finite mathematics schaum

SCHAUM’S OUTLlNE OF

THEORY AND PROBLEMS

OF

F I N I T E MATHE RIATIC S

0

BY

SEYMOUR LIPSCHUTZ, Ph.D. Associate Professor of Mathematics

Temple University

0

SC€IAUMfS OUTLINE SERIES McGRAW-HILL BOOK COMPANY

New York, St. Louis. San Francisco, Toronto, Sydne)

Copyright @ 1966 b j McGraw-Hill, Inc. All Rights Reserved. Printed in the United States of America. No part of this publication may be reproduced. stored in a rerrieval sy%tem, or transmitted, in any form or by any means. electronic, mechanical, photocopying, recording. or otherwise, without the prior written permissioii of the imblisher.

579hT

6 7 8 9 1 0 S H S H 7 5 4 3 2 1 0

Finite mathematics has in recent years become an integral part of the mathematical background necessary for such diverse fields as biology, chemistry, economics, psychology, sociology, education, political science, business and engineering. This book, in presenting the more essential material, is designed for use as a supplement to all current standard texts or as a textbook for a formal course in finite mathematics.

The material has been divided into twenty-five chapters, since the logical arrangement is thereby not disturbed while the usefulness as a text and reference book on any of several levels is greatly increased. logic; set theory; vectors and matrices: counting - permutations, combinations and partitions; probability and Markov chains; linear programming and game theory. The area on vectors and matrices includes a chapter on systems of linear equations; i t is in this context that the important concept of linear dependence and independence is introduced. The area on linear programming and game theory includes a chapter on inequalities and one on points, lines and hyperplanes; this is done to make this section self-contained. Furthermore, the simplex method is given for solving linear programming problems with more than two unknowns and for solving relatively large games. In using the book it is possible to change the order of many later chapters or even t o omit certain chapters without difficulty and without loss of continuity.

Each chapter begins with a clear statement of pertinent definitions, principles and theorems together with illustrative and other descriptive material. This is followed by graded sets of solved and supplementary problems. The solved problems serve to illustrate and amplify the theory, bring into sharp focus those fine points without which the student continually feels himself on unsafe ground, and provide the repetition of basic principles so vital t o effective learning. Proofs of theorems and derivations of basic results are included among the solved problems. The supplementary problems serve as a complete review of the material in each chapter.

More material has been included here than can be covered in most first courses. This has been done to make the book more flexible, to provide a more useful book of reference and to stimulate further interest in the topics.

I wish to thank many of my friends and colleagues, especially P. Hagis, J. Landman, B. Lide and T. Slook, for invaluable suggestions and critical review of the manuscript. I also wish to express my gratitude t o the staff of the Schaum Publishing Company, particularly to N. Monti, for their unfailing cooperation.

The basic areas covered are:

S. LIPSCHUTZ

Temple University June, 1966

C O N T E N T S

Page

Chapter 1 PROPOSITIONS AND TRUTH TABLES. . . . . . . . . . . . . . . . . . . . . . 1 Statements. Compound statements. Conjunction, p A q. Disjunction, p v q. Negation, -p. Propositions and t ruth tables.

Chapter 2 ALGEBRA OF PROPOSITIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Tautologies and contradictions. Logical equivalence. Algebra of propositions.

Chapter 3 CONDITIONAL STATEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Conditional, p + q. Biconditional, p ++ q. Conditional statements and variations.

Chapter 4 ARGUMENTS, LOGICAL IMPLICATION . . . . . . . . . . . . . . . . . . . . . 26 Arguments. Arguments and statements. Logical implication.

Chapter 5 SET THEORY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Sets and elements. Finite and infinite sets. Subsets. Universal and null sets. Set operations. Arguments and Venn diagrams.

Chapter 6 PRODUCT SETS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Ordered pairs. Product sets. Product sets in general. Truth sets of propositions.

Chapter 7 RELATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 Relations. Relations a s sets of ordered pairs. Inverse relation. Equivalence relations. Partitions. Equivalence relations and partitions.

Chapter 8 FUNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 Definition of a function. Graph of a function. Composition function. One-one and onto functions. Inverse and identity functions.

Chapter 9 VECTORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Column vectors. Vector addition. Scalar multiplication. Row vectors. Multi- plication of a row vector and a column vector.

Chapter 10 MATRICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Matrices. Matrix addition. Scalar multiplication. Matrix multiplication. Square matrices. Algebra of square matrices. Transpose.

C O N T E N T S

Page

Chapter I I LINEAR EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 Linear equation in two unknowns. Two linear equations in two unknowns. General linear equation. General system of linear equations. Homogeneous systems of linear equations.

Chapter 12 DETERMINANTS OF ORDER TWO AND THREE. . . . . . . . . . . . 127 Introduction. Determinants of order one. Determinants of order two. Linear equations in two unknowns and determinants. Determinants of order three. Linear equations in three unknowns and determinants. Invertible matrices. Invertible matrices and determinants.

Chapter 13 THE BINOMIAL COEFFICIENTS AND THEOREM . . . . . . . . . . 141 Factorial notation. Binomial coefficients. Binomial theorem. Pascal's triangle. Multinomial coefficients.

Chapter 14 PERMUTATIONS, ORDERED SAMPLES .................... 152 Fundamental principle of counting. Permutations. Permutations with repetitions. Ordered samples.

Chapter 15 COMBINATIONS, ORDERED PARTITIONS . . . . . . . . . . . . . . . . . 161 Combinations. Partitions and cross-partitions. Ordered partitions.

Chapter 16 TREE DIAGRAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 Tree diagrams.

Chapter 17 PROBABILITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Introduction. Sample space and events. Finite probability spaces. Equi- probable spaces. Theorems on finite probability spaces. Classical birthday problem.

Chapter 18 CONDITIONAL PROBABILITY. INDEPENDENCE . . . . . . . . . 199 Conditional probability. Multiplication theorem for conditional probability. Finite stochastic processes and tree diagrams. Independence.

Chapter 19 INDEPENDENT TRIALS, RANDOM VARIABLES. . . . . . . . . . . 216 Independent or repeated trials. Repeated trials with two outcomes. Random variables. Probability space and distribution of a random variable. Expected value. Gambling games.

Chapter 20 MARKOV CHAINS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Probability vectors. Stochastic and regular stochastic matrices. Fixed points of square matrices. Fixed points and .regular stochastic matrices. Markov chains. Higher transition probabilities. Stationary distribution of regular Markov chains. Absorbing states.

Chapter 21 INEQUALITIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 The real line. Positive numbers. Order. (Finite) Intervals. Infinite intervals. Linear inequalities in one unknown. Absolute value.

C O N T E N T S

Page

Chapter 22 POINTS, LINES AND HYPERPLANES.. . . . . . . . . . . . . . . . . . . . . . 266 Cartesian plane. Distance between points. Inclination and slope of a line. Lines and linear equations. Point-slope form. Slope-intercept form. Parallel lines. Distance between a point and a line. Euclidean nz-space. Bounded sets. Hyperplanes. Parallel hyperplanes. Distance between a point and a hyperplane.

Chapter 23 CONVEX SETS AND LINEAR INEQUALITIES. . . . . . . . . . . . . . 281 Line segments and convex sets. Linear inequalities. Polyhedral convex sets and extreme points. Polygonal convex sets and convex polygons. Linear functions on polyhedral convex sets.

Chapter 24 LINEAR PROGRAMJIIKG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 Linear programming problems. Dual problems. Matrix notation. Introduction t o the simplex method. Initial simplex tableau. Pivot entry of a simplex tableau. Calculating the new simplex tableau. Interpreting the terminal tableau. Algorithm of the simplex method.

Chapter 25 THEORY OF GAMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 Introduction to matrix games. Strategies. Optimum strategies and the value of a game. Strictly determined games. 2 X 2 matrix games. Recessive rows and columns. Solution of a matrix game by the simplex method. 2 X m and i i i x 2 matrix games. Summary.

INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

Chapter I

Propositions and Truth Tables STATEMENTS

true or false , but not both. Statements will usually be denoted by the letters

The truth or falsity of a statement is called its truth value.

A statement (or verbal assertion) is any collection of symbols (or sounds) which is either

P , (1, r , . . .

Example 1.1 : Consider the following expressions: (i) Par is is in England. (iii) Where a re you going?

(ii) 2 + 2 = 4 (iv) P u t the honiework on the blackboard.

The expressions (i) and 6) are statements; the first is false and the second is true. The expressions (iii) and [iv) a re not statements since neither is either t rue o r false.

COMPOUND STATEMENTS

connectives which we discuss subsequently. pound statements.

Some statements are composite, that is, composed of substatements and various logical Such composite statements are called com-

Example 2.1: “Roses a r e red and violets a re blue” is a compound statement with substatements “Roses a r e red” and “Violets a r e blue”.

Example 2.2: “He is intelligent o r studies every night” is, implicitly, a compound statement with substatements “He is intelligent” and “He studies every night”.

The fundamental property of a compound statement is that its truth value is completely determined by the truth values of its substatements together with the way in which they are connected to form the compound statement. We begin with a study of some of these connectives.

CONJUNCTION, p A Q

called the conjunction of the original statements. Symbolically, Any two statements can be combined by the word “and” to form a compound statement

P A 9 denotes the conjunction of the statements p and q, read “ p and q”.

Example 3.1: Let p be “It is raining” and let q be “The sun is shining”.

Then p A q denotes the statement “It is raining and the sun is shining”.

The truth value of the compound statement p ~ q satisfies the following property:

[TI] In other words, the conjunction of two statements is true only in the case when each sub- statement is true.

If p is true and Q is true, then p A q is true; otherwise, 11 A q is false.

1

2 PROPOSITIONS AND TRUTH TABLES [CHAP. 1

Example 3.2: Consider the following four statements:

(i) Paris is in France and 2 + 2 = 4.

(ii) Par is is in France and 2 + 2 = 5.

( i i i ) Par is is in England and 2 + 2 = 4.

(iv) Paris is in England and 2 - 2 = 5.

By property [Ti], only the first statement is true. is false since at least one of its substatements is false.

Each of the other statements

A convenient way to state property [Ti] is by means of a table as follows:

Here, the first line is a short way of saying that if p is true and q is true then p A 4 is true. The other lines have analogous meaning. We regard this table as defining precisely the truth value of the compound statement p A q as a function of the truth values of p and of q.

DISJUNCTION, 11 q

Any two statements can be combined by the word “or” (in the sense of “and/or”) to form a new statement which is called the disjunction of the original two statements. Symbolically,

P ” 4 denotes the disjunction of the statements p and q and is read ‘ip or q”

Example 4.1: Let p be “Marc studied French at the university”, and let q be “Marc lived in France”. Then p v q is the statement “Marc studied French at the university o r (Marc) lived in France”.

The truth value of the compound statement p v q satisfies the following property:

IT,] q is true; otherwise p v q

Accordingly, the disjunction of two statements is false only when both substatements are false. The property LT,] can also be written in the form of the table below, which we regard as defining 11 v q:

If p is true or q is true o r both p and q are true, then 13 is false.

Example 4.2: Consider the following four statements:

ii) Par is is in France o r 2 + 2 = 4.

( i i ) Par is is in France or 2 + 2 = 5. iiii) Par is is in England or 2 + 2 = 4.

iiv) Par is is in England o r 2 + 2 = 5.

By property [T2], only (iv) is false. least one of its substatements is true.

Each of the other statements is t rue since a t

CHAP. 11 PROPOSITIONS AND TRUTH TABLES 3

Remark: The English word “or” is commonly used in two distinct ways. Sometimes it is used in the sense of “ p o r (I or both”, i.e. a t least one of the two alternates occurs, as above, and sometimes it is used in the sense of “ p or 4 but not both”, i.e. exactly one of the two alternatives occurs. For example, the sentence “He ill go to Harvard o r to Yale” uses “or” in the latter sense, called the e,cclusice disjunction. Unless otherwise stated, “01”’ shall be used in the former sense. This discussion points out the precision we gain from o u r symbolic language: p b (I is defined by its truth table and always means “p andlor q”.

NEGATION, -11

Given any statement p , another statement, called the negatioii of 11, can be formed by writing “It is false tha t . . .” before p or , if possible, by inserting in p the word “not”. Symbolically,

-1, denotes the negation of p (read “not 71”).

Example 5.1 : Consider the following three statements: ( i ) Par is is in France. (ii, I t is false tha t Par is is in France. ( i i i ) Par is is not in France.

Then (ii) and (iii) a re each the negation of (i).

Consider the following statements: Example 5.2:

(i) 2 - 2 = 5 (ii) It is false t h a t 2 - 2 = 5 . (iii) 2 + 2 # 5

Then (ii) and (iii) a re each the negation of (i).

The truth value of the negation of a statement satisfies the following property:

[T,] Thus the truth value of the negation of any statement is always the opposite of the

If p is true, then -p is false; if 11 is false, then -p is true. truth

value of the original statement. The defining property IT,] of the connective can also be written in the form of a table:

P I -P

F T

Example 5.3: Consider the statements in Example 5.1. each its negation, a re false.

Consider the statements in Example 5.2. each its negation, a re true.

Obserre that t i ) is t rue and (ii) and [iii),

Example 5.4: Observe that ri) is false and (ii) and (iiii,

PROPOSITIONS AND TRUTH TABLES By repetitive use of the logical connectives (4, v, - and others discussed subsequently),

we can construct compound statements that are more involved. In the case where the substatements p , Q, . . . of a compound statement P(p, q, . . .) are variables, we call the compound statement a proposition.

Now the truth value of a proposition depends exclusively upon the t ruth values of its variables, that is, the truth value of a proposition is known once the t ruth values of its variables are known. A simple concise way to show this relationship is through a truth table. The truth table, for example, of the proposition - ( p A -q) is constructed as follows:

4

P Q

T T T F F I T F / F Step

PROPOSITIONS AND TRUTH TABLES

- ( P A - 4 )

[CHAP. 1

1’ 4 - ( P A - 4 ) P - (11 A - 4 T T T F F T T T T T F F T T F T T T F T F F T T T F F T F F F T F T T F F F T F F F F T F F F T F F T F

F F

Step

F F F

1 3 2 1 Step 4 1 3 2 1

CHAP. 11 PROPOSITIONS AND TRUTH TABLES 5

Solved Problems STATEMENTS 1.1. Let p be “It is cold” and let q be “It is raining.” Give a simple verbal sentence which

describes each of the following statements: (1) -21, ( 2 ) P A Q, (3) P ” 4, (4) cl ” -21, (5 ) ”P A -4, (6) - - Q.

In each case, translate A, v and - to read “and”, “or” and “I t is false that” o r “not”, respectively, and then simplify the English sentence.

(1) It is not cold. (41 It is raining or i t is not cold.

(2) It is cold and raining.

(3) It is cold or it is raining.

( 5 ) I t is not cold and i t is not raining.

( G I I t is not true tha t i t is not raining.

Let p be “He is tall” and let Q be “He is handsome.” statements in symbolic form using 11 and q.

(1) He is tall and handsome. (2) He is tall but not handsome. (3) It is false that he is short or handsome. (4) He is neither tall nor handsome. (5) He is tall, or he is short and handsome. (6) It is not true that he is short or not handsome.

Write each of the following

(Assume that “He is short” means “He is not tall”, i.e. -11.)

(1) P A q (3) -(-P v 4) ( 5 ) 1’ V (“P A 4)

(2) P A -!I (4) -P A @q (6) - ( - P ” -q)

TRUTH VALUES OF COMPOUND STATEMENTS 1.3. Determine the truth value of each of the following statements.

(i) 3 + 2 = 7 and 4 f 4 = 8. (ii) 2 + 1 = 3 and 7 + 2 = 9. (iii) 6 + 4 = 10 and 1 + 1 = 3 .

By property [TI], the compound statement “ p and q” is t rue only when p and q are both true. Hence: ti, False, iii) True, (iii) False.

1.4. Determine the truth value of each of the following statements.

(i) Paris is in England or 3 + 4 = 7. (ii) Paris is in France or 2 + 1 = 6. (iii) London is in France or 5 + 2 = 3 .

Hence: li) True, (ii) True, (iii) False. By property [ T L ] , the compound statement “ p or q” is false only when p and q a r e both false.

1.5. Determine the truth value of each of the following statements.

(i) It is not true that London is in France. (ii) It is not true that London is in England.

Hence: I i ) True, rii) False. By property IT,], the t ru th value of the negation of p is the opposite of the t ru th value of p .

6 PROPOSITIONS AND TRUTH TABLES

P a - P A P

T T F T F T T F F T F F F T T F T T F F T F F F

[CHAP. 1

Step

1.6. Determine the truth value of each of the following statements. (i) It is false that 2 + 2 = 4 and 1+ 1 = 5 . (ii) Copenhagen is in Denmark, and 1+1 = 5 or 2 f 2 = 4.

(iii) I t is false that 2 + 2 = 4 or London is in France.

2 1 3 1

(i) The conjunctive statement “2 + 2 = 4 and 1 + 1 = 5” is false since one of i ts substatements “I + 1 = 5” is false. Accordingly its negation, the given statement, is true.

(ii) The disjunctive statement “1 + 1 = 5 or 2 + 2 = 4” is t rue since one of i ts substatements Hence the given statement is t rue since it is the conjunction of two “2 + 2 = 4” is true.

t rue statements, “Copenhagen is in Denmark” and “1 + 1 = 5 o r 2 + 2 = 4”.

(iii) The disjunctive statement “ 2 + 2 = 4 o r London is in France” is t rue since one of its substatements “2 + 2 = 4” is true. Accordingly its negation, the given statement, is false.

T T F T F T F T T F F T F F T F F T T F

TRUTH TABLES OF PROPOSITIONS 1.7. Find the truth table of -13 ,. q .

9

T F T F

-

Method 1

1.8. Find the truth table of - ( ] I v 4).

T F T F T F F T

Method 1

1.9. Find the truth table of - ( p \d) -q) .

P I a I -9 I P V -a 1 - (?I” - 9 )

T F F

Step

p l q

F

Method 2

I - ( P v - a )

Method 1

1.10. Find the truth table of the following: (i) p A ( 4 v r ) , Since there are three variables, we will need 23 = 8 rows in the t ru th table.

(ii) ( p A 4) v ( p A T ) ~

C'HAP. I]

P q 7. P A ( q " r ) P q r

T T T T T T T T T T I F T T T T F T F T T T T F T T F F F F T F F F T T T F F , T T F T F T F F T F F F T ' T F F ~ F T F F F l F F F F F

PROPOSITIONS AND TRUTH TABLES

P A ? .

T T T F F T F F F F F F F ~F F i F

7

i

MISCELLANEOUS PROBLEMS 1.11. Let A p q denote p / q and let Sp denote -p. Rewrite the following propositions

using A and N instead of A and -. (9 P A -q (iii) -p A (-q A 1 )

(4 -(-P A Q) (iv) - ( p A -q) A (-(I A -7,)

(i) p A - q = p A iYq = A p N q

(ii) -(-p 4 q ) = -(.Vp A q ) = - ( A S p q ) = N A N p q

(iii) - p A ( - q A 1.) = N p A ( N q A r ) = S p A ( A N q r ) = ANpA'VqT

(iv) - ( p A -q) A ( - q \ -Y) = - ( A p X q ) A ( A N q S r ) = ( N A p N q ) A ( A S q S r ) = A N A p N q A N q X r

Observe tha t there a r e no parentheses in the final answer when A and .V are used instead of A and -. In fact, i t has been proved that parentheses a re never needed in any proposition using A and N .

1.12. Rewrite the following propositions using A and - instead of A and N. (i) N A m (iii) ApNq (v) Iv'AANpql.

(ii) ANpq (iv) ApAqr (vi) A S p A q N r

(i) NApq = S ( p A q ) = - ( p A q ) ( i i i ) ApNq = A p ( - q ) = p A -q (ii) A N p q = A ( - p ) q = - p A q (iv) A p A q r = A p ( q A 7,) = p A ( q A r )

(v) N A A S p q r = S A A ( - p ) q r = N A ( - p fi q)r = Nr(-p A q) / r1 = -:(-p A q) A r]

(vi) A N p A q S r = A N p A q ( - r ) = A N p ( q A -7) = A ( - p ) ( q A - 1 . ) = -lJ A ( Q A -r) Notice tha t the propositions involving A and N are unraveled from right to left.

Supplementary Problems STATEMENTS

1.13. Let p be "Marc is rich" and let q be "Marc is happy". Write each of the following in symbolic form.

(i) Marc is poor but happy.

(ii) Marc is neither rich nor happy.

(iii) Marc is either rich o r unhappy.

(iv) Marc is poor or else he is both rich and unhappy.

8 PROPOSITIONS AND TRUTH TABLES [CHAP. 1

1.14. Let p be “Erik reads Newsiccek”, let Q be “Erik reads Life” and le t r be “Erik reads Time”. Write each of the following in symbolic form.

( I ) Erik reads Newsweek or L i f e , but not Tima. iii) Erik reads Newsweek and Lit’c, o r he does not read Newsweek and Time.

( i i i ) I t is not t rue tha t Erik reads Newsweek but not Time.

(ivi I t is not t rue tha t Erik reads Time or Life but not Newsweek.

1.15. Let p be “Audrey speaks French” and let q be “Audrey speaks Danish”. Give a simple verbal sentence which describes each of the following.

(i) P v 4 (iii) p A -q (v) - - p

(ii) 11 A q (iv) - p v -q (vi) -(-P A -4

1.16. Determine the t ruth value of each of the following statements.

(i, 3 f 3 = 6 and 1 + 2 = 5. iiii I t is not t rue tha t 3 + 3 = 6 o r 1 f 2 = 3.

iiii) It is true t h a t 2 + 2 # 4 and 1 + 2 = 3.

iiv) It is not t rue tha t 3 + 3 # 6 o r 1 + 2 # 5.

TRUTH TABLES OF PROPOSITIONS

1.17. Find the t ruth table of each of the following.

ii) p v -q , iii) - p A -q, iiii) - ( - p A q ) , (iv) - ( -p v - q )

1.18. Find the t ruth table of each of the following.

( i ) ( p A - q ) v r , (ii) - p v ( q A - T ) , (iii) ( p v -v) A ( q v - T ) , (iv) - ( p v -9) A ( - p v 7 ) .

MISCELL-4NEOUS PROBLEMS

1.19. Let A p q denote P A q and let S p denote -p . (See Problem 1.11.) Rewrite the following propositions using A and N instead of n and -. (i) -p A q , (ii) - p A -q, iiii) -(p A -q ) , (iv) ( - p A q ) A -T.

1.20. Rewrite the following propositions using A and - instead of A and N . (i) N A p N q , (ii) A N A p q N y , (iii) A A p N r A q N p , (iv) A S A N p A N q r N p .

Answers to Supplementary Problems 1.13. [ i ) - p A q , (ii) -p A -q, ( i i i ) p v - q , (iv) - p v ( p A - q )

1.14. (i) (p v q ) A -7, (ii) ( p A q) v - ( p A T ) , (iii) - ( p A - T ) , (iv) -[(Y v q) A -p]

1.15. ( i ) Audrey speaks French o r Danish.

(ii) Audrey speaks French and Danish. (iii) Audrey speaks French but not Danish.

( iv) Audrey does not speak French or she does not speak Danish.

(v) I t is not t rue tha t Audrey does not speak French. (vi\ I t is not t rue tha t Audrey speaks neither French nor Danish.

CHAP. 11 PROPOSITIONS AND TRUTH TABLES

1.18. p q ) * ( i ) (ii) (iii)

T T T T F T F T T F F T T F T F T T F F F T F F T F T F F T T T T F T F T F F T T T F F T T T F F F F F F T T F

1.16. (i) F, ( i i ) F, (iiii F, (iv) F

(ivi

- p A -q - ( - p A 4 ) -(-pv - q )

F F T F F F F F F T T T F

1.19. (i) Ah'pq, (iii A.YpNq, (iii) N A p N q , (ivi A A S p q N r

9

1.20. (i) - ( p A - q ) , ( i i ) - ( p A q) A -Y, ( i i i ) ( p A -T ) A (q A - p ) , (iv) - [ - p n ( -p A T ) ] A - p

Chapter 2

Algebra of Propositions

TAUTOLOGIES AND CONTRADICTIONS Some propositions P(p ,q , . . . ) ca,ntain only T in the last column of their truth tables,

i.e. are true for any truth values of their variables. Such propositions are called tautologies. Similarly, a proposition P(p, (1, . . .) is called a cot i tmdic t ion if it contains only F in the last column of its truth table, i.e. is false for any truth values of its variables.

Example 1.1: The proposition " p o r not p", i.e. p v -11, is a tautology and the proposition " p and not p", i.e. p -11, is a contradiction. This is verified by constructing their t ruth tables.

1' 1 -11 1 P " - p ," I i l l ' P;-P

T F T

F ~ T ~ T F ~ T I F

Since a tautology is always true, the negation of a tautology is always false, i.e. is a contradiction, and vice versa. That is,

Theorem 2.1: If P(p,q , . . .) is a tautology then -Pip, (I, , . . ) is a contradiction, and

Now let P(p , (I, . . .) be a tautology, and let P,(p , q, . . . ), P,(p, q , . . .), . . . be any propositions. Since P(p, q , . . . ) does not depend upon the particular truth values of its variables p , 'I, . . ., we can substitute P , for p , P , for q, . . . in the tautology P(p, q, . . . ) and still have a tautology.

conversely.

In other words:

Theorem 2.2 (Principle of Substitution): If P(p, q , . . .) is a tautology, then P(P,, P, , . . . ) is a tautology f o r any propositions P,, P,~, , . . .

LOGICAL EQUIVALENCE

simply equicnleiit o r equal, denoted by

if they have identical truth tables.

Two propositions P(p, q, . . . ) and Q ( p , q , . . .) are said to be logically equivalent, or

P(l4 9 , . * * 1 = &(P, q, . . . )

Example 2.1: The t ruth tables of -(]I q i and - p v -q folio\\.

T F T T F T T

Accordingly, the propositions - ( p A q) and -11 v - q are logically equivalent:

" ( p A 9) - p V -q

10

CHAP. 21 ALGEBRA O F PROPOSITIONS 11

Example 2.2: The statement “It is false that roses a re red and violets a re blue”

can be written in the form - ( p A q ) where p is “Roses a re red” and q is “Violets a re blue”. By the preceding example, - ( p A q) is logically equivalent to - p v - q ; that is, the given statement is equivalent to the statement

“Either roses a re not red or violets a re not blue.”

ALGEBRA OF PROPOSITIONS

which are listed in Table 2.1 below. In fact, we formally state:

Theorem 2.3:

Propositions, under the relation of logical equivalence, satisfy various laws or identities

Propositions satisfy the laws of Table 2.1.

LAWS OF THE ALGEBRA OF PROPOSITIONS

Idempotent Laws

la . p v p = p l b . p ~ p = p

Associative Laws 2a. i p J q ) d T = p v (q v r ) 2b. ( p A q) A r = p /\ ( q A r )

Commutative Laws

3a. p 4 q = q v p 3b. p ~ q = q ~ p

Distributive Laws

4U. p J ( 4 / T ) ( p V 4 ) A ( p u‘ T ) 4 b . p A (q v r ) = ( p A q ) v ( p A r )

Identity Laws

5a. p v f = p 5b. p ~ t e p

6a . p ~ t = t 6b. P A f = f

Complement Laws

7a. p v - p 5 f 7b. p ~ - p f

8a. - - p = p 8b. --t 3 f , - f t

De Morgan’s Laws

9a. - ( p V q ) 3 -p A -q 9b. - ( p A q ) = - P ” - g

Table 2.1

In the above table, t and f denote variables which are restricted to the t ruth values true and false, respectively.

12 ALGEBRA O F PROPOSITIONS

p q p A 4 " ( P A 4 )

T T T F T F F T F T F T

F F F T

[CHAP. 2

P C " ( P /\ 4 )

T

T T

T

Solved Problems

T F F F

TAUTOLOGIES AND CONTRADICTIONS 2.1. Verify that the proposition p v - ( P A q) is a tautology.

T F T F T F F T

Construct the t ru th table of p v - ( p A 4):

Since the t ruth value of p v - ( p A q ) is T for all values of p and q, i t is a tautology.

2.2. Verify that the proposition ( p A q ) A - (p v q ) is a contradiction.

Construct the t ruth table of ( p A q ) A - ( p v q):

p 1 4 1 P A 4 I P V Q I " ( P V q ) 1 ( P A q ) * - ( p v Q )

T T F F

F F F F

Since the t ruth value of ( p + q ) A - ( p v q ) is F for all values of p and q, i t is a contradiction.

LOGICAL EQUIVALENCE 2.3. Prove the Associative Law: (11 A q ) A r = p A (4 A r ) .

Construct the required t ruth tables:

P

T T

T

T F F F

F

- 1'

Since the t ruth tables a re identical, the propositions are equivalent.

2.4. Prove that disjunction distributes over conjunction; tha t is, prove the Distributive Law: 11 v ( q A r ) = ( p v q) A ( p v Y).

Construct the required t ruth tables:

CHAP. 21 ALGEBRA O F PROPOSITIONS 13

Since the t ruth tables a r e identical, the propositions a re equivalent.

2.5. Prove that the operation of disjunction can be written in terms of the operations of conjunction and negation. Specifically, p v (r = -(-?I A -(I).

Construct the required t ru th tables:

Since the t ruth tables a r e identical, the propositions a re equivalent.

2.6. There are exactly four non-equivalent propositions of one variable; the t ruth tables of such propositions follow:

P 1 Pi(P0) 1 P,(?J) 1 p,(P) 1 pi(?,)

T F

Find four such propositions. Observe that ; ~ ; 1 PVT-P i P y P

F T T F

Hence PI (11 ) p v -P, P 2 h ) = p, P 3 ( p ) = - p , P4(p) = P A -11.

2.7. Determine the number of non-equivalent propositions of two variables p and 4.

appear as follows: The t ruth table of a proposition P ( p , q ) will contain 22 = 4 lines. In each line T or F can

In other words, there a re 24 = 16 non-equivalent propositions of two variables p and (I.

14 ALGEBRA O F PROPOSITIONS [CHAP. 2

2.8. Determine the number of non-equivalent propositions of: and 1’ ;

(i) three variables 11, q (ii) 72 variables pl, p , , . . . , p,].

( i ) The t ruth table of a proposition P ( p , q , r ) will contain 2? = 8 lines. can appear, there a r e 2 8 = 236 non-equivalent propositions of three variables.

The t ruth table of a proposition P ( p l , . . . , p,) will contain 2l’ lines; hence, as above, there are Z2” non-equivalent propositions of 71 variables.

Since in each line T o r F

(iir

N E GAT1 ON 2.9. Prove De Morgan’s Laws: (i) - ( p A (I) = -p v ”(7: (ii) - ( T I q ) = -p A -q.

In each case construct the required t ruth tables.

(i’ ~ l l -(?;Y) ip iq -”-‘

F T F T F F F F T T T

t t

F t

F T t

2.10. Verify: - - p = p . 1’ I -P I - - P

2.11. TJse the results of the preceding problems to simplify each of the following proposi tions: (i) -(p v -(I), (ii) -(-]I r a), (iii) -(-p v -a). (1) -[I1 V - q ) * p A - - q -1) * Y

(iii) -(-11 v -9) = - - p A - -q = p \ 4

(ii) -(-11 A q ) = - - p v -y = p v -y

2.12. Simplify each of the following statements.

(i) It is not true that his mother is English or his father is French. (ii) I t is not true that he studies physics but not mathematics. (iii) I t is not true that sales are decreasing and prices are rising. (iv) I t is not true that it is not cold or i t is raining.

(ii Let p denote “His mother is English” and let q denote “His father is French”. Then the given statement is - ( p v q ) . But - ( p v y) = - p A - q . Hence the given statement is logically equivalent to the statement “His mother is not English and his fa ther is not French”.

ALGEBRA O F PROPOSITIONS 15 CHAP. 21

(ii) Let p denote “He studies physics” and let y denote “He studies mathematics”. Then the given statement is -(PA -4). Hence the given statement is logically equivalent t o the statement “He does not study physics or he studies mathematics”

(iii) Since -()I (11 5 - p v -q, the given statement is logically equivalent to the statement

(iv) Since -(-p q ) = -q, the given statement is logically equivalent to the statement

But - ( p A - q ) = - p v - - y 5 - p v q.

“Sales a le increasing or prices are falling”.

“It is cold and it IS not raining”.

ALGEBRA OF PROPOSITIONS 2.13. Simplify the proposition ( p v q )

listed on Page 11. -11 by using the laws of the algebra of propositions

Statement Reason

(1) ( I ] \ 4) -11 =i - P A (PV 4 ) (1) Commutative law (2) = ( - P A P ) v ( - P (I) (2) Distributive law

(3) = f v ( - p / \ q ) (3) Complement law

(4) = - p A q (4) Identity law

2.14. Simplify the proposition p v ( p ~ q ) by using the laws of the algebra of propositions listed on Page 11.

Statement Reason

(1) (11 4) = ( P A t ) v ( P A 4 ) (1) Identity law

(2) = p A ( t v q ) (2) Distributive law

(3) E p A t (3) Identity law

(4) = p (4) Identity law

2.15. Simplify the proposition + v q ) ( - p ~ q ) by using the laws of the algebra of propositions listed on Page 11.

Statement Reason

(1) -(P \ 4 ) ‘4 (‘11 4 Q ) ( “ P A - q ) V (”p A 4 ) (1) De Morgan’s law (2) - p A ( - q v q ) (2) Distributive law

(3) - p A t (3) Complement law

(4) (4) Identity law - == - p

MISCELLANEOUS PROBLEMS 2.16. The propositional connective y is called the exclusive disjunction; p y q is read

“ p or q but not both”.

(i) Construct a truth table for I-, y q. (ii) Prove:

the original three connectives A , v and -. Now p 1 q is t rue if p is t rue or if q is t rue but not if both a re true: hence the t ru th table of p y q is as follows:

p y q = ( p v q ) ~ - ( p ~ q ) . Accordingly y can be written in terms of

(i)

16

T F T T F T T T F F F T F T T T T F F T F F F F F F T F F F

ALGEBRA OF PROPOSITIONS [CHAP. 2

Step

(iii

1 2 1 4 3 1 2 1

We construct the t ruth table of ( p v q ) A - ( p A q ) , by the second method,

P I Y / ( P ” 4 ) A - ( P A n)

as follows:

2.17. The propositional connective J, is called the joint denial; p J y is read “Neither p nor q”.

(i) Construct a truth table fo r p J q . (ii) Prove:

connective J, as follows: The three connectives V, A and - may be expressed in terms of the

(a) -13 = P J P , ( b ) P A C ? = (PJP)J,(YLY), ( c ) P J C ? = ( P L 4 ) J ( P J q ) *

( i ) Now p J q is t rue only in the case that p is not t rue and y is not true; hence the t ruth table of p 1 y is the following:

F

ciij Construct the appropriate t ruth tables:

(a ) ; 1; i ”%” ( b ) ; ; ”;‘ “ q (pyqLq)

F T F T F F F F F T T F

U f t

CHAP. 21 ALGEBRA OF PROPOSITIONS 17

Supplementary Problems LOGICAL EQUIVALENCE

2.18. Prove the associative law for disjunction: ( p v y ) v T = p v ( q v r )

2.19. Prove that conjunction distributes over disjunction: p A (q v r ) = (p A q ) v ( q A r) .

2.20. Prove ( p v y) A -p = -p A q by constructing the appropriate t ruth tables fsee Problem 2.13).

2.21. Prove p v ( p A q ) = p by constructing the appropriate t ruth tables (see Problem 2.14).

2.22. Pkove - ( p v q ) v ( - p q ) = - p by constructing the appropriate t ruth tables isee Problem 2.15).

2.23. (i) Express \i in ternis of A and -. (ii) Express A in terms of v and -.

2.25. Write the negation of each of the following statements a s simply as possible.

(i) He is tall but handsome.

(ii) He has blond hair o r blue eyes.

(iii) He is neither rich nor happy.

(iv) He lost his job o r he did not go to mork today.

(v) Neither Marc nor Erik is unhappy. (vi) Audrey speaks Spanish o r French, but not German.

ALGEBRA OF PROPOSITIOKS

2.26. Prove the following equivalences by using the l a ~ s of the algebra of propositions listed on Page 11:

(i) p ~ ( p v q ) = 1 1 , i i i ) ( p ~ q ) v - p = - p v q , f i i i ) p ~ ( - p v q ) 5 p A q .

Answers to supplementary Problems 2.23. (i) p v q = -(-p P - q ) , (ii) P A q = - ( -pv -4).

2.24. (i) - p v q, (ii) p - q , (iii) p v q.

2.25. (iii) He is rich or happy. (vi) Audrey speaks German but neither Spanish nor French.

2.26. (i) p ~ ( p v q ) = ( p v t ) r \ ( p v q ) = p v ( f (I) =- p f E p .

Chapter 3 Conditional Statements

CONDITIONAL, p -+ y

statements are called coriditioizal statements and are denoted by

The conditional p + Q can also be read:

Many statements, particularly in mathematics, are of the form “If p then q”. Such

p + q

(i) p implies q (iii) 23 is sufficient f o r (I (ii) p only if q (iv) (I is necessary f o r p .

The truth value of p + (I satisfies:

[T,] The truth table of the conditional statement follows:

The conditional p + y is true except in the case that 1-I is true and q is false.

Example 1.1 : Consider the following statements:

( i J If Paris is in France, then

(iij If Par is is in France, then

( i i i ) If Par is is in England, then

(iv) If Par is is in England, then

By the property I T J ] , only (ii) is a false statement;

21-2 = 4.

2 + 2 = 5. 2 + 2 = 4.

2 + 2 = 5.

the others are true. JVe emphasize that , by definition, (iv) is a t rue statement even though its substatements “Paris is in England” and “ 2 f 2 = 5” are false. It is a statement of the type: If monkeys are human, then the earth is flat.

Now consider the truth table of the proposition -11 J q:

T T

Observe that the above truth table is identical to the truth table of p-+ y. is logically equivalent to the proposition -p v q:

Hence 1 1 3 4

p’q = - p v q

In other words, the conditional statement “If p then g” is logically equivalent t o the statement “Not 11 or q” which only involves the connectives \ and - and thus was already a part of our language. We may regard p + q as an abbreviation for an oft-recurring statement.

18

CHAP. 31 CONDITIONAL STATEMENTS 19

T T T T F F F T F F F T

BICONDITIONAL, p * q

Such statements, denoted by

are called bicoudi f io / ia l statements. The truth value of the biconditional statement p e q satisfies the following property:

[T5J If p and Q have the same truth value, Then p - q is true; if p and (I have opposite truth values, then p - q is false.

The truth table of the biconditional follows:

Another common statement is of the form ‘‘p if and only if q” or , simply, “ p iff q“.

P f-$ (I

P 1 Q 1 P * 9

Conditional P q P - ’ 9

T T T T F F F T T F F T

Converse Inverse Contrapositive ( I + P -P + -9 - q + ”11

T T T T T F F F T T T T

Example 2.1 : Consider the following statements:

( i ) Paris is in France if and only if 2 2 = 4.

(ii) Paris is in France if and only if 2 2 = 5.

(iii) Par is is in England if and only if 2 - 2 = 4.

(iv) Paris is in England if and only if 2 + 2 = 5.

By property [T5], the statements (i) and (iv) a re true, and (ii) and (iii) a re false.

Recall that propositions P(p, (I, . . . ) and Q ( p , q, . . .) are logically equivalent if and only if they have the same truth table; but then, by property [Ti], the composite proposition P(p, (I, . . . ) cf Q ( p , Q, . . .) is always true, i.e. is a tautology. In other words,

Theorem 3.1: P ( p , (I, . . , ) = Q ( p , q, . . .) if and only if the proposition

is a tautology. PiPt (I, . f .) ‘3 &(P, q, . . . )

Theorem 3.2: A conditional statement p -+ q and its contrapositive -q + -p are logically equivalent.

20 CONDITIONAL STATEMENTS [CHAP. 3

Example 3.1: Consider the following statements about a triangle A :

p 3 q:

q - p :

If A is equilateral, then A is isosceles. If A is isosceles, then A is equilateral.

Note t h a t p + q is true, but q + p is false.

Example 3.2: Prove: ( p + q ) If ~2 is odd then x is odd. We show tha t the contrapositive - q -f -p, “If x is even then x2 is even”,

is true. Let J be even: then x = 2n where is an integer. Hence x2 = (2n)(2n) = 2 ( 2 4 ) is also even. Since the contrapositive statement -q + - p is true, the original conditional statement p --f q is also true.

Solved Problems

CONDITIONAL 3.1. Let p denote “It is cold” and let q denote “It rains”. Write the following statements

in symbolic form.

(i) I t rains only if i t is cold. (ii) A necessary condition for it to be cold is that it rain. jiii) A sufficient condition f o r it to be cold is that it rain. (iv) Whenever i t rains it is cold. (v) I t never rains when it is cold.

Recall that p + q can be read ‘‘p only if q”, “p is sufficient for q” or “q is necessary f o r p”.

(i) q + p i i i ) p + q ( i i i ) q - p

(iv) Now the statement “Whenever it rains i t is cold” is equivalent to “If i t rains then i t is cold”. That is, q + p.

(v ) The statement “It never rains when it is cold” is equivalent to “If i t i s cold then it does not rain”. That is, p + -q.

3.2. Rewrite the following statements without using the conditional.

( i) If i t is cold, he wears a hat. (ii) If productivity increases, then wages rise.

Recall tha t “If p then q” is equivalent to “Not p or q”.

It is not cold or he wears a hat.

Productivity does not increase or wages rise. ( i )

( i i )

3.3. Determine the truth table of ( p -$ q) * ( p A 4).

F T

C,HAP. 31 CONDITIONAL STATEMENTS

P Q ?“

T T T T T F T F T T F F F T T F T F F F T F F F

21

q “ r P + ( q f i T j P - + 9

T T T F F T F F F F F F T T T F T T F T T F T T

4

3.4. Determine the truth table of -p -+ ( q -+ p ) .

3.5. Verify that (p q ) -+ ( p v q ) is a tautology.

F T F T T F F F F T

3.6. Prove that the conditional operation distributes over conjunction:

P -+ ((I A 7 9 = ( P -+ (I) A (P -+ r )

T t

BICONDITIONAL 3.7. Show that “ p implies (I and (I implies 11’’ is logically equivalent to the biconditional

“71 if and only if 4”; that is, (21 + q ) A ( q .+ p ) = p cs q.

F T F T

F I F l T 1 I I 4

3.8. Determine the truth value of each statement. (i) 2 + 2 = 4 iff 3 + 6 = 9 (ii) 2 + 2 = 7 if and only if 5 + 1 = 2 (iii) 1 + 1 = 2 iff 3 + 2 = 8 (iv) 1 + 2 = 5 if and only if 3 + 1 = 4

Now p e q is true whenever p and q have the same t ru th value: hence (i) and (ii) a re t rue (Observe that (ii) is a t rue statement by definition of the statements. but (i i i) and (iv) a re false,

conditional, even though both substatements 2 + 2 = 7 and 5 + 1 = 2 are false.)


T T F F T F T T F T F T F F T F

3.9. Show that the biconditional p * q can be written in terms of the original three connectives v, A and -.

Now T I + q = - p v q and q + p = -qv p ; hence by Problem 3.7,

P * q ( P + Q ) A ( q + P ) ( “ P ‘ - ‘ q ) A ( - q V P )

F I T I T T F T T F I T I F F I F ~ T

3.10. Determine the truth value of (11 + 4 ) v - ( p H Wq).

T

F F

3.11. Determine the truth value of ( p t) -4) e (4 -+ p ) .

T T i :

F I T

T F F F T T T F F

Step I 1 1 3

NEGATION

T T F T

Method 1

- 4 )

Method 2

F T F F

4 1 1 ( 2 / 1

3.12. Verify by truth tables that the negation of the conditional and biconditional are as follows: (i) - ( p + 4) = 13 A -q , (ii) - (p t) 4 ) = 11 H - 4 = -p t) q.

ii) P 1 Q 1 P + q I “(P’4) 1 -q 1 P A - 4

F T F

R e m a r k : Since p + q E -p v q, we could have used De Morgan’s law t o verify (ii as follows: -(p + 4 ) -(-fJ V 4) 5 - - T I A -4 p A -q

CHAP. 31 C 0 N D IT I 0 N A L STATEMENTS 23

3.13. Simplify: (ii - ( / I c) -q ) , (ii) - ( - ] I H q ) , (iii) - ( -p + -(I).

(i) -01 - q ) p - -q = p tf q

(ii) - ( - ] I q ) = - - p * q = IJ tf (I

(iii) - ( - / I -i - q ) = - p A - -q = -11 ~ 11

3.14. Write the negation of each statement as simply as possible.

(i) If he studies, he will pass the exam. (ii) He sn-inis if and only if the water is warm. (iii) If it snows, then he does not drive the car.

(i) By Problem 3.12, - (p+ q) = 11 - q : hence the negation of ! i i is

He studies and he will not pass the exam.

(ii) By Problein 5.12, - ( p t) q ) = p - q = -p H q; hence the negation of (ii) is either of the f 011 ow in g :

He swims if and only if the water is not warm.

He does not swim if and only if the water is warm.

(iii) Note that - ( ] I + -q ) = p A - -q = p q. Hence the negation of (iiii is

I t sno\vs and he drives the car.

3.15. Write the negation of each statement in as simple a sentence as possible.

(i) I f it is cold, then he wears a coat but no sweater. (ii) If he studies, then he will go t o college or to art school.

(i) Let 11 he “I t is cold”, q be “He weals a coat” and r be “He n e a r s a sweater”. Then the given statement can be written as p + ( q - 7 ) . Now

-[p + ((I -?)I =_ 11 A - ( q A -?) = 11 ( - q v Y)

Hence the negation of (i) is

I t is cold and he wears a sweater 01 no coat.

(ii) The g i ~ e n statement is of the folm 1 1 - ( q v 1.). But - [p+ ((I 7 ) = p A - ( q v Y) = 11 - q I -r

Thus the negation of (ii) is

He studies and he does not go to college 01 t o a r t school.

CONDITIONAL STATEMENTS AKD VARIATIONS 3.16. Determine the contrapositive of each statement.

(i) If John is a poet, then he is poor. (ii) Only if Marc studies will he pass the test. (iii) It is necessary to have snow in order fo r Eric to ski. (iv) If ,I‘ is less than zero, then

(i)

is not positive.

The tontraposltive of 11- q is - q + -11. Hence the contrapositive of (i) is

If John is not poor, then he is not a poet.

(ii) The Riven statement is equivalent to “If Marc passes the test, then he studied”. contrapositive of (ii) is

Hence the

If Marc does not study, then he will not pass the test.

(iiij The given statement is equivalent to “If Eric skis, then it snowed”. Hence the contrapositive of iiii) is

If it did not snow, then Eric will not ski.

(iv) The contrapositive of p - -q is - -q + - p = q + -p. Hence the contrapositive of fiv) is

If n: is positive, then x is not less than zero.

3.17. Find and simplify: (i) Contrapositive of the contrapositive of p + q. (ii) Contra- positive of the converse of p + q.

(i, The contrapositive of p - t q is -q+ -p. The contrapositive of - q + - p is - -11 -t - -q

p + q ,

(ii, The converse of p + q is q + p . The contrapositive of q - t p is -p+ -q , which is the inverse of p - t q.

(iii) The inverse of p + q is - p i -q. The contrapositive of - p + -q is - -q + - -p = q + p, which is the converse of p-+ q.

In other words, the inverse and converse a re contrapositives of each other, and the conditional

(iii) Contrapositive of the inverse of p + q.

which is the original conditional proposition.

and contrapositive a re contrapositives of each other!

Supplementary Problems STATEMENTS 3.18. Let p denote “He is rich” and let q denote “He is happy”. Write each statement in symbolic form

using p and q.

(i) If he is rich then he is unhappy. (ii) He is neither rich nor happy. (iii, It is necessary to be poor in order to be happy.

(iv) To be poor is to be unhappy.

(v) Being rich is a sufficient condition to being happy. (vi) Being rich is a necessary condition t o being happy.

(vii) One is never happy when one is rich. (viii) He is poor only if he is happy.

(ix)

(x)

To be rich means the same as to be happy.

He is poor or else he is both rich and happy.

N o t e . Assume “He is poor” is equivalent to -p .

3.19. Determine the t ruth value of each statement.

(i) (ii) I t is not true t h a t 1 + 1 = 2 iff 3 + 4 = 5. (iii) A necessary condition tha t 1 + 2 = 3 is t h a t 4 + 4 = 4. (iv) I t is not t rue tha t 1 + 1 = 5 iff 3 + 3 = 1.

(vi If 3 < 5 , then -3<-5 .

(vi) A sufficient condition tha t 1 1. 2 = 3 is tha t 4 + 4 = 4.

If 5 < 3, then - 3 < -5.

3.20. Determine the t ruth value of each Statement.

(i) It is not t rue tha t if Z T 2 = 4, then 3 + 3 = 5 or 1 + 1 = 2.

(ii, If 2 1 . 2 = 4, then i t is not t rue that 2 + 1 = 3 and 5 + 5 = 10.

(iii) If 2 + 2 = 4, then 3 + 3 = 7 iff 1’1 = 4.

CHAP. 31 CONDITIONAL STATEMENTS 25

3.21. Write the negation of each statement in as simple a sentence as possible. (i) If stock prices fall, then unemployment rises. (ii) He has blond hair if and only if he has blue eyes. (iii) If Marc is rich, then both Eric and Audrey are happy. (iv) Betty smokes Kent or Salem only if she doesn't smoke Camcls. (v) Mary speaks Spanish or French if and only if she speaks Italian. (vi) If John reads Sewsweek then he reads neither Life nor Time.

TRUTH TABLES 3.22. Find the t ruth table of each proposition ii) ( - p v q ) + p , (ii, q e (-q ~ p ) .

3.23. Find the t ruth table of each proposition (i) ( P ++ -cl) + (-11 q ) , (ii) (-q v P ) ++ ( 4 + - P I .

3.24. Find the t ruth table of each proposition: (i) [ p r \ I - q + p ) ] A - ( p - - q ) + ( q v - p ) ] , in) q e ( r - + - p ) ] v [(-q+p)++r].

3.25. Prove: i i i ( p n q ) + v = ( p + r ) v ( q + ~ ) . ( i i i z ~ + ( q + r ) = ( P A - ? ) + -q.

CONDITIONAL XVD VARIATIONS 3.26. Determine the contrapositive of each statement.

(i) If he has courage he will win. (ii) It is necessary to be strong in order to be a sailor. (iii) Only if he does not tire will he win. (iv) It is sufficient for i t to be a square in order to be a rectangle.

3.27. Find: ii) Contrapositive of p + -q. (iii) Contrapositive of the converse of p + -q. ( i i ) Contrapositive of - p + q. iiv) Converse of the contrapositive of - p -+ -q.

3.18.

3.19.

3.20.

3.21.

3.22.

3.23.

3.24.

3.25.

3.26.

Answers to Supplementary Problems (i) P -+ -q (iii) q + -p iv) p q (vii) p 4 -q (ix) P ++ 4 (ii) -p A -q iiv) - p - -q (vii 4 -+ p (viii) -11 + q (x) -P" ( P A 4 )

(i) T, (ii) T, (iii) F , (iv) F, (v) F, (vi) T

(i) F, (ii, F , ( i i i ) T

(i) Stock prices fall and unemployment does not rise. (ii) He has blond hair but does not have blue eyes. (iii) Marc is rich and Eric or Audrey is unhappy. (iv) Betty smokes Kert or Salem, and Camels . (v) Mary speaks Spanish or French, but not Italian. (vi) John reads Sczcsweek, and Life o r T i m e .

(i) TTFF, (ii) F F F T

(i) TFTT, iii) F T F T

(i) F T F F , (ii) TTTFTTFT

Hint. Construct the appropriate t ruth tables.

(i) If he does not win, then he does not have courage. (ii) If he is not strong, then he is not a sailor. (iii) If he tires, then he will not win. (iv) If it is not a rectangle, then i t is not a square.

3.27. (i) q-f - p , iii) -q+ p, (iii) -p+ q, (iv) p + q

Chapter 4

T T T T F F F T T F F T

Arguments, Logical Implication

ARGUMENTS An aiyzrmerzt is an assertion that a given set of propositions P I , P,, . . ., P,I, called

premises, yields (has as a consequence) another proposition Q, called the conclz~sion. Such an argument is denoted by

The truth value of an argument is determined as follows:

[T,] An argument P I , P,, . . . , P,, - Q is true if Q is true whenever all the premises P,, P2, . . . , PII are true; otherwise the argument is false.

Thus an argument is a statement, i.e. has a truth value. If an argument is true it is called a valid argument; if an argument is false it is called a f a l lacy .

Example 1.1: The following argument is valid: 11, 11 + q + Q (Law of Detachment)

The proof of this rule follows from the following t ruth table.

P I Q j P ” 4

26

CHAP. 41 ARGUMEKTS, LOGICAL IMPLICATION

I-, 4 T

T T T

F T T

27

[ ( P + 4) ( n + r ) ] + ( P + r )

T T I T T T T T T T T T T T F T T T F T F F T T F F T F T T F F F F T T T T T T

T T F F T F T T

F T F F T T F T F F T F T F :~ T F F T F F F F T

F T T T ’ T T

F F T F T F T F T T T F T T F F F F T F T F T F T F T F

Example 1.4: The following argument is a fallacy:

P’q, -P t. -q

For the proposition table below.

l ( p + q ) 1 -p] .+ -q is not a tautology, as seen in the t ruth l l 2’;q ~ ; ( p + q ; A - l ’ l ; ~ [ ( p + q ) i - p l + - q

F T F F F T T T T T

Equivalently, the argument is a fallacy since, in Case (line) 3 of the t ruth table, p + q and - p are t rue but -q is false.

An argument can also be shown t o be valid by using previous results as illustrated in the next example.

Example 1.5: We prove tha t the argument p + -q, q t - p is valid:

Statement Reason

(1) q is true. (1) Given ( 2 ) p + - q is true. ( 2 ) Given (3) q + - p is true. (3) Contrapositive of (2) (4) -p is true. (4) Law of Detachment (Example 1.11

using (1) and (3)

ARGUMENTS AND STATEMENTS We now apply the theory of the preceding section to arguments involving specific

statements. We emphasize that the validity of a n argument does not depend upon the truth values nor the content of the statements appearing in the argument, but upon the particular form of the argument. This is illustrated in the following examples.

Example 2.1 : Consider the following argument:

S,: If a man is a bachelor, he is unhappy. S p : If a man is unhappy, he dies young.

S : Bachelors die young. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Here the statement S below the line denotes the conclusion of the argument, and the statements S1 and S, above the line denote the premises. We claim that the argument S1,S2 + S is valid. For the argument is of the form

p + q , q + r t. P + T

where p is “He is a bachelor”, q is “He is unhappy” and T is “He dies young”; and by Example 1.3 this argument (Law of Syllogism) is valid.

28 ARGUMENTS, LOGICAL IMPLICATION [CHAP. 4

Example 2.2: We claim tha t the following argument is not valid:

S1: If two sides of a triangle a re equal, then the opposite angles are equal. SB: Two sides of a triangle a r e not equal.

S: The opposite angles a r e not equal. ......................................................................

F o r the argument is of the form p - t q , - p c -q, where p is “Two sides of a triangle a r e equal” and q is “The opposite angles a re equal”; and by Example 1.4 this argument is a fallacy.

Although the conclusion S does follow from S2 and axioms of Euclidean geometry, the above argument does not constitute such a proof since the argument is a fallacy.

We claim tha t the following argument is valid: Example 2.3:

S1: If 5 is a prime number, then 5 does not divide 15. Sy: 5 divides 15.

............................................. S: 5 is not a prime number.

F o r the argument is of the form p + -q, q t- - p where p is “ 5 is a prime number” and q is “ 5 divides 15”; and we proved this argument is valid in Ex- ample 1.5.

We remark that although the conclusion here is obviously a false statement, the argument as given is still valid. It is because of the false premise Sl that we can logically arrive a t the false conclusion.

Determine the validity of the following argument: Example 2.4: S1: If 7 is less than 4, then 7 is not a prime number. Sy: 7 is not less than 4.

S: 7 is a prime number. ..................................................

We translate the argument into symbolic form. Let p be “7 is less than 4” and q be “7 is a prime number”. Then the argument is of the form

p - t - q , -P + 4 T T

The argument is a fallacy since in Case (line) T 4 of the adjacent t ruth table, p + -q and - p a r e t rue but q is false.

is irrelevant to the fact tha t the argument is a fallacy. The fact tha t the conclusion of the argument happens to be a t rue statement

LOGICAL IMPLICATION A proposition P(p, q, . . .) is said to logically imply a proposition Q ( p , q , . . . ) if

Q ( p , q, . . . ) is true whenever P(p, q, . . . ) is true.

Example 3.1: We claim tha t p logically implies p v q. For consider the t ru th tables of p and p v q in the adjacent table. Observe tha t p is t rue in Cases (lines) 1 and 2, and in these Cases p v q is also true. In other words, p logically implies P V Q . F

Now if Q ( p , 4, . . .) is true whenever P ( p , q, . . .) is true, then the argument

P(P, Q, . . .) I- Q@, 4, . . . ) is valid; and conversely. Furthermore, the argument P c Q is valid if and only if the conditional statement P+ Q is always true, i.e. a tautology. We state this result formally.

CHAP. 41 ARGUMENTS, LOGICAL IMPLICATION 29

P Q

T T

F F Step

Theorem 4.2: The proposition P(p, q , . . . ) logically implies the proposition & ( p , q, . . .) if and only if

(i) the argument P(p, q , . . . ) - Q ( p , q, . . . ) is valid or, equivalently,

(ii) the proposition P(p , q , . . . I + Q ( p , q, . . . ) is a tautology.

[ ( P - 4 ) A - 41 + - P

T T T F F T T ’ F T T F T F F F T F T I F T F T F T T F F T T T F

F T F T T F T I T F

1 2 1 3 2 1 4 2 1

Remark: The reader should be warned that logicians and many texts use the word “implies” in the same sense as we use “logically implies”, and so they distinguish between “implies” and “if . . . then”. These two distinct concepts are, of course, intimately related as seen in the above Theorem 4.2.

Solved Problems ARGUMENTS

F t F F T

4.1. Show that the following argument is valid: p - q , 4 t 11.

Method 1. Now p e p is t rue in Cases

(lines) 1 and 4, and Q is t rue in Cases 1 and 3: hence p w q and q are true simultaneously only in Case 1 where p is also true. Thus the argument p* q, q F p is valid.

Construct the t ruth table on the right.

Method 2. Construct the t ruth table of [ ( p t t q ) A 91 .+ p :

Since [ ( p w 4 ) A 91 --f p is a tautology, the argument is valid.

4.2.

4.3. Determine the validity of the argument -p+ q, p t -q. Construct the t ruth table of [(-]I+ q ) A py - q :

30 ARGUMENTS, LOGICAL IMPLICATION [CHAP. 4

F

T F F F T T F F T

T F F

F

~i

T T T

Since the proposition [ ( - p + q ) A ?I’ + -q is not a tautology, the argument -p + q, p + -q is a fallacy.

Observe that - p + q and p are both true in Case (line) 1 but in this Case -q is false.

4.4. Prove that the following argument is valid: p + -q, T + q , r I- y w . Method 1. Construct the following t ruth tables:

~ ?, 1 4 I r 1 p + - q 1 r + q 1 - p

T ‘ F T F F I F

; 1 ; T T F T

Now p + -y, 7 + q and r are true siinultaneously only in Case (line) 5 , where - p is also true; hence the given argument is valid.

Method 2. Construct the t ruth table for the proposition

[ ( P + - Q ) A (‘)“-’q) A r ] + -P

I t will be a tautology, and so the given argument is valid.

Method 3. Statement

(1) p + - y is true. (1)

( 2 ) r+q is true. ( 2 )

(3) -q + -7- is true. (3)

(4) p + - ~ is true. (4) ( 5 ) r + - p is true. ( 5 )

(6) r is true. (6)

(7) Hence -p is true. (7)

Reason

Given

Given

Contrapositive of (2)

Law of Syllogism, using (1) and (3)

Contrapositive of (4)

Given

Law of Detachment, using (5) and (6)

ARGUMENTS AND STATEMENTS 4.5. Test the validity of each argument:

(i) If it rains, Erik will be sick. It did not rain.

Erik was not sick.

(ii) If i t rains, Erik will be sick. Erik was not sick.

It did not rain.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * . , . . . . . . . . . . . . . .

Firs t translate the arguments into symbolic form:

(i) P + Y, -11 i- -q (ii) P -+ q, -q I- -P

where p is “It rains” and q is “Erik is sick”. By Example 1.4, the argument (i) is a fallacy; by Problem 4.2, the argument ( i i ) is valid.

CHAP. 41 ARGUMENTS, LOGICAL IMPLICATION

P 9 7, P’4

T T T T T T F T T F T F T F F F F T T T F T F T

31

9 “ r

T T T F T T T F

4.6. Test the validity of the following argument:

F F F F T T T T

If 6 is not even, then 5 is not prime. But 6 is even.

F T F T T T T T

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Therefore 5 is prime.

Translate the argument into symbolic form. Let p be Then the argument is “6 is even” and let q be “5 is prime.”

of the form - p - + - q , P t- 4

Now in the adjacent t ruth table, -11 + -q and p are both t rue in Case (line) 2; but in this Case q is false. Hence the argument is a fallacy.

F T ,

The argument can also be shown to be a fallacy by constructing the t ruth table of the [ ( - p + -9 ) ~ p / + q and observing tha t the proposition is not a tautology.

The fact that the conclusion is a true statement does not affect the fact tha t the argument proposition

is a fallacy.

4.7. Test the validity of the following argument: If I like mathematics, then I will study. Either I study or I fail.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . If I fail, then I do not like mathematics.

F i rs t translate the argument into synibolic form. Let p be “I like mathematics”, q be “I study” and r be “I fail”. Then the given argument is of the form

p + q , q v r F r + - p

To test the validity of the argument, construct the t ruth tables of the propositions p + q , q v r and ~ + - p :

-11 1 ? ’+ - p

Recall that an argument is valid if the conclusion is t rue whenever the premises are true. Now in Case (line) 1 of the above t ruth table, the premises p + q and qv 7‘ are both true but the conclusion 7.+ - p is false; hence the argument is a fallacy.

4.8. Test the validity of the following argument: If I study, then I ~7ill not fail mathematics. If I do not play basketball, then I will study. But I failed mathematics.

Therefore, I played basketball.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

32 ARGCRIENTS, LOGICAL IMPLICATION [CHAP. 4

p 1 q I r ’ -q

T ‘ T I F F

T F ~ F I T

F F ~ T ~ T

T T T I F

T I F I T I T

F T ~ T F

F I F F T

F I T F F

p + - q - r - r + P

F F T F T T T

F T T T T T F T F T T

LOGICAL IMPLICATION 4.9. Show that p 4 (I logically implies p cs 4.

Construct the t ruth table for ( p A q ) + (p q ) : l ; P i ‘ p i q ‘ ( p A q ) i ( p - q )

F F F T

Since ( I J q ) - ( p ++ q ) is a tautology, p q logically implies p tf q.

: I : F T

P I F

4.10. Show that p tf (r logically implies p + q. Consider the t ruth tables of p - g and p + q :

1’ I 4 I P - q I P - ) Q

T T F F F T T T

4.11. Prove: Let P(p,q, . . .) logically imply Q ( p , q , . . .). Then for any propositions P I , P,, . . . , P(P,, P,, . . .) logically implies Q(P,, P,, . , ,).

By Theorem 4.2, if P ( p , q , . . . ) logically implies Q ( p , q , . . . ) then the proposition P ( p , q , . . . ) -t Q ( p , q , , . . ) is a tautology. By the Principle of Substitution (Theorem 2.2), the proposition P(P, , P 2 , , . , I + Q(Pl,P,, . . .) is also a tautology. Accordingly, P(P, , P,, . . .) logically implies QIP,, P,, . . . ).

CHAP. 41 ARGUMEKTS. LOGICAL IMPLICATION 33

T ' T P F

A T F

T F T T F T I F T

4.12. Determine the number of nonequivalent propositions P ( p , q ) which logically imply the proposition 11 ++ q.

TI- F F F T F

Consdc,r the adjacent t ru th tahle of p ++ q . Now P ( p , (I) logically implies p t-t (J if IJ H q is t rue whenever P ( p , qi is true. q IS t rue only in Cases linesi 1 and 4; hence Pi7i, q ) c a n n o t be t rue in Cases 2 anti 3. There are foul, iuch propositions which a l e listed below.

But p

F T

T F T T

4.13. Show that pt-t-(1 does not logically imply T I + q.

Method I . Construct the t ruth tables of 21 cf -q and p + q : I

2' 1 r! I - r l1 p - - q I l , + q

Supplementary Problems ARGUMENTS

4.14. Test the validity of each argument: l i ) - ] I + q , p t- -q; (ii) -11- - q , q I- p .

4.15. Test the validity of each argument, ti1 11- 9, r ' + -q t- Y - - p ; ( i i ) p+ -q , - r+ -q t- p + - T .

4.16. Test the validity of each argument: r i I li+ - 4 , Y--' p , q t- - r : iii) p+ q , Y V -q , -r I- - p .

ARGUMENTS AND STATERIENTS

1.17. Test the validity of the argument:

If London is not in Denmark, then Paris is not in France.

But Paris is i n France. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Therefore, London is in Denmark.

4.18, Test the validity of the argument:

If I study, then I will not fail mathematics.

I did not study.

I failed mathematics.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

34 ARGTJPVIENTS, LOGICAL IMPLICATION [CHAP. 4

4.19. Translate into symbolic form and test the validity of the argument: icr ) If G is even, then 2 does not divide 7.

Either 5 is not prime or 2 divides 7 . But 5 is prime.

Therefore, 6 is odd (not even) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

( b ) On iiiy wife’s birthday, I bring her flowers. Either it’s iny wife’s birthday or I work late. I did not bring my wife flowers today.

Therefore, today I worked late.

If I woi.k, I cannot study. Either I work, o r I pass mathematics. I passed mathematics.

Therefore, I studied.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(c)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

i d ) If I work, I cannot study. Either I study, o r I pass mathematics. I worked. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ‘Therefore, I passed mathematics.

LOGICAL IXIPLICATION 4.20. Show tha t rii p A q logically implies p , (ii) p v q does not logically imply p .

4.21. Show tha t iil q logically implies p - q, (ii) - p logically implies p + q.

4.22. ( q v r ) logically implies ( p ? q ) v r.

4.23. Determine those propositions 15 hich logically imply (i) a tautology, (ii) a contradiction.

4.24.

Show t h a t p

Determine the number of nonequivalent propositions P ( p , q ) which logically imply the proposition p- q , and construct t ru th tables for such propositions (see Problern 4.12).

Answers to Supplementary Problems 4.14. ( I ) fallacy, (11) valid

4.15. ti) valid, (111 fallacy

4.16. (1) valid, i i i i valid

4.17. valid

4.18. fallacy

4.19. ( a ) p - -q, - 7 v q, r + -p ; valid. ( c ) p + - q , p v i , r F y, fallacy. ( b ) p + q , p v I * , - q + r ; valid. ( d ) p + - q , q v T , p + 7 , valid.

4.23. I ) E\ e iy pioposition logically implies a tautology. (ii, Only a contradiction logically implies a conti adiction.

Theie are eight such propositions 4.24.

F F T

F T

Chapter 5

Set Theory

SETS AND ELEMENTS The concept of a set appears in all branches of mathematics. Intuitively, a set is any

well-defined list or collection of objects, and will be denoted by capital letters A, B, X , Y , . . , , The objects comprising the set are called its elements or menzbem and will be denoted by lower case letters a, b , x , y , . . . . The statement ‘‘p is an element of A” or, equivalently, “ p belongs to A” is written

The negation of p E A is written p 6! A.

is to list its members. For example,

p E A

There are essentially two ways to specify a particular set. One way, if it is possible,

A = {a, e , i, 0, u)

denotes the set A whose elements are the letters a,e,i,o,u. Note that the elements are separated by commas and enclosed in braces { 1. The second way is to state those properties which characterize the elements in the set. For example,

B = {x: x i s a n integer, x > 0 }

which reads “B is the set of x such that x is an integer and x is greater than zero,” denotes the set B whose elements are the positive integers. A letter, usually x, is used to denote a typical member of the set; the colon is read as “such that” and the comma as “and”.

Example 1.1: The set B above can also be written as B = {1,2,3, . . .I. Observe tha t -6 4 B , 3 E B and P 4 B.

Example 1.2: The set A above can also be written as A = {Z : x is a letter in the English alphabet, E is a vowel}

Observe tha t b 6Z A , e E A and p fZ A .

Example 1.3: Let E = {x : xz-33x f 2 = 0 ) . In other words, E consists of those numbers which a r e solutions of the equation xz- 3% + 2 = 0, sometimes called the solution set of the given equation. Since the solutions of the equation a re 1 and 2, we could also write E = {1,2}.

Two sets A and B are equal, written A = B, if they consist of the same elements, i.e. if each member of A belongs to B and each member of B belongs to A. The negation of A = B is written A # B.

Example 1.4: Let E = {x : x2- 3x + 2 = 0 } , F = ( 2 , l} and G = {l, 2,2,1,6/3}. Then E = F = G. Observe that a set does not depend on the way in which

i ts elements are displayed. A set remains the same if its elements a r e repeated or rearranged.

35

36 S E T THEORY [CHAP. 5

FINITE AND INFINITE SETS Sets can be finite or infinite. X set is finite if it consists of exactly n different elements,

where 72 is some positive integer; otherwise i t is infinite.

Example 2.1: Let M be the set of the days of the week.

Then M is finite.

Let Y = { 2 , 4, 6, 8 , . , , } .

Let P = {x : s is a river on the earth}. number of rivers on the earth, P is a finite set.

In other words, M = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday]

Example 2.2:

Example 2.3:

Then Y is infinite.

Although it may be difficult to count the

SUBSETS A set A is a subset of a set E or, equivalently, B is a superset of A , written

A C E or B > A iff each element in A also belongs t o E ; that is, x E A implies x E B. A is contailzed in B or B contains A .

We also say that The negation of A c B is written A $! E or B $ A

and states that there is an x E A such that x 6! E .

Example 3.1: Consider the sets A = (1, 3, 5, 7, . . .}, C = {.c : 2 is prime,

Then C c A since every prime number B @ A since 10 E B but 10 @ A .

B = { 5 , 10, 15, 20, . . .I .T > 2 } = {3,5 ,7 ,11 , ...}

greater than 2 is odd. On the other hand,

Example 3.2: Let N denote the set of positive integers, 2 denote the set of integers, Q denote the set of rational numbers and R denote the set of real numbers.

N c Z c Q c R Then

Example 3.3: The set E = { 2 , 4 , 6 ] is a subset of the set F = {6,2,4} , since each number 2, 4 In fact, E = 8’. In a similar manner and 6 belonging to E also belongs to F .

it can be shown that every set is a subset of itself.

As noted in the preceding example, A C B does not exclude the possibility that A = B. In fact, we may restate the definition of equality of sets as follows:

I Definition: I Two sets A and E are equal if A c B and E c A .

In the case that A c B but A = E , we say that A is a proper subset of B or B contains A properly. The reader should be warned that some authors use the symbol C for a subset and the symbol c only for a proper subset.

The following theorem is a consequence of the preceding definitions:

Theorem 5.1: Let A , B and C be sets. Then: A = B; and (iii) if A C E and B C C, then A C C.

(i) A c A ; (ii) if A C B and B C A, then

UNIVERSAL AND NULL SETS In any application of the theory of sets, all sets under investigation are regarded as

subsets of a fixed set. We call this set the universal se t or universe of discourse and denote it (in this chapter) by C.

Example 1.1:

Example 4.2:

In plane geometry, the universal set consists of all the points in the plane.

In human population studies, the universal set consists of all the people in the world.

CHAP. 51 SET THEORY 37

It is also convenient to introduce the concept of the empty or null set, that is, a set This set, denoted by 8, is considered finite and a subset of which contains no elements.

every other set. Thus, fo r any set A , $8 c A C U.

Example 4.3: Let A = {x : x’ = 4, z is odd}. Then A is empty, i.e. A = @.

Example 4.4: Let B be the set of people in the world who a r e older than 200 years. According to known statistics, B is the null set.

CLASS, COLLECTION, FAMILY Frequently, the members of a set are sets themselves. For example, each line in a set

of lines is a set of points. To help clarify these situations, other words, such as “class”, “collection” and “family” are used. Usually we use class o r collection for a set of sets, and family fo r a set of classes. The words subclass, subcollection and subfamily have meanings analogous to subset.

Example 5.1: The members of the class {{2,3}, {2}, {5,6}} are the sets {2,3}, {2} and { 5 , S}.

Example 5.2: Consider any set A . The power set of A , denoted by ‘?‘(A) or Z A , is the class of all subsets of A . In particular, if A = {a, b , c}, then

T(A) = { A , {a, b } , {a, c>, ( 6 , c>, {a} , Cb), {c>, @I In general, if A is finite and has n elements, then “ ( A ) will have 2n elements.

SET OPERATIONS

belong to A o r to B:

Here “or” is used in the sense of andlor.

belong to both A and B:

If A n B = 0, that is, if A and B do not have any elements in common, then A and B are said to be disjoint or non-intersecting.

The relative co?nplernent of a set 5’ with respect to a set A or, simply, the difference of A and B, denoted by A\B, is the set of elements which belong to A but which do not belong to B:

A\B = { n : : x E A , x4B} Observe that A \B and B are disjoint, i.e. (A\B) n B = $3.

of elements which do not belong to A:

The union of two sets A and B, denoted by A U B, is the set of all elements which

A U B = { x : x E A or x E B }

The intersection of two sets A and B, denoted by A n B, is the set of elements which

A n B = { x : x E A and x E B }

The absolute complement or, simply, complement of a set A , denoted by Ac, is the set

A“ = { x : x E U , x 4 A )

That is, Ac is the difference of the universal set U and A .

Example 6.1 : The following diagrams, called Venn diagrams, illustrate the above set operations. Here sets are represented by simple plane areas and U , the universal set, by the area in the entire rectangle.

38 SET THEORY [CHAP. 5

A U B is shaded. A n B is shaded.

A\B is shaded. A c is shaded.

Example 6.2: Let A = {1,2, 3 ,4 ) and B = {3,4,5, 6} where 7,’ = {1,2,3, . . .}. Then:

A U B = (1, 2, 3, 4, 5, 6} A n B = (3, 4)

A\B = {1,2} Ac = (5, 6 , 7, . ..} Sets under the above operations satisfy various laws o r identities which are listed in

Table 5.1 below.

Theorem 5.2:

In fact we state:

Sets satisfy the laws in Table 5.1.

LAWS OF THE $LGEBRA OF SETS

Idempotent Laws l a . A U A = A lb. A n A = A

Associative Laws 2a. ( A u B ) u C = A u ( B U C ) 2b. ( A n B ) n C = A n ( B n C )

Commutative Laws 3a. A U B = B u A 3b. A n B = B n A

Distributive Laws 4a. A u ( B n C ) = ( A u B ) n ( A u C ) 4b. A n ( B u C ) = ( A n B ) u ( A n C )

5a. A u @ = A 6a. A U U = U

Identity Laws 5b. A n U = A 6b. A n @ = @

Complement Laws 7a. A u A c = U 7b. A n A c = @

8a. (Ac)c = A 8b. Uc = 13, @c = U

De Morgan’s Laws 9a. ( A u B ) ~ = AcnBc 9b. ( A nB)c = AcuBc

Table 5.1

Remark: Each of the above laws follows from the analogous logical law in Table 2.1, Page 11. For example,

A n B = { x : x E A and z E B } = { x : x E B a n d x E A } = B n A


Here me use the fact that if 11 is I E A and q is x E B, then p A q is logically equivalent to q ~ p : P A C ! = q ~ p .

Lastly we state the relationship between set inclusion and the above set operations:

Theorem 5.3: Each of the following conditions is equivalent t o A c B: (i) A n B = A (iii) Bc c Ac (v) B u A‘ = Lr (ii) A U B = B (iv) A n Br = $3

ARGUMENTS AND VENN DIAGRAMS Many verbal statements can be translated into equivalent statements about sets which

can be described by Venn diagrams. Hence Venn diagrams are very often used to determine the validity of an argument.

Example 7.1 : Consider the following argument:

S1: Babies are illogical.

Sp: Nobody is despised who can manage a crocodile. S,: Illogical people a r e despised.

S: Babies cannot manage crocodiles.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(The above argument is adapted from Lewis Carroll, Sumbolic Logic; he is also the author of Alice in WoizclerLa?zcZ.) Now by S1, the set of babies is a subset of the set of illogical people:

babies

By S,, the set of illogical people is contained in the set of despised people:

Furthermore, by S2 , the set of despised people and the set of people who can manage a crocodile a r e disjoint:

despised people

illogical people people who can manage

crocodiles

But by the above Venn diagram, the set of babies is disjoint f rom the set of people who can manage crocodiles, o r “Babies cannot manage crocodiles” is a consequence of S,, S , and S3.

is valid.

Thus the above argument,

s,, s,, s 3 + s

40 S E T THEORY [CHAP. 5

Solved Problems SETS, ELEMENTS 5.1. Let A = (x: 3x = 6) . Does A = 2 ?

A is the set which consists of the single element 2, that is, A = {2}. The number 2 belongs t o A ; it does not equal A . There is a basic difference between a n element p and the singleton set { p } .

5.2. Which of these sets are equal: { T , t , s } , {s, t, Y,S}, { t , s, t , r } , {s , r ,s , t } ? They are all equal. Order and repetition do not change a set.

5.3. M'hich of the following sets are finite? (i)

(iii) The number of people living on the earth. The first three sets a re finite; the last two sets a r e infinite.

The months of the year. (iv) {x : x is an even number) (ii) (1, 2, 3, . . ., 99, loo} (v) (1, 2, 3, * . .>

5.4. Determine which of the following sets are equal: 8, (01, {g} . Each is different from the other. The set ( 0 ) contains one element, the number zero. The

set 0 contains no elements; it is the empty set. The set (0) also contains one element, the null set.

5.5. Determine whether or not each set is the null set: (i) X = {x: x2 = 9, 2x = 4}, (ii) Y = {x: x # x}, (iii) 2 = {x: x + 8 = 8 ) .

(i) There is no number which satisfies both x2 = 9 and 2 2 = 4; hence X is empty, i.e. X = 0.

(ii) We assume tha t any object is itself, so Y is also empty. set as follows:

In fact, some texts define the null

$3 E { x : x # x )

(iii) The number zero satisfies z + 8 = 8; hence 2 = (0). Accordingly, 2 is not the empty set since i t contains 0. That is, Z # @ .

SUBSETS 5.6. Prove that A = {2,3,4,5} is not a subset of B = ( x : x is even}.

It is necessary to show that at least one element in A does not belong to B. Now 3 E A and, since B consists of even numbers, 3 6? B ; hence A is not a subset of B.

5.7. Prove Theorem 5.l(iii): We must show t h a t each element in A also belongs to C. Let x E A . Now A c B implies

x E B. But B c C; hence x E C. We have shown t h a t x E A implies x E C, t h a t is, that A c C.

If A C B and B c C , then A c C.

5.8. Find the power set "(8) of the set S = { 1,2,3} . {l}, {2}, (3) and the empty set 0. Hence

The power set T(S) of S is the class of all subsets of S; these a re {1,2,3), {1,2}, {1 ,3) , {2 ,3} ,

T(S) = { S , C1,3}, {2,3}, {1,2}, C1), {2>, { 3 ) ,

Note tha t there a re 23 = 8 subsets of S.


5.9. Let V = ( d ) , W = { c , d } , X = ( a , b , c ) , Y = { a , b } and 2 = ( a , b , d } . Determine whether each statement is true or false: (i) Y c X , ( i i ) W + Z , (iii) Z 3 17, ( iv) V c X , (v) X = W , ( ~ 7 i ) W c Y .

(i) Since each element in Y is a member of X , Y c X is true.

(ii) Now a E 2 but (-1 4 W ; hence W # 2 is true.

(iii) The only element in V is d and it also belongs to 2; hence Z > V is true.

(iv) V is not a subset of X since d E V but d 4 X ; hence V C X is false.

(v) Now u E X but a 4 W ; hence X = W is false.

(vi) W is not a subset of Y since c E 68 but c 4 Y ; hence W C Y is false.

5.10. Prove: If A is a subset of the empty set @, then A = @ . The null set 8 is a subset of every set; in particular, @ C A . But, by hypothesis, A c $3;

hence A = 0.

SET OPERATIONS 5.11. Let U = ( l , 2 , . . . , 8,9}, A = { l , 2 , 3 , 4 } , B = [2,4,6,8} and C={3,4,5,6}. Find:

(i) Ac, (ii) A n C, (iii) ( A n C ) [ , (iv) A u B, (v) B\C.

(i)

(ii)

(iii) (AnC)c consists of the elements in C that a re not in A n C . Now by iii), A n C = {3 ,4 ) and so (AnCC)c = {1,2,5,6,7,8,9}.

(iv) A U B consists of the elements in A or B (o r both): hence A LB = { 1 , 2 , 3 , 4 , 6 , 8}.

(v) B\C consists of the elements in B which a r e not in C ; hence B\C = {S,S}.

Ac consists of the elements in C that are not in A ; hence Ac = {5,6, 7,8,9}.

A n C consists of the elements in both A and C; hence A n C = {3 ,4} .

5.12. In each Venn diagram below, shade: [ i ) A U B, (ii) A n B.

(i) A L E consists of those elements which belong to A or B (or both); hence shade the area in A and in B as follows:

A U B is shaded.

(ii) A n B consists of the area tha t is common to both A and B . To compute A n B, first shade A with strokes slanting upward t o the right (////) and then shade B with strokes slanting downward t o the r ight (\\\\), as follows:

Then A n E consists of the cross-hatched area which is shaded below:

A n B is shaded.

Observe the following: ( a ) A nE is empty if A and E are disjoint. ( b ) A r E = B if B c A . i c ) A , - I B = A if A C E .

5.13. In the Venn diagram below, shade: (i) B", (ii) ( A C B ) < , (iii) (B\A)c, ( iv) A' n B c .

( i l Bc consists of the elements which do not belong to B: hence shade the area outside B a s follows:

EC is shaded.

(ii) Firs t shade A U B ; then iA1-Bj' is the area outside A I J B :

A U B is shaded. ( A U E)' is shaded.

(iii, Firs t shade B\A, the area in B which does not lie in A ; then (B\A)c is the area outside E\A.

B \A is shaded. ( E \ A)c is shaded.

(ivr Firs t shade Ac, the area outside of A , with strokes slanting upward to the right ( / / I / ) , and then shade Bc with strokes slanting downward t o the right ( \ \ \ \ I ; then Ac0iBc is the cross-hatched area:

CHAP. 51 S E T THEORY 43

Ac and Bc a r e shaded.

Observe that ( A LIB)" = ACnBc, a s expected by De Morgan's law.

AcnBc is shaded.

5.14. In the Venn diagram below, shade (i) A n ( B U C), (ii) ( A n B ) U ( A n C).

(i) Firs t shade A with upward slanted strokes, and then shade B U C with downward slanted strokes; now A n (BU C) is the cross-hatched area:

A and BUC a r e shaded. A n ( B u C ) is shaded.

(ii) Firs t shade A nB with upward slanted strokes, and then shade A n C with downward slanted strokes; now ( A n B ) U ( A n C) is the total area shaded:

A n B and A 17 C are shaded. ( A n B) U ( A n C) is shaded.

Notice tha t A n ( B u C) = ( A n B ) u (A n C), as expected by the distributive law.

5.15. Prove: B\A = B n A". Thus the set operation of difference can be written in terms of the operations of intersection and complementation.

B\A = {x: x E B , . , ;@A) = {x: x E B , x E A C } = BnAc


5.16. Prove the Distributive Law: A n ( B U C) = ( A n B) U ( A n C). A n ( B u C ) 17 {x : x € A ; x E B u C }

= { x : x E A ; x € B o r x E C }

=x {x: % € A , x E B ; o r x E A , x E C } = { r e : x E A n B or a E A n C } = ( A n B ) u ( A n C )

Observe tha t in the third step above we used the analogous logical law

P 4 ( Q " 4 = (P A 4) " ( P A 4

5.17. Prove: ( A \B) n B = @. ( A \ B ) n B = {x: xEA\B, x E B }

= {x: x E A , x4B; z E B }

= ! a The last step follows from the fact tha t there is no element 2 satisfying z E B and z 48.

5.18. Prove De Morgan's Law: (AUB)c = AcnBc. ( A u B ) ~ = {x: x 4 A u B }

= { x : x 4 A , x 4 B } = {x: x E A c , x E B C }

= AcnBc

Observe that in the second step above we used the analogous logical law

" ( p V 4) -p A -q

5.19. Prove: For any sets A and B, A nB C A C A U B .

Let z E A n B ; then x E A and x E B. In particular, x E A . Since x E A n B implies x € A , A n B c A. Furthermore, if z E A , then x E A or x E B , i.e. 3: E A u B . Hence A c AUB. In other words, A n B c A c A u B .

5.20. Prove Theorem 5.3(i): A c B if and onlyif A n B = A . Suppose A cB. Let x E A ; then by hypothesis, x E B. Hence x € A and x E B, i.e. x E A n B.

Accordingly, A c A nB. On the other hand, i t i s always t rue (Problem 5.19) tha t A n B C A. Thus A n B = A .

Now suppose t h a t A n B = A . Then in particular, A C A nB. But i t is always t rue tha t A nB c B. Thus A c A n B c B and so, by Theorem 5.1, A c B.

ARGUMENTS AND VENN DIAGRAMS 5.21. Show that the following argument is not valid by constructing a Venn diagram in

which the premises hold but the conclusion does not hold: Some students are lazy. All males are lazy.

Some students are males. .....................

Consider the following Venn diagram:


lazy people

students males

Notice tha t both premises hold, but the conclusion does not hold. For an argument to be valid, the conclusion must always be t rue whenever the premises a re true. Since the diagram represents a case in which the conclusion is false, even though the premises are true, the argument is false.

It is possible to construct a Venn diagram in which the premises and conclusion hold, such as

lazy people

students males

5.22. Show that the following argument is not valid: All students are lazy. Nobody who is wealthy is a student.

Lazy people are not wealthy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Consider the following Venn diagram:

wealthy people

Now the premises hold in the above diagram, but the conclusion does not hold; hence the argument is not valid.

5.23. For the following set of premises, find a conclusion such that the argument is valid: S,: All lawyers are wealthy. S,: Poets are temperamental. S,: No temperamental person is wealthy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S:

By S,, the set of lawyers is a subset of the set of wealthy people: and by S3, the set of wealthy people and the set of temperamental people are disjoint. Thus

temperamental people

46 S E T THEORY CHAP. 5

By Sy, the set of poets is a subset of the set of temperamental people; hence

Thus the statement “No poet is a lawyer’’ o r equivalently “No lawyer is a poet” is a valid

The statements “NO poet is wealthy” and “No lawyer is temperamental” are also valid conclusion.

conclusions which do not make use of all the premises.

MISCELLANEOUS PROBLEMS 5.24. Let A = (2, (4,5}, 4). Which statements are incorrect and why?

(i) (4 ,5) C A , (ii) {4,5} E A , (iii) ((4,5}} C A. The elements of A are 2, 1 and the set {4,5}. Therefore (iii is correct, but (i) is an incorrect

statement. Furthermore, (iii) is also a correct statement since the set consisting of the single element {4, 5 ) which belongs to A is a subset of A .

5.25. Let A = (2, {4,5}, 4}. Which statements are incorrect and why? (il 5 E A, (ii) (5) E A , jiii) {5) c A.

Each statement is incorrect. The elements of A are 2 , 4 and the set {4,5}; hence ( i i and (ii) are incorrect. There a re eight subsets of A and (5) is not one of them; so (iii) is also incorrect.

5.26. Find the power set T(S) of the set S = (3, {1 ,4)} . Note first tha t S contains two elements, 3 and the set {1,4). Therefore T(S) contains

22 = 3 elements: S itself, the empty set (3, and the two singleton sets which contain the elements 3 and { 1 , 4 } respectively, i.e. {3} and {{1,4}}. In other words,

Tu(S) = {S, {3}, {{L4}}> c)}

Supplementary Problems

SETS, SUBSETS 5.27. Let A = {1 ,2 ,..., 8,9}, B = {2,4,6,8}, C = {1,3,5,7,9] , D = {3,4,5} and E = { 3 , 5 } .

(iv) x C c but x # A . Which sets can equal X if we are given the following information? (i) X and E are disjoint. (ii) X c D but X # B. (iii) X C A but x’ C.

5.28. State whether each statement is t rue or false: (ii Every subset of a finite set is finite. rii) Every subset of an infinite set is infinite.

5.29. Find the power set “ ( A ) of A = (1 ,2 ,3 ,4} and the power set T(i3) of B = {l, { 2 , 3 ) , 4 ) .

CHAP. 51 S E T THEORY 47

5.30. State whether each set is finite or infinite:

(i) The set of lines parallel to the 7 axis. (ii) The set of letters in the English alphabet. (iii) The set of numbers which a re multiples of 5. (iv) The set of animals living on the earth. (v) The set of numbers which are solutions of the equation x 2 i + 2 6 ~ 1 ~ - 17x11 + 7x3 - 10 = 0. (vi) The set of circles through the origin (0,O).

5.31. State whether each statement is t rue o r false:

(i) {1,4,3] = { 3 , 4 , 1 ] (iv) C4) E CC4)) (ii) {1,3,1,2, 3 ,2 ) c {I, 2,3} (v) {4} c CC4)) (iii) {1 ,2 } = { 2 , 1 , 1 , 2 , 1 ] (vi) 8 c CC43)

SET OPERATIONS 5.32. Let U = { u , b , c , d , e , f , g } , A = { a , b , c , d , e ) , B = { a , c , e , g } and C = { b , e , f , g } . Find:

( i ] AUC (iii) C \ B (v) C c n A ivii) ( A \ B c ) c

iiir B n A (iv) B C ~ C (vi) ( A \ C P iviii) ( A n A c ) c

5.33. In the Venn diagrams below, shade ( i ) W\V (ii) VcUW (iiii V n W c (iv) Vc\Wc.

5.34. Draw a Venn diagram of three non-empty sets A , B and C so tha t A , B and C have the following properties:

( i ) A c B , C c B , AnC = 8 (iii) A c C, A # C, B n C = $3 (ii, A C E , CQB, AnC # 0 (iv) A c ( B n C ) , B c C , C I B , A Z C

5.35. The formula A \ B = A nBc defines the difference operation in terms of the operations of Find a formula that defines the union of two sets, A U B, in terms of intersection and complement.

the operations of intersection and complement.

5.36. Prove Theorem 5.3iiil: A c B if and only if A U B = B.

5.37. Prove: If A T E = (3, then A c B c .

5.38. Prove: Ac \ Bc = b' \A.

5.39. Prove: A c B implies A u ( B \A) = B.

5.40. (i) Prove: An(B\C) = ( A n B ) \(A?C).

(ii) Give an example to show tha t A ( B \ C) f ( A U B ) \ ( A J C ) .

-ARGUMENTS AND VENS DIAGRAMS 5.41. Determine the validity of each argument for each proposed conclusion.

No college professor is wealthy. Some poets a re wealthy.

(i) Some poets a re college professors. (ii) Some poets a r e not college professors.

SET THEORY [CHAP. 5 48

5.42.

5.43.

5.44.

Determine the validity of each arg~iment for each proposed conclusion.

All poets a r e interesting people. Audrey is an interesting person.

i i I Audrey is a poet. i i i i Audrey is not a poet.

Determine the validity of the argument fo r each proposed conclusion. All poets a re poor. In order to be a teacher, one must graduate from college. No college graduate is poor.

(i) Teachers a r e not poor. (ii) Poets are not teachers.

(iii) If Marc is a college graduate, then he is not a poet.

Determine the validity of the argument fo r each proposed conclusion. A11 mathematicians are interesting people. Some teachers sell insurance. Only uninteresting people become insurance salesmen.

Insurance salesnien a re not mathematicians. Some interesting people a r e not teachers. Some teachers are not interesting people. Some mathematicians a r e teachers. Some teachers a re not mathematicians. If Eric is a mathematician, then he does not sell insurance.

Answers to Supplementary Problems 5.27. ( i ) C and E , (ii) D and E , [iii) A , E and D, (iv) None

5.28. ( i ) T, (ii, F

5.30. (i) infinite, (ii) finite, (iii) infinite, (iv) finite, (v) finite, (v i ) infinite

5.31. (i) T, (ii) T, (iii) T, (iv) T, (v) F, (vi) T

5.32. (i) A U C = U (vj CcnA = {a,c ,d: = CL (ii) B n A = { a , c , e } (vi) ( A \ C). = { b , e , f, d (iii) C \ B = {b , f} (vii) ( A \Bc)c = { b , c I , t’, gl

(iv) BCUC = { b , d , e , f , g } (viii) (AnAc)c = U

5.33. ( a )


W\V

Observe that V c u W = U and V n W C = 0 in case ( b ) where V C VV.

5.34. (i) @ (ii)

(iii) @@

5.35. A u B = (AcnBc)C

5.40. (ii)

A L ( E \ C) is shaded. ( A u B ) \ ( A u C) is shaded.

5.41. (i) fallacy, (iii valid

5.42. (i) fallacy, (iij fallacy

5.43. (i) valid, (ii, valid, iiii) valid

5.44. (i) valid, (ii) fallacy, riii) valid, (iv) fallacy, ivi valid, (vi) valid

The following Venn diagrams show why (ii) and (iv) a r e fallacies:

interesting

insurance

mathematicians 0 teachers u

Chapter 6 Product Sets

ORDERED PAIRS

is designated as the first element and the other as the second element. pair is written

Two ordered pairs (a, b) and (c, d ) are equal if and only if a = c and b = d.

An ordeyed pair consists of two elements, say a and 0 , in which one of them, say a, Such an ordered

(a, b )

Example 1.1:

Example 1.2:

The ordered pairs ( 2 , 3 ) and (3 ,2 ) are different.

The points in the Cartesian plane in Fig. 6-1 below represent ordered pairs of real numbers.

The set {2 ,3} is not an ordered pair since the elements 2 and 3 are not distinguished.

Ordered pairs can have the same first and second elements, such as (1, l), (4,4) and

Example 1.3:

Example 1.4: (5,5).

Remark: An ordered pair (u ,b) can be defined rigorously as follows:

the key here being that {a} c { a , b J which we use to distinguish a first element.

From this definition, the fundamental property of ordered pairs

(a, b) = { @ I t @ , b ) } as the

can be proven:

(a, b ) = (c, d ) if and only if u = c and b = d

PRODUCT SETS

A X B, consists of all ordered pairs (a, b ) where a E A and b E B:

The product of a set with itself, say A X A, is sometimes denoted by A2.

Let A and B be two sets. The iwoduct set (or Cartesian product ) of A and B, written

A x B = ( (a ,b ) : a E A, b E B }

Example 2.1: The reader is familiar with the Cartesian plane R2 = R X R (Fig. 6-1 below). Here each point P represents a n ordered pair (u , b ) of real numbers and vice versa; the vertical line through P meets the x axis a t a , and the horizontal line through P meets the 7~ axis at b.

t -1

J::

B P b

a

A 1 2 8

Fig. 6-1 Fig. 6-2

50

CHAP. 61 PRODUCT SETS

T I T T T F T F T T F F F T T F T F F F T F F F

51

T

Since A and B do not contain many elements, it is possible to represent A X B by a coordinate diagram as shown in Fig. 6-2 above. Here the vertical lines through the points of A and the horizontal lines through the points of B meet in 6 points which represent A X B in the obvious way. The point P is the ordered pair (2, b) .

In general, if a finite set A has s elements and a finite set B has t elements, then A x B has s times t elements. If either A or B is empty, then A X B is empty. Lastly, if either A or B is infinite, and the other is not empty, then A x B is also infinite.

PRODUCT SETS I N GENERAL The concept of a product set can be extended to more than two sets in a natural way.

The Cartesian product of sets A , B, C, denoted by A X B X C, consists of all ordered triplets (a, b, c ) where u E A, 2, E B and c E C:

A x B x C = ( ( c t , b , c ) : u E A , b E B , c E C ]

Analogously, the Cartesian product of n sets A,, A,, . . . , A n , denoted by A, X A, X consists of all ordered n-tuples (a,, a,, . . ., uil) where a, E A,, . . . , an E An:

- X An,

A , x A, x * x An = {(cL,, . . .,an) : a, E A,, . . ., an E An} Here an ordered ?i-tuple has the obvious intuitive meaning, that is, it consists of n elements, not necessarily distinct, in which one of them is designated as the first element, another as the second element, etc. Furthermore,

(a,, . . . ,an) = ( b l , . . ., b J iff a, = b,, . . ., ail = biz

Example 3.1 : In three dimensional Euclidean geometry each point represents a n ordered triplet: its x-component, its y-component and its z-component.

TRUTH SETS OF PROPOSITIONS

value to each of the eight cases below: Recall that any proposition P containing, say, three variables p, (r and r , assigns a truth

P I q l r

52 PRODUCT SETS [CHAP. 6

p + q

T T F F T T T T

-1 The truth set of a proposition P, written S ( P ) , consists of those elements of U for which the proposition P is true.

Example 1.1: Following is the t ruth table of ( p + q ) A ( q + T ) :

q + r ( P + q ) A \ ( Q + r )

T T F F T F T F T T F F T T T T

F T

T

Accordingly, the t ruth set of ( p -+ q) A ( q + T ) is

T ( ( p + q) A ( q + Y)) = {TTT, FTT, FFT, FFF}

The next theorem shows the intimate relationship between the set operations and the logical connectives.

Theorem 6.1: Let P and Q be propositions. Then:

(i) S(P + Q ) = 7 ( P ) n S(Q)

(ii) T ( P \ Q) = T ( P ) U T ( Q )

(iii) T(-P) = (T(P))"

(iv) P logically implies Q if and only if S(P) c 7(Q)

The proof of this theorem follows directly from the definitions of the logical connectives and the set operations.

Solved Problems

ORDERED PAIRS

6.1. Let W = {John, Jim, Tom) and let V = {Betty, Mary}. Find W X V . TV x V consists of all ordered pairs (a, b ) where a E IT' and b E V . Hence,

{(John, Betty), (John, Mary), (Jim, Betty), (Jim, ilIaryi, (Tom, Betty), (Tom, Rlary)} W X V =

6.2. Suppose (x+ y, 1) = (3, n.- y). Find r and y. If ( x - u, 1) = (3, x - y) then, by the fundamental property of ordered pairs,

. r + r / = 3 and l = x - y

The solution of these simultaneous equations is giverrby .2: = 2, y = 1.

CHAP. 61 PRODUCT SETS 53

6.3. Let A = {a, b , c , d , e , f } and B = (a, e , i , o , u } . Determine B the ordered pairs corresponding to the points PI, P,, P , and P, which appear in the coordinate diagram of A X B on the right.

The vertical line through P I crosses the A axis at b and the horizontal line through PI crosses the B axis a t i; hence PI corresponds to the ordered pair ( b , i). Similarly, P , = (a , a) , P , = (d, u) and P, = ( e , e ) .

0

i

a A

a b c d e f

PRODUCT SETS 6.1. Let A = {1,2,3}, B = {2,4} and C = {3,4,5}. Find AX B X C.

A convenient method of finding A X B >; C is through the so-called “tree diagram” shown below:

The “tree” is constructed from the left to the right. listed to the r ight of the “tree”.

A X B X C consists of the ordered triples

6.5. Let A = {a, b ) , B = {2,3) and C = {3,4} . Find: (i) A x ( B u C), (ii) ( A x B ) u ( A x C ) , (iii) A x ( B n C), (iv) ( A X B ) fl ( A X C).

(i) Firs t compute B u C = {2,3,4}. Then A X (BUG‘) = { (a , 21, (a, 3), (a, 4), ( b , 2 ) , ( b , 3), ( b , 4):

(ii) Firs t find A X B and A X C: A x B = { ( a , 2), (a, 3), ( b , 2 ) , ( b , 3 ) ) A x C = { ( a , 31, (a , 4), ( b , 3), ( b , 4))

{ ( a , 21, (a , 3), ( b , 2), ( b , 31, ( a , 41, ( b , 4))

Then compute the union of the two sets:

( A x B ) u ( A x C) = Observe, from (i) and (ii), tha t

A x ( B u C ) = ( A X B ) U ( A X C )

(iii) Firs t compute B r C = {3}. Then A x ( B r, C) = { (a , 31, ( b , 3))

54 PRODUCT SETS [CHAP. 6

(iv) Now 4 X B and A X C were computed above. The intersection of A X B and A X C consists of those ordered pairs which belong t o both sets:

( A x B ) n ( A x C ) = { ( a , 31, ( h 3))

A x ( B n C ) = ( A x B ) n ( A x C ) Observe from (iii) and (iv) that

6.6. Prove: A x (B n C) = ( A x B ) n ( A x C). A x ( B n C ) = {(x,y) : x € A , y E BnC}

= {(x,y) : x E A , y E B , y E C} = {(x, Y) : (x, 4 E A x B , (2, Y) E A x c> = ( A x B) n (A x C)

6.7. Let S = { a , b } , W = { l , 2 , 3 , 4 , 5 , 6 } and V = { 3 , 5 , 7 , 9 } . Find ( S x W ) n ( S X V ) . The product set (S X W ) n ( S K V ) can be found by first computing S X W and S X V , and

then computing the intersection of these sets. On the other hand, by the preceding problem, (S X W ) n (S X V ) = S X ( W n V ) . NOW W n V = {3,5}, and so

(S x W ) n (S x V ) = S x (Wn V ) = { ( a , 31, ( a , 5 ) , ( b , 3), ( b , 5))

6.8. Prove: Let A C B and C C D; then ( A X C) C ( B X D). Let (x,y) be any arbi t rary element in A X C; then x € A and y EC. By hypothesis, A c B and

C c D ; hence x E B and y E D. Accordingly, (x, y) belongs t o B X D. We have shown tha t (x ,y ) € A X C implies (z,y) E B X D ; hence ( A X C) C ( B X D).

TRUTH SETS O F PROPOSITIONS 6.9. Find the truth set of p A -q.

Firs t construct the t ruth table of P A -q:

F

Note tha t p A -q is t rue only in the case t h a t p is t rue and q is false; hence

PA -q) = {TF)

6.10. Find the truth set of - p + q. Firs t construct the t ruth table of - p + q:

F

Now - p + q is t rue in the first three cases; hence

T ( - p + q) = {TT, T F , F T ]

CHAP. 61 PRODUCT SETS

P

55

rl 1 ?’ 1 P ” 4 I ( P V d A Y

T T T T T T T F T F T F T T T T F F T F F T T T T F T F T F F F T F F F F F F F

6.12. Suppose the proposition P = P(p , q, , , . ) is a tautology. Determine the t ru th set T ( P ) of the proposition P.

of P is the universal set: T ( P ) = U . If P is a tautology, then P is true for any t ruth values of its variables. Hence the t ruth set

6.13. Suppose the proposition P = P ( p , g , , . . ) is a contradiction. Determine the t ru th set T ( P ) of the proposition P.

If P is a contradiction, then there is no case in which P is true, i.e. P is false fo r any t ruth values of its variables. Hence the t ru th set of P is empty: T ( P ) = $3,

6.14. Let P = P(p, q , . , .) and Q = Q ( p , q, . . , ) be propositions such tha t P A Q is a contradiction. Show tha t the t ru th sets T ( P ) and T(Q) a re disjoint.

If P A Q is a contradiction, then its t ruth set is empty: T ( P A Q) = 0. Hence, by Theorem 6.1(i),

T ( P ) n 7 ( Q ) = T ( P A Q ) = $3

6.15. Suppose tha t the proposition P = P(p, g, . . .) logically implies the proposition Q = Q ( p , g, . . . ) . Show tha t the t ru th sets T ( P ) and T(-Q) a re disjoint.

Since P logically implies Q, 7 ( P ) is a subset of T(Q). But by Theorem 5.3,

7 ( P ) c T(Q) is equivalent to T ( P ) n ( 7 ( Q ) ) c = @

Furthermore, by Theorem 6.l(iii), ( T ( Q ) ) c = 7(-Q). Hence

7 ( P ) n T ( - Q ) = @

56 PRODUCT S E T S ‘CHAP. 6

Supplementary Problems ORDERED P-IIIRS, PRODUCT SETS 6.16. Suppose (u -2, 2 x t 1) = ( . I . - 1, ~ ~ 2 ) . Find x and y.

6.17. Find the ordered pairs corresponding to the points P I , P,, P , and P, which appear below in the coordinate diagram of {l, 2, 3 , 4 > Y (2 ,4 ,6 , S}. :m

4

2

1 2 3 4

6.18. Let W = {Mark,Eric, Paul: and let V = {Eric,David]. Find: t i ) IT’ v T’, (ii) V X W , (iii) 1’1 = T’X V.

6.19. Let A = {2,3}, B = { 1 , 3 , 5 ) and C = {3,4}. Construct the “tree diagram” of A X E X C, as in Problem 6.4, and then find A X B X C.

6.20. Let S = {a , b , c}, T = { b , c, cl) and W = {a, d } . Construct the tree diagram of S X ?’ A TI.‘ and then find S X T X W .

6.21. Suppose tha t the sets V , W and Z have 3, 4 and 5 elements respectively. Determine the number of elenients in (i) V x W x 2, iii) Z x V X W , (iii) W X 2 X V .

6.22. Let A = B r: C. Determine if either statement is true: rii A A A = ( B x B ) n ( ( C x C ) , (i i) A x A = ( B x C ) ? ( C X B ) .

6.23. Prove: A X (BUG‘) = ( A X B ) u ( A X C).

TRUTH SETS OF PROPOSITIONS 6.2-1. Find the t ruth set of p t) -q ,

6.25. Find the t ru th set of - p v -q.

6.26. Find the t ru th set of ( p v Q ) i -r.

6.27. Find the t ruth set of ( p .+ q ) A ( p ++ 9.).

6.28. Let P = P ( p , q, . . .) and Q = Q ( p , q , . . .) be propositions such tha t P v Q is a tautology. Show that the union of the t ru th sets T ( P ) and T(Q) is the universal set: T ( P ) U T(Q) = 7J.

6.29. Let the proposition P = P ( p , q , . . . ) logically imply the proposition Q = Q ( p , q, . . . ) . Show t h a t the union of the t ru th sets T ( - P ) and T ( Q ) is the universal set: T( -P) u T(Q) = L‘.

Answers to Supplementary Problems 6.16. .I. = 2, u = 3.

6.17. P I = ( 1 , 3 ) , P, = (2 ,8) , P; = ( 4 6 ) and P4 = (3 ,2) .

6.18. i i ) V A V = {(Mark, Eric), (Mark, David), (Eric, Eric), (Eric, David), (Paul, Eric) , (Paul, David)) V X W = {(Eric, Mark) , (David, Nark), (Eric, Eric), (David, Eric), (Eric, Paul), iDavid, Paul)} (ii)

(iii) V X V = ((Eric, Eric), (Eric, David), (David, Eric), (David, David)}

CHAP. 61

6.19.

PRODUCT SETS 57

The elements of A x E x C are the ordered triplets to the right

3, 3) 3, 4)

5 , 3) 5 . 4)

1, 3 ) 1, 4)

3, 3 ) 3 , 4)

5 , 3 ) 5 , 4)

the tree diagram above.

6.21. Each has 60 elements.

6.22. Both are true: A X A = ( B X B ) n (C X C) = ( B X C) n (C X 6’).

6.23. A X (BuC) = { ( Y , y) : 2 E A , y EBUC} = { ( x , ~ ) : % € A ; y E B o r y E C }

= {(x,~) : x E A , ZJ E B: or .T E A , y E C} =

=

Observe tha t the logical law p A ( q \ T ) E ( p 1, q ) v ( p A T ) was used in the third step above.

{(x, y) : (x, y) E A X B ; o r (s, u) E A X C} ( A x B ) u ( A X C)

6.24. T(p ts - q ) = {TF. FT}

6.25. T(--p v - q ) = {TF, FT, F F }

6.26. T ( ( p v q) -t - T ) = {TTF, T F F , FTF, F F T , FFF]

Chapter 7

Relations

RELATIONS

(a, b) in A Y E exactly one of the following statements: A binuisu i7elotion or, simply, re la t ion R from a set A to a set B assigns to each pair

(i) “u is related to b”, written a R b (ii) “u is not related to b”, written a$ Z,

A relation from a set A to the same set A is called a relation in A .

Example 1.1: Set inclusion is a relation in any class of sets. For, given any pair of sets A and B, either A C B or A d B .

Marriage is a relation from the set M of men t o the set W of women. For, given any man m E -11 and any woman w E TV, either ?ii is married to ~ t ‘ or nz is not married to w.

Order, symbolized by “<”, or, equivalently, the sentence “x is less than y”, is a relation in any set of real numbers. For, given any ordered pair ( u , b ) of real numbers, either

Perpendicularity is a relation in the set of lines in the plane. of lines a and b , either a is perpendicular t o b or a is not perpendicular t o b.

Example 1.2:

Example 1.3:

a < b or a Q b

Example 1.4: For, given any pair

RELATIOKS AS SETS OF ORDERED PAIRS

follows: €2::: = { (a , b ) : a is related to b } = {(a, b ) : a R b }

On the other hand, any subset R,’ of A x B defines a relation R from A to B as follows:

Now any relation R from a set A t o a set B uniquely defines a subset R“ of A X B as

n I: 2, iff (a ,b ) E

In view of the correspondence between relations R from A t o B and subsets of A X B, we redefine a relation by

I Definition: 1 A relation R from A to B is a subset of A x B. I l l 1

a

Example 2.1: Let R be the following relation from A = {1,2 , 3) to B = {a, b } :

R = CCL 4, (1, b ) , (370)) Then 1 R a, 2 @ b , 3 R a, and 3 $ b. The relation R is displayed on the coordinate diagram of A X B on the right. 1 2 3

Example 2 . 2 : Let R be the following relation in W = {a , b, c]:

R = { ( a , b ) , (a , c), (c, 4, (c, b ) ) Then a$a, u R b , c R c and c $ a .

58

CHAP. 71 RELATIONS 59

Example 2.3: Let A be any set. The i d e x t i t y relation in A , denoted by 1 or P A , is the set of all pairs in A X A with equal coordinates:

A.4 = { (a ,a) : a E A } The identity relation is also called the diagonal by virtue of i ts position in a coordinate diagram of A X A .

INVERSE RELATION

from B to A which consists of those ordered pairs which when reversed belong to R: Let R be a relation from A to B. The inverse of €2, denoted by R-l, is the relation

R-’ = { ( b , a ) : (a , b ) E R }

Example 3.1 : Consider the relation R = {(1,2), (1,3), (2 ,3 ) )

in A = {1,2,3}. Then R-’ = ((2, I), (3,1), (3,2)$

Observe tha t R and R-’ are identical, respectively, to the relations < and > in A , i.e.,

(a , b ) E R iff a b

Example 3.2: The inverses of the relations defined by

“x is the husband of y” and ‘‘x is taller than y”

a r e respectively

“x is the wife of y” and “3: is shorter than y”

EQUIVALENCE RELATIONS A relation R in a set A is called .i.e.flexive if a R a , i.e. (a, a) E R, for every a E A .

Then Example 4.1: ( i ) Let R be the relation of similarity in the set of triangles in the plane. R is reflexive since every triangle is similar to itself.

(ii) Let R be the relation < in any set of real numbers, i.e. (a, b ) E R iff a < b. Then R is not reflexive since a Q a for any real number a.

A relation R in a set A is called symmetric if whenever a R b then b Ra, i.e. if (a, b ) E R implies ( b , a ) E R.

Example 4.2: ii) The relation R of similarity of triangles is symmetric. similar to triangle /3, then ,8 is similar to 01.

For if triangle 01 is

(ii) The relation R = {(l, 3), (2,3), (2,2), (3, l ) } in A = {l, 2,3} is not symmetric since (2,3) E R but (3 ,2 ) 4 R.

A relation R in a set A is called t ~ a n s i t i v e if whenever aR b and b R c then a R c , i.e. if (a, 6) E R and ( b , c ) E R implies (a, c) E R.

Example 4.3: (i) The relation R of similarity of triangles is transitive since if triangle 01 is similar to ,8 and /3 is similar to y, then a is similar to y.

(ii) The relation R of perpendicularity of lines in the plane is not transitive. Since if line a is perpendicular to line b and line b is perpendicular to line c, then a is parallel and not perpendicular to c.

A relation R is an equivalence r e l a t i m if R is (i) reflexive, (ii) symmetric, and (iii) transitive.

Example 4.4: By the three preceding examples, the relation R of similarity of triangles is a n equivalence relation since it is reflexive, symmetric and transitive.

60 RELATIONS [CHAP. 7

PARTITIONS

union is X , i.e. such that each a € X belongs to one and only one of the subsets. subsets in a partition are called cells.

A part i t ion of a set X is a subdivision of X into subsets which are disjoint and whose The

Thus the collection { A l , A,, . . . , of subsets of X is a partition of X iff: (ii) for any Ai ,A, , either A , = A j or A , n A j = (2 (ij X = A,UA,U - a + UA,,,;

Example 5.1 : Consider the following classes of subsets of X = { l , 2 , . . . ,8,9}:

(i) [{I, 3,5}, {2 ,6 ) , {4,8,931

!ii) [{I, 3,5}, {2,4,6,83, {5,7,9}]

(iii) [{I, 3,51, {2,4,6,8}, (7,911

Then (i) is not a partition of X since 7 E X but 7 does not belong to any of the cells. Furthermore, (ii) is not a partition of X since 5 E X and 5 belongs t o both { l , 3,5} and {5,7,9). On the other hand, ( 5 ) is a partition of X since each element of x belongs to exactly one cell.

EQUIVALENCE RELATIONS AND PARTITIONS

equivalence class of A , be the set of elements to which a is related: Let R be an equivalence relation in a set A and, for each a E A , let [a], called the

[a] = {X : ( a , ~ ) E R }

The collection of equivalence classes of A , denoted by A I R, is called the quotient of A by R: A I R = { [ a ] : a E A )

The fundamental property of a quotient set is contained in the following theorem.

Theorem 7.1: Let R be an equivalence relation in a set A . is a partition of A . Specifically,

Then the quotient set A I R

(i) (iij [a] = [b] if and only if (a, b ) E R; (iii) if [a] # [b] , then [a] and [ b ] are disjoint.

a E [a] , for every a E A ;

Example 6.1: Let R, be the relation in Z, the set of integers, defined by x = y (mod 5 )

which reads “z is congruent to y modulo 5” and which means tha t the difference x - y is divisible by 5 . Then R5 is an equivalence relation in Z. There a r e exactly five distinct equivalence classes in Z I Rj :

A , zz . . .} A , = { ..., -9, -4, 1, 6, 11, ...}

A , = { ..., -8, -3, 2 , 7 , 12, ...}

A , = { ..., -7, -2, 3, 8, 13, . . .}

A4 = { ..., -6, -1, 4, 9, 14, ...}

{. . ., -10, -5 , 0, 5 , 10,

Observe tha t each integer x which is uniquely expressible in the form x = 5 q + T

where 0 5 r < 5 is a member of the equivalence A , where r is the remainder. Note t h a t the equivalence classes a r e pairwise disjoint and t h a t

2 = A , U A , U A , U A , U A ,

Solved Problems RELATIONS 7.1. Let El be the relation < from A = (1 ,2 ,3,4) t o B = (1,3,5}, i.e. defined by

“x is less than y”.

(i) Write R as it set of ordered pairs. (ii) Plot R on a coordinate diagram of ‘4 x B. (iii) Find the inverse relation l?*.

(i) R consists of those ordered pairs (a, b ) E A X B for

1 IiEEE which a < b ; hence

R = h l , 3 ) , (1 ,5 ) , (2,31, 12, 51, 13 , 51, (4 ,5))

(ii) R is sketched on the coordinate diagram of A X E as shown in the figure.

(iii) The inverse of R consists of the same pairs as a re in R but in the reverse order; hence

R-’ = ( (3 ,1) , (5 ,1 ) , (3 ,21, ( 5 , 2 ) , 15 ,3 ) , (5 ,4))

Observe that R-1 is the relation >, i.e. defined by ‘‘x is greater than y”.

1 2 3 4

7.2. Let €2 be the relation from E = [ 2 , 3 , 4 , 5 ) to F = {3,6,7,10) defined by “x divides y”.

(i) Write R as a set of ordered pairs. (ii) Plot R on a coordinate diagram of E X F . (iii) Find the inverse relation R-I.

(i) Choose from the sixteen ordered pairs in E X F those in which the first element divides the second; then

R = { ( 2 , 6), ( 2 , lo) , (3 , 31, 13, 61, 1 5 , l O ) )

(ii) R is sketched on the coordinate diagram of E X F as shown in the figure.

(iii) To find the inverse of R, write the elements of R but in reverse order.

R-l = {(6,2) , (10,2), ( 3 , 3 ) , f6,3i, 110,5)}

7.3. Let d l = (a, b , c, d } and let R be the relation in M consisting of those points which are displayed on the d

coordinate diagram of M x M on the right.

(i) Find all the elements in A l which are related to b, that is, ( J . (x, b ) E R } . b

(ii) Find all those elements in J l to which d is re- a lated, that is, {x : (d,x) € R ) .

(iii) Find the inverse relation R-l.

(i) The horizontal line through b contains all points of R in xh ich b appears as the second element (cr, b ) , ( b , b ) and (d, b ) . Hence the desired set is { a , b , d ] .

(ii) The vertical line through d contains all the points of R in Nhich ( I appears as the first element (d,a) and ( t l , b ) . Hence { u , b \ i b the desired set.

(iii) F i i s t n r i t e R as a set of ordered pails, and then write the pairs in reverse order

C

a b c d

R = ( ( a , b ) , ( b , a) , (6, b) , (b , 4, (c, c ) , (d , (0, (d , b ) ) R-’ = ( ( b , a), (a , b ) , ( b , b ) , (6 b ) , (c , c), (a , 4, ( b , 4 )


EQUIVALENCE RELATIONS 7.4. Let R be the relation 4 in N = {1,2,3, . . .}, i.e. (a, b ) E R iff a L b. Determine

whether R is (i) reflexive, (ii) symmetric, (iii) transitive, (iv) an equivalence relation. (i) (ii) (iii) R is transitive since a 5 b and b 6 c implies a 6 c.

(iv) R is not an equivalence relation since i t is not symmetric.

R is reflexive since a f a for every a € N . R is not symmetric since, for example, 3 f 5 but 5 a 3, i.e. (3,5) E R but (5 ,3) 4 R.

7.5. Let R be the relation ( 1 (parallel) in the set of lines in the plane. Determine whether R is (i) reflexive, (ii) symmetric, (iii) transitive, (iv) an equivalence relation. (Assume that every line is parallel to itself.) (i) (ii)

iiii) R is transitive since if Q ) j /3 and /3 1 ) y then a I) y.

(iv) R is an equivalence relation since it is reflexive, symmetric and transitive.

R is reflexive since, by assumption, (Y / I a f o r every line a. R is symmetric since if a ( 1 p then p 1 1 a, i.e. if the line a is parallel to the line /3 then /3 is parallel to a.

7.6. Let W = {1,2,3,4}. Consider the following relations in W : R, = ((1, 2 ) , (4,317 (2,2), (2,1), (3 ,1)) R, = ((2,219 (2,3), (392)) R, = {(1,3))

Determine whether each relation is (i) symmetric, (ii) transitive. (i) Now a relation R is not symmetric if there exists an ordered pair (a , b ) E R such that

(b , a ) 4 R. Hence: R, is not symmetric since (4,3) E Rl but (3,4) 4 R, , R3 is not symmetric since (1,3) E R3 but ( 3 , l ) 4 R3.

On the other hand, R, is symmetric.

A relation R is not transitive if there exist elements a, b and c , not necessarily distinct, such tha t

(a , b ) E R and ( b , c) E R but (a , c ) 4 R Hence R , is not transitive since

(ii)

(4 ,3) E R1 and ( 3 , l ) E Rl but ( 4 , l ) 4 Rl Furthermore, R, is not transitive since

(3 ,2) E R, and (2,3) E R, but (3 ,3) 4 R,

On the other hand, R3 is transitive.

7.7. Let R be a relation in A . Show that: (i) R is reflexive iff A C R; (ii) R is symmetric iff R = R-l.

(i) Recall tha t A = {(a,a) : a E A } . Thus R is reflexive iff (a,a) E R for every a € A iff

(ii) Suppose R is symmetric. Let (a , b ) E R, then (b ,a) E R by symmetry. Hence (a , b ) E R-1, and so R c R-1. On the other hand, let (a, b) E R-1; then ( b , a) E R and, by symmetry, ( a , b ) E R. Thus R-1 c R, and so R = R-1.

Now suppose R = R-1. Let (a, b) E R; then ( b , a ) E R-1 = R. Accordingly R is symmetric.

A c R.


7.8. Consider the relation R = ((1, l), (2 ,3) , (3,2)} in X = {1 ,2 ,3} . Determine whether or not El is ( i ) reflexive, (ii) symmetric, (iii) transitive. (ij R is not reflexive since 2 E X but 12,2j 4 R. (ii) R is symmetric since R-1 = ((1, l), (3,2), (2,3)} = R. (iii) R is not transitive since (3 ,2 ) E R and 12,3) E R but (3,3) 4 R.

7.9. Let N = [l, 2,3 , . . . }, and let R be the relation = in N X N defined by

Prove that R is an equivalence relation. (a, b ) (c, d ) iff ad = bc

For every (a, b ) E N X N , (a, b ) -- (a , b ) since a6 = ba; hence R is reflexive. Suppose (a, b ) = (c, d). Then ad = bc, which implies that cb = da. Hence (c, d ) (a , b )

Now suppose (a, b ) (e ,d) and (c, d ) = ( e , f ) . Then ad = bc and cf = de. Thus and so R is symmetric.

iatZ)(cf) = (be)(&) and, by cancelling from both sides, at’ = b e . Accordingly, (a, b ) = (e , f ) and so R is transitive.

Since R is reflexive, symmetric and transitive, R is an equivalence relation.

Observe that if the ordered pair (a, b ) is written as a fraction j then the above relation R is, b a c .

d in fact, the usual definition of equality between fractions, i.e. 7 = - 1ff ad = bc.

7.10. Prove Theorem 7.1: Let R be an equivalence relation in a set A . Then the quotient set A 1 R i s a partition of A . Specifically, (i) a E [ a ] , for every a € A ; (ii) [a] = rbl if and only if (a, b ) E R ; (iii) if [ a ] + Ib l , then la] and [bl are disjoint.

Proot o f ( I ) , Since R is reflexive, (a, a ) E R for every a € A and therefore a € [a] . Proof o t (11). Suppose (a , b ) E R. We want to show tha t [ a ] = [ b ] . Let x E [b] ; then ( b , x) E R.

But by hypothesis ( a , b ) E R and so, by transitivity, (a , a ) E R. Accordingly x € [a]. Thus [bl C [ a ] . To prove that a c , b ] , we observe that (a , b ) E R implies, by symmetry, tha t ( b , a) E R. Then by a similar argument, we obtain [a] ~ [ b l . Consequently, [a] = [ b ] .

On the other hand, if [a] = [ b ] , then, by ( I ) , b E [b] = [a] ; hence ( a , b ) E R. Proof o t (1111. We prove the equivalent contrapositive statement

if ‘a1 bl # @ then [a] = [bl

If (b, x) E R.

[a] r b] P $3, then there exists an element x E A with .c E [a ] n [ b l . Hence (a,x) E R and Consequently by (ii), [a ] = lbl. By symmetry, (x, b) E R and by transitivity, (a , b ) E R.

PARTITIONS 7.11. Let X = {a , b , c , d , e , f , g } , and let:

(i) A , = { a , c , e ) , A , = { b } , A , = { d , ~ } ; (ii) B , = (a , e , Y), B, = { c , d } , B, = { b , e, f } ; (iii) C, = (a , b , e , g } , C, = {c } , C, = { d , j ) ;

Which of (i) { A , , A , , A , ] is not a partition of X since t E X but f does not belong to either A l , A > , o r A , . (ii) { B , , E 2 , E , ] is not a partition of X since e E X belongs to both B , and B,. (iii) {C,, C2, C,] is a partition of X since each element in X belongs t o exactly one cell, i.e.

X = C, d C 2 u C3 and the sets are pairwise disjoint. (iv) {D,; is a partition of X .

(iv) D, = {a , b , c, d, e , f , SI. {A,, A2, A 3 } , {B l , B2, BJ), {C,, C,, C3}, {D,} are partitions of X ?

7.12.

Supplementary Problems RELATIONS 7.13. Let R be the relation in A = ( 2 , 3 ,4 , 5 ) defined by “x and ?/ are relatively prime”, i.e. “the only

comnion divisor of x and y is 1”.

(i) Write R as a set of ordered pairs. (ii) Plot R on a coordinate diagram of A X A . (iii) Find R-1.

7.14. Let N = {l, 2 , 3 , . . . ] and let R be the relation in N defined by r + 2y = 8, i.e., R = { ( r r , y ) : x , y E N , r + 2 y = 8 }

(i) Write R as a set of ordered pairs. (ii) Find R - l .

7.15. Let C = { 1 , 2 , 3 , 4 , 5 } and let R be the relation in C consisting of the points displayed in the following coordinate diagram of C X C:

(i)

(ii)

State whether each is t rue or false:

Find the elements of each of the following subsets of C: (a) {x : 3 R x}, ( b ) {x : (4, x ) E R } , ( c ) {x : ( x , 2 ) 62 R } , ( d ) {Z : x R 5).

(a) 1 R 4, ( b ) 2 R 5, (c) 3 $ 1, (d) 5 $ 3.

7.16. Consider the relations < and 5 in N = {l, 2,3, . . .}. Show tha t < U A = f where 1 is the diagonal relation.

EQUIVALENCE RELAATIONS 7.17. Let TY = {l, 2 ,3 ,4} . Consider the following relations in W :

Rl = {(1,1), (1 ,2 ) } R4 = ((1, I), (2,217 (3,311

R , = {(1,3), (2 ,4 ) ) R , = 8 R , = {(1,1), (2 ,3 ) , (431)) R, = T.V X W

Determine whether or not each relation is reflexive.


7.18. Determine whether o r not each relation in Problem 7.17 is symmetric.

7.19. Determine whether o r not each relation in Problem 7.17 is transitive.

7.20. Let R be the relation I of perpendicularity in the set of lines in the plane. Determine whether R is Ii) reflexive, (ii) symmetric, ( i i i ) transitive, (iv) a n equivalence relation.

7.21. Let N = {l, 2,3, . . . } and let = be the relation in N X N defined by

( a , b ) ( c , d ) iff a + d = b + c

(i) Prore r is an equivalence relation. (ii) Find the equivalence class of (2,5), i.e. [ (2 ,5) ] .

7.22. Prove: If R and S a r e equivalence relations in a set X , then R n S is also a n equivalence relation in X .

7.23. (i) Show tha t if relations R and S are each reflexive and symmetric, then R U S is also reflexive and symmetric.

Give an example of transitive relations R and S f o r which R U S is not transitive. (ii)

7.25. Find all partitions of V = {l,Z, 3) .

7.26. Let [ A , , A , , . . ., A,,,1 and [B,, B,, . . ., B,] be partitions of a set X . Show tha t the collection of sets

[ A , n B , : i = l , ..., m , j = l , . . . , n] is also a partition (called the CTOSS p ~ ~ , t i t i ~ i l ) of X .

7.13.

7.14.

7.15.

7.17.

7.18.

7.19.

7.20.

7.21.

1.23.

7.24.

7.25.

R,, R, and R, are the only symmetric relations.

All the relations are transitive.

(i) no, ( i i ) yes, i i i i ) no, (iv) no

(ii) [(Z, 5)l = {(a, b ) : a + 5 = b + 2, a , b E N )

= {(a, a + 3) : a E NI = {(I, 4), (2,5), (3,6), (4 ,7 ) , . . .I (ii) R = { \ l , 2 ) ) and S = {(2,3)}

(i) no, (iii no, ( i i i ) yes, (iv) yes

[ {1 ,2 ,3) ] , [{l:, {Z, 3)], [C2>, C1,3$], [{3), {1,2)1 and [{I}, { z : , (311

Chapter 8

Functions

DEFINITION OF A FUNCTION Suppose that to each element of a set A there is assigned a unique element of a set B ;

the collection, f , of such assignments is called a functioyz (or mapping) from (or on) A into B and is written

The unique element in B assigned to a E A by f is denoted by f ’ (a) , and called the i m a g e of a under f o r the value of f at a. The domain of f is A , the co-domain B. The m x g e of f, denoted by f [ A ] is the set of images, i.e.,

f l A ] = { f (a ) : a € A }

f : ~ + B or A & B

Example 1.1: Let f assign to each real number its square, tha t is, for every real number 2: let f ( x ) = x2. Then the image of -3 is 9 and so we may write f ( - 3 ) = 9 or f : -3 + 9.

Example 1.2: Let f assign to each country in the world its capital city. Here the domain of f is the set of countries in the world; the co-domain is the list of cities of the world. The image of France is Paris, t h a t is, f \France) = Paris.

Example 1.3: Let A = { a , b , c , d ) and B = {a, b, c } . The assignments a - b , b - t c , c + c and d + b

define a function f from A into B. Here f(a) = b, f(b) = c , f(c) = c and f ( c1 ) = b. The range of f is { b , c ) , that is, f [ A ] = { b , c } .

Example 1.4: Let R be the set of real numbers, and let f : R + R assign to each rational number the number 1, and to each irrational number the number -1. Thus

1 if 3: is rational -1 if x is irrational f ( x ) =

The range of f consists of 1 and -1: f [ R ] = {I,-1).

Example 1.5: Let A = {a, b , c , d ) and B = {x,y,z}. The following diagram defines a function f : A + B .

Here f (a) = y, f(b) = N , f(c) = x and f ( d ) = y. and the co-domain a re identical.

Also f [ A ] = B, tha t is, the range

66

CHAP. S] FCNCTIONS 67

GRAPH OF A FUNCTION To each function J : A -+ B there corresponds the relation in A X B given by

((a, / '(a)) : a E A ) We call this set the g r a p h of f . Two functions f : A + B and g : A -+ B are defined to be equal, written f = y, if ?(a) = g(a) for every a E A , that is, if they have the same graph. Accordingly, we do not distinguish between a function and its graph. The negation of J = g is written i = y and is the statement:

there exists an ci E A f o r which f ( a ) - g ( a )

A subset f of A x B, i.e. a relation from A to B, is a function if it possesses the folloving

Each a E A appears as the first coordinate in exactly one ordered pair (a, b) in f . Accordingly, if (a , b) E f, then t ( a ) = b ,

property:

[F]

Example 2.1: Let f : A -+ B be the function defined by the diagram in Example 1.5. Then the graph of f is the relation

{ ( a , Y), ( b , x), (c, x ) , ( d . L/)S

Example 2.2: Consider the following relations in A = {1,2,3] :

f is a function from A into A since each member of A appears as the first coordinate in exactly one ordered pair in f ; here f(1) = 3, f ( 2 ) = 3 and f(3) = 1. g is not a function from A into A since 2 € A is not the first coordinate of any pair in g and so g does not assign any image to 2. Also h is not a function from A into A since 1 € A appears as the first coordinate of two distinct ordered pairs in h , (1,3) and (1,2). If h is t o be a function i t cannot assign both 3 and 2 to the element 1 E A .

Example 2.3: A real-valued function ,f : R - R of the form f ( x ) = ax t b (or: defined by y = a.c 4- b )

is called a linear f u x c t i o x : its graph is a line in the Cartesian plane RZ, The graph of a linear function can be obtained by plotting ( a t least) two of its points. For example, to obtain the graph of f j x ) = 2% - 1, set up a table with a t least t w o values of x and the corresponding values of f ( x ) as in the adjoining table. The line through these points, (-2, -5), (0, -1) and (2,3), is the graph of f as shown in the diagram.

Graph of f(x) = 2 x - 1

68 FUNCTIONS [CHAP. 8

- 2 -1

0 1 2

Esaiiiple 2.4:

5 0

-3 --1 -3

3 0 4 5

A real-valued function f’ : R + R of the form

f ( l ) = q)T’* + alX,--l + . . . + a,,-,:c + a,

is called a p o l y z o m i u l f m c t i o n . The graph of such a function is sketched by plotting various points and drawing a smooth continuous curve through them.

Consider, for example, the function f i x ) = .c’ - 2.1; - 3. Set up a table of values for s and then find the corresponding values f o r f ( x ) as in the adjoining table. The diagram shows these points and the sketch of the graph.

t Graph of f (x) = - 2.1: - 3

COMPOSITION FUNCTION Consider now functions f : A + E and g : B + C illustrated below:

:c I f (x)

Let a E A ; then its image f ( a ) is in E , the domain of g . Hence we can find the image of f ( a ) under the function g, i.e. g ( f ( a ) ) . The function from A into C which assigns t o each a E A the element g(f(a)) E C is called the composition o r product of f and g and is denoted by g 0 f . Hence, by definition,

( g o f ) ( a ) = !ma))

Example 3.1: Let f : A .+ E and g : B 3 C be defined by the following diagrams:

A f B g C

We compute ( g 0 t ’ ) : A + C by its definition.

( g o f ) ( 4 = s( f (a)) = d u ) = t ( 9 0 f N b ) = g ( f ( b ) ) = g ( z ) = ?“

iu O f ) ( C ) = g ( f ( c ) ) = u ( u ) = t

Notice tha t the composition function g 0 t‘ is equivalent to “following the arrows” from A to C in the diagrams of the functions j and g.

CHAP. 81 FUNCTIONS 69

Example 3.2: Let R be the set of real numbers, and let f : R + R and g : R -+ R be defined a s follows:

Then ( f o g ) ( 2 ) = f ( d 2 ) ) = f(5) = 25 (g Of)@) = d f ( 2 ) ) = g(4) = 7

Observe tha t the product functions f o g and g o f are not the same function. We compute a general formula for these functions:

( f o g ) ( x ) = f ( g ( z ) ) = f ( x t 3 ) = ( x + 3 ) 2 = x2 + 62 + 9

( g o f ) ( x ) = g ( f ( x ) ) = g(22) = + 3

f(2) = 2 2 and g ( x ) = z + 3

ONE-ONE AND ONTO FUNCTIONS A function f : A + B is said to be one-to-one (or: one-one or 1-1) if different elements

in the domain have distinct images. Equivalently, f : A + B is one-one if

f ( a ) = f(a') implies a = a'

Example 4.1: Consider the functions f : A - B, g : B + C and h : C + D defined by the following diagram:

A f B B C h D

a

W Now f is not one-one since the two elements a and c in its domain have the same image 1. Also, g is not one-one since 1 and 3 have the same image y. On the other hand, h is one-one since the elements in the domain, x, y and x , have distinct images.

A function f : A + B is said to be onto (or: f is a function from A onto B or f maps A on to B ) if every Z, E B is the image of some a E A . Hence f : A + B is onto iff the range of f is the entire co-domain, i.e. f [ A ] = B.

Example 4.2: Consider the functions f, g and h in the preceding example. since 2 € B is not the image of any element in the domain A . both g and h a r e onto functions.

Then f is not onto On the other hand,

Example 4.3: Let R be the set of real numbers and let f : R + R, g : R + R and h : R + R be defined as follows:

f ( x ) = 22, g(z) = z3 - x and h ( z ) = x2

The graphs of these functions follow:

f ( x ) = 2" g(2) = 23 - x h(x) = 22


The function f is one-one; geometrically, this means tha t each horizontal line does not contain more than one point of f. The function g is onto; geometrically, this means t h a t each horizontal line contains a t least one point of g. The function h is neither one-one or onto; for h(2) = h ( - Z ) = 4, i.e. the two elements 2 and -2 have the same image 4, and h[R] is a proper subset of R ; for example, -16 4 h[R] .

INVERSE AND IDENTITY FUNCTIONS In general, the inverse relation f - I of a function f C A X B need not be a function.

However, if f is both one-one and onto, then f - ' is a function from B onto A and is called t,he inverse function.

Example 5.1: Let A = { a , b, c ) and B = { r , s, t}. Then

f = { (a , s), ( b , t ) , ( c , 41 is a function from A into B which is both one-one and onto. seen by the following diagram of f :

This can easily be

Hence the inverse relation f-1 = ( ( 5 , a ) , ( 4 JJ), ( r , 4)

is a function from B into A . The diagram of f-' follows:

Observe tha t the diagram of f-' can be obtained from the diagram of f by reversing the arrows.

For any set A, the function f : A + A defined by f ( x ) = x, i.e. which assigns to each element in A itself, is called the i d e n t i t y function on A and is usually denoted by 1, or simply 1. Note that the identity function 1, is the same as the diagonal relation: 1, = A,. The identity function satisfies the following properties:

Theorem 8.1: For any function f : A + B, lBOf = f = f o l ,

Theorem 8.2: If f : A + B is both one-one and onto, and so has an inverse function f- ' : B + A , then

f - l o f = 1, and f o f - ' = 1,

The converse of the previous theorem is also true:

Theorem 8.3: Let f : A -+ B and y : B + A satisfy

g o f = 1, and f o g = 1,

Then f is both one-one and onto, and y = f - ' .

CHAP. 81 F U I\: C T I ONS 71

Solved Problems FUNCTIONS 8.1. State whether or not each diagram defines a function from A = ( a , b , c } into

B = { r , y , z ] , m@gJm C z C z C 2

(i) (ii) (iii)

(i) No. There is nothing assigned t o the element b € A . (ii) No. Two elements, x and x , are assigned t o c € A . (iii) Yes.

8.2. Rewrite each of the following functions using a formula: (i) To each number let f assign its cube. (ii) To each number let g assign the number 5. (iii) To each positive number let li assign its square, and to each non-positive number

let 11 assign the number 4.

(i) Since f assigns l o any number ?. its cube .v3, f can be defined by f ( . c ) = x3, (ii) Since g assigns 5 to any number s, we can define g by g(..) = 5. (iii) Two different rules a re used to define k as follows:

$2 if x > 0 4 i f x " 0

li (?.) =

8.3. Let f , g and 11 be the functions of the preceding problem. Find: (i) f(4), f ' ( -2) , f(@); (ii) ~~(41 , g ( - 2 ) , ~(0); (iii) h(4), h(-2), 4 0 ) .

(i) Now ,t'(:cs) = x3 for every number 2:; hence f(4) = 43 = 64. J ( - 2 ) = ( -2)s = -8, f(0) O3 = 0. (ii) Since y(x) = 5 for every number x , y(4) = 5, g(-2) = 5 and g(0) = 5. (iii) If 3: > 0, then h(x) = x2; hence h(4) = 42 = 16. On the other hand, if x f 0, then

h(a) = 4: thus h(-2) = 4 and h i O ) = 4.

8.4. Let A = (1 ,2 ,3 ,4 ,5} and let J:,4 + A be the function defined in the diagram:

(i) Find the range of f . (ii) Find the graph of f , i.e. write f as a set of ordered pairs.

( i ) The range f[AI of the function f consists of all the image points. Kow only 2, 3 and 5 appear as the image of any elements of A ; hence f[A] = {2 ,3 ,5 ) .

(ii) The ordered pairs ( a , f ( a ) ) , where a € A , form the graph of f. Now f(1) = 3, f ( 2 ) = 5, f(3) = 5, f(4) = 2 and f(5) = 3; hence f = {(1,3), ( 2 , 5 ) , (3,5), (4,2), (5,3)}.


8.5. Let X = (1,2,3,4>. Determine whether o r not each relation is a function from X into X. ti) f = ii2,3), (1,4), (2, I), ( 3 , 2 ) , 14,4)} i i i ) g = “3, l), (4,2), (1, 1)) iiii) h = j(2, l ) , (3,4), (1,4), j2,11, (4,4)}

Recall that a subset f of X X X is a function f : X --f X if and only if each a E X appears

No. Two different ordered pairs ( 2 , 3 ) and ( 2 , l ) in f’ have the same number 2 as their first coordinate. No. The element 2 E X does not nppear as the first coordinate in any ordered pair in 8 .

ordered pairs a re equal.

as the first coordinate in exactly one ordered pair in f . iir

i i i l i i i i i Yes. Although 2 E X appears as the first coordinate in two ordered pairs in h, these two

8.6. Find the geometric conditions under which a set f of points on the coordinate diagram of A x B defines a function f : A + B.

The requirement t h a t each u E A appear a s the first coordinate in exactly one ordered pair in f is equivalent to the geometric condition tha t each vertical line contains exactly one point in f .

8.7. Let TV = { a , b, c, d } . Determine whether the set of points in each coordinate diagram of T I ’ x IT’ is a function froin I T ’ into W . tM b :m- fff!Ff

a a a

a b c d a b c d a b c d

(i) (ii) (iii)

( i )

( i i )

(iiir Yes. Each vertical line contains exactly one point of the set.

No. The vertical line through b contains two points of the set, i.e. two different ordered pairs ( b , b ) and ( b , d ) contain the same first element b. KO. The vertical line through c contains no point of the set, i.e. c E W does not appear as the first element in any ordered pair.

GRAPHS OF REAL-VALUED FUNCTIONS 8.8. Sketch the graph of f ( x ) = 3 x - 2 .

This is a linear function: only two points (three as a check) a re needed to sketch its graph. Set up a table with three values of s, say, x = -2, 0, 2 and find the corresponding values of f(r):

t’(-2) = 3(-2) - 2 = -8, ((0) = 3(0) - 2 = -2, f(2) = 3(2) - 2 = 4

Drau the line through these points as in the diagram.

X

10, - 2 )

7 3 CHAP. 81 FUNCTIONS

-2 -1

2

4

8.9. Sketch the graph of (i) f ( x ) = 2 t .L‘ - 6 , (ii) g(x) = x? - 3x2 - x + 3. In each case, set up a table of values for x and then find the corresponding values of t ’ ( ~ ) .

Plot the points In a coordinate diagram, and then draw a smooth continuous curve through the points

-15 0

0 3 1 0

-3 3 0

15

Graph of f .

(ii) x I gis)

Graph of g.

COMPOSITION OF FUNCTIONS 8.10. Let the functions f : A -+ B and y : B -+ C be defined by the diagram

(i) Find the composition function y o I: : A .+ C. (ii) Find the ranges of f , g and y o f .

(i) We use the definition of the composition function to compute.

(Y {)(a) = M a ) ) = g(u) = t ( g o f ) ( b ) = g ( f ( b ) ) = g ( r ) = s

(Y t ’ ) ( C ) = g ( f ( c ) ) = s(l/) = t

Note that we arrive at the same answer if we “follow the arrows” in the diagram: a --t u + t , b + x + s , c + y + t

(ii) By the diagram, the images under the function f are x and y, and the images under g a re T , s and t : hence

range of f = {.c, g } and range of g = { Y , s, t }

By i ~ , the images under the composition function a re t and s. hence range of g o f = {s, t )

Note that the ranges of g and y c t’ are different.

8.11. Consider the functions f = { (1 ,3 ) , ( 2 , 5 ) , (3,317 (4,1), (592)) g = {(I , 4), (2, I), (3911, (4, a, 15,311

from X = j l , 2 , 3 , 4 , 5 } into X. (i) Determine the range of f and of g. (ii) Find the composition functions g o f and f o g .


(ij The range of a function is the set of image values, i.e. the set of second coordinates: hence

range o f f = ( 3 , 5 , 1 , 2 } and range of g = { 4 , 1 , 2 , 3 }

8.12. Let the functions f and g be defined by f(x) = 2 x f l and g(x) = x 2 - 2 . Find formulas defining the composition functions (i) g 0 t' and (ii) f o g .

(i) Compute g 0 f as follows: (g 0 f i ( r j = g ( f ( z ) ) = g(2.2. + 1) = ( 2 ~ + 1)2 - 2 = 422 + 42 - 1.

Observe tha t the same answer can be found by writing

y = f ( ~ ) = 2.t + 1 and z = g(y) = y2 - 2

and then eliminating y from both equations: x = u 2 - 2 = (2% + 1)2 - 2 = 4 x 2 + 4.1 - 1

(iil Compute f o g as follows: ( f o g ) ( x ) = f ( g ( x ) ) = f(x'- 2 ) = 2 ( n 2 - 2 ) + 1 = 2%'- 3.

Note tha t f o g # g of.

8.13. Prove the associative law for composition of functions: if A A B % C % D

then ( J z o g ) o f = h o ( g o f ) .

For every a E A , ( ( h 0s) 0 = ( h o g ) ( f ( a ) ) = h ( u ( f ( a j ) )

( { l (Y f ) j ( a ) = h((s O f)(a)) = h ( g ( f ( a ) ) )

Hence ( h 0 9 ) 0 f = h o ( g o f ) , since they each assign the same image to every a E A . Accordingly, we may simply write the composition function without parentheses: h 0 g 0 f .

ONE-ONE AND ONTO FUNCTIONS 8.14. Let A = ( a , b , c , d , e J , and let B be the set of letters in the alphabet. Let the

functions f , g and h from A into B be defined as follows: f 9 h

a+ r a+ x a+ a b + a b + Y b + c C + S C + X c + e d - + r d + Y d - + r

(9 (ii) (iii) e - + e e + x e + s

Are any of these functions one-one?


Recall tha t a function is one-one if it assigns distinct image values t o distinct elements in the domain. (i) No. (ii) No. (iii) Yes.

For f assigns T to both a and cl. F o r g assigns x to both a and e . F o r h assigns distinct images to different elements in the domain.

8.15. Determine if each function is one-one. (i) To each person on the earth assign the number which corresponds to his age. (ii) To each country in the world assign the latitude and longitude of its capital. (iii) To each book written by only one author assign the author. (iv) To each country in the world which has a prime minister assign its prime

minister.

(i) No. (ii) Yes. (iii) No. There a re different books with the same author. (iv) Yes.

Many people in the world have the same age.

Different countries in the world have different prime ministers.

8.16. Prove: If f : A + B and g : B + C are one-one functions, then the composition

Let (g o f ) ( a ) = ( g o f ) ( a ’ ) ; i.e. &“(a)) = g( j (a ’ ) ) . Then f(a) = f(a’) since g is one-one.

function g 0 f : A + C is also one-one.

Furthermore, a = a‘ since f is one-one. Accordingly, g o f is also one-one.

8.17. Let the functions f : A + B, g : B + C and h : C + D be defined by the diagram.

A i B C h D

(i) Determine if each function is onto.

(i) The function f : A + B is not onto since 3 E B is not the image of any element in A. The function g : B --f C is not onto since z E C is not the image of any element in B. The function h : C + D is onto since each element in D is the image of some element of C.

(ii) Find the composition function h o g o f .

(ii) Now a + Z + x + 4 , b + l + y + 6 , c + Z + x + 4 . Hence h o g o f = { ( a , 4 ) , ( b , 6 ) , ( ~ , 4 ) } .

8.18. Prove: If f : A + B and g : B -+ C are onto functions, then the composition function

Let e be any arbi t rary element of C. Since g is onto, there exists a b E B such t h a t g ( b ) = c. But then

g o f : A + C is also onto.

Since f is onto, there exists a n a E A such tha t f ( a ) = b . ( g o f ) ( 4 = g(f(a\) = g ( b ) = c

Hence each c E C is the image of some element a € A . Accordingly, g 0 f is a n onto function.

76 FUNCTIONS ;CHAP. 8

INVERSE FUNCTIONS 8.19. Let W = { 1 , 2 , 3 , 4 , 5 } a n d l e t f : W - + W , g : W + W and h : W - + W bedefinedby

the following diagrams:

Determine whether each function has an inverse function.

Only h is one-one and onto; hence h , and only h, has a n inverse function. In order for a function to have an inverse, the function must be both one-one and onto.

8.20. Let f : R -+ R be defined by f ( x ) = 2x - 3. Now f is one-one and onto; hence f has an inverse function f - l . Find a formula for fF1.

Let y be the image of x under the function f : y = f ( x ) = 2x - 3

Consequently, 2; will be the image of y under the inverse function f-1. Solve for x in terms of y in the above equation:

x = (y + 3)/2

Then J-W = ( ~ + 3 ) / 2 is a formula defining the inverse function.

8.21. Let A = (a, b, c, d } . Then f = ((a, b) , ( b , d ) , (c ,a) , (d , c)} is a one-one, onto function from A into A . Find the inverse function f - ' .

in reverse order: To find the inverse function f - 1 , which is the inverse relation, simply write each ordered pair

f-' = { ( b , 4, (4 b) , (a , c ) , ( c , 4)

The function assigns to any element the square of the element minus 3 times the element plus 2.

(a) t ( -3) = (-3)2 - 3(-3) + 2 = 9 + 9 + 2 = 20 ( b ) f(2) = (2)2 - 3(2) + 2 = 0, f(-4) = (-4)2 - 3(-4) + 2 = 30. Then

f (2) - f(-4) = 0 - 30 = -30

(c) 1(y) = (L/)2 - 3(y) + 2 = y2 - 3y + 2 (d ) f(a2) = (a2)2 - 3(a*) + 2 = a1 - 3a* + 2 (t) t(.') = ( X 2 ) Z - 3(x2) + 2 = a4 - 32' + 2

( t ) f ( y - x )

(9) f(.c + / I )

= (y - z)2 - 3(y - 2 ) + 2 (x + h)2 - 3(a + h ) + 2

= y2 - 2yz + 22 - 3y + 32 + 2 x2 + 2xh + h' - 3% - 3h + 2 = =


(h) f ( x + 3) = ( X + 3)' - 3(x + 3) + 2 = (N? + 62 + 9) - 3% - 9 + 2 = x2 + 3% + 2

(i) f ( 2 x - 3 ) = ( 2 2 - 3)2 - 3(2x - 3) + 2 = 4r2 - 122 + 9 - 6x + 9 + 2 = 4x2 - 18% + 20

( j ) Using ( h ) and (i), we have f ( 2 ~ - 3) + f ( ~ + 3) = (429 - 18s + 20) + (6 + 3% + 2) = 5 ~ ' - 15% + 22

(k) f (6- 3~ + 2 ) = (9- 3% + 2)2 - 3 ( z * - 3% + 2) + 2 = ~4 - 6x3 + 1 0 ~ ' - 3% (1) f(f(x)) = f ( d - 3~ + 2) = x4 - 623 + 10x2 - 3%

(m) f(f(x + 1))

(n) By (g), f (z + h ) = x2 + 2xh + h2 - 3% - 3h + 2. Hence

= f([(x + 1)2 - 3 ( x + 1) + 21) = f([x2 + 22 + 1 - 3 x - 3 + 21) = f(z' - x) = ( 2 2 - x)' - 3(s'- x) + 2 = x4 - 2x3 - 2x2 + 3 2 + 2

f (2 I h ) - t ( x ) = ( ~ 2 + 2 ~ h + h' - 3.2: - 3 h + 2) - ( $ 2 - 3~ i- 2 ) = 2xh + h2 - 3 h

(0) Using ( n ) , we have [ f ( x + h) - f ( z ) ] / h = (2rh + h2 - 3 h ) h = 2r + h - 3

8.23. Prove Theorem 8.1: For any function f : A + B, l n o f = f = f~ 1,. Let a be any arbi t rary element in A ; then

( 1 ~ 0!)(4 = 1 ~ ( f ( a ) ) = f ( a ) and ( f o l A ) ( a ) = f(la(a)) = f ( a ) Hence 1,o f = f = f 0 1, since they each assign f (a) to every element a E A .

Supplementary Problems FUNCTIONS

8.24. State whether each diagram defines a function from { l , 2 , 3 } into { 4 , 5 , 6 ) .

8.25. Define each function by a formula:

(i) To each number let f assign its square plus 3.

(ii) To each number let g assign its cube plus twice the number.

(iii) To each number greater than o r equal to 3 let h assign the number squared, and to each number less than 3 let h assign the number -2.

8.26. Determine the number of different functions from {a , b } into {l, 2 , 3 ) .

8.27. Let f : R + R be defined by f(x) = z2 - 4 2 + 3. Find (i) f ( 4 ) , (ii) f(-3), (iii) f ( y - 2 x ) , (iv) f ( x - 2 ) .

8.28. Let g : R + R be defined by g(x) =


8.30. Let the function g assign t o each name in the set {Betty, Martin, David, Alan, Rebecca} the number of different letters needed t o spell the name. Find the graph of g , i.e. write g a s a set of ordered pairs.

8.31. Let i t ’ = [ 1 , 2 , 3 , 4 } and let g : W + T I T be defined by the diagram

( i i Write g a s a set of ordered pairs. the range of g.

Let V = {1 .2 ,3 ,4} . is a function from V into 1’.

iii) Plot g on the coordinate diagram of W X TT’. (iii) Find

8.32. Determine whether the set of points in each coordinate diagram of V X V :m fffF ## 4

3

2 2

1 1

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

(i) (ii) (iii) (iv)

GRAPHS OF REhL-VALUED FUNCTIONS

8.33. Sketch the graph of each function: (i) f(z) = 2, (ii) g(x) = 4% - 1, ( i i i ) k ( z ) 1 2%’- 4% - 3.

8.34. Sketch the graph of each function: 0 if x = O

- i f x # O ( i ) f ( x ) = :c3 - 3% -I- 2, (ii) g(z) = :ci - 10x2 + 9, (iii) h ( x ) =

COMPOSITION OF FUNCTIONS 8.35, The functions f : A + B, g : B + A, h : C + B, F : B + C, G : A + C a r e pictured in the diagram

below.

Determine whether each of the folloLving defines a product function and if i t does, find its domain and co-domain: (i) g o f , iii) h o f , ( i i i ) F o f , (iv) G o f , (v) g c h , (vi) F o h , (vii) h o G o g , rviiil l i 0 G.

8.36. The following diagrams define functions .f, g and h which map the set {1,2,3,4} into itself. @Q@-$)m 4 4 4 4 4 4

[i t ( i i ) Find the composition functions (1) f o g , (2) h o j , (3) g2, i.e. gog.

Find the ranges of f, y and h .

CHAP. 8 ) F U K C T I O N S 79

--

8.37. Let f : R 3 R and g R + R be defined by Ji.r) = z2 + 3% + 1 and g(r) = 2s - 3. Find fomiulas defining the product functions (i) f c g , i i i ) g 3 f , (iii) g o g , (iv) fof.

4

-- 2

ONE-ONE, ONTO . iKD INVERSE FUNCTIONS 8.38 Let f : X - 17. Which conditions define a one-one function:

l i ) J ( n l = f ( b ) implies a = b (iii) f ( a ) # J ( b ) iniplies a # b iiir n = b implies f(n) = f(b) (iv) a f b implies J ( a ) # f ( b )

8.39. ( i ) State whether o r not each function i n Problem 8.36 is one-one (ii) State whether or not each function in Problem 8.36 is onto.

8.10. Prove Theorem 8.2: If f : A -f B is one-one and onto, and so has an inverse function f - 1 , then (i) f - 1 o . f = 1, and (ii) f o f - 1 = 1,.

8.41. Prove: If t : A 4 E and g : B + A satisfy y o f = lA, then f is one-one and g is onto.

8.42. Let f : R + R be defined by f ( x ) = 3x - 7 . Find a formula for the inverse function f - 1 : R .+ R.

8.43. Let g : R + R be defined by g(z) = x? -t 2. Find a formula for the inverse function g-1 : R 3 R.

8.14. from A into the quotient Show tha t 17 is an onto function.

8.45. Prove Theorem 8.3: Let f : A + B and y : E .+ A satisfy g 0 t' = 1,4 and f o g = lg. Then

Let R be an equivalence relation in a non-empty set A . The function set AIR is defined by ~ ( a ) = [ a ] , the equivalence class of a.

f is one-one and onto, and g = f - 1 .

Answers to Supplementary Problems 8.24.

8.25.

8.26.

8.27.

8.28.

8.29.

8.30.

8.31.

8.32.

8.33.

(i) No, i i i ) Yes, iiii) No

Nine.

(i) 3, (iil 24, ( i i i ) y2 - 4xy + 4x2 - 4y + 8x + 3, (iv) x2 - 8% 4- 15

(i) g ( 5 ) = 10. ( i i ) y(0) = 2, (iii) g(-2) = 0

(i) Yes, iii) No, i i i i ) Yes, (iv) No

g = {(Betty, 4, (RIartin, 6), (David, 4) , (Alan, 3 ) , (Rebecca, 5))

(i) g = C(1,2), (2,3), (3, I), (4,3)), (iii) :2 ,3 ,1 )

(i) No, (i i \ No, (iii) Yes, (iv) No + -2 -3

Graph of f

F C N C T I 0 N S 80 #CHAP. 8

8.34.

( i i i )

-3 -5 -3

Graph of h

#- - 4 - 6

Graph o f f

-15

Graph of g

' - - - 4

Graph of h

8.35. ( i ) g o f : A + A , (ii) No, (iii) F o f : A + C, 8.37. lii ( , f o g ) ( x ) = 4 2 2 - 6 ~ + 1

(i i) (g 0 f ) (x) = 2x2 + 6 r - 1

( i i i ) (y 0 g ) ( x ) = 4x - 9 (ivl (f o f ) ( x ) = x4 + 6x3 + 14x2 + 152 + 5

(iv) No, (vii) h 0 G 0 g : B + B,

range of g = {1 ,2 ,3 ,4} , range of h 1 {1,3}

(v) g 0 h : C + A , (vi) F 0 h : C -t C, (viii) h 0 G : A + B

8.36. (i) range of f = {1,2,4},

8.38. ( i ) Yes, Iii) No, (iii) No, (iv) Yes

8.39.

8.42.

8.43. g-l (3:) = -2

iii) ; 1 to;^^) I (I/(Z) ,I 9;) ti) Only g is one-one.

I-1 ( x ) = (x + 7)/3

(ii, Only g is onto.

4

Chapter 9

Vectors

COLUMN VECTOKS A column rector ~r is a set of numbers u l , u,, . . . , zcn written in a column:

ZG =

The numbers t i , are called the components of the vector u.

Example 1.1: The following are column vectors:

The first two vectors have two components; whereas the last two vectors have three components.

Two column vectors 2 i and v are equal, written u = v, if they have the same number of components and if corresponding components are equal. The vectors

\3i' are not equal since corresponding elements are not equal.

Example 1.2: Let

Then, by definition of equality of column vectors,

x - y = 4 z + y = 2 x - l = 3

Solving the above system of equations gives x = 3, y = -1, and x = 4.

Remark: In this chapter we shall frequently refer to numbers as scalars.

VECTOR ADDITION The sum of ZL

and v, denoted by ti -t v , is the column vector obtained by adding corresponding components: Let u and 2' be column vectors with the same number of components.

81

82 VECTORS [CHAP. 9

Note that ii + 1: has the same number of components as 11 and 2'. The sum of two vectors with different numbers of components is not defined.

Example 2.1:

Example 2.2: The sum (-a; + f ,-5 4)l //

is not defined since the vectors have different numbers of components.

A column vector whose components are all zero is called a zero vector and is also denoted by 0. The next Example shows that the zero vector is similar to the number zero in that, for any vector 11, u f O = u.

SCALAR RIULTIPLICATION

column vector obtained by multiplying each component of 21 by It: The product of a scalar lc and a column vector u, denoted by k - u or simply kx, is the

k * 11 = k r) = p un 4/

Observe that ZJ and 1c.u have the same number of components. We also define:

-u = -1.21 and u- 2j = z! t (-v)

Example 3.1:

The main properties of the column vectors under the operations of vector addition and scalar multiplication are contained in the following theorem:

CHAP. 9 VECTORS 83

Theorem 9.1: Let VTL be the set of all 1%-component column vectors, and let ZI, I - , 20 E V , and let k , k r be scalars. Then:

(i) (ii) I' t 2%

(iii) (ivj

(vi) (vii) (kk'jrc = k(lc'u); (viii) l u = 21.

(21 t 1 9 ) -L w = u + (c t 2c), i.e. addition is associative;

ZI -t 0 = 0 + u = u; LI T ( - 1 1 ) = (-u) + z/ = 0: l;(u t 2') = ku + kv; ( k + ~ ' ) I I = ku + lc'u;

v + u, i.e. addition is commutative:

(v)

The properties listed in the above theorem are those which are used to define an abstract niathematical system called a linear spuce 3r rector space. Accordingly, the theorem can be restated as follon-s:

Theorem 9.1: The set of all ,i-component column vectors under the operations of vector addition and scalar multiplication is a vector space.

ROM' VECTORS Analogously, a row vector 2) is a set of numbers i f I , u,, , . .,u, written in a row:

ZL = (ul, u,, . . . , 21,)

The numbers ut are called the coiiiyoilents of the vector 21. Two row vectors are equal if they have the same number of components and if corresponding components are equal. Observe that a row vector is simply an ordered ri-tuple of numbers.

The sun1 of two row vectors ?( and 21 with the same number of components, denoted by u + r , is the row vector obtained by adding corresponding components from 11 and u:

1' T 2' = (ul, u,, . . . , u,,) + (vl, v2, . . ., w,,) = (211 + w l , u2+v2, . . . , 2 0 , + 2yJ

The product of a scalar lc and a row vector u, denoted by l;*u or simply kri, is the row vector obtained by multiplying each component of 21 by k :

k - u = k(/,ll , u , , . . . ) un) = (lW1, kLI,, . . . ) rCuJ

We also define - R = -1.u and u-zt = u + ( - w ) as v e did f o r column vectors.

Example 4.1: (1, -2, 3, -4) + (0 , 5, -7, 11) = (I, 3, -4, 7 ) 5 * (1, -2, 3, -4) = (5, -10, 15, -20)

We also have a theorem f o r row vectors which corresponds to Theorem 9.1.

Theorem 9.2: The set of all 11-component row vectors under the operations of vector addition and scalar multipIication satisfies the properties listed in Theorem 9.1, that is, is a vector space.

MULTIPLICATION O F A ROW VECTOR AND A COLUMN VECTOR If a row vector u and a column vector v have the same number of components, then

their product, denoted by 21 * 1' or simply uv, is the scalar obtained by multiplying corresponding elements and adding the resulting products:

86 VECTORS [CHAP. 9

ROW VECTORS 9.6. Let 11. = (2,-7,1), 2: = (-3,0,4) and IL' = (0,5,-8). Find: (i) u + v , (ii) ZI fw,

iiii) -3u, (iv) -w. 1 i 1 Add corresponding components:

u + 2, = 12, - 7 , 1) + (-3, 0, 4) = (2- 3, - 7 + 0 , 1 + 4 ) = (-1, - 7 , 5 )

I i i ) Add corresponding components:

2, + 20 = ( - 3 , 0, 4) + (0 , 5, -8 ) = (-3 f 0, 0 + 5 , 4 - 8 ) = (-3, 5, -4)

iiiii Multiply each component of u by the scalar -3. -3u = -3(2,-7, l ) = (-6,21,-3).

iiv) Xultiply each component of w by -1, i.e. change the sign of each component:

-W = - (O, 5, - 8 ) = (0 , -5, 8)

9.7. Let u, v and w be the row vectors of the preceding problem. ( i i ) 22~ + 3v - 5w.

Find (i) 3u-4v,

Firs t perform the scalar multiplication and then the vector addition.

(i) 3% - 42, = 3(2, - 7 , l ) - 4(-3,0,4) = (6, -21.3) + (12,0, -16) = (18, -21, -13)

i i i ) 2 x + 3v - 5w = 2(2, -7, 1) + 3(-3, 0, 4) - 5(0, 5, -8 ) = =

(4, -14, 2 ) + (-9, 0, 12) + (0, -25, 40) (4 - 9 i- 0, -14 + 0 - 25, 2 + 12 4 40) = (-5, -39, 54)

9.8. Find x, y and x if (2x, 3, v) = (4, x + x , 22).

Set corresponding components equal to each other to obtain the system of equations

2x = 4 3 = x + x y = 2 2

Then x = 2, x = 1 and y = 2 .

9.9. Find x and ?-J if s(1,l) -C y(2,-1) = (1,4). First multiply by the scalars x and y and then add:

x ( l , 1) f y(2, -1) = (z, x) + (2y, -y) = (x + 2y, x - y) = (1, 4)

x t 2 y = 1 x - - l / = 4

Solve the system of etpations to find x = 3 and y = -1.

Now set corresponding components equal to each other to obtain

RilULTIPLICATION OF A ROW VECTOR AND A COLUMN VECTOR

( i ) Multiply corresponding coinponents and add:

(2, -3,6)(-!) = 2 . 8 + ( - 3 ) * 2 + 6*(-3) = 16 - 6 - 18 = -8

CHAP. 9J \' E CT 0 R S 87

(ii) The product is not defined since the vectors have different numbers of components.

(iii) Multiply corresponding components and add:

/ 4 ( 3 , - 5 , 2 , 1 ) = 3 . 4 + (-51.1 + 2 * ( - 2 ) 4- 1 . 5 = 1 2 - 5 - 4 + 5 = 8

f 2' 9.11. Let u = (1, li., -3) and v = \-!I . Determine k so that Z L * U = 0.

Compute z( * r and set i t equal to 0:

2 1 . 2 , = ( l , k , - 3 ) -5 = 1 - 2 T ( - 5 l . k - ( - 3 ) . 4 = 2 - 5k - 12 = -10 - 5k = 0 :i The solution t o the equation -10-5k = 0 is k = -2.

RIISCELLANEOUS PROBLEMS 9.12. Prove Theorem 9,1(i): For any vectors 11, v and w, (u Au) + zv = u + (v +w).

Let i t , , uZ and zc, be the ith components of I ! , 21 and w respectively. Then u,+ v, is the 7th component of z( - 2: and so (24% + 71,) - ti', is the zth component of ( I ( - 2:) - z ~ . On the other hand, v,+ w, is the zth component of v + ZL' and so z(, + ( v , f w,) is the ith component of u + ( t , + zc). But u,, T~ and u', are numbers for which the associative law holds. tha t is,

(u, + vt) A w, = i(, + (v, + w J , for each i

Accordingly, ( z ( - 1%) - w = u + (v + 1 1 ) since all their corresponding components a re equal.

9.13. Let 0 be the number zero and 0 the zero vector. Show that, for any vector u, O I L = 8. Method I :

component of 0(( is zero, Ozt is the zero vector, i.e. Ozt = e. Let ? I , be the ith component of u. Then 0 i t i = O is the ith component of Ozi. Since every

Method 2: We use Theorem 9.1:

0 u = (0 + 0)zc = ou + ou Now adding -0" to both sides gives

e = Ozt + (-010 = 021 + [Oa+ (-Ou)] = O H e = O I (

9.14. Prove Theorem 9.l(vi): ( k + k')u = kzi + k'u. Observe that the first plus sign refers to the addition of the two scalars k and k' whereas

the second plus sign refers to the vector addition of the two vectors k u and k'u.

Let u, be the ith component of u . Then (k+k')z i1 is the ith component of ( k + k')u. On the other hand, kii , and k'u, a r e the ith components of the vectors kzi and k'", respectively, and so ku, + k'u, is the ith component of the vector kit - k'u. But k , k' and i!, are numbers; hence

( k + k')zc, = k z ~ , + k 'ul , for each z

as corresponding components a re equal. Thus (k - k')zi = / ; I ( + k'zi,

9.15. The norm 1 u 1 of a vector u = ( u l , I [ , , . . . , u,J is defined by

IIull = du; +u; + . * . +v;

88 V E C T O R S [CHAP. 9

Supplementary COLUMN VECTORS

9.16. Compute: (i) (-!) + (ii) f

-1 -7

Problems

9.17. Compute: (i) 3 I-:) + 5 (1;) (ii) 3 - 2 (iii) - (1;) - 4

-4

9.19. Determine x and y if: (i) (. + = ("6 (ii) x (i) = 2

9.20. Determine 2, y and x if z + y (-t]i + .(8) = (-:) .

9.21. Determine r and y if x (t) = 2~ ( 2 2 ) .

ROW VECTORS 9.22. Let u = (2,-1,0,-3), w = (1, -1, -1,3) and w = (1,3, - 2 , Z ) . Find:

(iI 32(, (iij I L + 71, (iii) 2u - 3c, (ivj 5 2 4 - 3V - 4W, (v) --2l + 2V - 22c.

9.23. Find 2 and y if x(1,Z) = -4(y ,3) .

9.24. Find 2, u and x if x(l,l,O) + y(2,0, -1) + z ( O , l , l ) = (-1,3,3)

MULTIPLIC.-ITION OF A ROW VECTOR AND A COLUMN VECTOR

CHAP. 91 VECTORS 89

9.26. Determine lc so that (1, k , -2 , -3)

9.27. Prove: If u = ( u I , . . . , u J i s a row vector such tha t u . 2 ) = 0 for every column vector v with n components, then u is the zero vector.

MISCELLANEOUS PROBLEMS

9.28. Prove Theorem 9 . l (v) : For any vectors u and v and scalar li, k(ic T I S ) = k u + kv.

9.29. Prove Theorem 9.1tvii): For any vector zi and scalars k,k ' , ( k k ' ) ~ = k (k 'u ) .

9.30. Find ii) 1 1 ( 3 , 5 , -4) 11, (ii) I / (2, -3, 6, 4) 1

Answers to Supplementary Problems

9.16.

9.17.

9.18.

9.19.

9.20.

9.21.

9.22.

9.23.

9.24.

9.25.

9.26.

9.30.

(i) x = 2, y = 4 (ii) x = -1, y = - 3 / 2

z = 2 , y = 3 , x = - 2

x = O , y = 0 or x = - 2 , y = - 4

3% = (6, -3,0, -9), u + ZI = (3, -2, -1,O), 22t - 3~ = (1,1,3,-15), 52t - 3~ - 4~ = (3, -14,11, - 3 2 ) , -u + 2v - 2w = ( - 2 , - 7 , 2 , 5 )

x = l , y = - l , 2 = 2

(i) 7 (ii) Not defined (iii) 0

Chapter 10

Matrices MATRI C E S

and n columns is A rnatyiy is a rectangular array of numbers; the general form of a matrix with Tn rows

all a13 * * * , a21 a22 a23 * * * %::: '\ 1 \ . . . . . . . . . . . . . . . . . . . . .

UnLl am2 a m 3 * . ' ainn

We denote such a matrix by (a,,),,,, o r simply (a,])

and call it an n z x "rz matrix (read ( 'nz by n"). Note that the row and column of the element atj is indicated by its first and second subscript respectively.

(: -; 3 Example 1.1: Consider the 2 X 3 matrix

Its rows are (1, -3,4) and (0,5, -2) and its columns a r e

In this chapter, capital letters A, B, . . . denote matrices whereas lower case letters a, b, . . . denote numbers, which we call scalars. Two matrices A and E are equal, written A = E, if they have the same shape, i.e. the same number of rows and the same number of columns, and if corresponding elements are equal. Hence the equality of two m X n matrices is equivalent to a system of mn equalities, one for each pair of elements.

'I = (; :) ( x + y x - y x - l . 0

22 + w Example 1.2: The statement

is equivalent to the system of equations

x + u = 3 x - y = 1

2 i + w = 5 -I 2 - 2 0 = 4

The solution of the system of equations is x = 2, u = 1, 2 = 3 , w = -1.

Remark: A matrix with one row is simply a row vector, and a matrix with one column is simply a column vector. Hence vectors are a special case of matrices.

MATRIX ADDITION Let A and B be two matrices with the same shape, i.e. the same number of rows and

of columns. The sum of A and B, written A + E , is the matrix obtained by adding corresponding elements from A and B:

90

CHAP. 101 i\I A T R I CE S 9 1

all + bll a12 + 2112 * * aln + bll l \ all a12 . + bll b12 " '

. . . . . . I

a m n b m l bmz * anti + b,i,l an,? + bi7,2 amn + b,,,,/

azl + bzl a22 + b.2 aan + 212, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . be1 bz2

Note that A + B has the same shape as A and B. The sum of two matrices with different shapes is not defined.

3 0 -6) = 1 + 3 - 2 + 0 3+ f -6 ) 2 -3 11 0 + 2 4 + ( - 3 ) 5'1

Example 2.1:

Example 2.2: The sum

is not defined since the matrices have different shapes.

A matrix whose elements are all zero is called a zero matrix and is also denoted by 0. Part (iii) of the following theorem shows the similarity between the zero matrix and the scalar zero.

Theorem 10.1: For matrices A , B and C (with the same shape),

(i) ( A + B) + C = A + ( B + C), i.e. addition is associative;

(ii) A + B = B + A , i.e. addition is commutative; (iii) A + 0 = 0 + A = A.

SCALAR MULTIPLICATION

by multiplying each element of A by lz: The product of a scalar lc and a matrix A, written k A or Ak, is the matrix obtained

all a12 kall kaln 1 . . ka In

kazl ka22 9 + . k U 2 ,

G n 1 am2 * - a ka,l kam2 * ' ka,,,, . . . . . . . . . . . . . . . . . ....................

Note that A and /<A have the same shape.

/1 -2 0) = (::: 3.(--2) 3 ( 4 3 -5 3.3 3.(-5) Example 3.1:

We also introduce the following notation: -A = (-l)A and A - B = A + (-B)

The next theorem follows directly from the above definition of scalar multiplication.

Theorem 10.2: For any scalars k , and lz., and any matrices A and B (with the same shape), (i) (klk2)A = kl(kpA) (iv) 1 . A = A , and OA = 0

(ii) kl(A + B ) = k1A + k1B (v) A + (-A) = ( - A ) + A = 0.

(iii) (kl + k2)A = klA + kzA

92 RI AT RICE S [CHAP. 10

Using (iii) and (iv) above, we also have that

A + A = 2A, A + A + A = 3A, , . .

MATRIX PIULTIPLICATION Let A and B be matrices such that the number of columns of A is equal to the number

of rows of B. Then the product of A and B, written AB, is the matrix with the same number of rows as A and of columns as B and whose element in the ith row and j th column is obtained by multiplying the ith row of A by the j th column of B:

where ciJ = aiibij + a22b23 + - * * + atpbpJ. In other words, if A = (at j ) is an m x p matrix and B = ( b t l ) is a p X n matrix, then

A B = ( c i J ) is the nz X n matrix f o r which czj = atlblj + &2b& f * ' . + a z p b p j = $ a z k b k j

k = l

If the number of columns of A is not equal to the number of rows of B, say A is m x p and B is q X 71 where p # q, then the product A 6 is not defined.

val + sb , m2 + sb, ra3 + sb , t a l + x b l ta2 + ub, ta, + t tb,

Example 4.1:

Example 4.2:

1 . 1 + 1 . 3 1 * 2 + 1 ° 4 ) = (' ;) (t :)(: :) = ( 0 * 1 + 2 * 3 0 . 2 + 2 * 4

We see by the preceding example that matrices under the operation of matrix multiplication do not satisfy the commutative law, i.e. the products A B and B A of matrices need not be equal.

Matrix multiplication does, however, satisfy the following properties:

Theorem 10.3: (i) (AB)C = A(BC)

(ii) A ( B + C ) = A B + AC (iii) ( B + C ) A = B A + CA

(iv) k(AB) = (kA)B = A(kB) , where k is a scalar.

We assume that the sums and products in the above theorem are defined.

Remark: In the special case where one of the factors of A B is a vector, then the product is also a vector:

CHAP. 101 h I 4 T RI CE S 93

Example 4.3:

(2, - 3 , 4 ) -;I = (2 * 1 T (-3) * 5 + 4 . (-2), 2 . (-3) t ( - 3 ) * 0 + 4.4) = (-21,lO)

1 -3 1 * (-1) + (-3) * 2

1-2) (-1) + 4 * 2 5 * ( - 1 ) + 0 * 2 ) = “I)

Example 4.4:

A system of linear equations, such as

is equivalent t o the matrix equation

That is, any solution to the system of equations is also a solution to the matrix equation, and vice versa.

SQUARE MATRICES A matrix with the same number of rows as columns is called a square matrix. A

square matrix with 71 rows and n columns is said to be of order 11, and is called a n n-square matrix. The main diagonal, or simply diagonal, of a square matrix A = (at,) is the numbers a,,, aZ2, . . ., ~ l , ~ ~ .

Example 5.1:

The matrix

is a square matrix of order 3. The numbers along the main diagonal a re 1, -4 and 2.

The n-square matrix with 1’s along the main diagonal and 0’s elsewhere, e.g.,

1 0 0 1 0 \,o 0 0 1

I

is called the unit matrix and will be denoted by I . The unit matrix I plays the same role in matrix multiplication as the number 1 does in the usual multiplication of numbers. Specifically,

Theorem 10.4: For any square matrix A , AI = I A = A

94 RIA TRI CE S [CHAP. 10

ALGEBRA OF SQUARE MATRICES However, if we only

consider square matrices of some given order n, then this inconvenience disappears. Specifically, any 12 x n matrix can be added to or multiplied by another n x n matrix, or multiplied by a scalar, and the result is again an n X n matrix.

Recall that not every two matrices can be added or multiplied.

In particular, if A is any n-square matrix, we can form powers of A : A 2 = A A , A 3 = A 2 A , . . . and A n = Z

We can also form polynomials in A . That is, for any polynomial

f(x) = Uo + Uln : + u2x2 + * . . + U n X n

we define f ( A ) to be the matrix f ( A ) zz aoZ + alA + uzA2 + * a * + anAn

In the case that J ( A ) is the zero matrix, then A is said to be a zero or root of the polynomial J(.T).

/1 '); then A2 = Example 6.1: Let A = ( 3 -4

If f ( x ) = 2x9 - 3~ + 5, then

On the other hand, if g(x) = x2 + 3x - 10 then

g ( A ) = (-: ii) + 3(' 3 -4 ') - lo( : ;) = (' 0 0 '1 Thus A is a zero of the polynomial g(z).

TRANSPOSE

of A , in order, as columns: The transpose of a matrix A , written At, is the matrix obtained by writing the rows

a1 u2 a1 bl * * *

b2

Note that if A is an m x n matrix, then .4t is an n X nz matrix.

Example 7.1:

The transpose operation on matrices satisfies the following properties.

Theorem 10.5: (i) ( A + B)t = At + Bt (ii) (At)t = A (iii) ( ~ c A ) ~ = kAt, for k a scalar (iv) (AB)t = F A t .

CHAP. 101 MATRICES 95

Solved Problems

MATRIX ADDITION AND SCALAR MULTIPLICATION 10.1. Compute:

1 2 -3

11 -1

1 -2

(i) Add corresponding elements:

4 + 0 5 + 3 6 + ( - 5 ) 1 + 1 2 + ( - 1 )

4 5 6

(ii)

(iii) Add corresponding elements:

The sum is not defined since the matrices have different shapes.

0 + 2 (-5) + 0 1 + (-2) (-1) + (-3) 1 + 3 2 + (-5) (-3) + 6

2 -5 -1 -4 4 -3 3

3 -5 6 -1) - 0 -5 1 -1 2 0 -2 -3) - 1 2 -3

(i) Multiply each element of the matrix by the scalar 3: 3.2 3 6 12‘

3(-: ;) = (3.(-3) 3:;) = (-9 3)

(ii) Multiply each element of the matrix by the scalar -2: (-2) 1 (-2) 7 1-2 -14

-Z(i 0 -”) -1 = ( ( - 2 ) * 2 (-2) * 0 (-2)*(-3)) (-2) ’ (-1) = (,,-; ;) (iii) Multiply each element of the matrix by -1, or equivalently change the sign of each element

in the matrix: 2 -3

1 -2 -3) + q o 1 -2) 10.3. Compute: 3 % ; -: -:) - 2 \ o ’ -1 5 1 -1 -1

Firs t perform the scalar multiplication, and then the matrix addition:

3(23 -: -:> - 2 ( 0 -1 5 / ’ {1 -1 -1 -2) 1 -2 -3) 2

6 -15 3 ) + ( - 2 4 6) + (0 4 - 8 ) 9 0 -12 0 2 -10 4 -4 -4

6 + (-2) + 0 -15 + 4 + 4 3 + 6 + (-8) 4 -7 9 + 0 + 4 0 + 2 + (-4) -12 + (-10) + (-4)

96 $1 A TRI CE S -CHAP. 10

10.4. Find x, y, x and w if: x 6 4 X + Z J

\-1 2,) + j z + w 3

First write each side as a single matrix:

($ :;,) = ( x + 4 X S W - 1 2 x + 3 .?: + y + 6

Set corresponding elements equal t o each other to obtain the four linear equations 32 = 2 + 4 22 = 4

3y = z + u + 6 2 y = 6 f x

3X = z t ? C - l 22 = Zt' - 1 3w = 220 + 3 zc: = 3

or

The solution of this system of equations is z = 2, ?J = 4, z = I , w = 3.

MATRIX NIULTIPLICATION 10.5.

10.6.

10.7.

Let (Y x s) denote a matrix with shape r X s. When will the product of t w o matrices (r x s ) ( t X ZL) be defined and what will be its shape?

The product (r X s ) ( t X u) of an Y X s matrix and a t X u matrix is defined if the inner numbers s and t are equal, i.e. s = t . The shape of the product will then be the outer numbers in the given order. i.e. T X ZL.

Let (Y x s) denote a matrix with shape r x s. Find the shape of the following products if the product is defined: (i) (2 x 3)(3 x 4) (iv) ( 5 x 2)(2 X 3)

(iii) (1 X 2)(3 x 1) (vi) (2 x 2)(2 X 4) (ii) (4 x 1)(1 x 2) (v) (4 x 4)(3 x 3)

In each case the product is defined if the inner numbers are equal, and then the product will have the shape of the outer numbers in the given order.

(i) The product is a 2 X 4 matrix. (ii) The product is a 4 X 2 matrix.

(iii) The product is not defined since the inner numbers 2 and 3 are not equal.

(iv) The product is a 5 X 3 matrix.

(v) The product is not defined since the innar numbers 4 and 3 a re not equal. (vi) The product is a 2 X 4 matrix.

" -') Find (i) AB, (ii) BA. Let A = I\' 2 -1 ") and B = ( 3 -2 6 '

(i) Now A is 2 X 2 and B is 2 x 3, so the product matrix A B is defined and is a 2 x 3 matrix. To obtain the elements in the first row of the product matrix AB, multiply the first row

( l , 3 ) of A by the columns , (-:) and (-$ of B , respectively:

-* 6 :I 1 . 2 - 3 . 3 1 . 0 + 3.(-2) 1 * ( - 4 ) + 3 . 6 ) , = (11 -6

14) = ( To obtain the elements in the second row of the product matrix AB, multiply the second row (2,-1) of A by the columns of 6, respectively:

CHAP. 101

(ii)

10.8. Let

(i)

(ii)

10.9. Let

(i)

MATRICES 97

-6 14

1 3 \ /

11 = ( 2 * 2 4- (-1).3 2 . 0 -C (-1)*(-2) 2 . ( -4 ) (-1)*6

Thus 11 -6 14) 1 2 -14) A B = (

Now B is 2 X 3 and A is 2 X 2. Since the inner numbers 3 and 2 are not equal, the product BA is not defined.

Find (i) AB, (ii) BA.

Now A is 1 X 2 and B is 2 X 3, so the product A B is defined and is a 1 X 3 matrix, i.e. a row vector with 3 components.

To obtain the elements of the product AB, multiply the row of A by each column of B:

A B = (2,1) (i -: -:) = ( 2 . 1 + 1.4, 2. ( -2) + 1 . 5 , 2 . 0 + 1.(-3)) = (6,1,-3)

Now B is 2 x 3 and A is 1 X 2. Since the inner numbers 3 and 1 are not equal, the product BA is not defined.

1 2 -1 A = i 1 0 ) a n d B = (i -f -:I. Find (i) AB, (ii) BA.

\-3 4

Now A is 3 X 2 and B is 2 X 3, so the product A B is defined and is a 3 X 3 matrix. To obtain the first row of the product matrix AB, multiply the first row of A by each column of B, respectively:

\

= r -8 -I0?

2 . 1 + (-1)*3 2 . ( -2) + ( -1)*4 2 . ( -5 ) + ( - 1 ) * O =I : To obtain the second row of the product matrix AB, multiply the second row of A by each column of B , respectively:

-8 -10 -1 -8 -10 1 . l t 0 . 3 1 * ( - 2 ) + 0 * 4 1 * ( - 5 ) & 0 - 0 ) = 1 -2 - 5 )

To obtain the third row of the product matrix AB, multiply the third row of A by each column of B, respectively:

-8 -10 -1 -8 -10 -3 4 / \

-1 1 -2 -5 =I : (-3) 1 + 4 * 3 (-3) * (-2) + 4 4 (-3) * ( - 5 ) + 4 * 0 9 22

Thus -1 -8 -10

A B = i/ --: ---I (ii) Now B is 2 x 3 and A is 3 X 2, so the product BA is defined and is a 2 X 2 matrix. To obtain

the first row of the product matrix BA, multiply the first row of B by each column of A , respectively:

98 AT AT RICE S [CHAP. 10

= ;" 1 ' 2 + (-2) ' 1 + (-5) * (-3) 1 * (-1) + (-2) * 0 + (-5) 4

= ( To obtain the second row of the product matrix B A , multiply the second row of B by each column of A :

15 -21 ) = (10 -3) 15 -21

= ( 3 . 2 + 4 . 1 + 0*(-3) 3 * ( - 1 ) + 4 . 0 + 0 . 4

Thus 15 -21 = (10 -3)

Remark: Observe t h a t in this case both A B and B A are defined, but A B and B A are not equal; in fact, they do not even have the same shapes.

(i) Determine the shape of AB. j t h column of the product matrix AB, that is, AB = (el!).

(ii) Let cij denote the element in the ith row and Find: c23, c14, CPI and CIZ.

(ii

(ii)

Since A is 2 X 3 and B is 3 X 4, the product A B is a 2 X 4 matrix.

Now c I j is defined as the product of the ith row of A by the j th column of B . Hence:

/ o \

/ 1\ C14 = (2,-1,O) -1 = 2 ~ 1 + ( - 1 ) ~ ( - 1 ) + 0 ~ 0 = 2 + 1 + 0 = 3

0:'

c21 = = 1 . 1 + 0 . 2 + ( - 3 ) * 4 = 1 + 0 - 12 = -11

~ 1 2 1 (2, -1, ,,(Tb) = 2 * ( - 4 ) + ( - l )*( -1) + 0 . 0 = -8 + 1 + 0 = -7

(ii) ( ' \ -3 5 / - 7 )

(i) The first factor is 2 X 2 and the second is 2 X 2, so the product is defined and is a 2 X 2 matrix:

1 * 4 + 6 * 2 1 * 0 + 6 * ( - l ) (-3) 4 + 5 2 (-3) 0 + 5 ' (-1) -2 -5

CHAP. 101 RIA TRICES 99

(ii) The first factor is 2 X 2 and the second is 2 X 1, so the product is defined and is a 2 X 1 matrix

/ 1 . 2 + 6 . ( - 7 ) ) -

\ ( - 3 ) * 2 + 5 * ( - 7 ) / -

(iii) Now the first factor is 2 X 1 and the second is 2 X 2. Since the inner numbers 1 and 2 are distinct, the product is not defined..

(iv) Here the first factor is 2 X 1 and the second is 1 X 2, so the product is defined and is a 2 Y, 2 matrix:

(v) The first factor is 1 X 2 and the second is 2 j: 1, so the product is defined and is a 1 X 1 matrix which we frequently write as a scalar.

(2, -1) (-:) = ( 2 . 1 + (-1). (-6)) = 181 = 8 /

SQUARE MATRICES

. Find (i) A', [ i i ) As, (iii) f ( A ) , where J (x ) = 2x3 - 42, + 5 . \ 4 -3 10.12. Let A = I

(iv) Show tha t A is a zero of t he polynomial g(x) = M? 2s - 11.

(i) A2 = A A = (: -;)(: -$ 9 -41

\-8 IT, 1 . 1 + 2 . 4 1 . 2 + 2 . ( -3 )

4 . 1 + ( - 3 ) * 4 4 . 2 ( -3) . ( -3)

1 G O -67 1 . 9 $- 2.(-8) 1 . ( - 4 ) f 2 . 1 7

4 * (-4) + (-3) * 17 4 9 + (-3) * (-8)

(iii) To find J ( A ) , first substitute A for x and 51 for the constant 5 in the given polynomial m = 2x3 - 4 x - 5 :

Then multiply each matrix by its respective scalar:

Lastly, add the corresponding elements in the matrices:

-13 52') 104 -117

- / - 1 4 - 4 + 5 6 0 - 8 + 0 ) -133 + 1 2 - 5, - 1

\ l 2 0 - 16 + 0

(iv) Now A is a zero of g(x) if the matrix g ( A ) is the zero matrix. for J ( A ) , i.e. first substitute A for .z and 111 fo r the constant 11 in g ( x ) = x2+ 22 - 11:

Compute g ( A ) as was done

Then multiply each matrix by the scalar preceding it:

100 >I AT RI CE S ;CHAP. 10

Lastly, add the corresponding elements in the matrices.

/ 9 + 2 - 11 -4 + 4 i- o:, - s ( A ) = (-8 t 8 + 0 17 - 6 - 111

Since g iA) = 0, A is a zero of the polynomial g(z).

10.13. Let B = [' 3 -4" 0 -1

Firs t coinpute B2:

and let j ( x ) = x2 - 4x - 3 . Findf(B).

2 2 BL = B B =

1.1 + 2 . 3 + 0 . 0 3 . 1 + ( - 4 ) . 3 + 5 . 0 3 . 2 + (-4).(-4) - 5*(-1) 3 . 0 + ( - 4 ) * 5 + 5 . 2 0 * 1 + (-1) * 3 + 2 * 0 0 - 2 + (-1) * (-4) - 2 * (-1) 0 . 0 + ( -1)*5 + 2 . 2

1 . 2 + 2 . ( - 4 ) T 0 * ( - 1 ) 1 . 0 + 2 . 5 + 0 . 2

-3 2 -1/

Then

2 f ( B ) = 6' - 4B + 31 =

0 -1 0 0 1

-6

-3 -1 - - (-: 1; -;:) +

7 - 4 + 3 - 6 - 8 + 0 1 0 - 0 + 0 6 -14 = (-9 - 1 2 - 0 1 7 + 1 6 + 3 -10 - 20 + O ) = (-21 3; --;:I

- 3 - 0 + 0 2 + 4 + 0 - 1 - 8 + 3 -3 -6

. Find a non-zero column vector 2~ = such that AZL = 3u. j 4 -3

10.14. Let -4 = /1

Firs t set up the matrix equation Au = 3u:

Write each side as a single matrix (column vector):

Set corresponding elements equal to each other to obtain the system of equations

x + 321 = 3 2 -2.1. + 321 = 0 or

4x - 3y = 3y 4~ - 621 = 0

The two linear equations a re the same (see Chapter 11) and there exist a n infinite number of

solutions. One such solution is N = 3, = 2. In other words, the vector u = ("2> is non-zero and has the property t h a t A u = 32~.

CHAP. 101 X A T R I C E S 101

1 0 1 0 \ 10.15. Find the transpose At of the matrix A =

0 3 4 "\

TRANSPOSE

Rewrite the rows of A a s the columns of At: A t = [l . 0 5 4

10.16.Let A be an arbitrary matrix. Under what conditions is the product AAt defined? Suppose A is an m X n matrix: then At is n x m . Thus the product AAt is always defined.

Observe tha t AtA is also defined. Here AAt is an m X m matrix, whereas AtA is a n n X n matrix.

/ 10.17. Let A = ") Find (i) AAt , (ii) AtA. j 3 -1 4 *

1 3\ To obtain At, rewrite the rows of A as columns: At = (i -:). Then

- 1.1 + 2 * 2 + 0 . 0 1.3 + 2*(-1) + 0 . 4 - (3-1 + (-1)*2 + 4 . 0 3.3 + (-1)*(-1) -I- 4 . 4 ) = ( n 2;)

1.1 + 3 . 3 1 . 2 + 3*(-1) 10 -1 12

0.1 + 4 - 3 0 . 2 + 4*(-1) 0 . 0 + 4 . 4 12 -4 16 = 2.1 + (-1).3 2 . 2 + (-l)*(-l) 2 . 0 + ( - 1 ) . 4 i

PROOFS 10.18. Prove Theorem 10.3(i): (AB)C = A(BC) .

Let A = (a , ] ) , B = ( b j k ) and C = (chi). Furthermore, let A B = S = ( s , k ) and BC = T = (t,J. Then m

s ,k = a l l b l k a12brk -k f aambm, = x a,,bJh

t j i = b j i c i i + b@c,i + " ' + bjncnL = 2 b,hChl

Now multiplying S by C, i.e. (AB) by C, the element in the i th row and lth column of the matrix

S,1C11 + S,$21 + ..' f s l , I C n ~ = 2 S l k c k [ = 2 ( a L I b & ? d

On the other hand, multiplying A by T , i.e. A by BC, the element in the ath row and Zth column of the matrix A(BC) is

1 = 1 n

h = l

(AB)G is n n m

k = l ) = 1 k = l

Since the above sums a r e equal, the theorem is proven.

102 MATRICES [CHAP. 10

10.19. Prove Theorem 10.3(ii): A ( B + C) = AB + AC. Let A = ( a i j ) , B = (bik) and C = (cjk). Furthermore, let D = B + C = (djk), E = A B = (eiJ

and F = AC = ( f i k ) . Then

d j k = b j k + m

j = 1 e i k = ailbl , + a i i b z k + * . * + a imbn ,k = CLijbjk

a l l C l k a l a c a k + . ' * + a i ~ r l c i n k = x a i j c j k j = 1

fiirc

Hence the element in the ith row and kth column of the matrix A B + AC is ??1 m m

j = 1 j = 1 j = 1 e i k + f i k = aijbJi, + 2 CLijCjk = x a i j ( b j k + Cjk)

On the other hand, the element in the ith row and kth column of the matrix A D = A ( B + C ) is m in

i = 1 j = 1 a i l d l k f CLi2dzk f ' ' ' + a i l n d m k = 2 a,,dj, = 3 U, j (b jk + Cjk)

Thus A ( B + C) = A B + AC since the corresponding elements are equal.

10.20. Prove Theorem 10.5(iv): (AB) t = P A t .

Let A = (a, ] ) and B = ( b J . Then the element in the ith row and j t h column of the matrix A B is

allb,, + a22b23 + . * * + a n n b m , (1 1 Thus (1) is the element which appears in the j th row and ith column of the transpose matrix (AB)t .

On the other hand, the j th row of Bf consists of the elements from the j t h column of B:

( b l J bzt ". b W I J ) (21

Furthermore, the ith column of A f consists of the elements from the i th row of A :

i'i a i m

( 3 )

Consequently, the element appearing in the j t h row and ith column of the matrix BtAt is the product of ( 2 ) by (3 ) which gives (1) . Thus (AB)t = BtAt.

Supplementary Problems MATRIX OPERATIONS

For Problems 10.21-23, let 2 -3 0 -:), C = ( 5 -1 -4 ") and D = (,--!I

- 1 0 0 3

10.21. Find: (i) A + B , (ii) A + C, (iii) 3A - 4B.

10.22. Find: i i ) A B , (ii) AC, (iii) A D , (iv) BC. (v) BD, (vi) CD.

10.23. Find: ( i ) At , (ii) AtC, (iii) DtAt, (iv) BtA, (v) DtD, (vi) DDt.

CHAP. 101 $1 AT R I CE S 103

SQUARE MATRICES l 2 10.24. Let A = ') (i) Find A2 and A3, (ii) If f ( x ) = 2 3 - 3 9 - 2 r + 4, find f ( A ) . (iii) If j 3 -1 .

g ( ~ ) z .(.? - .C - 8, find g ( A ) .

(i) If f ( x ) = 2.c2 - 4.c + 3, find f ( B ) . (ii) If g ( r ) = r 2 - 42 - 12, find g ( B ) . j 5 3 . f " such t h a t Bu = 6 2 4 . (iii) Find a non-zero column vector ti = \ Y)

f 1 "> 10.25. Let B =

10.26. Matrices A and E are said to commute if A B = B A . Find all matrices (E s> which commute

with (t t ) .

2\ , Find An. b 1 ) 10.27. Let A =

PROOFS 10.28. Prove Theorem 10.2iiii): ( k , + k,)A = klA + k,A.

10.29. Prove Theorem 10.4[iii): k(AB) = ( k A ) B = A ( k B ) .

10.30. Prove Theorem 10.5(i): ( A + B)t = A + + Bt .


10.21. ( i ) (-! -; (ii) Not defined. iiiii ( 17

10.22. (i) Not defined. (iii) (') - 5 ) (vi) Not defined. 11 -12 5 -2 4

(ii) (T1 -3 -12 18

1 0 4 -7 4 -2

2 4 -3 12 10.23. (i) (-1 3) (ii) Not defined. (iii) ( 9 , 9 ) ( i d ( 0 -6 -1) (v) 14 (vi) (-: -3 1 -!) 10.24. (i) A2 = (Io 2 ) , A3 = (26 Is\ (ii, f ( A ) = ( y : -1$ (iii) g ( A ) =

3 1 ) 27 -1)

10.26. Only matrices of the form ( 8 i) commute with

Chapter 11

linear Equations

LINEAR EQUATION IN TWO UNKNOWNS A linear equation in two unknowns x and y is of the form

ax + by = c

where a, b , c are real numbers. the equation, that is, for which

is a true statement. to x and solving fo r y (or vice versa).

We seek a pair of numbers N = (k,, k2) which satisfies

ak, + bk, = c Solutions of the equation can be found by assigning arbitrary values

Example 1.1 : Consider the equation

If we substitute :c = -2 in the equation, we obtain

2 r + y = 4

2 2 0 T 3 -2

2 * ( - 2 ) + y = 4 or - 4 + y = 4 o r y = 8

Hence (-2,8) is a solution. If we substitute N = 3 in the equation, we obtain

2 * 3 + y = 4 or 6 + y = 4 o r y = -2

Hence (3, -2) is a solution. The table on the right lists six possible values for z and the corresponding values for y, i.e. six solutions of the equation.

Now any solution a = (k, ,k,) of the linear equation ax + b y = c determines a point in the Cartesian plane R'. If a, b are not both zero, the solutions of the linear equation correspond precisely to the points on a straight line (whence the name linear equation).

Example 1.2: Consider the equation 2x + y = 4 of the preceding example. On the r ight we have plotted the six solutions of the equation which appear in the table above. Note tha t they all lie on the same line. We call this line the graph of the equation since i t corresponds precisely to the solution set of the equation. Graph of 2 x + y = 4

TWO LINEAR EQUATIONS IN w o UNKNOWNS We now consider a system of two linear equations in two unknowns x and p:

U,X + b,y = c,

uPx + b2y = c2

104

CHAP. 111 LINEAR EQUATIONS 105

A pair of numbers which satisfies both equations is called a simultaneous solution of the given equations o r a solution of the system of equations, There are three cases which can be described geometrically. (Here we assume that the coefficients of x and y in each equation are not both zero.)

t z - c y = 1 x + y = 1 x - y = -3

x + 2 y = 3 2 2 f 2y = 6 3x + 3y = 3

(a ) ( b ) (4

Fig. 10-1

T h e system has exactly one solution. Here the lines corresponding to the linear equations intersect in one point as shown in Fig. 1 0 - l ( ~ ) above.

T h e system has no solutions. Here the lines corresponding to the linear equations are parallel as shown in Fig. lO- l (b ) above.

T h e systenz has an infinite number o f solictions. linear equations coincide as shown in Fig. 10-l(c) above.

Now the special cases 2 and 3 can occur if and only if the coefficients of x and the

1.

2.

3. Here the lines corresponding to the

bl a2 b2

coefficients of y are proportional: - - - -

There are then two possibilities:

(i)

(ii)

If the constant terms of the equations are in the same proportion, i.e.

Cl

a, b2 c2

- - _ - - b 1 - -

then the lines are coincident. Hence the system has an infinite number of solutions which correspond to the solutions of either equation.

If the constant terms of the equations are not in the same proportion, i.e. C1 f - b1 a,

a, b2 CZ _ - - -

then the lines are parallel. Hence the system has no solution.

Example 2.1: Consider the system 3% + 6y = 9 2% + 4y = 6

Note tha t 3/2 = 6/4 = 9/6. Hence the lines are coincident and the system has an infinite number of solutions which correspond to the solutions of either equation.

106 LINEAR EQUATIONS [CHAP. 11

Particular solutions of the first equation and hence of the system can be obtained as shown in Example 1.1. For example, substitute x = 1 in the first equation t o obtain

3 * 1 + 6 2 / = 9 or 3 f 6 y = 9 o r 6y = 6 or y = 1

Thus x = 1 and y = 1 or, in other words, the pair of numbers a = (1,l) is a solution of the system. By substituting other values fo r x (or y) in either equation, we obtain other solutions of the system.

Example 2.2: Consider the equations 3x + 62/ = 9

2x -t. 4y = 5

Note tha t 312 = 614 but 312 = 614 # 915. Hence the lines are parallel and the system has no solution.

On the other hand, if the coefficients of the unknowns are not proportional, i.e.

bl a2 b2

f - a, -

then case 1 occurs; that is, the system has a unique solution. This solution can be obtained by a process known as elimination, i.e. by reducing the equations in two unknowns to an equation in only one unknown. This is accomplished by:

(i) multiplying each of the given equations by numbers such that the coefficients of one of the unknowns in the resulting equations are negatives of each other;

(ii) adding the resulting equations.

We illustrate this method in the following example.

Example 2.3 : Consider the equations ( 1 ) 32 + 2y = 8

(2 ) 22 - 5y = -1

Note t h a t 3 / 2 i. 21-5, so the system has a unique solution. and (2 ) by -3 and then add in order to “eliminate” 2:

Multiply ( 1 ) by 2

2 X ( 1 ) : 6% + 4y = 16

-3 X (2 ) : - 6 ~ + 1571 = 3

Addition: 19y = 19 o r y = 1

Substitute y = 1 in ( 1 ) to obtain 3 x + 2 * 1 = 8 o r 3 x + 2 = 8 o r 3% = 6 or 2 = 2

Thus x = 2 and y = 1 or, in other words, the pair 01 = ( 2 , l ) is the unique solution to the given system.

GENERAL LINEAR EQUATION

xl, x,, . . ., xI1 .

where a,,a,, . . .,a,,, b are real numbers. The numbers at are called the coefficients of xi , and b is called the constant of the equation. An n-tuple of real numbers (Y = (kl, k,, , , . , k,), i.e. an n-component row vector, is a solution of the above equation if, on substituting ki for xl, the statement

a,k, + a2k2 + . - . + a,kI1 = b

is true. We then say that N satisfies the equation.

We now consider a linear equation in any arbitrary number of unknowns, say Such an equation is of the form

a,x, + a p , + * . . + a?txll = b


Example 3.1 : Consider the equation x + 2 y - 4 x + w = 3 Now 01 = (3,2,1,0) is a solution of the equation since

3 + 2 . 2 - 4 * 1 + 0 = 3 or 3 + 4 - 4 + 0 = 3 or 3 = 3 is a t rue statement. equation since

On the other hand, /3 = (1 ,1 ,2 ,2) is not a solution of the

1 + 2 * 1 - 4 * 2 + 2 = 3 o r 1 + 2 - 8 + 2 = 3 or -3 = 3 is not a t rue statement.

A linear equation is said to be d e g e n e m t e if the coefficients of the unknowns are all zero. There are two cases:

(i)

(ii)

The constant is not zero, i.e. the equation is of the form

Then there is no solution to the linear equation.

The constant is also zero, i.e. the equation is of the form

Then every n-tuple of real numbers is a solution of the equation.

0x1 + 0x2 + . . . + Oxn = b, with b i 0

0x1 + 0x2 + * * * + O X , = 0

GENERAL SYSTEM O F LINEAR EQUATIONS A system of nz equations in n unknowns n.,, x2, . . . , xn is of the form

all x1 + U 1 z X 2 + * ' ' + alnxn = Z;I

a21 x1 + U 2 2 X Z + * ' ' + ~ 2 n x n = bz

am1 XI + am2 ~2 + * ' + a m n x n = bm

where the ail, b, are real numbers. satisfies all the equations is called a solution of the system.

where every a,, = 0. There are two cases:

4 n n-tuple of numbers O( = (kl, k,, . . . , kJ which

We first consider the special case where all the above equations are degenerate, i.e.

(i) If one of the constants b, # 0, i.e. the system has an equation of the form

then this equation, and hence the system, has no solution. 0x1 + 0x2 + . - - + Ox, = b,, with b, # 0

(ii) If every constant bt= 0, i.e. every equation in the system is of the form 0x1 + 0x2 + . * + Oxn = 0

then each equation, and hence the system, has every n-tuple of real numbers as a solution.

Example 4.1: The system ox + o y + o x + ow = 0 ox + oy + o x + ow = 4 O x + O y + O x + O w = -1

has no solution since one of the constants on the right is not zero.

On the other hand, the system ox+ o y + o x + o w = 0 ox + oy + o x + ow = 0 ox + oy + o x + ow = 0

has every 4-tuple a = (kl , k,, k,,k,) as a solution since all the constants on the right a r e zero.


In the usual case where the equations are not all degenerate, i.e. where one of the a,,#O, we reduce the system ( 1 ) to a simpler system which is equivalent to the original system, i.e. has the same solutions. This process of reduction, known as (Gauss) elimination, is as follows:

(i) Interchange equations and the position of the unknowns so that all#O. (ii) Multiply the first equation by the appropriate non-zero constant so that a,, = 1. (iiij For each i > 1, multiply the first equation by -ai, and add i t to the ith equation

so that the first unknown is eliminated.

Then the system (1) is replaced by an equivalent system of the form XI + a:;"& -+ ( g 3 X 3 + . . . + al,x, b l

aZ2x2 -+ a2353 + - . . + a&x,, = b;

uLx.2 + am353 + * . * + am,5,, = b:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Example 4.2: Consider the system 2 ~ + 6 y - x = 4

3 x - 2 2 y - x = 1

5% + 9y - 22 = 12

Multiply the first equation by 4 so that the system is replaced by

x + 3 y - * x = 2

3 % - 2 2 y - x = 1

5x + 9y - 2x = 12

Multiply the new first equation by -3 and add it to the second equation so that the system is replaced by

x + 3 y - * x = 2

-1ly + 3 2 = -5

52 + 9y - 2x = 12

Lastly multiply the first equation by -5 and add it to the third equation so that the system is replaced by the following which is in the desired form:

x + 3 y - + z = 2

-1ly + 42 = -5

-6y + $2 = 2

Example 4.3: Consider the system of the preceding example. simpler form without first changing the leading coefficient to 1. the advantage of minimizing the number of fractions appearing.

nate x from the second equation:

Here we reduce the system to a This method has

Multiply the first equation by 3 and the second by -2 and then add to elimi-

3 X first: G X + 1 8 ~ - 3x = 12

-2 X second: -6.2: + 4x1 + 22 = -2

Addition: 22r/ - x = 10

Now multiply the first equation by 5 and the third equation by -2 to eliminate 3 from the third equation:

5 X first: lox T 30y - 52 = 20

-2 X third: -lox - 18y + 4x = -24

Addition: 12y - 2 = -4

CHAP. 111 LIKEAR EQUATIONS 109

Thus the system is equivalent t o

2x + 61/ - x = 4 r + 3 u - 3 z = 2 22y - z = 10 o r 22y - z = 10 122) - z = -4 12y - z = -4

Continuing the above process, we obtain the following fundamental result:

Theorem 11.1: Every system of m equations in n unknowns can be reduced t o an equivalent system of the following form:

x1 f c12x2 + C 1 3 s 3 + * ' ' f C ~ T ~ T f ' ' ' + ClnT,, = dl, ?" ??%

x 2 + c.323 + . . . + czrx7 + * . . + c2,zxn = dz s j + ' . + c3rrr + * * * t C?,,Xn = dS

X r + . . . + c77,xn = a, 0 = d T + l

o = d ,

Remark: A system of linear equations in the above form, where the leading coefficient is not zero, is said to be in echelon form.

The solutions of a system of equations in the echelon form ( 2 ) can easily be described

. . . , d m are not all zero, i.e. there is an

and found.

(i) Inconsisteizt Equutiotis. If the numbers equation of the form

then the system is said to be inconsistent, and there a re no solutions. Oxr + Oxz + . . . + Ox,, = dk, with d h = 0

(ii) Consistent Equations. If the numbers d,,,, . . . , d m are all zero, then the system is

Then we can successively solve uniquely for the unknowns xr, x r p l , . . ., x, and so there exists a unique solution for the system.

Then we can arbitrarily assign values to the unknowns x,,,, , , .,xn and then solve uniquely for the unknowns xr, , . . , x,. L4ccordingly there exists an infinite number of solutions.

said to be consistent and there does exist a solution. (a) r=n, that is, there are as many non-zero equations as unknowns.

There are two cases:

( b ) r<n, that is, there are more unknowns then there are non-zero equations.

The following diagram shows the various cases:

System of linear equations

I Inconsistent

I I I I

solution solution


Example 4.4:

Example 4.5:

Example 4.6:

Consider the system

x + 2 y - O x + y +

ox + Oy +

3x = 4 x+2y-332 = 4 2x = 5 o r y + 2 x = 5 ox = 3 0 = 3

Since the system has an equation of the form 0 = c with c # 0, the system is inconsistent and has no solution.

o x + y + 2 z = 5 y + 2 x = 5 O x + O y + x = 1 z = 1

o r o r y t 2 x = 5

x = l ox + oy + 02 = 0 o = o

Since there is no equation of the form 0 = c, with c f 0, the system is consistent. Furthermore, since there a r e three unknowns and three non-zero equations, the system has a unique solution. Substituting x = 1 in the second equation we obtain

y + 2 . 1 = 5 or y + 2 = 5 or y = 3

Substituting z = 1 and y = 3 in the first equation, we obtain

x f 2 . 3 - 3 . 1 = 4 or x + 6 - 3 = 4 o r x + 3 = 4 or s = l

Thus x = 1, y = 3 and x = 1 or, in other words, the ordered triple (1 ,3 ,1) is the unique solution of the system.

Consider the system z + 2 y - 3 x + w = 4 2 + 2 y - 3 x + w = 4

y + 2 x + 3 w = 5 0 2 + y + 2 2 + 3 w = 5 or 02 + o y + O x + Ow = 0 o = o

The system is consistent, and since there a re more unknowns than non-zero equations the system has an infinite number of solutions. In fact, we can arbitrarily give values to z and zu and solve for x and y. To obtain a particular solution substitute, say, x = 1 and w = 2 in the second equation to obtain

y + 2 * 1 + 3 * 2 = 5 or y + 2 + 6 = 5 o r y + 8 = 5 or y = - 3

Substitute y = -3, z = 1 and w = 2 in the first equation to obtain

x + 2 * ( - 3 ) - 3 * 1 + 2 = 4 or x - 6 - 3 + 2 = 4 or x - 7 = 4 o r x = 11

Thus x = 11, y = -3, x = 1 and w = 2 or, in other words, the row vector (11, -3,1,2) is a specific solution to the system.

To obtain the general solution to the system substitute, say, z = a and zu = b in the second equation t o obtain

y + 2 u + 3 b = 5 o r = 5 - 2 a - 3 b

Substitute y = 5 - 2a - 3b, x = u and t*! = b into the first equation to obtain

x + 2 ( 5 - 2 a - 3 b ) - 3 a + b = 4 or ~ + 1 0 - 4 ~ - 6 b - 3 3 a + b = 4

o r z + 1 0 - 7 u - 5 b = 4 o r s = 7 a + 5 b - 6

Thus the general solution of the system is

(7a + 5b - 6, 5 - 2a - 3b, a, b), where a and b are real numbers

Frequently, the general solution is left in terms of z and w (instead of a and b ) as follows:

or, ( 7 ~ + 5 ~ - 6 , 5 - 2 ~ - 3 ~ , 2, W)

x = 72 f 5 % ~ - 6

y = 5 - 2x - 3w

The previous remarks give us the next theorem.

Theorem 11.2: A system of linear equations belongs to exactly one of the following cases: 1. 2 . 3.

Recall that, in the special case of two equations in two unknowns, the above three cases correspond to the following cases described geometrically:

1. 2. 3.

The system has no solution. The system has a unique solution. The system has an infinite number of solutions.

The two lines are parallel. The two lines intersect in exactly one point. The two lines are coincident.

HOMOGENEOUS SYSTEMS OF EQUATIONS A system of linear equations is called homogeneous if i t is of the form

allxl + u12x2 + ' * . + u1nxn = 0 a21 x1 + u22x2 -t ' ' ' + uznx, = 0

am1 x1 + urn222 + . * + am, x n = 0

( 3 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

that is, if all the constants on the right are zero. equivalent to a system of the form

By Theorem 11.1, the above system is

X1 + ClZX. + * * * + C1,Zr + * . * + C l n X n = 0 , 7' L m

X 2 + - . * + C 2 , X r + * * * + C z n X n = 0

o = o Clearly, the above system is consistent since the constants are all zero; in fact, the system ( 3 ) always has the zero solution (0, 0, , , . 0) which is called the trivial solution. It has a non-trivial solution only if r < n. Hence if we originally begin with fewer equations than unknowns in ( 3 ) , then T < n and so the system always has a non-trivial solution. That is,

Theorem 11.3: A homogeneous system of linear equations with more unknowns than equations has a non-zero solution.

Example 5.1 : The homogeneous system

has

Example 5.2: We

x + z y - 3 x + w = 0 x - s y + x-zw = 0

2 x + y - 3 x + 5 w = 0 a non-trivial solution since there a r e more unknowns than equations.

reduce the following system to echelon form: z + y - x = 0 x + y - z = 0 x + y - x = 0

y-"x = 0 2 % - 3 y - t x = 0 t o -5y + 3x = 0 t o 5

x - 4 y + z 2 = 0 -5y + 32 = 0 o = o

112 LINEAR EQUATIONS ‘CHAP. 11

The system has a non-trivial solution since we finally have only two non-zero equations in three unknowns. For example, choose r = 5 ; then y = 3 and TC = 2. In other words ( 2 , 3 , 5 1 is a particular solution.

We reduce the following system to echelon form: Example 5.3: x + y - r = 0 x + u - r = 0 x + y - 2 = 0

2 x + 4 y - 2 = 0 to 2 y + r = 0 to y + + z = 0 3 x + 2 y + 2 2 = 0 - y - z = 0 - 3 2 = 0

z + y - z = 0 to U A * % = 0

r = O Since in echelon form there a r e three non-zero equations in the three unknowns, the system has a unique solution, the zero solution ( 0 , 0 , 0).

MATRICES AND LINEAR EQUATIONS The system (1 ) of linear equations is equivalent t o the matrix equation

. . . . . . . . . . . . . . . . . am1 d m 2 * * . a m n

in that every solution to (1) is a solution to the above matrix equation and vice versa. In other words, if A = ( a , , ) is the matrix of coefficients, X = (x,) is the column vector of unknowns, and B = ( b J is the column vector of constants, then the system ( I ) is equivalent to the simple matrix equation

A X = B In particular, the homogeneous system of linear equations ( 3 ) is equivalent to the matrix equation

A X = 0

Example 6.1: The following system of linear equations and the matrix equation are equivalent:

2 2 + 3 y - 4 x x - 2 y - 5 5 x = = 3 ’ 7 . (; -; I;)(:) = (;I In studying linear equations, it is usually simpler t o use the language and theory of

The general solution of a non-homogeneous system of linear equations A X = B is obtained by adding the general solution of the homogeneous system AX = 0 to a particular solution of the non-homogeneous system A X = B.

Proof. Let a , be a fixed solution of A X = B. If p is any solution of the homogeneous

matrices.

Theorem 11.4:

system A X = 0, then

That is, the sum nl + p is a solution to the non-homogeneous system A X = B. A(a,+/3) = A a , + A P = B + O = B

On the other hand, suppose a, is a solution of AX = B distinct from a,. Then A(a2 - a,) = Aa, - Aa, = B - B = 0

That is, the difference N., - a1 is a solution of the homogeneous system A X = 0.

hence every solution of A X = B can be obtained by adding a solution of AX=O to the particular solution a1 of A X = B.

But n2 = a, + ( a 2 - a,)


Theorem 11.5: If n l , a?, , . . , all are solutions to the homogeneous system of linear equations A X = 0, then every linear combination of the of the form

k,a, + k p , + . . . + li,,a,, where the k L are scalars, is also a solution of the homogeneous system AX = 0.

Proof. We are given that Aa, = 0, . . ,, ila,, = 0. Hence

A ( k , n , - . . . + k,{a,,) = klAnl + . . + k,Aa, = k,O + . . . + k,,O = 0

Accordingly, k,a, + . . . T k p r l is also a solution of the homogeneous system A X = 0. In particular, every multiple ke of any solution a of A X = 0 is also a solution of AX = 0.

Example 6.2: Consider the homogeneous system in Example 5.2: . r + y - x = 0

2.r - 3y + x = 0

. r - 4 y + 2 x = 0

As was noted, (Y = 12,3 ,5) is a solution of the above. such as (4 ,6 ,10 ) and i -6, -9, -15), is also a solution.

Hence every multiple of a ,

We now restate Theorem 11.2 using the language of matrix theory, and give an independent proof of the theorem.

non-trivial solution to the homogeneous system A X =

A ( a - P ) =z An -- AP = B Hence every multiple k ( ~ - p ) of a - p is a solution scalar k,

is a distinct solution to A X = B. Thus AX = B has claimed.

n + k ( N - p)

Theorem 11.2:

Proof.

If A X = B has more than one solution, then it has an infinite number of solutions.

Then the difference N - ,5 is a Let a and 13 be distinct solutions to AX = B. 0:

- B = 0 of A X = O . Accordingly, fo r each

a n infinite number of solutions as

VECTORS AND LINEAR EQUATIONS The system ( I ) of linear equations is also equivalent to the vector equation

X1 + x2 r)+ . . . + x s t) =

an, 1 am2 &n bln

In other words, if ( Y ~ ,

the system (1) is equivalent to the vector equation . . . , an and p denote the above column vectors respectively, then

X p , + S p 2 + . ’ ’ A X n N n = p If the above equation has a solution, then p is said to be a linear combination of the vectors

or is said to depend upon the vectors We restate this concept formally:

The vector p is a linear combination of the vectors al, . . . , N~ if there exist scalars k, , . . . , kll such that

p = k l N l 4- k2a2 + * * . + k,a,

114 L I N E A R EQUATIONS [CHAP. 11

i.e. if there exists a solution to

p = X,a, + X2a2 + . . . + 5"an

where the X, are unknown scalars.

Remark: The above definition applies to both column vectors and row vectors, although our illustration was in terms of column vectors.

ExampIe 7.1: Let

/ 3 = I; i ) , (ul = (i), a2 = (a) and a3 = ($ :-4

Then /3 does depend upon the vectors a,, ue and a3 since

p = -4q + 7a2 - a3

that is, the equation (or system)

2 = x + y + x 3 = x + y

-4 = x

has a solution ( -4 ,7 , -1).

VECTORS AND HOMOGENEOUS LINEAR EQUATIONS

The system ( 3 ) of homogeneous linear equations is equivalent to the vector equation

XI :"u + x2 [I+... + Xn f) = [I a m i anlz Clmn

That is, if a,, a2, . . . , an denote the above column vectors respectively, then the homogeneous system ( 3 ) is equivaIent to the vector equation

X,a1 + x2a2 + * . . + Xnafl = 0

If the above equation has a non-zero solution, then the vectors al, . . . ,an are said to be dependent; on the other hand, if the above equation has only the trivial (zero) solution, then the vectors are said to be independent. We restate this concept formally:

1 Definition: I The vectors al, a*, . . . , a l l are dependent if there exists scalars k,, k,, . . . , kn, not all zero, such that

k p , + kznz + * * ' + k p , = 0

Z,a, + X2a2 + ' * + X,Ian = 0

i.e. if there exists a non-trivial solution to

where the xi are unknown scalars. in dependent.

Otherwise, the vectors are said to be

Remark: The above definition applies to both column vectors and row vectors although our illustration was in terms of column vectors,


Example 8.1: The only solution to

is the trivial solution s = 0, ?/ = 0 and x = 0. pendent.

: c + y + z = 0 o r . c + y = o

x = o Hence the three vectors are inde-

Example 8.2: The vector equation lor ' system of linear equations)

x + 2 y + x = 0

z f 3 y t 3 x = 0

has the non-trivial solution ( 3 , - 2 , l ) . Accordingly, the three vectors a r e dependent.

Solved Problems LINEAR EQUATIONS IN TWO UNKNOWNS 11.1. Determine three distinct solutions of 2n. - 3y = 14 and plot its graph.

Choose any value for either unknown, say 2: = -2. Substitute z = -2 into the equation to obtain

2 . ( - 2 ) - 3y = 14 or -4 - 3u = 14 or -3y = 18 o r y = -6

Thus x = -2 and y = -6 or, in other words, the pair (-2, -6) is a solution.

Now substitute, say, z = 0 into the equation t o obtain

2.0 - 3y = 14 o r -3u = 14 or y = -14/3

Thus (0, -14/3) is another solution.

Lastly, substitute. say, y=O into the equation to obtain

2 x - 3 . 0 = 14 o r 2x = 14 or z = 7

Hence (7,O) is still another solution.

Now plot the three solutions on the Cartesian plane R2 as shown below; the line passing through these three points is the graph of the equation.

Y t5

3 x - 2 y = 7 11.2. Solve the system: x + 2 y = 1 *

Since 311 # -212, the system has a unique solution. the negatives of each other: hence a d d both equations:

32 - 2y = 7

x + 2 y = 1

Now the coefficients of y are already

Addition: 4x = 8 or . 1 : = 2

Substitute .r = 2 into the second equation to obtain

2 + 2 y = 1 or 2y = -1 or u = -$ Thus 2: = 2 and y = -4 or, in other words, the pair (Y = ( 2 , -3) is the solution of the system.

Check pour answer by substituting the solution back into both original equations:

3 * 2 - 2 * ( - $ ) = 7 or 6 + 1 = 7 or 7 = 7

2 + 2 * ( - 3 ) = 1 or 2 - 1 = 1 o r 1 = 1 and

(I) 2 x + 5 y = 8 11.3. Solve the system:

(2) 3.2: - 2y = -7 Note that 213 # 51-2: hence the system has a unique solution. To eliminate x, multiply

( I ) by 3 and (2) by -2 and then add:

3 X ( 1 ) : 62 + 15y = 24

-2 X ( 2 ) : -6% + 4~ = 14

Addition: 19y = 38 o r y = 2

Substitute y = 2 into one of the original equations, say ( I ) , t o obtain

2 x l - 5 . 2 = 8 o r 23:+10 = 8 or 22 = -2 or x = -1

Hence .c = --1 and y = 2 or, in other words, the pair ( - 1 , Z j is the unique solution to the system. Check your answer by substituting the solution back into both original equations:

( 1 ) : 2*(--1) 4- 5 . 2 = 8 or -2 f 10 = 8 or 8 = 8

(2): 3 . ( - 1 ) - 2 * 2 = -7 or - 3 - 4 = -7 or -7 = -7

We could also solve the system by first eliminating y as follows. Multiply ( I ) by 2 and ( 2 ) by 5 and then add:

2 i< (1) : 4% + 1Oy = 16

5 X (2): 15% - 1Oy = -35

Addition: 19x = -19 or x = -1

Substitute 2 = -1 in ( 1 ) to obtain

2*( -1 ) + 5y = 8 o r - 2 + 5 y = 8 or 6~ = 10 or y = 2

Again we get (-1,Z) as a solution.

( I ) x - 2 y = 5 11.4. Solve the system:

( 2 ) -32 + 6y = -10 '

Note tha t 1/-3 = -216: hence the system does not have a unique solution. But -2 5 - + - - - 1 -

-3 6 -10 Hence the lines a r e parallel and the system has no solution.

- _


(I) 5 x - 2 ~ = 8 11.5. Solve the system:

(2) 3 x + 4 y = 10

Note t h a t 5/3 f -2/4; hence the sybtem has a unique solution. To eliminate y, multiply (1) by 2 and add it t o (2 ) :

2 X (1): 10s - 4y = 16

(2 ) : 32 + 4y = 10

Addition: 13x = 2 6 or x = 2

Substitute x = 2 in either original equation, say (2 ) , to obtain

3 . 2 +4y = 10 or 6 + 4y = 10 or 4y = 4 o r y = 1 Thus the pair 12 , l ) is the unique solution t o the system.

Check the answer by substituting the solution back into both original equations:

( 1 ) : 5 . 2 - 2 - 1 = 8 o r 1 0 - 2 = 8 o r 8 = 8

( 2 ) : 3 . 2 + 4 . 1 = 10 or 6 + 4 = 10 or 10 = 10

( I ) x - 2 v = 5 11.6. Solve the system:

( 2 ) - 3 ~ + 6 2 j = -15 *

Note that

Accordingly, the two lines a r e coincident. Hence the system has an infinite number of solutions which correspond to the solutions of either equation.

Particular solutions can be found as follows: Let y = 1 and substitute in (1) to obtain

x - 2 * 1 = 5 or x - 2 = 5 or x = 7

Hence ( 7 , l ) is a particular solution. Let 11 = 2 and substitute in ( 1 ) t o obtain

x - 2 * 2 = 5 or x - 4 = 5 or x = 9

Then (9 ,21 is another specific solution of the system. And so forth.

The general solution to the system is obtained as follows. Let y = a and substitute in ( 1 ) t o obtain

Accordingly, the general solution to the system is (5 + 2a, a ) , where a is any real number.

x - 2 y = 5 or z = 5 + 2 a

GENERAL SPSTERIS OF LINEAR EQUATIONS 11.7. Solve the following system in echelon form:

x + 2 y - 3 2 + 4 w = 5 O x + y f 5 x I - 2 ~ = 1 or ?/ + 5x +22u = 1 0 2 + O y + O 2 + O w = 2

x + zy - 32 + 4w = 5

0 = 2

The system is inconsistent since i t has an equation of the form 0 = c with c # 0. Hence the system has no solution.

11.8. Solve the following system in echelon form: x+3y-22X= 4 x + 3 2 j - 2 2 = 4

on:+ y - & = 2 y - 5 2 = 2 o x + o y + x = -1 x = -1 ox + o y + ox = 0 o = 0

or


The system is consistent since it has no equation of the form O = c with c f 0 ; hence the system has a solution. Furthermore, since there a r e three unknowns and three non-zero equations, the system has a unique solution.

Substitute x = -1 into the second equation to obtain y - 5 * ( - 1 ) = 2 or y + 5 = 2 o r y = -3

Then substitute y = -3 and x = -1 into the first equation to obtain

x + 3 . ( - 3 ) - 2 * ( - 1 ) = 4 or z - 9 + 2 = 4 o r 2 - 7 = 4 or x = 11 Thus x = 11, y = -3 and z = -1 or, in other words, the ordered triple (11, -3, -1) solution t o the system.

is the unique

11.9. Solve the following system in echelon form: x + 2 y - 3 2 + 4 w = 5 x + 2 y - % + 4 w = 5

ox + y + 5 x + 2242 = 1 or y+55x+2w = 1 o x + o y + o z + o w = 0 o = o

The system is consistent and so has a solution. Furthermore, since there a r e four unknowns and only two non-zero equations, the system has a n infinite number of solutions which can be obtained by assigning arbi t rary values to the unknowns x and tu.

second equation to obtain

Now substitute x = a, w = b and y = 1 - 5a - 2b into the first equation to obtain x + 2 ( 1 - 5 a - 2 b ) - 3 a + 4 b = 5 or z + 2 - 1 0 a - 4 b - 3 ~ + 4 b = 5

o r x + 2 - 1 3 a = 5 or x = 3 + 1 3 a

To find the general solution to the system let, say, x = a and w = b, and substitute into the

y + 5 a + 2 b = 1 or y = 1 - 5 a - 2 b

Thus the general solution is

(3 + 13a, 1 - 5a - 2b, a, b) where a and b are real numbers

To find a particular solution t o the system let, say, a = 1 and b = 2 to obtain

(16, -8, 1, 2) (3 + 13 * 1, 1 - 5 - 1 - 2 * 2, 1, 2) or

N o t e . Some texts leave the general solution in terms of x and w as follows:

x = 3 + 13x y = 1 - 5 x - 2 w

(3 + 132, 1 - 5x - 2w, x , w) where z and w are any real numbers

(I) z + ~ Y - ~ x = -4 11.10. Solve the following system: (2) 5x - 3y - 72 = 6 .

( 3 ) 3 x - 2 y + 3 2 = 11

Reduce the system to echelon form by first eliminating x from the second and third equations. Multiply (I) by -5 and add to ( 2 ) to eliminate z from the second equation:

-5 x (I): -5x - 1oy + 202 = 20

( 2 ) : 5% - 3y - '72 = 6

Addition: -13y + 13x = 26 or y - x = -2

Now multiply ( 1 ) by -3 and add to (3) to eliminate x from the third equation:

-3 X (I): - 3 ~ - 6y t 122 = 12

(3): 3x - 2y + 32 = 11 Addition: -8y 4- 152 = 23


Thus the original system is equivalent to the system

x + 2y - 4x = -4 y - x = - 2

-8y i- 152 = 23

Next multiply the second equation by 8 and add to the third equation to eliminate y from the third equation:

8 X second: 8y - 82 = -16 third: -8y + 15x = 23

Addition: 7 x = 7 or x = l

Thus the original system is equivalent t o the following system in echelon form:

x t 2 y - 4 x = -4 y - - z = -2

x = 1

The system is consistent since i t has no equation of the form 0 = c with c # 0; hence the system has a solution. Furthermore, since there a re three unknowns and also three non-zero equations, the system has a unique solution.

Substitute z = 1 into the second equation to obtain

y - 1 = -2 o r y = -1

Now substitute z = 1 and y = -1 into the first equation to obtain

x + 2 . ( - 1 ) - 4 . 1 = -4 or x - 2 - 4 = -4 or 2 - 6 = -4 or z = 2

Thus x = 2 , y 2 -1 and z = 1 or, in other words, the ordered triple ( 2 , -1, 1) is the unique solution to the system.

(1) + 2 y - % = -1

11.11. Solve the following system: (2) -3x + y - 2x = -7 . (3) 5x 4-3y-42 = 2

Reduce to echelon form by first eliminating x from the second and third equations. Multiply ( 1 ) by 3 and add t o ( 2 ) to eliminate z from the second equation:

3 x (1 ) : 3s + 6y - 9~ = -3

(2): -3x + y - 2x = -7

Addition: 7y - 112 = -10

Now multiply ( 1 ) by -5 and add to (3) to eliminate x from the third equation:

-5 X ( 1 ) : - 5 ~ - 1Oy + 1 5 ~ = 5

(3): 5 2 + 3y - 4x = 2 Addition: -7y + llx = 7

Thus the system is equivalent to the system

x + 2y - 32 = -1 7y - 11z = -10

-7y + 11z = 7

However, if we add the second equation t o the third equation, we obtain

Ox+ Oy + O x = -3 or 0 = -3

Thus the system is inconsistent since it gives rise to an equation 0 = c with c f 0; accordingly, the system has no solution.

120 LINEAR EQUATIONS 'CHAP. 11

( I ) ~ + 2 y - 3 ~ = 6 11.12. Solve the following system: ( 2 ) 2x - y + 4x = 2 .

(3) 4 x + 3 1 ~ - 2 ~ = 14

Reduce the system to echelon form by first eliminating :c from the second and third equations. Multiply ( 1 ) by -2 and add t o (2) '

-2 X (1): -2a - 4y f 6.2. = -12

( 2 ) : 22 - y f 42 = 2

Addition: -5y + 102 = -10 or y - 22 = 2

Multiply (1) by -4 and add i t t o (3):

-4 X ( 1 ) : -4.2. - 8y + 122 = -24

( 3 ) : 42 + 3y - 22 = 14

11.13.

-5y f 102 = -10 or y - 22 = 2 Addition:

Thus the system is equivalent to x + 2 y - 3 ~ = 6

~ f 2 u - 3 ~ 6 y - 2 2 = 2 or simply

y - 2 2 = 2 y - 2 2 = 2

( S o t c . Since the second and third equations a re identical, we can disregard one of them. In fact, if we multiply the second equation by -1 and add to the third equation we obtain 0 = 0.)

The system is now in echelon form. The system i s consistent and since there a r e three unknowns but only two non-zero equations, the system has an infinite number of solutions which can be obtained by assigning arbi t rary values to 2.

Then substitute z = a into the second equation to obtain

Now substitute z = a and y = 2 + 2a into the first equation t o obtain

To obtain the general solution t o the system let, say, z = a .

y - 2 2 a = 2 or y = 2 + 2 a

n + 2 ( 2 + 2 a ) - 3 a = 6 or . ? + 4 - 4 4 a - 3 ~ = 6 o r r + 4 + a = 6 or x = 2 - a

Thus the general solution t o the system is

(2 - a , 2 + 2a, a) where a is any real number.

To obtain a particular solution of the system let, say, a = l and substitute into the general solution t o obtain ( 2 - 1 , 2 + 2 . 1 , 1) or (1, 4, 1).

(1) 2x + 2/ - 3x = 1 Solve the system: (2) 32 - 9 - 42 = 7 .

(3) 5~ + 21~ - 62 1 5

Method 1.

( I ) by -3 and (2) by 2 and then add to eliminate Reduce to echelon form by first eliminating x from the second and third equations. Multiply

from the second equation: -3 X (1): -62 - 3y + 92 = -3

2 X ( 2 ) : 6x - 22/- 8 2 = 14

Addition: -5y + 2 = 11

Multiply ( 1 ) by -5 and (5) by 2 and then add to eliminate z from the third equation:

-5 X (1) : -10% - 51/ + 152 = -5

2 x (3): 10% i 4y - 122 = 10

Addition: -1/+ 32 = 5 or y - 32 = -5


Thus the system is equivalent to the system

2.1- + y - 3x = 1 2 2 + 2 / - & = 1

-5u + x = 11 or , interchanging equations, y - 3 2 = -5

y - 3 x = -5 -5y + x = 11

Now niultiply the second equation by 5 and add to the third equation t o eliminate y from the third equation:

5 X second: 5y - 152 = -25

third: -5y + x = 11 -14x = -14 o r x = 1 Addition:

Thus the original system is equivalent t o the system

23: + - 3x = 1

y - 3 % = -5

x = 1

Substitute z = 1 into the second equation to obtain

y - 3 . 1 = -5 o r y - 3 = -5 or y = - - 2

Substitute 2 = 1 and y = - 2 into the first equation to obtain

2 2 + ( - 2 ) - 3 3 1 = 1 or 2 x - 2 - 3 = 1 or 2 2 - 5 = 1 o r 22 = 6 o r z = 3

Thus 1: = 3, y = -2 and x = 1 is the unique solution to the system.

Method 2.

( 1 ) to ( 2 ) to obtain

Multiply ( 1 ) by -2 and add to (3):

Consider y as the first unknown and eliminate y from the second and third equation. Add

5 ~ - 7 2 = 8

-2 X ( I ) : -4% - 2y + 62 = -2

(3): 5x + 2y - 62 = 5 Addition: x = 3

Thus the system is equivalent to

2 2 + y - 3 x = 1 y - 3 2 + 2 ~ = 1

5% - 7 x = 8 or -7% + 5~ = 8

X = 3 r = 3

Substitute 1: = 3 into the second equation t o obtain

- 7 2 + 1 5 = 8 o r -72 = -7 or x = 1

Substitute x = 3 and z = 1 into the first equation to obtain

y - 3 + 6 = 1 o r y + 3 = 1 o r y = - 2

Thus N = 3, y = -2 and z = 1 is the unique solution of the system.

VECTORS AND LINEAR 11.14. Reduce the following

and solve:

EQUATIONS vector equation to a n equivalent system of linear equations


Multiply the vectors on the right by the unknown scalars and then add:

Set the corresponding components of the vectors equal to each other to form the system

x + y + x = 2 z T y + x = 2 x + y = - 4 or y + x = -4 X = 7 x = 7

Substitute r = 7 into the second equation to obtain

y f 7 = -4 o r y = -11

Substitute -c = 7 and y = -11 into the first equation to obtain

x - l 1 + 7 = 2 o r x - 4 = 2 or z = 6

Thus x = 7, y = -11 and x = 6 is the solution of the vector equation.

11.15. Write the vector v = (1, -2 ,5) as a linear combination of the vectors e l = (1,1, l),

We wish t o express ZI as v = x e l + ye, + xe3, with x, y and z as yet unknown. Thus we require:

e, = (1 ,2 ,3 ) and e3 = (2, -1,l).

(1, -2, 5) = x(1, 1, 1) + y ( l , 2 , 3) + 4 2 , -1, 1)

(2, x, x) + (Y, 2Y, 3Y) + (22, -2 , 4 = ( x + y + 2 x , x + 2 y - x , x f 3 y i z )

=

Hence s + y - 2 2 = 1

x + 2 y - x = -2

.L:+31/- x = 5 or

x + y + 2 x = 1

2 y - z = 4 y - 3 2 = -3 or

x f y + 2 x = 1

or y - 3 2 = -3

x = 2

Substitute z = 2 into the second equation t o obtain

y - 3 . 2 = -3 o r y - 6 = -3 o r y = 3

Substitute z = 2 and y = 3 into the first equation to obtain

s + 3 + 2 . 2 = 1 o r x + 3 + 4 = 1 or x + 7 = 1 or x = - 6

Consequently, u = -6e, + 3e2 + 2e3

11.16. Write 2' = as a linear combination of el = [-:I, e2 = (1;) and e3 = [-:I. z + 2 y + z

-32 - 4y - 52

\-5 Set I' as a linear combination of the e, using unknowns x, y and z : w %el + y e 2 + ~ e 3 .

(-%) = x {-:) f 2 + y (1;) + (-:) = (-:) + (::) f (-5) = ( 2 x - Y f 7 2 )


Hence x t 2 y f 2 = 2 z + 2 y + x = 2 x + 2 y + x = 2

22- y + 7 % = 3 or - 5 y + 5 x = -1 or y - z = g - 3 X - 41/ - 52 = -5 2 y - 22 = 1 - 5 y + 5 x = -1

z t 2 y + x = 2 or y - - x = 1

o = "

The system is inconsistent and so has no solution. Accordingly, u cannot be written as a linear combination of the vectors e l , e2 and e 3 .

DEPENDENCE AND INDEPENDENCE 11.17. Determine whether the vectors il,1, -l), (2, -3,l) and (8, - 7 , l ) are dependent or

independent. Firs t set a linear combination of the vectors equal to the zero vector:

(0, 0, 0) = ~ ( 1 , 1, -1) + ~ ( 2 , -3, 1) + ~ ( 8 , -7, 1)

Then reduce the vector equation to an equivalent system of homogeneous linear equations:

(0, 0, 0) = ( r , 2, - x ) + ( 2 ~ , - 3 ~ , II) + (82, -72, Z )

= ( X + 2u 4- 8x, x - 3y - 72, -2 + y + X )

01, r + 2 y + 8 x = 0 2 - 3 y - 7 x = 0

- x + y + 2 = 0

Lastly, reduce the system to echelon form. Eliminate 5 from the second and third equation to obtain

z + 2 y + 8x = 0 r + 2 y + 8 x = 0 z + 2 y + 8 z = 0

-5y - 152 = 0 or y + 3 2 = 0 or y + 3 x = 0

3y + 92 = 0 y + 3 x = 0 o = o Since there a re three unknowns but only two non-zero equations, the system has a non-trivial solution.

Remark: We do not need to solve the system to determine dependence or independence; we

Thus the original vectors are dependent.

only need to know if a non-zero solution exists.

(-:I, and [:) are dependent or independent. 11.18. Determine whether the vectors \-3 \ -1

Firs t set a linear combination of the vectors equal to the zero vector; and then obtain an equivalent system of homogeneous linear equations:

-1 -32 -u/ -3% - + Z or I x i 2 y + 3 x = 0

-22 + 3y + 2x = 0 -32- y + 2 = 0

Eliminate z f r o m the second and third equations to obtain z + 2 y + 32 = 0 x + 2 y + 3 2 = 0

7y + 82 = 0 5y + 10% = 0


Eliminate y from the third equation to obtain

z + 2 y + 3 x = 0 s f 2 y i - 3 2 = 0

y + 2 2 = 0 or y + 2 x = 0

-62 = 0 z = o

In echelon form the system has three unknowns and three non-zero equations; hence the system has the unique solution x=O, y = 0, z = 0, i.e. the zero solution. Hence the original vectors are independent.

Reduce the problem to a homogeneous system of equations by setting a linear combination of the vectors equal to the zero vector:

x + y + 32 f 2v - 8~ -2% 4- 2y - 72 f UJ

3 2 + y - 6v - 7~ -42 + 5y + 22 - 5v + 4w

X + y + 3 2 + 2 ~ - 8 ~ = 0

- 2 2 + 2 y - 7 2 $. W = 0

3 ~ + y - 6 ~ - ' 7 ~ = 0

- 4 2 f 5 2 / + 2 2 - 5 5 v + 4 w = 0

Now the system of homogeneous equations has five unknowns but only four equations. Hence, by Theorem 11.3, the system has a non-trivial solution and so the vectors a r e dependent.

11.20. Let vl,v2, . . .,vm be n-component vectors. Show that if the number of vectors is greater than the number of components, i.e. m > 12, then the vectors are dependent.

Note, as in the preceding problem, t h a t the vector equation

x1v1 + 22v2 + ... + XmWm = 0

gives rise to a homogeneous system of linear equations in which the number of unknowns equals the number of vectors and the number of equations equals the number of components. Hence if the number of vectors is greater than the number of components, then the system has a non-zero solution by Theorem 11.3 and so the vectors a re dependent.


Supplementary Problems LINEAR EQUATIONS IN TWO UNKNOWNS 11.21. Solve:

32 - 2y = 7 2 + 31/ = -16

2 s + 3y = 1 5 s + 7y = 3

(iii) 22 - 5y = 1 32 + 2y = 11 (ii)

11.22. Solve: 22 - 3y = 5t i 32 + y = 2t

4s - 2y = 5 -62 + 3y = 1 (iii) (ii) { 22 + 4y = 10

32 + 6y = 15

11.23. Solve:

GENERAL SYSTEMS OF LINEAR EQUATIONS 11.24. Solve:

2 2 + y - 3 % = 5 32 - 2y + 22 = 5 5 2 - 3y - z = 16

11.25. Solve: 22 + 3y = 3 2 - 2 y = 5

32 + 2y = 7

11.26. Solve: z + 2 y + 2 x = 2

3 2 - 2 y - 2 = 5 (i) 22 - 5y + 3% = -4

x + 4 y + 6 2 = 0

x + 2 y + 3x = 3 2x + 3y + 8x = 4 3x + 2y + 172 = 1

2 2 + 3y - 2x = 5 z - 2y + 3% = 2

4 2 - y + 4 x = 1

Z + 5 y + 4 ~ - 1 3 ~ = 3 (ii) 32 - y + 22 + 5w = 2

2 2 + 2 y + 3 x - 4w = 1

VECTORS AND EQUATIONS 11.27. Write v = (9 ,5 ,7) as a linear combination of the vectors el = (1 ,1 ,1) , e2 = (1,-1,0) and

e3 = (2 ,0 ,1) .

11.28. Determine whether the vectors (1, -2, 3, l), (3, 2, 1, -2) and (1, 6, -5, -4) are dependent o r independent.

11.29. Determine whether the vectors (-i), (i) and (-!) are dependent or independent.

11.30. Prove: If the vectors el, e2 and e3 are independent, then any vector v can be written in a t most one way as a linear combination of the vectors e l , e2 and e3.

11.21. ( i ) 13, 11, o r 5 = 3 , y = 1 (ii) ( 2 , -l), or x = 2 , y = -1 (iii) (-1, -5) , or z = -1, y = -5

11.22. (ii (5 - 2 a , a), or x = 5 - 2 a , y = a iii) no solution (iii) ( t , -t), or z = t , y = --t

11.23. (ii 16, 8) [ii) (5, 2 ) (iii) ( 5 , 2 )

x = -1-77x y = 2 + 2 x 11.24. ( i ) (1, - 3 , -2) (ii) no solution (iii) (-1 - 7% 2 + 2a, a), or

z = -x+2w y = 1 + 2 x - 2 w

11.25. (ii (3 , -1) (ii) (-a+ 2b, 1 + 2u - 2 b , a , b ) , or

( 5 = 6 - 2 y - 5 2 (iii) (6 - 2a - 5b, a, b, 2b - l), or < w = 2 2 - 1

11.26. (ii (2, 1, -1) (ii) no solution

11.27. w = 2e, - 3e2 + 5e3

11.28. Dependent

11.29. Independent

21.30. Proof: Suppose w = alel + a2e2 + u3e3 and w = b,e , + b,ez + b3e3. Subtracting, we obtain

0 = w - v = (al - bl)el + (a2 - b,)e , + (a3 - ba)ea

Since the vectors a re independent, the coefficients must be zero, i.e.

al - b , = 0 , a, - b2 = 0 and u3 - b3 = 0

Accordingly, a, = b, , u2 = b , and u3 = b,.

Chapter 12

Determinants of Order Two and Three

INTRODUCTION

the matrix. of linear equations, has many interesting properties in its own right.

two and three, including its application to linear equations. of higher order are beyond the scope of this text.

To every square matrix there is assigned a specific number called the determinant of This determinant function, first discovered in the investigation of systems

In this chapter we define and investigate the determinant of square matrices of order Determinants of matrices

DETERMINANTS OF ORDER ONE

square matrices only. Write det(A) or 'A( for the determinant of the matrix A ; it is a number assigned to

we i.e.

The determinant of a 1 x 1 matrix ( a ) is the number a itself: det(a) = a. Observe that in the linear equation in one unknown x,

ax = b

can view a as a 1 x 1 matrix (a) applied to the one-component vector x. if a # 0, then the equation has the unique solution x = ;b.

If det(a) # 0, However, if det(a) = 0, i.e. 1

w

if a = 0, then the equation has no solution if b f 0, and every number is a solution if b = 0; hence the solution is not unique.

As we shall subsequently see, the preceding result holds true in general, i.e. det(A) # 0 is a necessary and sufficient condition f o r the linear equation A x = b to have a unique solution.

DETERMINANTS OF ORDER TWO

The determinant of the 2 x 2 matrix (: :) is denoted and defined as follows:

Example 1.1:

We emphasize that a square array of numbers enclosed by straight lines is not a matrix but denotes the number that the determinant function assigns to the enclosed array of numbers, i.e. t o the enclosed square matrix.

127

D E T E R M I N A N T S O F ORDER TWO AND T H R E E [CHAP. 12 128

A very important property of the determinant function is that i t is multiplicative; that is,

Theorem 12.1: The determinant of a product of matrices is the product of the determinants of the matrices: det(AB) = det(A) det(B).

Remark: The diagram on the right may help the reader obtain the determinant of a 2 x 2 matrix. Here the arrow slanting downward connects the two numbers in the plus term of the determinant ad - be, and the arrow slanting upward connects the two numbers in the minus term of the -

determinant ad - be.

(X +

LINEAR EQUATIONS IN TWO UNKNOWNS AND DETERMINANTS Consider two linear equations in two unknowns:

a1x + b1y = c1 a2,r + bey = cz

Let us solve the system by eliminating y. equation by -bl, and then add:

Multiply the first equation by b, and the second

bp X first: albzx + b1bey = b 2 C 1

-bl x second: -arblx - blb2y = -blcz

Addition: (a1bzra2b1)x = b 2 C 1 - blcz

Now the system has a unique solution if and only if the coefficient of x in this last equation is not zero. i.e.

Observe that D is the determinant of the matrix of coefficients of the system of equations. In this case, where D # 0, we can uniquely solve for x and y as quotients of determinants as follows:

Here D, the determinant of the matrix of coefficients, appears in the denominator of both quotients. The numerators N Z and N y of the quotients for x and y, respectively, can be obtained by substituting the column of constant terms in place of the column of coefficients of the given unknown in the matrix of coefficients.

Example 2.1: Solve using determinants: 2x - 3y = 7

3x + 5y = 1

The determinant D of the matr ix of coefficients is

-3 = 2.5 - 3.(-3) = 10 f 9 = 19 D = ) 3 5 1

CHAP. 12)

a1 bl c1

a3 bs c3

a2 b2 CP

DETERMINANTS O F ORDER TWO AND THREE 129

= alb2c3 + a2bscl + a3blcZ - alb3c2 - aZblc3 -- a3bZcl

Since D f 0, the system has a unique solution. To obtain the numerator LV, replace, in the matrix of coefficients, the coefficients of x by the constant terms:

To obtain the numerator N, replace, in the matrix of coefficients, the coefficients of y by the constant terms:

2 7 = 2 . 1 - 3 . 7 = 2 - 21 = -19 = l 3

Thus the unique solution of the system is

, 2 x = 5 + y Example 2.2: Solve using determinants: \I

3 + 2 y + 3 % = 0

Firs t arrange the equations in standard form: 2 2 - y = 5 3 x f 2 y = -3

The determinant D of the matrix of coefficients is

Since D # 0, the system has a unique solution. Now,

= 5 . 2 - (-3)(-1) = 10 - 3 = 7 NZ =

Thus the unique solution of the system is

Example 2.3: Solve using determinants: {::I:; 1 ; The determinant D of the matrix of coefficients is

-' = 2- ( -6 ) - 3 - ( - 4 ) = -12 + 12 = 0 = 1 3 -61

Since D = 0, the system does not have a unique solution, and we cannot solve the system by determinants. (The discussion of the previous chapter shows that the system has no solution since 2/3 = -41-6 f 7/5.)

130 DETERMINANTS O F ORDER TWO AND THREE

+ (-4)

[CHAP. 1 2

0 -4

or

a1 bl c1

D = a2 b2 cz a3 b3 c3

which is a linear combination of three determinants of order two whose coefficients (with alternating signs) are the first row of the given matrix. Note that each 2 x 2 matrix can be obtained by deleting, in the original matrix, the row and column containing its coefficient:

a1

# 0

= 2( -20+2) - 3(0 -2 ) - 4 ( 0 + 4 ) = -36 + 6 - 16 = -46

Example 3.2:

' = 2(-12-00) - ( 1 0 + 4 ) + 3(0-24) = -24 - 14 - 72 = -110

Remark: Problems 12.8 and 12.18 show that the determinant of a 3 x 3 matrix can be expressed as a linear combination of three determinants of order two with coefficients from any row or from any column, and not just from the first row as exhibited here.

CHAP. 121 DETERMINANTS O F ORDER TWO AND T H R E E 131

N , 1

In this case, the unique solution of the system can be expressed as quotients of determinants,

bi dt uz bz dz

~ a3 u1 b3 d3

d l bl c1 a1 dl c1

d:! bz cz , NU = a2 dz CZ , N , = ds b3 c3 a3 d3 c3

Example 4.1:

Solve the following system by determinants: 2 r + y - x = 3 2 + u f x = 1 2 - 2 2 y - 3 x = 4

The determinant D of the matrix of coefficients is obtained as follows:

= 2( -3+2) - 1(-3-1) - l ( - 2 - 1 ) = -2 + 4 + 3 = 5

Since D # 0, the system has a unique solution. for x, y and x respectively:

We evaluate S,, S, and N,, the numerators

3 1 -1

-2 -3 N , = 1 ; 1 1 1 = 3( -3+2) - 1(-3-4) - 1( -2 -4 ) = -3 + 7 + 6 = 10

N?, = = 2(-3-4) - 3(-3-1) - l ( 4 - 1 ) = -14 + 12 - 3 = -5

N , = = 2 ( 4 + 2 ) - 1 ( 4 - 1 ) + 3 ( - 2 - 1 ) = 1 2 - 3 - 9 = 0

Thus the unique solution is

= 2 , = - = - = N , -5 -1, x = ? = - = o N O D 5

w, - 10 D 5 2 = - - - D 5

INVERTIBLE MATRICES

that

Such a matrix B is unique; fo r

AB1= B1A = I and ABz = BZA = Z

A square matrix A is said to be invertible if there exists a matrix B with the property

A B = B A = I , the identitymatrix

implies BI = BIZ = BI(ABP) = (B1A)BZ = IBa B2.

132 DETERMINANTS O F ORDER TWO AND THREE [CHAP. 12

We call such a matrix B the iwaerse of A and denote it by A-l . Observe that the above relation is symmetric; that is, if B is the inverse of A , then A is also the inverse of B.

Example 5.1:

3 -5 2 . 3 + S*(-l) 2*(-5) + 5 . 2 1 0 (f :)(-1 2) = ( 1 * 3 + 3 . ( - 1 ) 1 * ( - 5 ) + 3 * 2 ) = (0 1)

3 - 5 2 5 3 . 2 + ( -5)al 3 . 5 + ( -5)*3 (-1 2 ) ( 1 3) = ( ( - 1 ) * 2 + 2 . 1 ( - 1 ) * 5 + 2 . 3 ) = (i

It is known that AB = Z if and only if BA = Z (Problem 12.17); hence it is necessary to test only one product to determine whether two given matrices are inverses, as in the next example.

Example 5.2:

0

6 -1

1 * ( - 1 1 ) + 0 * ( - 4 ) + 2 . 6 1 . 2 + 0 * 0 + 2 . ( - 1 ) 1 * 2 + 0 . 1 + 2 * ( - 1 )

4 * ( - 1 1 ) f 1 . ( - 4 ) + 8 . 6 4 * 2 + 1 * 0 + 8 . ( - 1 ) 4 * 2 + 1 * 1 + 8 * ( - 1 ) 2.(-11) + (-1)*(-4) + 3.6 2 . 2 + ( - l ) . O + 3*(-1) 2 . 2 + ( -1) ' l + 3*(-1)

Thus the two matrices a r e invertible and a r e inverses of each other.

Example 5.3:

Find the inverse of (i . We seek scalars x, y, z and w fo r which

or which satisfy y + 2 w = 0

3 y + 4 w = 1

The solutions of the systems a r e 2: = -2, z = 312 and y = 1, u) = -112. Thus the inverse of the

given matrix is (pi -:) . We test our answer:

INVERTIBLE MATRICES AND DETERMINANTS

We now calculate the inverse of a general 2 X 2 matrix A = (: i ) . We seek scalars x , y , x and w such that

ax + bx a y f bw ex + dx cy + dw

whjch reduces to solving the following two systems of linear equations in two unknowns: ax + bx = 1 cx + dx = 0

a y + b w = 0 c y + d w = 1

CHAP. 121 DETERMINANTS OF ORDER TWO AND THREE 133

Note first that one equation in each system is homogeneous and has (0,O) as a solution, whereas the other equation does not have (0 ,O) as a solution. In other words, the lines in each system are not coincident and so each system has either a unique solution or no solution. Furthermore, the matrix of coefficients in each system is identical to the original matrix A . Accordingly, the systems have a solution (and so A has an inverse) if and only if the determinant of A is not zero. We formally state this result (which holds for matrices of all orders):

Theorem 12.2:

otherwise. Hence the preceding theorem can be restated as follows:

A matrix is invertible if and only if its determinant is not zero. Usually a matrix is said to be singular if its determinant is zero, and nonsingular

Theorem 12.3:

for the unknowns 2 , y, x and w as follows:

A matrix is invertible if and only if it is nonsingular. Now if the determinant of the above matrix A is not zero, then we can uniquely solve

In other words, we can obtain the inverse of a 2 x 2 matrix, with determinant not zero, by (i) interchanging the elements on the main diagonal, (ii) taking the negative of the other elements, and (iii) dividing each element by the determinant of the original matrix.

Example 6.1: Find the inverse of the matrix A = (i i ) . Now det(A) = IAI = 2 * 5 - 4 3 = 10 - 12 = -2. Hence the matrix does

To obtain A-1, interchange the elements 2 and 5 on the main have an inverse. diagonal and take the negative of the other elements to obtain the matrix

(-: -;I Now divide each element by -2, the determinant of A , t o obtain A-I:

Solved Problems

DETERMINANTS OF ORDER TWO 12.1. Compute the determinant of each matrix:

134 D E T E R M I N A N T S O F ORDER TWO A N D THREE [CHAP. 12

12.2. Solve for R' and y, using determinants:

2 x + y = 7 a x - 2 b y = c 3 x - 5 y = 4 3ax-5by = 2c

(ii) , where a b # 0 (i,

1: , " ,=o* 12.3. Determine those values of k for which

Hence 2k=O o r k = 0 , and k - 2 = 0 o r k = 2 . That is, if k = O or k = 2 , the determinant is zero.

12.4. Prove Theorem 12.1 (in the case of 2 x 2 matrices): The determinant of a product of matrices is the product of the determinants of the matrices, i.e. det(AB) = det(A) det(B).

s \ a n d s o AB = and B = \ t Z t / ' Let A = \

det(AB) = (ar + b t ) ( c s + d u ) - (as -t bu)(cr + d t ) = acrs + adrzc. + bcst + bdtu - acrs - adst - bcru - bdtu

= a d m + bcst - a d s t - bcru

and det(A) det(B) = ( a d - b c ) ( ~ u - s t ) = adrzc - adst - bcru + best

Thus d N A B ) = det(A) det(B).

CHAP. 121 DETERMINANTS O F ORDER TWO AND THREE

(i)

135

1 2 3 1 2 $ 3 4 -2

0 5 -1 0 5 4 -2 3 = 11

DETERMINANTS OF ORDER THREE 12.5. Find the determinant of each matrix:

= l (2-15) - 2 ( - 4 + 0 ) + 3(20+O) = -13 + 8 + 60 = 55

= 4(2+6) + l ( 0 + 15) - 2(0-10) = 32 + 15 + 20 1 67

= 2(10-6) + 3(5-3) + 4(-2+2) = 8 + 6 + 0 = 14

3y+2x = x + l 12.6. Solve, using determinants: 3% + 2 2 = 8 - 5y

3 2 - 1 x x - 2 ~

2.1:+3y- 2 = 1 Firs t arrange the equation in standard f o r m with the unknowns appearing in columns:

3% + 5y + 22 = 8 :c - 2y - 32 = -1

Compute the determinant D of the matrix of coefficients:

3 -1 5 2 1 = 2(-15+4) - 3(-9-2) - (-6-5) = -22 + 33 + 11 = 22

1 -2 -3 D =

Since D # 0, the system has a unique solution, To compute N,, N, and N,, replace the coefficients of the unknown in the matrix of coefficients by the constant terms:

N , = ~

N , = ~ 3 8 2 1 = 2(-24+2) - 1(-9 -2) - 1(-3-8) = -44 + 11 + 11 = -22

3 -1 5 2 ~ = 1 ( - 1 5 r 4 ) - 3(-24+2) - 1(-16*5) = -11 + 66 + 11 = 66

-1 -2 -3

1 2 1 -1

1 -1 -3


a1 b l c1

a3 b3 c3

a2 b2 c2

3

1 -2 -1 N, = I 5 1 = 2(-5+16) - 3(-3-8) + 1 ( - 6 - 5 ) = 22 + 33 - 11 = 44

= alh,c3 + h1~2a3 + cla2b3 - a3b2cl - b3c2al - c3a2bl

12.7. Consider the 3 x 3 matrix (Et !: ::). Show that the diagram appearing below,

where the first two columns are rewritten to the right of the matrix, can be used to obtain the determinant of the given matrix:

a3 b3 c3

+ + + - - -

Form the product of each of t,he three numbers joined by an arrow slanting downward, and precede each product by a plus sign as follows:

+ albZc3 + blcza3 + Clazb,

Now form the product of each of the three numbers joined by an arrow slanting upward, and precede each product by a minus sign as follows:

- a,b2cl - b3c2al - c3a2b,

12.8. We represented the determinant of a 3 x 3 matrix A as a linear combination of determinants of order two with coefficients from the first row of A . Show that det(A) can also be represented as a linear combination of determinants of order two with coefficients from (i) the second row of A, (ii) the third row of A .

a1 bl c1

(i) a2 b2 c2 = alb2c3 + u2h3cl + a3blc2 - alb3c, - a2hlc3 - a3b2c, ' a3 b3 c3 i = - ~ Z ( b , c 3 - b 3 ~ 1 ) + b Z ( ~ 1 ~ 3 - ~ 3 ~ 1 ) - cz(alb3 - aab,)

(ii)

CHAP. 12j DETERMINANTS O F ORDER TWO AND THREE 137

Observe tha t the 2 X 2 matrix accompanying any coefficient can be obtained by deleting, in the original matrix, the row and column containing the coefficient. Furthermore, the signs accompanying the coefficients form a checkerboard pattern in the original matrix:

Remayk: We can also expand the above determinant in terms of the elements of any column in a n analogous way (Problem 12.18).

INVERTIBLE MATRICES

12.9. Find the inverse of (i 35) . Method 1.

We seek scalars 2, y, x and w for which

1 = (: ;) ( 2 x + 32 2y + 3w 32 + 52 3y + 5w (; :)(: :) = (; ;) Or

o r which satisfy 3 y + 5w = 0 2 y + 3 w = 1

and { 32 + 52 = 1 22 + 32 = 0

To solve the first system, multiply the first equation by 2 and the second equation by -3 and then add:

2 X first: 62 + 102 = 2

-3 X second: -62 - 9x = 0

Addition: x = 2

Substitute z = 2 into the first equation t o obtain

3 x + 5 * 2 = 1 or 3 2 + 1 0 = 1 or 32 = -9 or z = -3

To solve the second system, multiply the first equation by 2 and the second equation by -3 and then add:

2 X first: 6y + low = 0

-3 X second: -6y - 9w = -3

Addition: w = -3

Substitute ?L' = -3 in the first equation to obtain

3 y f 5 * ( - 3 ) = 0 or 3 y - 15 = 0 or 31,' = 15 or y = 5

Thus the inverse of the given matrix is

Method 2.

the given matrix: We use the general formula for the inverse of a 2 X 2 matrix. F i r s t find the determinant of


Kow interchange the elements on the main diagonal of the given matrix and take the negative of the other elements to obtain

(-2” -35)

(-; -35) Lastly. divide each element of this matrix by the determinant of the given matrix, tha t is, by -1:

The above is the required inverse.

o r which satisfy

2x3 + 3y3 - 23 = 0 x3 + 2y3 + 23 =Z 0

-23 - 763 f 323 = 1

2x2 + 3y2 - Xg = 0 22 + 2y2 + 22 = 1

-22 - yg + 3x2 = 0

2x1 f 3y1 - X i = 1 2 1 + 2y, + 2, = 0 I -71 - y1 + 32, = 0

We solve each system by determinants. the original matrix. Its determinant D is

Observe tha t the matrix of coefficients in each system is

2 3 -1 D = 1 1 2 f ( = 2 ( 6 + 1 ) - 3(3+1) - 1 ( - l f 2 ) = 14 - 12 - 1 = 1

-1 -1

Since D # 0, each system has a solution and so the given matrix has a n inverse. each of the three systems follows:

The solution of

x1 = 7, y1 = -4, x1 = 1; x2 = -8, 2/2 = 5, zg = -1; xg = 5, y3 = -3, 2 3 1 1

Thus the inverse of the given matrix is

CHAP. 121 DETERMINANTS OF ORDER TWO AND THREE 139

12.11.

12.12.

12.13.

12.14.

12.15.

12.16.

12.17.

12.18.


Find the determinant of each matrix:

Solve each system using determinants:

5x = 2y - 7 3y = 4 + 4z

22 - 3 y = -1 4 2 + 7y = -1

(iii) (32 + 5y = 8

( ii) (i) 1 4 x - 2y = 1

Find the determinant of each matrix:

1 3 -2 -4 -2 -1

1 -3

Solve each system using determinants:

2 2 f 3 = yJ-322

3 y + 2 = 2-222 z + 2y - 42 = 3 (ii)

2% - 5y + 22 = 7

3x - 4y - 6 2 = 5

2 -3 Find the inverse of each matrix: (i) (; i) (ii) ( 3)

Find the inverse of [-: , f S o t e : There will be 9 unknowns.) 4 -2

Prove: A B = Z if and only if B A =I, where A and B are square matrices and I is the identity matrix.

Express the determinant of a 3 X 3 matrix as a linear combination of determinants of order two with coefficients from ti) the first column, (ii) the second column, (iii) the third column.


12.11. (i) -18, iii) -15, (iii) 8, (iv) 1, (v) 44, (vi) -57

21 29 5 1 13 8 12.12. (i) x = -, y = - (ii) x = - - y = - (iii) = -- Y = -- 26 26 13' 13 7 ' 7

140 DETERMINANTS O F ORDER TWO AND T H R E E [CHAP. 12

12.13. (i) 21, (ii) -11, (iii) 100, (iv) 0

12.14. ii) x = 5 , y = 1, x = 1 (ii) z = z + y, y = -2. The determinant of this system = 0.

Chapter 13

The Binomial Coefficients and Theorem

FACTORIAL NOTATION The product of the positive integers from 1 to n inclusive is denoted by n! (read

‘‘R factorial”): n ! = 1 . 2 . 3 . . . . . ( n - 2 ) ( n - l ) n

We can also define n ! recursively as follows: l! = 1 and n! = n - ( n - l ) !

It is also convenient to define O! = 1.

Example 1.1: 2 ! = 1 . 2 = 2, 3 ! = 1 0 2 . 3 = 6, 4! = 1 . 2 . 3 . 4 = 24,

5 ! = 5 * 4 ! = 5.24 = 120, 6 ! = 6*5! = 6.120 = 720

1 2 * 1 1 . 1 0 * 9 ! 12! 9! 9 !

- B ! 8 . 7 * 6 ! 6 ! 12.11.10 = 6 ! = 8 . 7 = 56 _ - - Example 1.2:

Example 1.3: n(12 - 1). . .(n - T + l ) ( n - r ) ( n - T - 1). * a 3 ’ 2 1 - ?Z(n - 1). * *(?Z - T + 1) = - n!

(n - r ) ! (n - ~ ) ( n - r - q . a . 3 2 * 1

n ! -._ - - no? - 1)’ ‘ ‘(7%- r + 1) 1 n ! 1 = n ( n - - l ) . . . ( n - r + l ) - = 1 * 2 * 3 ’ . . ( r - l ) r r ! ( n - T ) ! T ! r ! ( n - v ) !

BINOMIAL COEFFICIENTS

The symbol (:) (read “n Cr ” ) , where r and n are positive integers with r 6 n, is

n(n - l ) (n - 2) * - -(n - r + 1) defined as follows:

(:) = 1 . 2 . 3 . * * ( T - l)r

These numbers are called the binomial coe,tficients in view of the theorem in the following section.

Example 2.1:

Note that (:) has exactly r factors in both the numerator and the denominator.

Observe, by Example 1.3, that n! - n(n - 1) * - . (n - T + 1)

- 1 . 2 ’ 3 . * . ( r - l ) r r ! (n - ?.) !

141

142 T H E BINOMIAL COEFFICIENTS AND THEOREM [CHAP. 13

But n - (R - r ) = 1'; hence we have the following important relation:

Theorem 13.1 : (n ") = ( ::I1 or, in other words, if a t Z, = ?a then

Example 2.2:

Compute (lo). By definition, \ 7

On the other hand, 10 - 7 = 8 and so we can also compute (lo' as follows: \ 7 /

Observe t h a t the second method saves space and time.

Example 2.3:

Compute 'I4) and (20' l . Now 1 4 - 1 2 = 2 and 2 0 - 1 7 = 3; hence !\ 12 17 //

2 0 . 1 9 - 1 8 = 1140 1 . 2 . 3

and the fact that 0 ! = 1, we can extend the Remark: Using the above formula for

definition of (:) to the following cases:

= 1 n ! O ! O ! O ! = 1 and, in particular, (a) = O!n!

BINOMIAL THEOREM

general expression for the expansion of (a + b)qt.

Theorem 13.2 (Binomial Theorem) :

The Binomial Theorem, which can be proved by mathematical induction, gives the

a n - 3 b 3 + . . , ?2(72 - 1) n(n - l)(n - 2) (a + b)" = + na"-'l, + , 2 anP2b2 + 1 e 2 , 3

n(n - 1) + 1.2a'Z,"-' + ?2abn-l + b"

CHAP. 131 THE BINOMIAL COEFFICIENTS AND THEOREM 143

The following properties of the expansion of (a + b)" should be observed:

(i) There are 11 + 1 terms.

(ii) The sum of the exponents of a and b in every term is n.

(iii) The exponents of n decrease term by term from n to 0; the exponents of b increase

(iv) The coefficient of any term is ( y , ) where k is the exponent of either a or b. (This

term by term from 0 to n.

property follows from Theorem 13.1.)

(v) The coefficients of terms equidistant from the ends are equal.

Example 3.2:

Expand (2.c + 3 y 3 5 and simplify.

Observe tha t a and b in the Blnom1al Theorem correspond in this case to 22 and 3y2 respec-

5 . 4 5 . 4 tively. Thus

(2x+ 3y2)5 = (2 . r ) ; + 5 ( 2 ~ ) ~ ( 3 y 2 ) - ~ ( 2 r ) ? ( 3 y ~ ) ~ + 1 . 2 ( 2 . ~ ) ' ( 3 y ~ ) ~ + 5 ( 2 ~ ) ( 3 y ~ ) ~ + ( 3 ~ / ~ ) ; * 2

= 2 j . 2 + 5 . 2 4 ~ 4 . 3 ! / ~ + 10 * 2 3 r s * 3 2 y 4 + 10.22xL.33y6 + 5 * 2 ~ * 3 ~ y 8 + 351/l0

= 3 2 ~ : + 240x4~2 + 7 2 0 . ~ ' ? ( ~ + 1 0 8 0 ~ ~ ~ ~ + 810.c$ - 2432/'O

PASCAL'S TRIANGLE The coefficients of the successive powers of a + b can be arranged in a triangular array

of numbers, called Pascal's triangle, as follows: ( a + b)O = 1 1

(a -L b)' = a + b 1 1

(a + b)' = a2 + 2 a b 1 b2

(a T b ) 3 = a3 + 3a2b + 3ab' + b3

1 2 1

1 3 3 1

( a + b)4 = a4 t 4a3b + 6a2b2 + 4ab3 -C b4

(a + b)5 = a5 + 5a4b + lOa3b2 + 10aQb3 + 5ub4 + b5 1 5 'g l o 4 5 1

(a+ b)G = a6 - 6a5b 1_ 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 1 6 8 8 1 5 6 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................

Pascal's triangle has the following interesting properties.

(i) The first number and the last number in each row is 1.

(ii) Every other number in the array can be obtained by adding the two numbers appearing directly above it. For example: 10 = 4 f 6 , 15 = 5 + 10, 20 = 1 0 f 10.

Now since the numbers appearing in Pascal's triangle are the binomial coefficients, we can rewrite the triangular array of numbers as follows:

144 T H E BINOMIAL COEFFICIENTS A N D THEOREM [CHAP. 13

3 j, (30 ) (;) (:) !,3 '

(fi) (;).$1) (:) ' . . (nill\l (:) ( n i l ) ( % ; I ) . . . :": l) (:;;)

(0") i') (;) (i' . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

In order t o establish the above property (ii) of Pascal's triangle, i t is necessary to prove the following theorem.

Theorem 13.3: ( n + l ) = (I/L1) +

In the next example we prove a special case of this theorem; the general proof of Theorem 13.3 is given as a solved problem.

Example 4.1:

'11 We prove (7) = ,/ + (I;) which is Theorem 13.3 when n = l l and T = 7. Now

7 5 7 Multiply the first fraction by - and the second fraction by 5 to obtain the same denominator,

and then add:

5 - 1 1 ! - 7 - 1 1 ! 5 '11! m+- 7 ! 5! -

7 - 1 1 ! + - 7 * 6 ! 5 ! 7 ! * 5 . 4 !

- 7 - 1 1 ! + 5 - 1 1 ! = 1 2 - 1 1 ! - 12! ( 7 + 5 ) * 1 1 ! - - - - - 7 ! 5! 7! 5! 7 ! 5 ! 7! 5!

The general proof of Theorem 13.3 is very similar t o the proof given above.

MULTINOMIAL COEFFICIENTS

Let n 1 , 1 1 2 , . . . . nr be non-negative integers such that ? i l + n2 + . . . + n, = n. Then the

is defined as follows: . . . . n n !

nl, n2, . . . . n, i = nl!na! * . . n,!

n n,

CHAP. 131 THE BINOMIAL COEFFICIENTS A N D THEOREM 145

These numbers are called the ?nultiiioniial coefficients in view of the following theorem which generalizes the binomial theorem.

al"1aF.. . n Theorem 13.4: ( u ~ t a z f ~ ~ ~ t a , ) ~ =

Example 5.1: 7 * 6 . 5 . 4 . 3 * 2 * 1 = 210 2 - 1 . 3 . 2 . 1 . 2 . 1

- - - 7!

( 2 3 - - 2! 3! 2!

Example 5.2:

nl + n2 + n3 = 3: Expand (a - b A c ) ~ . Consider all triplets of non-negative integers (nl ,nz,n,) for which

= a3 + 3a'b + 3a2c + Gnbc b3 + 3bzc + 3ub2 + c3 3ac2 + 3bc2

Solved Problems FACTORIAL 13.1. Compute 4! , 5!, 6!, 7! and 8!.

4! = 1 ' 2 ' 3 ' 4 = 2.1

5! = 1 . 2 . 3 . 4 - 5 = 5 . 4 ! = 5 - 2 4 = 120

6 ! = 1 - 2 . 3 . 4 . 5 . 6 = 6 * 5 ! = 6.120 = 720

7! = 7 * 6 ! = 7.720 = 5040

8! = 8 * 7 ! = 8.5040 = 40,320

13 ! 7 ! 11 ! 13.2. Compute: (i) -, (ii) m,

(iii) 56. (i) 27.26, (ii) 14. 13. 1 2 , 1 13.3. Write in terms of factorials:

2 7 * 2 6 * 2 5 ! - 3 25!

- (i) 27.26 = 25!

56-55! - 56! (iii) 56 = __ - - 55! 55!

- ll! - 11! - - - 1 14 * 13 * 12

(ii) 14 * 13 * 12 * 11 ! 14!

146 THE BINORIIAL COEFFICIENTS A N D THEOREM [CHAP. 13

( n - 1) ! (iv) ____ ( I 2 + 2) ! *

( n - l ) ! ( n - l ) ! - %

n ! (n + 1) ! ( n - I ) ! ' 13.4. Simplify: (i) m, iii i

n(n- l ) (n - 2 ) . ' . 3 * 2 ' 1 72 ! -

11 . ? L ! - n(n- l ) ! - _ - - - n or, simply, ___ - ( i ) __

( J 1 - I)! (72 - 1)(n - 2 ) ' ' ' 3 * 2 * 1

(71 + I ) ! (n+ l)n(n- 1)()~- 2 ) . * . 3 * 2 * 1 - - ( n + l ) * n = ? " + n ~~ - (iii) -__ ( n - l ) ! - ( n - l ) ( ? z - 2 ) . ' ~ 3 ' 2 * 1

( n T l ) ! - ( n + l ) . n . ( n - l ) ! - or, simply, m- (n- l)! - (n + 1) - n = n2 + n

( n - l ) ! - (n - 1) ! - 1 iivi ~~ (n + 2) !

- 1 - - - (n + 2)(n + 1) ' IZ * (11 - 1) ! (n + 2)(n 1) * n 123 + 3n2 + 2%

BINOMIAL COEFFICIENTS

13.5. Compute: (i) ( y ), (ii) ' 12'1 , (iii) ( y ) . '1 4 / '

Recall tha t there a re as many factors in the numerator a s in the denominator.

13.6. Compute: (i) (:), (ii) (;), (iii) ( y ) . 1 8 ) - 8 . 7 . 6 . 5 . 4 = 56 \ 5 / 1 . 2 . 3 . 4 - 5

Note that 8 - 5 = 3; hence we could also compute ( 8 ) 1 a s follows:

- ti)

\ 5 / 8 . 7 . 6

( i i ) Now 9 - 7 = 2; hence

1 0 . 9 . 8 ' 7 = 210. (iii) Now 1 0 - 6 = 4; hence (16p) = (140) = 1 . 2 . 3 . 4

13.7. Prove:

16! 6 16! Multiply the first fraction by 3 and the second by 11 3 to obtain the same denominator in both fractions; and then add:

11*16! - 6 - 1 6 ! 11-16! - + - (?) + (I:) = 6*5!.11! 6!.11.10! G!*11! G ! e l l !

(6+11)*1G! - 17 * 16!

6 - 1 6 !

17! - - -- 6 - 1 6 ! + 1 1 * 1 6 ! -

- - - 6 ! * 1 1 ! 6 ! *11! 6! 11! 6!*11!


(The technique in this proof is similar t o that of the preceding problem and Example 4.1.)

11 ! n! To obtain the same denomi- n - r + l Hence n - - r + l '

( r - 1): * (ti - r + I)! + r ! - (n- r ) ! \ T - 1) I/ )+(:) = - Now I

nator in both fractions, multiply the first fraction by? and the second fraction by r

T. * 1 2 ! ( n - T + 1) n! ~~- - - r ! ( n - r + l ) ! + ? - ! ( n - y + l ) !

Ir -- (12 - r - l)] * n! - r - n ! t ( n - r + l ) * n ! - - -

Y ! ( n - r + l ) ! r ! (n -rTf l ) !

- ( i L t l ) ? i ! - (n + 1) ! __ Y ! oz-r+m - T ! (n - ?' + 1 7

BINOMIAL THEOREM 13.9. Expand and simplify: (z + 3 7 ~ ) ~ .

13.13. Given (a - b)6 = a6 + 6a5b + 15a4bbL + 20a3b3 + 15a2b4 -L 6ab5 + b6. Expand (a + b)'

We could expand (u+ b)7 and (a T b ) k using the binomial theorem. However, given the expansion of (a - b ) F , it is simpler to use Pascal's triangle to obtain the coefficients of (a+ b)' and ( a + b ) h . Firs t n r i t e down the coefficients of (a + b)6 and then compute the next two rows of the triangle

and (a - b)'.

148 THE BINOMIAL COEFFICIENTS AND THEOREM [CHAP. 13

1 6 20 15 6 1

Recall that each number is obtained by adding the two numbers above, e.g. 35 = 1 5 + 2 0 , 28 = 7 + 21, 56 = 35 + 21. From the above diagram we obtain

(a + b)7 = a7 + 7a6b + 21ajbz + 35a4b3 + 35a3b4 + Zla'b5 + 7ab6 + b7

(a + b)8 = a6 + 8a7b + 28a6bz + 56a5b3 + 70a4b* + 56a3b5 + 28a2b6 + 8ab7 + b8

13.14. (i) How does (a - b)" differ from (a + b)"? (ii) Use Problem 13.11 to obtain (2% - y2)5.

The expansion of ( a - b)n will be identical to the expansion of (a+ b)" except tha t the signs in ( a - b)" will alternate beginning with plus, i.e. terms with even powers of b will be plus and terms with odd powers of b will be minus. This follows from the fact that ( - b ) r = br if T is even and ( - b ) r = -br if T is odd.

( i i ) By Problem 13.11,

i i )

(2% + Y ' ) ~ = 3225 + 80x4~2 + 80x3~4 + 40x2y6 + 10xy6 + y10

To obtain (2% - y2)5, just alternate the signs in (2s + y*)?

( 2 ~ - = 32x5 - 80&42/2 + 8 0 ~ ~ 9 - 4Ox2y6 + 1 0 % ~ ~ - y'O

13.15. Obtain and simplify the term in the expansion of (239 - y3)8 which contains do.

Nethod 1. The general term in the expansion of (2&-y3)8 is

(;) ( 2 2 2 ) s - r (-y3)' (:) 26-r %16-2r (- Y ) 3 7

Hence the term containing 210 has ~ 1 6 - 2 " = do or 16 - 27 = 10 or T = 3, t h a t is, is the term

Method 2.

hence first write down the following: Each term in the expansion contains a binomial coefficient, a power of 2x2 and a power of -y3;

To obtain x10, the power of 2x2 must be 5; hence insert the exponent 5 to obtain

( ) (2x215 ( 3 3 )

Now the sum of the exponents must add up to 8, the exponent of (222-y3)8; hence the exponent of (-us) must be 3. Insert the 3 to obtain

( ) (2x95 (-1/3)3

Last, the binomial coefficient must contain 8 above and either exponent below. Thus the required term is (:) (2x2)s (-y3)3 = - 1792x10~9

CHAP. 13! THE BINOMIAL COEFFICIENTS AND THEOREM 149

13.16. Prove: 24 = 16 =

Expand (1 + 1)4 using the binomial theorern:

71 In\

13.17. Prove the Binomial Theorem 13.2: (a+b)" = \a"-'@. r = ~ \y)

The theorem is true f o r n = 1, since

We assume the theorem holds for ( a + b)n and prove it is t rue for (a+ b)n+'.

(a + b ) n r l = ( a + b)(a + b).

= ( a + b ) [ a n + ( Y ) a n - l b + . . . + (Tnl) a n - r t l b r - l

3 + ( : ) a n - r b ~ + . . . + ( 7 ) a bn-' + bil

N o w the term in the product which contains br is obtained from

[(Tn J + (91 br

But, by Theorem 13.3, (?'

Note that (a + b ) ( a + b)n is a polynomial of degree n + 1 in b. Consequently,

which was to be proved.

MULTINOMIAL COEFFICIENTS

13.18. Compute:

8! - 8 * 7 * 6 * 6 * 4 * 3 * 2 * 1 = 420 iii) ( 4 , 2 8 2 , O ) = 4 ! 2 ! 2 ! 0 ! - 4 * 3 * 2 * 1 * 2 * 1 * 2 . 1 * 1

(iii) The expression ( 5 , ;\, 2 > has no meaning since 5 + 3 + 2 + 2 # 10.

150 THE BINOMIAL C O E F F I C I E N T S AND T H E O R E M [CHAP. 13

13.19. Show that

implicitly implies that n, i- n2 = n or n2 = n - n,.

n ! n l ! n,!

- - n! n, ! (n - n,) !

Observe that the expression , Hence

13.20. Find the term in the expansion of ( 2 ~ ~ - 3 x y ~ + z ~ ) ~ which contains xll and y4.

(a, b , c /

- ( a , b , c /

The general term of the expansion is

'1 2a3;:3a (-3)b~byZb 2%

- 2 a ( - 3 ) b % 3 a C b @ b z2c

Thus the term containing xll and ~4 has 3a+ b = 11 and 2 b = 4 or, b = 2 and a = 3. Also, since a - b + c = 6, we have c = 1. Substituting in the above gives

Supplementary Problems FACTORIAL

13.21. Compute: ( I ) 9!, (ii) l o ! , iiii) 11!

16' 14 ! 8 ' l o ! 14! ll!' l o ! 13! 13.22. Compute Ii) :, (ii) - iiii) :, (iv) - .

(iii) 42. 1 1 0 . 1 1 - 1 2 ' 13.23. Write in terms of factorials: ii) 2 1 * 2 3 - 2 2 - 2 1 , (ii)

(nt l ) ! 7 2 ! (n - 1) ! ( n - r + l ) ! 13.24. Simplify, ( i ) 7, (ii, __ (iii) (iv) (?z - - I ) ! . n. ( T i - 2 ) ! '

BINOMIAL AND MULTINOMIAL COEFFICIENTS

13.26. Show iwithout using Theorem 13.3) that (7) = (7) + (;)a

BINOMIAL 1 S D MULTINOMIAL THEOREMS

13.28. Expand and simplify: (i) 12x y"3, iii) (z2 - 3y)4, (iii) (+a + 2b)5, (iv) (2az - b)6.

13.29. The eighth row of Pascal's triangle is as follows:

1 8 28 56 70 56 28 8 1

Compute the ninth and tenth rows of the triangle.


13.30. Show that [ n ) + (7 ) + (;) + / 7 1 \ + ... + (E) = 2 n . !\ 0 , 3 /

13.31. Show that (c; - (;) + (;) - (;I T 3 . . t (;) = 0.

13.32. Find the term in the expansion of (2x2 - +2/3)6 which contains x*.

Find the term in the expansion of ( 3 ~ ~ 2 + 22) : which contains yG.

Find the term in the expansion of (2.j.' - y? + 32)' which contains T~ and z4.

Find the term in the expression of ( ru - ~2 - 2 z ) 6 which contains ~3 and us.

13.33.

13.34.

13.35.

Answers to Supplementary Problems 13.21. (i) 362,880 iii 3,628,800 (iii) 39,916,800

13.22. (i) 240 iiii 2184 (iii) 1/90 (iv) 111716

13.23. (i) 24!/20! rii) 9! /12! (iii) 42!/41! or 7 ! / 5 !

13.24. (i) n + 1 iiii x(x - 1) = n 2 - n iiii) l /[?t(n + l)(n + 2)] (iv) (12 - r)(n - r 4- 1)

13.25. (ij 10 ( i i ) 35 iiiii 91 (iv) 15 I V ) 1140 (vi) 816

13.27. (i) 504 i i i i 210 iiii) 180

13.28. (i) 8 d + 1 2 ~ ~ ; 1 / ~ 7 b y 4 + y6

(ii) x8 - 1 2 . ~ 6 ~ - 5 4 x 4 ~ ~ - 108x2y3 + 81yq

(iii) a"/32 + 5 d b / 8 + 5u3b2 + 20a2b3 + 40ab4 + 32b5

(iv) 64a" - 192a"Jb A 240a8b2 - 160a6b3 - 60a4b4 - l W b 5 + b6

13.29. 1 8 28 56 70 5 6 28 8 1

1 9 36 84 126 126 84 36 9 1

1 10 45 120 210 252 210 120 45 10 1

13.30. Hint. Expand (1 + l)'].

13.31. Hint. Expand (1 - l ) n .

13.32. 7OxS.y"

13.33. 945x3yGz8

13.34. - ( 1 0 5 / 4 ) ~ 4 ~ ' ~ 4

13.35. - 2 4 0 ~ 3 ~ 5 ~ '

154 PE R 31 U T A T I 0 N S, 0 RD E R E D S A 11 P L E S [CHAP. 14

Theorem 14.4: The number of permutations of n objects of which n, are alike, n2 are alike, . . ., n,. are alike is

n ! nl! nz! * . * n,!

We indicate the proof of the above theorem by a particular example. Suppose we want to form all possible 5 letter words using the letters from the word “DADDY”. Now there are 5 ! = 120 permutations of the objects DI, A, D2, Di, Y, where the three D’s are distinguished.

D1DrD3AY, D2DD1D8AY, D3D1D2AY, D1D3DzAY, D2D3D1AY, and D3D2D1AY

produce the same word when the subscripts are removed. The 6 comes from the fact that there are 3 ! = 3 . 2 . 1 = 6 different ways of placing the three D’s in the first three positions in the permutation. This is true for each of the three possible positions in which the D’s can appear.

Observe that the following six permutations

Accordingly there are

- ~ = 20 5 ! - 120

3 ! - 6 different 5 letter words that can be formed using the letters from the word “DADDY”.

Example 3.1: How many 7 letter words can be formed using the letters of the word “BENZENE”? We seek the number of permutations of 7 objects of which 3 a r e alike (the three E’s), and 2 a r e alike ( the t w o N’s). By Theorem 14.4, there a r e

7 . 6 . 5 . 4 - 3 . 2 . 1 ~ = 420 7! - - 3 ! 2! 3 . 2 . 1 - 2 . 1

such words.

Example 3.2: How many different signals, each consisting of 8 flags hung in a vertical line, can be formed from a set of 4 indistinguishable red flags, 3 indistinguishable white flags, and a blue flag? We seek the number of permutations of 8 objects of which 4 are alike and 3 a re alike. There a re

8 . 7 - 6 . 5 . 4 . 3 . 2 . 1 = 280 L- - 1! 3! 4 . 3 * 2 . 1 * 3 . 2 * 1 different signals.

ORDERED SAMPLES Many problems in combinatorial analysie and, in particular, probability, are concerned

with choosing a ball from an urn containing n balls (or a card from a deck, or a person from a population). When we choose one ball after the other from the urn, say T times, we call each choice an ordered sample of size r. We consider two cases:

(i) Sampling with replacewzeizt. Here the ball is replaced in the urn before the next Now since there are n different ways to choose each ball, by the ball is chosen.

fundamental principle of counting there are r times & n - n - n . . . n = r z l

different ordered samples with replacement of size T .

(ii) Sampliizg without replucemcizt. Here the ball is not replaced in the urn after it is chosen. In other words, an ordered sample of size 7- without replacement is simply an r-permutation of the objects in the urn.

Thus there are no repetitions in the ordered sample.

Accordingly, there a re

:HAP. 141 PERMUTATIOSS. ORDERED SAMPLES 155

12 ! ( 1 2 - r ) !

P(n, r ) = njn - 1 ) ( 1 2 - 2) . ’ .(n - r + 1) = ___

different ordered samples of size r without replacement from a population of n objects.

Example 4.1: In how many ways can one choose three cards in succession from a deck of 52 cards iii with replacement, lii’l without replacement? If each card is replaced in the deck before the next card is chosen, then each card can be chosen in 52 different ways. Hence there are

5 2 * . 5 2 . 5 2 = 523 = 140,608

different ordered samples of size 3 with replacement. In case there is no replacement, then the first card can be chosen in 52 different

ways; the second card can be chosen in 51 different ways; and, lastly, the third card can be chosen in 50 different ways.

5 2 - 5 1 - 5 0 = 132,600

Thus there a re

different ordered samples of size 3 without replacement.

Solved Problems 14.1. There are 4 bus lines between A and B ; and 3 bus lines between B and C.

(i) In how many ways can a man travel by bus from A to C by way of B? (ii) In how many ways can a man travel roundtrip by bus from A to C by way of B ? (iii) In how many ways can a man travel roundtrip by bus from A to C by way of B,

if he doesn’t want to use a bus line more than once?

(i) There are 4 ways to go from A to B and 3 ways to go from E to C; hence there are 4 * 3 = 1 2 ways to go from A t o C by way of B.

(ii) Method 1. There are 12 ways to go from A to C by way of B , and 1 2 ways to return. Hence there

are 12 * 12 = 144 ways to travel roundtrip.

Method 2.

A + E + C + B + A arrows as follows:

Above each arrow write the nuinher of v a y s tha t par t of the t r ip can be traveled; 4 ways from A t o E . 3 ways from B to C , 3 ways from C to B , and 4 ways from B to A :

A%B$C-%’B%A

The man d l travel from A to B t o C to B to A . Enter these letters with connecting

Thus there a re 4 * 3 * 3 . 4 = 144 ways to travel roundtrip.

(iii) The man will travel from A t o E to C to E to A . Enter these letters with connecting arrows

A + E + C + B + A as follows:

The man can travel 4 ways from A t o E and 3 ways from E t o C; but he can only travel 2 ways from C to B and 3 ways from R t o A since he doesn’t want to use a bus line more than once.

A h E % C % B & A

Enter these numbers above the corresponding arrows as follows:

Thus there a r e 4 - 3 . 2 . 3 = 72 ways t o travel roundtrip without using the same bus line more than once.

156

14.2.

14.3.

14.4.

PERJIUTATIONS, ORDERED SANPLES [CHAP. 14

If repetitions are not permitted, (i) how many 3 digit numbers can be formed from the six digits 2, 3, 5 , 6, 7 and 9 ? (ii) How many of these are less than 400? (iii) How many are even? (iv) How many are odd? (v) How many are multiples of 5 ?

In each case draw three boxes 0 to represent an arbi t rary number, and then

write in each box the number of digits tha t can be placed there.

(ij The box on the left can be filled in 6 ways; following this, the middle box can be filled in --- 5 ways: and, lastly, the box on the r ight can be filled in 4 ways: I 6 I I 5 I I 4 I . a re 6 . 5 . 4 = 120 numbers.

Thus there

(ii) The box on the left can be filled in only 2 ways, by 2 or 3, since each number must be less than 400: the middle box can be filled in 5 ways; and, lastly, the box on the right can be filled in 4 ways: ]?I

(iii, The box on the right can be filled in only 2 ways, by 2 or 6, since the numbers must be even; the box on the left can then be filled in 5 ways; and, lastly, the middle box can be filled in

4 ways: Fl

T&i . Thus there a re 2 5 . 4 = 40 numbers.

PI. Thus there a r e 5 4 * 2 = 40 numbers.

(ivj The box on the right can be filled in only 4 ways, by 3, 5 , 7 or 9, since the numbers must be odd: the box on the left can then be filled in 5 ways; and, lastly, the box in the middle can

be filled in 4 ways: FI . Thus there a re 5 * 4 * 4 = 80 numbers.

(v) The box on the r ight can be filled in only 1 way, by 5 , since the numbers must be multiples of 5 : the box on the left can then be filled in 5 ways; and, lastly, the box in the middle can

be fiIled in 4 ways: ,

Thus there a r e 5 * 4 1 = 20 numbers.

Solve the preceding problem if repetitions are permitted.

(ij Each box can be filled in 6 ways: [q . Thus there a re 6 6 6 = 216 numbers.

(ii) The box on the left can still be filled in only two ways, by 2 or 3, and each of the others in

6 ways:

(iii) The box on the r ight can still be filled in only 2 ways, by 2 or 6, and each of the others in 6 ways: F1 . Thus there a r e 6 * 6 - 2 = 72 numbers.

(iv) The box on the right can still be filled in only 4 ways, by 3, 5, 7 or 9, and each of the other boxes in 6 ways:

The box on the left can still be filled in only 1 way, by 5, and each of the other boxes in

6 ways: Fl .

Lf._l . Thus there a r e 2 * 6 6 = 72 numbers.

El (41 . Thus there a r e 6 * 6 * 4 = 144 numbers.

(vj Thus there a r e 6 * 6 1 = 36 numbers.

In how many ways can a party of 7 persons arrange themselves 7 chairs? (ii) around a circular table?

(i) in a row of

(ij

(ii)

The seven persons can arrange themselves in a row in

One person can s i t at any place in the circular table. The other six persons can then arrange themselves in 6 * 5 4 3 * 2 1 = 6! ways around the table.

This is a n example of a circular permutation. In general, n objects can be arranged in a circle in (n - l)(n- 2) - . . 3 2 9 1 = (n- l)! ways.

7 6 . 5 4 * 3 2 1 = 7 ! ways.

CHAP. 141 PERMUTATIOKS, ORDERED SAMPLES 157

14.5.

14.6.

14.7.

14.8.

14.9.

(i) In how many ways can 3 boys and 2 girls sit in a row? can they sit in a row if the boys and girls are each to sit together? many ways can they sit in a row if just the girls are to sit together?

(ii) In how many ways (iii) In how

(9 (ii)

(iii)

The five persons can sit in a row in

There a re 2 ways to distribute them according to sex: BBBGG o r GGBBB. In each case the boys can sit in 3 - 2 . 1 = 3! = 6 ways, and the girls can sit in 2 * 1 = 2! = 2 ways. Thus, altogether, there a r e 2 * 3! - 2 ! = 2 - 6 * 2 = 24 ways.

There a re 4 ways to distribute them according to sex: GGBBB, BGGBB, BBGGB, BBBGG. Note that each way corresponds t o the number, 0, 1, 2 or 3, of boys sitting to the left of the girls. In each case, the boys can sit in 3! ways, and the girls in 2! ways. Thus, altogether, there a re 4 * 3 ! * 2 ! = 4 . 6 - 2 = 48 ways.

5 ~ 4 . 3 . 2 . 1 = 5! = 120 ways.

Solve the preceding problem in the case of r boys and s girls. left in factorials.)

(i)

(ii)

(Answers are to be

The P + s persons can s i t in a row in ( T + s) ! ways.

There are still 2 ways to distribute them according to sex, the boys on the left or the girls on the left. In each case the boys can sit in r ! ways and the girls in s! ways. Thus, altogether, there a re 2 * r ! - s! ways.

(iii) There a re r + 1 ways to distribute them according to sex, each way corresponding t o the number, 0,1, . . . , r , of boys sitting to the left of the girls. In each case the boys can sit in r ! ways and the girls in s ! ways. Thus, altogether, there a re ( r + 1) * r ! * s! ways.

How many distinct permutations can be formed from all the letters of each word: (i) them, (ii) that, (iii) radar, (iv) unusual, (v) sociological.

(i)

(ii) 4r = 12, since there a re 4 letters of which 2 a r e t .

(iii) 5! = 30, since there a re 5 letters of which 2 a r e r and 2 are a,

(iv)

4 ! = 24, since there a re 4 letters and no repetition.

2!

2 ! 2 ! 7 ' 3! = 840, since there a re 7 letters of which 3 are u.

1 2 ! 3 ! 2 ! 2 ! 2 ! ' since there a re 12 letters of which 3 are 0, 2 are c, 2 are i, and 2 a r e 1. (")

How many different signals, each consisting of 6 flags hung in a vertical line, can be formed from 4 identical red flags and 2 identical blue flags?

6! 4! 2 !

there a r e 6 flags of which 4 a r e red and 2 are blue. This problem concerns permutations with repetitions. There a re - = 15 signals since

In how many ways can 4 mathematics books, 3 history books, 3 chemistry books and 2 sociology books be arranged on a shelf so that all books of the same subject are together?

Firs t the books must be arranged on the shelf in 4 units according to subject matter:

0 a. The box on the left can be filled by any of the four subjects; the next by

any 3 subjects remaining; the next by any 2 subjects remaining; and the box on the right by the

las t subject: . Thus there a r e 4 . 3 - 2 - 1 = 4! ways to arrange the books

on the shelf according to subject matter.

books in 3 ! ways, the chemistry books in 3 ! ways, and the sociology books in 2 ! ways. altogether, there a re 4 ! 4 ! 3 ! 3 ! 2 ! = 41,472 arrangements.

a In each of the above cases, the mathematics books can be arranged in 4 ! ways, the history

Thus,

158 PERM U TAT I 0 N S, ORDERED SAM PL E S [CHAP. 14

14.10. Solve the preceding problem if there are r subjects, A,,A,, . . . ,A7 , containing s,, s2, . . . , s , books, respectively.

Since there a re r subjects, the books can be arranged on the shelf in r ! ways according to subject matter. In each case, the books in subject A , can be arranged in sl! ways, the books in subject A? in s,! ways, . . ., the books in subject A , in s,! ways. Thus, altogether, there are r ! s1 ! s,! . . . s, ! arrangements.

14.11. (i) In how many ways can 3 -4mericans, 4 Frenchmen, 4 Danes and 2 Italians be seated in a row so that those of the same nationality sit together?

(ii) Solve the same problem if they sit a t a round table.

(i) The 4 nationalities can be arranged in a row in 4! ways. In each case, the 3 Americans can be seated in 3 ! ways, the 4 Frenchmen in 4! ways, the 4 Danes in 4! ways, and the 2 Italians in 2 ! ways. Thus, altogether, there a re 4! 3 ! 4! 4! 2 ! = 165,888 arrangements.

( i i ) The 4 nationalities can be arranged in a circle in 3! ways (see Problem 14.4 on circular permutations). In each case, the 3 Americans can be seated in 3! ways, the 4 Frenchmen in 4! ways, the 4 Danes in 4! wags, and the 2 Italians in 2! ways. Thus, altogether, there are 3 ! 3 ! 4 ! 4 ! 2 ! = 41,472 arrangements.

14.12.Find the total number of positive integers that can be formed from the digits

Let sl, sq, ss and s4 denote the number of integers containing 1 , 2, 3 and 4 digits respectively.

Also, since there a re 4 digits, there a re 4 * 3 = 12 integers containing two digits, i.e. s2 = 12. Similarly, there a r e 4 * 3 - 2 = 24 integers containing three digits and 4 * 3 * 2 * 1 = 24 integers containing four digits, i.e. s:] = 24 and s,$ = 24. Thus, altogether, there are s1 + s2 + sg + sq = 4 + 1 2 + 24 + 24 = 64 integers.

1, 2, 3 and 4 if no digit is repeated in any one integer. Note tha t no integer can contain more than 4 digits.

Since there a re 4 digits, there a re 4 integers containing exactly one digit, i.e. s1 = 4 .

We compute each si separately.

14.13. Suppose an urn contains 8 balls. Find the number of ordered samples of size 3 (i) with replacement, (ii) without replacement.

(i) Each ball in the ordered sample can be chosen in 8 ways; hence there are 8 8 8 = 83 = 512 samples with replacement.

The first ball in the ordered sample can be chosen in 8 ways, the next in 7 ways, and the last in 6 ways. Thus there a re 8 * 7 * 6 = 496 samples without replacement.

(ii,

14.14. Find n if (i) P(n, 2) = 72, (ii) P(n, 4) = 42P(?z, 2), (iii) 2P(n, 2) + 50 = P(2n, 2).

(i) P ( n , 2 ) = n(n-1) = n2--n; hence n2-n = 72 or n * - n - 7 2 = 0 or ( n - 9 ) ( n + 8 ) = 0.

Since n must be positive, the only answer is n = 9.

(ii) P ( n , 4) = n(n- l)(n - 2)(n - 3) and P(n, 2) = n(n- 1). Hence

n(n - l)(n - 2)(n - 3) = 42n(n - 1) or, if n # 0, # 1, (n - 2)(n - 3) = 42

o r n2 - 5% + 6 = 42 or n 2 - 5% - 36 = 0 or ( n - 9 ) ( n + 4) = 0


(iii) P(Jz , 2 ) = ?z(n - 1) = n2 - ?z and P ( h , 2) = 2742% - 1) = 4n2 - 2%. 2 ( n Z - n) t 50 = 4nz - 212 or 2x2 - 2% + 50 = 4n* - 2% or 50 = 2n2 or n2 = 25

Hence


CHAP. 141 PERMUTATIONS, ORDERED SAMPLES 159

f 12 and p(n, r ) . 14.15. Find the relationship between

Recall that (I) = ,).! ( n - r ) !

?'

and P(?i, T ) = n ! nl Multiply P(n,r ) by $ t o obtain

12 ! = r ! ( : )

( n - r ) ! .

- r ! r ! n! - n ! - P ( n , T ) = ________ ~

(n - r ) ! T ! ( n - r ) ! r ! (?Z--T)! -.-

Supplementary Problems 14.16. (i) How many automobile license plates can be made if each plate contains 2 different letters

followed by 3 different digits? (ii) Solve the problem if the first digit cannot be 0.

14.17. There are 6 roads between A and B and 4 roads between B and C.

(i) In how many ways can one drive from A to C by way of B? (ii) In how many ways can one drive roundtrip from A to C by way of B? (iii) In how many ways can one drive roundtrip from A to C without using the same road more

than once?

14.18. Find the number of ways in which 6 people can ride a toboggan if one of three must drive.

14.19. (i) Find the number of ways in which five persons can sit in a row. How many ways a re there if two of the persons insist on sitting next to one another? (ii)

14.20. Solve the preceding problem if they sit around a circular table.

14.21. Find the number of ways in which a judge can award first, second and third places in a contest with ten contestants.

14.22. (i) Find the number of four letter words that can be formed from the letters of the word HISTORY. (ii) How many of them contain only consonants? (iii) How many of them begin and end in a consonant? (ivj How many of them begin with a vowel? (vj How many contain the letter Y ? (vi) How many begin with T and end in a vowel? (vii) How many begin with T and also contain S? (viii) How many contain both vowels?

14.23. How many different signals, each consisting of 8 flags hung in a vertical line, can be formed from 4 red flags, 2 blue flags and 2 green flags?

14.24. Find the number of permutations that can be formed from all the letters of each word: (ii) committee, (iii) proposition, (iv) baseball.

(i) queue,

14.25. (i) Find the number of ways in which 4 boys and 4 girls can be seated in a row if the boys and girls are to have alternate seats.

(ii) Find the number of ways if they sit alternately and if one boy and one girl a re t o sit in adjacent seats.

(iii) Find the number of ways if they sit alternately and if one boy and one girl must not sit in adjacent seats.

14.26. Solve the preceding problem if they sit around a circular table.

160 PERM U T A T I 0 N S, ORDERED SARI PLE S [CHAP. 14

14.27. An urn contains 10 balls. ( i ) of size 3 with replacement, (ii, of size 3 without replacement, fiii) of size 4 with replacement, (iv) of size 5 without replacement.

Find the number of ordered samples

14.28. Find the number of ways in which 5 large books, 4 medium-size books and 3 small books can be placed on a shelf so tha t all books of the same size a re together.

14.29. Consider all positive integers with 3 different digits. (Note tha t 0 cannot be the first digit.) (iv) How (i, How many are greater than 7001

many are divisible by 5? i i i ) How many are odd? (iii) How many are even?

14.30. (i) Find the number of distinct permutations tha t can be formed from all of the letters of the word ELEVEN. fii) How many of them begin and end with E l f i i i ) How many of them have the 3 E's together? (iv) How many begin with E and end with N ?


14.17.

14.18.

14.19.

14.20.

14.21.

14.22.

14.23.

14.24.

14.25.

14.26.

14.27.

14.28.

14.29.

(i) 2 6 . 2 5 . 1 0 . 9 - 8 = 468,000 iii) 2 6 . 2 5 . 9 . 9 . 8 = 421,200

3 . 5 * 4 . 3 . 2 * 1 = 360

(i) 5 ! = 120 (ii) 4 . 2 ! * 3 ! = 48

(i) 4! = 24 (ii) 2!3! = 12

1 0 . 9 . 8 = 720

9 ' - (ii) - 2 ! 2! 2! 45,360 (iii) ~ = 1,663,200 2! 3! 2!

8 ! 2! 2! 2! (iv) __ = 5040

(i) 2 * 4 ! * 4 ! = 1152 (ii) 2 . 7 * 3 ! . 3 ! = 504 (iii) 1152-504 = 648

(i) 3 ! * 4 ! = 144 (ii) 2 * 3 ! * 3 ! = 72 fiii) 144-72 = 72

(i) 10 * 10 10 = 1000 (iii) 10 * 10 10 10 = 10,000

(ii, 1 0 ' 9 . 8 = 720 (iv) 10 * 9 * 8 7 * 6 = 30,240

3 ! 5 ! 4 ! 3 ! = 103,680

(i) 3 ' 9 . 8 = 216 (ii) 8 ' 8 . 5 = 320

(iii, 9 ' 8 . 1 = 72 end in 0, and 8 ' 8 . 4 = 256 end in the other even digits; hence, altogether,

(iv) 9.8.1 = 72 end in 0, and 8.8.1 = 64 end in 5; hence, altogether, 72+64 = 136 are

72+256 = 328 a r e even.

divisible by 5.

G ! 4 ! 3 . 2 !

14.30. (i) 7 = 120 (ii) 4! = 24 iiii) 4 * 3! = 24 (iv) - = 12

Chapter 15

abc

abd

acd

bcd

Combinations, Ordered Partitions

a b c , a c b , bac, bca , c u b , c b a

a b d , a d b , b a d , bda , d a b , dba

a c d , a d c , c a d , c d a , d u e , dca

bcd, bdc , c b d , c d b , d b c , c k b

P(n, r ) = n ! Theorem 15.1: C(n, T ) = ~ r ! T ! (n - T ) !

161

162 COMBINATIOKS, ORDERED PARTITIONS [CHAP. 15

n! Recall that the binomial coefficient . hence

I I

We shall use C(n,r) and (:) interchangeably.

Example 1.3: How many committees of 3 can be formed from 8 people? essentially, a combination of the 8 people taken 3 a t a time.

Each committee is, Thus

different committees can be formed.

Example 1.4: A farmer buys 3 cows, 2 pigs and 4 hens from a man who has 6 cows, 5 pigs and 8 hens.

The farmer can choose the cows in /6' ways, the pigs in (i ways, and

How many choices does the farmer have?

\3 1 the hens in ways. Hence altogether he can choose the animals in

PARTITIONS AND CROSS-PARTITIONS Recall (Page 60) that a part i t ion of a set X is a subdivision of X into subsets which

are disjoint and whose union is X, i.e. such that each a E X belongs to one and only one of the subsets. In other words, the collection {A,,A,, . , . , A m } of subsets of X is a partition of X iff

(i) X = A , u A 2 u 9 . . UA,; (ii) foranyA,,A,, either A l = A 3 or A t n A j = @ The subsets of a partition are called cells.

Example 2.1: Consider the following classes of subsets of X = {1 ,2 , . . ., 8,9}:

6) [{I, 3,51, {2 ,6} , {4,8,931

(ii) [{I, 3,5), { 2 , 4 , 6 , 8 ) , {5 ,7 ,9 ) ]

(iii) [{I, 3,51, { 2 , 4 , 6 , 8 } , (7,911

Then (i) is not a partition of X , since 7 6 X but 7 does not belong to any cell in (i). Also (ii) is not a partition of X since 5 € X and 5 belongs t o both { 1 , 3 , 5 } and {5,7,9}. On the other hand, (iii) is a partition of X since each element of X belongs to exactly one cell.

Suppose {A,,A,, ..., A,.} and {B,,B,, , . .,Bs} are both partitions of the same set X. Then the class of intersections {At n B,} form a new partition of X called the cross-partition.

Example 2.2: Let { A , B, C, D} be the partition of the undergraduate students at a university into freshmen, sophomores, juniors and seniors, respectively; and let {M, F } be the partition of the students into maies and females, respectively. The cross- partition consists of the following:

A n M : male freshmen A n F : female freshmen

B n M : male sophomores B n F : female sophomores

CnM: male juniors C n F : female juniors

D n M : male seniors D n F: female seniors

CHAP. 151 COMBINATIONS, ORDERED PARTITIONS 163

ORDERED PARTITIONS Suppose an urn A contains seven balls numbered 1 through 7. We compute the

number of ways we can draw, first, 2 balls from the urn, then 3 balls from the urn, and lastly 2 balls from the urn. In other words. we want to compute the number of ordered partitions

(AI, A2, A3)

of the set of 7 balls into cells A , containing 2 balls, A, containing 3 balls and A, containing 2 balls. We call these ordered partitions since we distinguish between

({I, 21, {3,4,5), ( 6 7 ) ) and (C6,7), {3,4,51, {1,21)

each of which determines the same partition of A .

Now we begin with 7 balls in the urn, so there are ways of drawing the first 2 balls, (3 i.e. of determining the first cell A, ; follon.ing this, there are 5 balls left in the urn and so

mays of drawing the 3 balls, i.e. of determining the second cell A , ; finally, /c, \

there are 2 balls left in the urn and so there are Hence there are

ways of determining the last cell A , .

7 . 6 5 . 4 . 3 2 . 1 - 210

different ordered partitions of A into cells A, containing 2 balls, A, containing 3 balls, and A, containing 2 balls.

We state the above result in its general form as a theorem.

Theorem 15.2: Let A contain n elements and let n,, n2, . . . , nr be positive integers with 1 2 , 1 ? 1 , 4 . . - + n? = 12. Then there exist

1 (;$n i 2 n 1 )( n - nz ) . . . ( n - nl - n2 - * * - -nr-l

nr

different ordered partitions of A of the form (A,,A,, . . . , A r ) where A, contains n, elements, A, contains n, elements, . . ., and A , contains nT elements.

Now observe that

since each numerator after the first of the previous factor. Similarly,

( n )( n - nl ) ( n - nl - nz nl i n2 n3

n!

- 7 ! - 7 ! 5! 2! -.-.- 2!5! 3!2! 2 ! 0 ! 2!3!2!

is cancelled by the second term in the denominator

. . . ~

-, - - n,! (n- n l ) ! * nz! ( n - n1 - n2)! . n3! (n- n, - n2 - n3) !

nl ! n,! ns! . - n,!

n,! n! - -

The above computation together with the preceding theorem gives us our next theorem.

164 CO R I B I N A T I 0 N S , 0 RD E R E D PART IT I 0 K S [CHAP. 15

Theorem 15.3: Let A contain 71 elements and let nl, 712, . . . , n, be positive integers with 111 + n2 + . . . + n, = n.

n ! nl! na! ns! . . . n,!

different ordered partitions of A of the form (A1, Az, . . . , A,) where A1 contains nl elements, A> contains n2 elements, . . . , and Ar contains nr elements.

In how many ways can 9 toys be divided between 4 children if the youngest child is to receive 3 toys and each of the others 2 toys?

We wish to find the number of ordered partitions of the 9 toys into 4 cells containing 3, 2, 2 and 2 toys respectively.

Then there exist

Example 3.1:

By Theorem 15.3, there a re

= 2520 3! 2! 2! 2! such ordered partitions.

Solved Problems COMBINATIONS 15.1. In how many ways can a committee consisting of 3 men and 2 women be chosen

from 7 men and 5 women? The 3 men can be chosen from the 7 men in ways, and the 2 women can be chosen from the

7 . 6 . 5 5 - 4 5 women in(') ways. Hence the committee can be chosen in 350 ways.

2)

3.5.2. A bag contains 6 white balls and 5 black balls. Find the number of ways 4 balls can be drawn from the bag if ( i) they can be of any color, (ii) 2 must be white and 2 black, (iii) they must all be of the same color.

(ii The 4 balls (of any color) can be chosen from the 11 balls in the urn in 330 ways.

( i i ) The 2 white balls can be chosen in

Thus there a re

ways, and the 2 black balls can be chosen in

' 150 ways of drawing 2 white balls and 2 black balls.

(:) = 15 ways of drawing 4 white balls, and ") = 5 ways of drawing 4 black (iii) There are

balls. Thus there a re 15 + 5 = 20 ways of drawing 4 balls of the same color. (4,

15.3. A delegation of 4 students is selected each year from a college to attend the National Student Association annual meeting. (i) In how many ways can the delegation be chosen if there are 12 eligible students? (ii) In how many ways if two of the eligible students will not attend the meeting together? (iii) In how many ways if two of the eligible students are married and will only attend the meeting together?

12 11 ' 10 * 9 = 495 ways. ( y ) = 1 . 2 . 3 - 4 (i, The 4 students can be chosen from the 12 students in

(ii, Let A and B denote the students who will not attend the meeting together.


Method 1. 1 0 - 9 * 8 . 7 -

- (140) = 1 . 2 . 3 . 4 If neither A nor B is included, then the delegation can be chosen in

210 ways. If either A or B, but not both, is included, then the delegation can be chosen in

2 * ( y ) = 2 . 1 . 2 . 3 10'9.8 1 240 ways. Thus, altogether, the delegation can be chosen in

210 + 240 = 450 ways.

Method 2. If A and B are both included, then the other 2 members of the delegation can be chosen in

Thus there a r e 495 - 45 = 450 ways the delegation can be chosen if "'1 = 45 wags. ( 2 , A and B are not both included.

(iii) Let C and ll denote the married students.

/"" ~ 210 ways. chosen in

(120\' = 45 wags. Altogether, the delegation can be chosen in 210 t 45 = 255 wags.

If C and D do not go, then the delegation can be

If both C and D go, then the delegation can be chosen in \ 4 : ~

15.4.

15.5.

There are 12 points A, B, . . . in a given plane, no three on the same line. (i) How many lines are determined by the points? (ii) How many of these lines pass through the point A ? (iii) How many triangles are determined by the points? (iv) How many of these triangles contain the point A as a vertex?

12 * 11 Since two points determine a line, there are ('2") = 1.2 = 66 lines.

To determine a line through A , one other point must be chosen: hence there a re 11 lines through A .

Since three points determine a triangle, there a r e ( 3 ) - 1 . 2 . 3 = 220 triangles.

Method 1.

('2' ) =

12 - 1 2 * 1 1 * 1 0

To determine a triangle with vertex A , two other points must be chosen; hence there a re

= 55 triangles with A as a vertex.

Method 2. There are (131 ) = = 165 triangles without A a s a vertex. Thus 220 - 165 = 55

, I

of the triangles do have A as a vertex.

A student is to answer 8 out of 10 questions on an exam. has he? if he must answer at least 4 of the first 5 questions?

(i) How many choices (iii) How many (ii) How many if he must answer the first 3 questions?

(i)

(ii)

(iii)

The 8 questions can be selected in lo - lo - = 45 ways. ( 8 ) - ( 2 ) -

If he answers the first 3 questions, then he can choose the other 5 questions from the last

7 questions in '7 - 7 - 7.6 = 21 ways. (5) - ( 2 ) - 1 . 2

If he answers all the first 5 questions, then he can choose the other 3 questions from the last

5 in (i) = 10 ways. On the other hand, if he answers only 4 of the first 5 questions, then

he can choose these 4 in and he can choose the other 4 questions

from the last 5 in (i) = (f) = 5 ways; hence he can choose the 8 questions in 5 . 5 =

25 ways.

4) = = 5 ways,

(5i ( 5 )

Thus he has a total of 35 choices.

166 COMBINATIOXS, ORDERED PARTITIONS [CHAP. 15

15.6. How many diagonals has an octagon? An octagon has 8 sides and 8 vertices. Any two vertices determine either a side or a diagonal.

= 28 sides plus diagonals. But there are 8 sides; hence there are

28 - 8 = 20 diagonals.

15.7. How many diagonals has a regular polygon with ?z sides? The regular polygon with 71 sides has, also, n vertices. Any two vertices determine either a

- - sides plus diagonals. But there ’n\ n(n-1) n(n- 1) :2 / 1 .2 side or a diagonal. Thus there are 1

are 71 sides: hence there a r e

722 - n 2n - n2 - 3n - - - ’3 diagonals n(n - 1)

2 ~-

2 2 2 2 n =

15.8. Which regular polygon has the same number of diagonals as sides? nz - 3 n

The regular polygon with 71 sides has, by the preceding problem, 7 diagonals. Thus we seek the polygon of n sides for which

n2 - 3 n - - n o r n1 - 3 7 ~ = 2n or n2 - 5n = 0 or n(n-5) = 0 2

Since 71 must be a positive integer, the onIy answer is n= 5. In other words, the pentagon is the only regular polygon with the same number of diagonals a s sides.

15.9. A man is dealt a poker hand (5 cards) from an ordinary playing deck. In how many ways can he be dealt (i) a spade flush (5 spades), (ii) an ace high spade flush (5 spades with the ace), (iii) a flush ( 5 of the same suit), (iv) an ace high full house (3 aces with another pair), ( ~ 7 ) a full house (3 of one kind and 2 of another), (vi) 3 aces (without the fourth ace or another pair), (vii) 3 of a kind (without another pair), (viii) two pair?

13) - 13 * 1 2 - 1 1 1 0 - 9 = 1287 ways. (i) ( 5 ,/ - 1 . 2 * 3 * 4 . 5 The 5 spades can be dealt from the 13 spades in

(ii) If he receives the spade ace, then he can be dealt the other 4 spades from the remaining

(iii) There a re 4 suits, and a flush from each suit can be dealt (see (i)) in (153) ways; hence a

flush can be dealt in 4 . (7) = 5148 ways.

(iv) The 3 aces can be dealt from the i aces in (i) = (:> = 4 ways. The pair can be selected

in 12 ways, and the 2 cards of this pair can be dealt in ways. Thus the ace high full

house can be dealt in 4 12 {4) = 4 12 0- 4 * 3 = 288 ways. \2 1 . 2

(v) One kind can be selected in 13 ways and 3 of this kind can be dealt in ways; another

ways. Thus a full c 1

kind can be selected in 12 ways and 2 of this kind can be dealt in house can be dealt in

13 ($ * 12 * (i) = 3744 ways

fobserve, by (iii), tha t a flush is more common than a full house.)

CHAP. 151 COMBINATIOKS, ORDERED PARTITIOKS 167

(vi) The 3 aces can be selected from 4 aces in ways. The 2 other cards can be selected from

the remaining 12 kinds in ( y ) ways; and 1 card in each kind can be dealt in 4 ways.

Thus a poker hand of 3 aces can be’ dealt in

(:) /

(vii) One kind can be selected in 13 ways: and 3 of this kind can be dealt in (43) ways. The

2 other kinds can be selected from the remaining 12 kinds in I”~‘ ways; and 1 card in each

kind can be dealt in 4 ways. Thus a poker hand of 3 of a kind can be dealt in \ 2 /

(viii) Two kinds (for the pairs) can be selected from the 13 kinds in 13‘ ways; and 2 of each kind

ways. Another kind can be selected in 11 mays; and 1 of this kind in \ 2

can be dealt in

4 ways. Thus a poker hand of t w o pair can be dealt in (:I

15.10. Consider 4 vowels (including a) and 8 consonants (including b ) .

(i) How many 5 letter “words” containing 2 different von~els and 3 different consonants can be formed from the letters?

(ii) How many of them contain b ?

(iii) How many of them begin with b?

(iv) How many of them begin with a ?

(v) How many of them begin with a and contain b?

(i) The 2 vov,els can be selected from the 4 vowels in (;) ways, and the 3 consonants can be

ways. Furthermore, each 5 letters can be arranged /8‘ \ 31

selected from the 8 consonants in

in a row fas a “word”) in 5 ! ways. Thus we can form

(:) (:; * 5 ! = 6 56 * 120 = 40,320 words

i \ (ii) The 2 vowels can still be selected in ways. However, since b is one of the consonants, the

‘ 2 other 2 consonants can be selected from the remaining 7 consonants in ways. Again each 5 letters can be arranged as a word in 5! ways. Thus we can form

(i) * (2‘) * 5 ! = 6 * 21 * 120 = 15,120 Ivords containing b

(iii) The 2 vowels can be selected in ( 4 ) ways, and the other 2 consonants can be selected in (z) ways. The 4 letters can be arranged, following b, in 4 ! ways. Thus we can form \ 2 /

(i) (2’) - 4! = 6 - 21 * 24 = 3024 words beginning with b

168 COMBINATIONS, ORDERED PARTITIONS [CHAP. 15

( i d The other vowel can be chosen in 3 ways, and the 3 consonants can be chosen in ways.

The 4 letters can be arranged, following a, in 4! ways. Thus we form 3 - / 8 ) * 4 ! = 3 . 5 6 . 2 4 = 4032 words beginning with a.

(:) \ 3 /

(vi The other vowel can be chosen in 3 ways, and the 2 other consonants can be chosen in (z) x a y s . The 4 letters can be arranged, following a, in 4! ways. Thus we can form

3 ( 7 ) - 4! = 3 - 2 1 - 2 4 = 1612 words beginning with a and containing b . 2 ,

15.11. How many committees of 5 Tvith a given chairman can be selected from 12 persons? The chairman can be chosen in 12 ways and, following this, the other 4 on the committee can

be chosen from the 11 remaining in ( y ) ways. Thus there are 1 2 . (121) = 12.330 = 3960 such committees.

15.12. Find the number of subsets of a set X containing ?z elements.

Method 1. The number of subsets of X with T

subsets of X. The above sum (see Problem 13.30, Page 151) is equal to 2n, i.e. there a re 2" subsets of X.

Method 2.

hence there are There are two possibilities for each element of X : either it belongs to the subset o r it doesn't;

times

2 . 2 . . . . a 2 = 2n

ways t o form a subset of X , i.e. there a re 2n different subsets of X.

15.13. In how many ways can a teacher choose one or more students from six eligible students ?

Method 1. By the preceding problem, there are 2s = 64 subsets of the set consisting of the six students.

However, the empty set must be deleted since one or more students a re chosen. Accordingly, there are 26 - 1 = 64 - 1 = 63 ways to choose the students.

Method 2. Either 1 , 2 . 3 , 4 , 5 or 6 students are chosen. Hence the number of choices is

(F) + 4- (i) + (:) + (i) + (z) = 6 + 15 + 20 + 15 + 6 i- 1 = 63

15.14. In how many ways can three or more persons be selected from 12 persons? There are 21a - 1 = 4096 - 1 = 4096 ways of choosing one o r more of the twelve persons.

Now there a re ( y ) + (7) = 12 + 66 = 78 ways of choosing one or two of the twelve persons.

Hence there are 4096-78 = 4018 ways of choosing three or more.


PARTITIONS AND CROSS-PARTITIONS 15.15. Which of the following are partitions of X = {a, b, c, d , e, f , g } ?

(i) [{a. 0, c:, "Zj, i f ,9)]

(ii) [{a, 0, c ) , {c, d , e}, { f , s } ] (iii) [{a, b , c ) , ( d , e ) , { f , 9 } ]

(i) This is not a partition of X since e E X, but e does not belong to any cell.

(ii) This is not a partition of X since c E A' and belongs to both { a , 6, c) and {c, d, e } .

(iii) This is a partition of X . Each element of T belongs to exactly one cell.

15.16. Let cA = { A , , A , , . . . , A , , L } and 49 = (B, , B,, . . . , B J be partitions of the same is also a set X .

partition of X. Let z E X.

Show that the cross-partition C = {Ai n B I : Ai E d, Bj E CB}

Then x belongs to some dLo and to some Bi,, since 04 and '73 are partitions of X . Thus x E A , o n E , o and so belongs t o a member of the cross-partition.

On the other hand, suppose 3: E A ; , , I ~ I B , ~ and AilnB,il . Then z E Aio and A , , , whence A . = A i since d is a partition of X . Similarly, B . = B j l . Accordingly, AiOnBj, = A , l n B , l and so C is a partition of X . '0 1 30

13.17. Let X = (1,2,3,4,5,6,7,8}. Find the cross-partition of each pair of partitions of X : (i) [{I, 2,3,4}, .(5,6,7,8>1 and [{I, 2 , 8 ) , (3 ,4 ,5 ,6 ,7 ) ] (ii) [{I, 2 , 3 , 4 ) , {5 ,6 ,7 ,8} ] and [il, 2 ) , {3 ,4 ,5} , {6,7,8)]

(i) {1,2,3,4] n {l, 2 , 8 } = {1,2} ; C5, 6,7,8> n {1,2, 8) = {8)

To obtain the cross-paTAition, intersect each set of one partition with each set of the other.

{ 1 , 2 , 3 , 4 ) n{3,4,5,6,7} = {3,4}; {5 ,6,7,8}n{3,4,5,6,7] = {5,6,7}

Thus the cross-partition is [{l, Z}, {3 ,4} , {8), {5,6,7}].

(ii) (1,2,3,4$ n {l, 2 ) = {1,2} ; {5,6,7,8) n U , 2 } = '2) { 1 , 2 , 3 , 4 } n ( 3 , 4 , 5 } = {3,4}; {5 ,6 ,7 ,8 )n {3 ,4 ,5 } = (5)

{1,2,3,4: n {6,7,8> = P ; ( 5 , 6, 7, 8) n {6,7,8} = (6, 7,8}

Thus the cross-partition is

[{l, 2 } , {3,4}, {5}, {6,7,8), 81 or, simply, [{I, a} , {3,4), (51, {6,7,8}]

Remark: We do not distinguish partitions which differ by the empty set.

15.18. Let X denote the domain of a proposition P(p, q, 7') of three variables: X = {TTT, TTF, TFT, TFF, FTT, FTF, FFT, FFF}

(i) (ii) For the proposition P

(iii) For .the proposition Q = - p + ( q A T ) , partition X into 7(Q) and ('T(Q))c.

(iv) Find the cross-partition of the partitions of X in (ii) and (iii).

Partition X according to the number of T's appearing in each element.

that 7 ( P ) is the truth set of P.) (11- q ) A r , partition X into T ( P ) and (T(P))c . (Recall

(i) Combine the elements of X with the same number of T's t o form the cells:

[{FFF}, (TFF, F T F , FFT}, {TTF, TFT, FTT) , {TTT}]

COMBIKATIONS, ORDERED PARTITIONS [CHAP. 15 170

Q r

T T

T F

F T F F T T

T F

F T

F F

i i i ) and (iii). Construct the t ruth tables of P = ( p -+ q ) P T and Q = - p + (q A r):

P + 4 ( P + Q ) A T

T T

T F F F F F T T

T F T T T F

f F T T

T I B I T F

T 1 , ' l Note tha t P is t rue in Cases (lines) 1 , 5 and 7, and Q is t rue in Cases 1 , 2 , 3 , 4 and 5. Thus

T(P) = {TTT, FTT, FFT) and I(Q) = {TTT, TTF, TFT, TFF , FTT}

and so the required partitions a re

[{TTT, FTT, FFT}, {TTF, TFT, TFF , FTF, FFF}]

[{TTT, TTF, TFT, TFF , FTT}, {FTF, FFT, FFF}] and

(iv) The cross-partition is obtained by intersecting each set of one partition with each set of the other:

{TTT, FTT, FFT} n {TTT, TTF, TFT, TFF, FTT} = {TTT, FTT}

{TTT, FTT, FFT} n {FTF, FFT, F F F } = {FFT}

{TTF, TFT, TFF , FTF , F F F ) n {TTT, TTF, TFT, TFF , FTT} = {TTF, TFT, TFF}

{TTF, TFT, TFF , FTF, FFF] n {FTF, FFT, F F F ) = {FTF, F F F }

Thus the cross-partition is

[{TTT, FTT}, {FFT}, {TTF, TFT, TFF}, {FTF, FFF}]

ORDERED AND UNORDERED PARTITIONS 15.19.In how many ways can 7 toys be divided among 3 chiIdren if the youngest gets

3 toys and each of the others gets 2? We seek the number of ordered partitions of 7 objects into cells containing 3, 2 and 2 objects,

respectively. By Theorem 15.3, there a re - "' = 210 such partitions. 3! 2 ! 2!

15.20.There are 12 students in a class. In how many ways can the 12 students take 3 different tests if 4 students are to take each test?

Method 1. We seek the number of ordered partitions of the 1 2 students into cells containing 4 students

1 2 ! each. By Theorem 15.3, there a re = 34,650 such partitions.

Method 2. There are (7) ways to choose 4 students to take the first test; following this, there are

''1 ways t o choose 4 students to take the second test. The remaining students take the third test.

Thus. altogether, there a re ( y ) * ( t ) = 495 70 = 34,650 ways for the students to take the tests.


15.21. In how many ways can 12 students be partitioned into 3 teams, A , , A , and A,, so that each team contains 4 students?

Method 1. Observe tha t each partition { A l , A 2 , A 3 } of the students can be arranged in 3! = 6 ways

as a n ordered partition. Since (see the preceding problem) there a re = 34,650 such ordered partitions, there a re 34,650/6 = 5775 (unordered) partitions.

Method 2.

1 2 !

Let A denote one of the students. Then there a re (131) ways to choose 3 other students to be

on the same team as A . Now let B denote a student who is not on the same team as A ; then there

a re (:) ways t o choose 3 students of the remaining students t o be on the same team as B. The

remaining 4 students constitute the third team. Thus, altogether, there a re 165.35 = 5775 ways t o partition the students. ( Y ) *(;I =

15.22. In how many ways can 6 students be partitioned into (i) 2 teams containing 3 students each, (ii) 3 teams containing 2 students each?

(i) Method 1. 61

3! 3! There are - = 20 ordered partitions into 2 cells containing 3 students each. Since

each unordered partition determines 2 ! = 2 ordered partitions, there a r e 20/2 = 10 unordered partitions.

Method 2. Let A denote one of the students; then there a re ([) = 10 ways to choose 2 other students

to be on the same team as A . The other 3 students constitute the other team. In other words, there are 10 ways to partition the students.

(ii) Method 1. 61

There a re & = 90 ordered partitions into 3 cells containing 2 students each. Since each unordered partition determines 3! = 6 ordered partitions, there a re 90/6 = 15 unordered partitions.

Method 2. Let A denote one of the students, Then there a r e 5 ways to choose the other student to

be on the same team as A . Let B denote a student who isn't on the same team as A ; then there a re 3 ways to choose another student to be on the same team as B. The remaining 2 students constitute the other team. Thus, altogether, there a re 5 - 3 = 15 ways to partition the students.

15.23.In how many ways can a class X with 10 students be partitioned into 4 teams A,, A,, B,, B,, where A , and A, contain 2 students each and B, and B, contain 3 students each ?

Method 1. l o !

There are 2 ! 2 ! 3 ! 3 ! 3 students, respectively.

= 25,200 ordered partitions of X into 4 cells containing 2,2,3 and

However, each unordered partition { A , , A,, B1, B,} of X determines 2! 2! = 4 ordered partitions of X . Thus, altogether, there a r e 25,20014 = 6300 unordered partitions.

Method 2. There a re ( y ) ways to choose 4 students who will be on the teams A , and A,, and there are

3 ways in which each 4 students can be partitioned into 2 teams of 2 students each. On the other hand, there are 10 ways (see preceding problem) t o partition the remaining 6 students into 2 teams

containing 3 students each. Thus, altogether, there a re (:)*3*10 = 2 1 0 * 3 * 1 0 = 6300 ways to partition the students.

172 COMBINATIONS, ORDERED PARTITIONS ;CHAP. 15

15.24.In how many ways can 9 students be partitioned into three teams containing 4, 3

Since all the cells contain different numbers of students, the number of unordered partitions

and 2 students, respectively?

equals the number of ordered partitions which, by Theorem 15.3, is & - - 1260.

15.25. (i) In how many ways can a set X containing 10 elements be partitioned into two cells?

(ii) In how many ways can 10 students be partitioned into two teams?

(i, Method 1. Each subset A of X determines an ordered partition ( A , A c ) of X (where Ac is the com-

plement of A ) , and so there are 210 = 1024 such ordered partitions. However, each unordered partition { A , B } determines two ordered partitions, ( A , B ) and ( B , A ) , and so there a re l024/2 = 512 unordered partitions.

Method 2.

2q = 512 ways of choosing a subset of X - {x} to be in the same cell as x. there are 512 such partitions. Here we assume tha t each team must contain at least one student, hence-we do not allow the partition for which one team contains 10 students, and the other team none. Accordingly, there a re 5 1 2 - 1 = 511 possible partitions.

Let j . denote one of the elements in X . Now X - { x } contains 9 elements, and so there a re In other words,

(Observe t h a t { X , @ ) is one of these partitions.)

( i i i

Supplementary Problems COMBINATIONS 15.26.

15.27.

15.28.

15.29.

15.30.

15.31.

A class contains 9 boys and 3 girls. (i) In how many ways can the teacher choose a committee of 4? (iii How many of them will contain at least one girl? (iii) How many of them will contain exactly one girl?

A woman has 11 close friends. ( i ) In how many ways can she invite 5 of them to dinner? (ii) In how many ways if two of the friends are married and will not attend separately? iiii) In how many ways if two of them are not on speaking terms and will not attend together?

A woman has 11 close friends of whom 6 are also women. ii) In how many ways can she invite 3 o r more of them to a par ty? (ii) In how many ways can she invite 3 or more of them if she wants the same number of men as women fincluding herself)?

There are 10 points A , R , . , . in a plane, no three on the same line. (i) How many lines a r e determined by the points? (ii) How many of these lines do not pass through A or E ? (iii) How many triangles a r e determined by the points? (iv) How many of these triangles contain the point A ? IW) How many of these triangles contain the side A B ?

A student is t o answer 10 out of 13 questions on a n exam. (i) How many choices has he? iii) How many if he must answer the first two questions? (iii) How many if he must answer the first or second question but not both? (iv) How many if he must answer exactly 3 of the first 5 questions? (v) How many if he must answer a t least 3 of the first 5 questions?

How inany diagonals has a ( i ) hexagon, (ii) decagon?


15.32. Which regular polygon has (i) twice as many diagonals as sides, (ii) three times as many diagonals a s sides?

15.33. (i)

(ii)

How many triangles are determined by the vertices of a regular polygon with 12 sides if the sides of the polygon are not to be sides of any triangle?

A man is dealt a poker hand (5 cards) from an ordinary playing deck. In how many ways can he be dealt (i) a straight flush, (ii) four of a kind, (iii) a straight, (iv) a pair of aces, (v) two of a kind (a pair)?

The English alphabet has 26 letters of which 5 are vowels. (i) How many 5 letter words containing 3 different consonants and 2 different vowels can be

formed?

(ii) How many of them contain the letter b? (iii) How many of them contain the letters b and c?

(iv) How many of them begin with b and contain the letter c?

(v) How many of them begin with b and end with c?

(vi) How many of them contain the letters a and b? (vii) How many of them begin with a and contain b? (viii) How many of them begin with b and contain a? (ix) How many of them begin with a and end with b? (x) How many of them contain the letters a , b and c?

How many triangles a r e determined by the vertices of a n octagon? How many if the sides of the octagon are not to be sides of any triangle?

15.34.

15.35.

15.36.

PARTITIONS AND CROSS-PARTITIONS 15.37. Which of the following are partitions of X = (a,e, i ,o,u}?

(i) [{a, e , Q, Co), {u)] (iv) [{a , e l , Ci), (01, {a, e l , {ull (ii) [{a,i, u>, {e l , (0, ull (v) [{a , e , 4, (0, u1, @I (iii) [{a, e ) , Co), Czt)] (vi) [@, { a ) , Ce, u.1, @, {i, 011

35.38. Find the cross-partition of each pair of partitions of X = { 1 , 2 , 3 , 4 , 5 , 6 ) :

(i) [{l, 2 ,3 ) , {4 ,5 ,6 ) ] and [{I, 3,5) , {2 ,4 ,6 ) ] . (ii) [{I, 2 ,3) , {4 ,5 ,6 ) ] and [{I, 21, {3,4, 5) , (611 . Let X denote the domain of a proposition P ( p , q , ~ ) of three variables: 15.39.

X = {TTT, TTF, TFT, T F F , FTT, F T F , F F T , FFF)

(i) Partition X according to the number of F's appearing in each element. (ii) For the proposition P = ( p v q) A T , partition X into T ( P ) and ( 7 ( P ) ) c . (iii) Partition X into T(q) and T(-q ) . (iv) Find the cross-partition of the partitions in (ii) and (iii).

15.40. For any proposition P = P(p, q, . . .), show that [ T ( P ) , T( -P)] forms a partition of the domain of P .

15.41. Let the partition { A , , A , , . . . ,A,.} be a refinement of the partition {B,, B,, . . .,Bs}, that is, each Ai is a subset of some Bj. Describe the cross-partition.

Show that a cross-partition is a refinement (see preceding problem) of each of the original partitions. 15.42.

ORDERED AND UNORDERED PARTITIONS 15.43.

15.44.

In how many ways can 9 toys be divided evenly among 3 children?

In how many ways can 9 students be evenly divided into three teams?

174 COMBINATIONS, ORDERED PARTITIONS 'CHAP. 15

15.45.

15.46.

15.47.

15.48.

15.49.

15.50.

15.51.

15.52.

In how many ways can 1 0 students be divided into three teams, one containing 4 students and the others 3 1

There a re 12 balls in a n urn. In how many ways can 3 balls be drawn from the urn, four times in succession?

In how many ways can a club with 12 members be partitioned into three committees containing 5,4 and 3 members respectively?

In how many ways can n students be partitioned into two teams containing at least one student?

In how many ways can 14 men be partitioned into 6 committees where 2 of the committees contain 3 men and the others 2?

In how many ways can a set X with 3 elements be partitioned into (i) three ordered cells, (ii, three (unordered) cells?

In how many ways can a set X with 4 elements be partitioned into (i) three ordered cells, (ii) three (unordered) cells?

In bridge, 13 cards a re dealt to each of four players who are called North, South, Eas t and West. The distribution of the cards is called a bridge hand. ii) How many bridge hands a re there? (ii, In how many of them will one player be dealt all four aces? (iii) In how many of them will each player be dealt a n ace? (iv) In how many of them will North be dealt 8 spades and South the other 5 spades? (v) In how many of them will North and South have, together, all four aces? (Rrmark. Leave answers in factorial notation.)


15.26. (i, (7) = 495, (ii) ( y ) - (:) = 369, (iii) 3 * (:) = 252

15.27. (i, ) = 462, (ii) ( 9 3 ) + = 210, (iii) 2 *(:I = 252

( y ) - (';) = 1981 o r (t:> + ('41) + . * . + (") = 1981 15.28. (i) 211 - 1 - \ 11

15.29. (i) ('2", = 45, (ii) (i) = 28, (iii) (130) = 120, (iv) = 36, (v) 8

15.30. (i) = (7) = 286

(ii) ( y ) = ( y ) = 165

15.31. (i) (i) - 6 = 9, (ii) (Y) - 10 = 35

15.32. Hl?zt. By Problem 15.7, the regular polygon with n sides has ( n 2 - 3n)/2 diagonals. ( i ) (?zL - 3 n ) / 2 = 2% or n = 7, ( i i ) (nL - 3n)/2 = 3n or ?z = 9

15.33. (i) = 56, (ii) (:) - 8 (t) - 8 = 16


15.35. (i) 4 - 1 0 = 40, iii) 1 3 - 4 8 = 624, (iii) 10.45 - 40 = 10,200. (We subtract the number of straight flushes.) (iv) ( 4 ) /12 ) * 43 = 84,480, (v) 13 (:)(?) 43 = 1,098,240

2 \ 3

15.36. (i) (:)(:) - 5 ! = 1,596,000 (v) 19 * (i) * 3! = 1140 (ix) 4 * (T) * 3! = 4456

(ii) ti)(:) * 5 ! = 228,000

(iii) 19 (i) 5! = 22,800

(vi) 4 * ( y ) * 5! = 91,200

(vii) 4 * (7) 4 ! = 18,240

(x) 4 * 1 9 . 5 ! = 9120

(iv) 19-(:) 4 ! = 4560 fviii) 18,240

15.37. (i) yes, (ii) no, (iii) no, (iv) yes, (v) yes, (vi) yes

15.38. (i) [{1,31, { 2 } , C51, (4, Q], (ii) [C1,2}, {3}, @, C4,5}, (611

15.39. (i) [{TTT}, {TTF, TFT, FTT}, {TFF, FTF, FFT}, {FFF}]

(ii) [{TTT, TFT, FTT}, {TTF, TFF , FTF , FFT, FFF}] (iii) [{TTT, TTF, FTT, FTF}, {TFT, TFF, FFT, FFF}] (iv) [{TTT, FTT}, {TFT}, {TTF, FTF) , {TFF, FFT, FFF}]

{ A , , A , , . . . , A , } , i.e. the refinement partition. 15.41.

15.42. A , n B , c A , and A , n B , c B , .

15.43. - '! = 1680 3! 3! 3!

(:)(;) = 280 15.44. 1680/3! = 280 or

15.45. - lo! 0 - - - 2100 o r ti)(;) = 2100 4!3!3! 2!

12! = 369,600 15.46* 3! 31 3! 3!

12! 5! 4! 3! 15.47. - = 27,720

15.48. 2n-1 - 1

.- - - 3,153,150 14! 15'49' 3! 3! 2! 2! 2! 2! 2 ! 4!

15.50. (i) 33 = 27 (Each element can be placed in any of the three cells.) The number of elements in the three cells can be distributed as follows: (ii)

(4 [{31, (01, { O } ] , ( b ) [{2}, {1>, {Oil, (c) [{I}, c11, (111 Thus the number of partitions is 1 + 3 + 1 = 5.

15.51. (i) 3 4 = 81. (ii) The number of elements in the three cells can be distributed a s follows:

is 1 + 4 + 3 + 6 = 14.

( a ) [{4}, (01, {O}], ( b ) [{3$, {I}, {O}], (4 [@I, C2}, { O ) ] , (4 [@I, {I}, U}l. Thus the number of partitions

48! 48! iiv) (13) 39! 9! 13! 13! 13! (iii) 4! 12! 121 12! 12! 8 5!8!13!13! (ii) 4 52! (i) 13! 13! 13! 13! 15.52.

48! 48! 48! - 48! (v) 2 * 9! 13! 13! 13! + 2 ' 4 * l o ! 12! 13! 13! + 2 * 3 * ll! ll! 13! 13!

- 2300 11! 13! 13! 13!

Chapter 16 Tree Diagrams

TREE DIAGRAMS

of events where each event can occur in a finite number of ways. tree diagrams is illustrated in the following examples.

A tree diagram is a device used to enumerate all the logical possibilities of a sequence The construction of

Example 1: Find the product set A X B X C where A = (1 ,2} , B = {a, b , c} and C = {3 ,4} .

The tree diagram follows: (1, a, 3)

(1, a, 4)

3 (1, b, 3)

' < A (1, b, 4)

'<: (1, c, 4)

'<: (2, a, 4)

'<4 (2, b, 4)

(1, c, 3)

(2, a, 3)

3 (2, b , 3)

(2, c, 3)

Observe that << the tree is constructed from c<: left t o right, and (2, that c, 4) the number of

branches at each point corresponds to the number of ways the next event can occur.

Mark and Eric a re to play a tennis tournament. The first person to win two games in a row or who wins a total of three games wins the tournament. The following diagram gives the various ways the tournament can occur.

Example 2:

Observe tha t there a re 10 endpoints which correspond to the 10 ways that the tournament can occur:

MM, MEMM, MEMEM, MEMEE, MEE, EhIM, EMEMM, EMEME, EMEE, E E

The path from the beginning of the tree to the endpoint describes who won which game in the individual tournament.

176

CHAP. 161 TREE DIAGRAMS 177

Example 3: A man has time to play roulette at most five times. A t each play he wins or loses a dollar. The man begins with one dollar and will stop playing before the five times if he loses all his money or if he wins three dollars, i.e. if he has four dollars. The tree diagram describes the way the betting can occur. Each number in the diagram denotes the number of dollars he has at t h a t point.

The betting can occur in 11 different ways. Note t h a t he will stop betting before the five times a r e up in only three of the cases.

3-

Solved Problems 16.1. Construct the tree diagram for the number of permutations of {a, b, c } .

abc

acb c bac

a bca

The six permutations a r e listed on the right of the diagram.

16.2. Find the product set { 1,2,3} x (2 ,4} x {2,3,4} by constructing the appropriate tree diagram.

(1, 2, 2) (1, 2, 3) (1, 2, 4)

2 6 2 3 1 4

, , / ' 5 4 < : (1, (1, (1, 4, 4, 4, 2) 4) 3)

(2, 2, 2)

(2, 2, 4) 245$: (2, 2, 3)

4

(3, 2, 2) \ /2e! (3,2, (3, 2, 4) 3)

-3\4+23 (3, (3, 4, 4, 2) 3) 4 (3, 4, 4)

The eighteen elements of the product set a re listed to the r ight of the tree diagram.

178 TREE DIAGRAMS [CHAP. 16

16.3. Teams A and B play in a basketball tournament. The team that first wins 3 games wins the tournament. Find the number of possible ways in which the tournament can occur.

Construct the appropriate tree diagram:

The tournament can occur in twenty ways: AAA, AABA, AABBA, AABBB, ABAA, ABABA, ABABB, ABBAA, ABBAB, ABBB

BAAA, BAABA, BAABB, BABAA, BABAB, BABB, BBAAA, BBAAB, BBAB, BBB

16.4. A man is a t the origin on the x-axis and takes a one unit step either to the left or to the right. He stops if he reaches 3 or -3, or if he occupies any position, other than the origin, more than once. Find the number of different paths the man can travel.

Construct the appropriate tree diagram. There a re 14 different paths, each path

corresponding to a n end point of the tree diagram.


16.5. Teams A and B play a hockey game consisting of three periods. goals (points) scored in each period satisfies the following conditions:

The number of

(i j The total number of points scored by both teams in any one period is one 01' two.

(iij The teams are never tied at the end of any period. (iii) Neither team is ahead by more than 2 points a t the end of any period.

(iv) Team A is leading a t the end of the first period.

Find the number of ways the scoring can occur if the above conditions are satisfied. In how many ways will team B win?

Note first, by condition (i), tha t the scoring in any one period can occur in only five ways:

B j Bj 1jR.l The number a t the top is the number of points scored by team A, and the number at the bottom by team B.


Observe that condition (iv) allows only two of the above five scores to occur in the first period. By the tree diagram, the scoring can occur in only 13 ways, and in 4 of these which a re exhibited on the right of the diagram, Team B will win.

T R E E DIAGRAMS [CHAP. 16 180

16.6. Teams A and B play in baseball's world series. (Here, the team that first wins 4 games wins the series.) Find the number of ways the series can occur if (i) A wins the first game and (ii) the team that wins the second game also wins the fourth game.


Observe tha t there is only one choice in the fourth game, the winner of the second game. The series can occur in 15 ways.

16.7. In the following diagram let A , B , . . ., F denote islands, and the lines connecting them bridges. A man begins at A and walks from island to island. He stops for lunch when he cannot continue to walk without crossing the same bridge twice. Find the number of ways that he can take his walk before eating lunch.


C D

D B D F E

A

F

There are eleven ways t o take his walk. Observe tha t he must ea t his lunch at either B, D or E.


A-

1 1-1 I-1-1

16.8. Consider the adjacent diagram with nine points A,B,C,R, S, T, X, Y , 2. A man begins at X and is allowed to move horizontally o r vertically, one step at a time. He stops when he cannot continue to walk without reaching the same point more than once. Find the number of ways he can take his walk, if he first moves from X to R. (By symmetry, the total number of ways is twice this.) X-Y- z


Y z

Y s

/ ‘ s / ‘z Y

\Y z T C

/ \

‘ Y z T C B A

There are 10 different trips. (Note that in only 4 of them are all nine points covered.)

Supplementary Problems 16.9. Construct the tree diagram for the number of permutations of { a , b , c , d } .

16.10.

16.11.

Find the product set {a, b , c } X {a, c } X { b } X { b , c} by constructing the appropriate tree diagram.

Teams A and B play in a basketball tournament, The first team that wins two games in a row o r a total of four games wins the tournament. Find the number of ways the tournament can occur.

A man has time t o play roulette five times. He wins or loses a dollar a t each play. The man begins with two dollars and will stop playing before the five times if he loses all his money or wins three dollars (i.e. has five dollars). Find the number of ways the playing can occur.

A man is at the origin on the x-axis and takes a unit step either t o the left or to the right. He stops after 5 steps or if he reaches 3 or -2. Construct the tree diagram to describe all possible paths the man can travel.

Teams A and B play in baseball’s world series. (The team that first wins 4 games wins the series.) Find the number of ways the series can occur if the team tha t wins the first game also wins the third game, and the team tha t wins the second game also wins the fourth game.

16-12.

16.13.

16.14.

16.15. Solve Problem 16.7 if the man (i) begins a t E , (ii) begins at F .

16.16. Solve Problem 16.8 if the man moves to 2.

Solve Problem 16.5 if condition (ii) is replaced by the following: (ii) The game does not end in a tie.

(i) begins a t Y and first moves to S, (ii) begins at Y and first

16.17.

182 TREE DIAGRAMS [CHAP. 16


16.13. Hint . The tree is essentially the same as the tree of the preceding problem.

16.15. (i) (ii)

E- C- D D

F- C- D

F \

D C- F-

E

TREE DIAGRAMS

- - 2 2 0 0

1 1 0 0 1

0 1 1

0 0 2 2

- - __ __ - - - - __

- __ - - - 0 0

and __ ~

__ - - - __ ~

- d

183 CHAP. 161

16.16. (i)

(ii)

/ R<y ‘4 B C T z R X

I’ sqB<; T z R X

B A T-=====Z

Chapter 17

Probability

INTRODUCTION If a die is tossed

in the air, then i t is certain that the die will come down, but i t is not certain that, say, a 6 will appear. However, suppose we repeat this experiment of tossing a die; let s be the number of successes, i.e. the number of times a 6 appears, and let n be the number of tosses. Then it has been empirically observed that the ratio f = sln, called the relative f r equency , becomes stable in the long ru;, i.e. approaches a limit. This stability is the basis of probability theory.

In probability theory, we define a mathematical model of the above phenomenon by assigning “probabilities” (or: the limit values of the relative frequencies) to each possible outcome of an experiment. Furthermore, since the relative frequency of each outcome is non-negative and the sum of the relative frequencies of all possible outcomes is unity, we require that our assigned “probabilities” also satisfy these two properties. The reliability of our mathematical model for a given experiment depends upon the closeness of the assigned probabilities to the actual relative frequency. This then gives rise to problems of testing and reliability which form the subject matter of statistics.

Historically, probability theory began with the study of games of chance, such as roulette and cards. The probability 11 of an event A was defined as follows: if A can occur in s ways out of a total of n equally likely ways, then

Probability is the study of random or nondeterministic experiments.

s p = P(A) = - n

For example, in tossing a die an even number can occur in 3 ways out of 6 “equally likely” ways; hence p = f = 4. This classical definition of probability is essentially circular since the idea of “equally likely” is the same as that of “with equal probability” which has not been defined. The modern treatment of probability theory is purely axiomatic. In particular, we require that the probabilities of our events satisfy the three axioms listed in Theorem 17.1. However, those spaces consisting of n equally likely elementary events, called equiprobable spaces, are an important class of finite probability spaces and shall provide us with many examples.

SAMPLE SPACE AND EVENTS The set S of all possible outcomes of some given experiment is called the sample space.

A particular outcome, i.e. an element in S, is called a sample point or sample. An event A is a set of outcomes or, in other words, a subset of the sample space S. In particular, the set { a } consisting of a single sample a E S is an event and called an eZewzentury event . Furthermore, the empty set $2 and S itself are subsets of S and so are events; $3 is sometimes called the impossible event .

184

CHAP. 171 PROBABILITY 185

Since an event is a set, we can combine events to form new events using the various

(i) A U B is the event that occurs iff A occurs o r B occurs (or both);

(ii) A n B is the event that occurs iff A occurs and B occurs; (iii) Ac, the complement of A, also written A, is the event that occurs iff A does not

occur. Two events A and B are called mutually exclusive if they are disjoint, i.e. A n B = 8. In other words. A and B are mutually exclusive iff they cannot occur simultaneously.

set operations:

Example 1.1:

Example 1.2:

Example 1.3:

Example 1.4:

Experiment: Toss a die and observe the number tha t appears on top. sample space consists of the six possible numbers:

S = (1, 2, 3, 4, 5, 6)

Let A be the event t h a t an even number occurs, B tha t an odd number occurs and C that a prime number occurs:

Then the

A = { 2 , 4 , 6 } , B = { 1 , 3 , 5 } , C = { 2 , 3 , 5 } Then:

A u C = { Z , 3 , 4 , 5, 6}

B n C = (3, 5) is the event t h a t a n odd prime number occurs;

Cc = (1, 4, S} is the event tha t a prime number does not occur.

is the event tha t an even or a prime number occurs;

Note tha t A and B are mutually exclusive: A n B = 0; in other words, an even number and a n odd number cannot occur simultaneously.

Experiment Toss a coin 3 times and observe the sequence of heads (H) and tails (T) tha t appears. The sample space S consists of eight elements:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Let A be the event tha t two or more heads appear consecutively, and B tha t all the tosses are the same:

A = { H H H , H H T , T H H } and B = {HHH,TTT}

Then A n B = (HHH} is the elementary event in which only heads appear. The event t h a t 5 heads appear is the empty set 8.

Experiment: the coin was tossed. Here infinite number of times, infinite.

Toss a coin until a head appears and then count the number of times The sample space of this experiment is S = (1 ,2 ,3 , . . ., -}.

refers to the case when a head never appears and so the coin is tossed an This is a n example of a sample space which is co~72tabZy

Experiment: Let a pencil drop, head first, into a rectangular box and note the point on the bottom of the box that the pencil first touches. Here S consists of all the points on the bottom of the box. Let the rectangular area on the r ight represent these points. Let A and B be the events t h a t the penciI drops into the corresponding areas illustrated on the right. This is a n example of a sample space which is not finite nor even countably infinite, i.e. which is non-denumerable. For theoretical reasons every subset of this space cannot be considered to be an event.

S I The sample spaces in the preceding two examples, as noted, are not finite. The theory

concerning such sample spaces, especially the non-denumerable case, lies beyond the scope of this text. Thus, unless otherwise stated, all our sample spaces S shall be finite.

186 PROBABILITY [CHAP. 17

FINITE PROBABILITY SPACES Let S be a finite sample space: A finite probability space is

obtained by assigning to each point a7 E S a real number TI,, called the probabili ty of at, satisfying the following properties:

S = {u,,a,, . . ., a , l ] .

(i) each p l is non-negative, p i 1 0 (ii) the sum of the p l is one, p , + p 2 + * * . + p n = 1.

The probabili ty of any event A , written P ( A ) , is then defined to be the sum of the probabilities of the points in A . For notational convenience we write P(u,) for P({a , } ) .

Example 2.1: Let three coins be tossed and the number of heads observed; then the sample space is S = { 0 , 1 , 2 , 3 ) . We obtain a probability space by the following assignment

P(0) = 4, P ( 1 ) = #, P(2) = Q and P ( 3 ) = 8 since each probability is non-negative and the sum of the probabilities is 1. Let A be the event that a t least one head appears and let B be the event tha t all heads or all tails appear’

A = (1, 2, 3) 2nd B = (0, 3) Then, by definition,

P ( A ) = P ( 1 ) + P ( 2 ) P ( 3 ) = 3 + 3 + 1 = 2 8 8 8 8

and P ( B ) = P(0) + P(3) = & + 4 = * Example 2.2: Three horses A , B and C a r e in a race; A is twice as likely to win as B and 13 is

twice as likely to win as C. What a r e their respective probabilities of winning, i.e. P ( A ) , P(B) and P(C)?

Let P ( C ) = p ; since B is twice as likely t o win as C, P(B) = 2 p ; and since A is twice a s likely to win as B, P(A) = 2 P ( B ) = 2 ( 2 p ) = 4p. Now the sum of the probabilities must be 1: hence

Accordingly, p + Z p f 4 p = l or 7 p = 1 or p = +

P ( A ) = 4 p = 4, P(B) = 2 p = +, P(C) = p = 3 Question: What is the probability tha t B or C wins, i.e. P({B, C})? By definition

P({B, C}) = P(B) + P ( C ) = + + 3 = +

EQUIPROBABLE SPACES Frequently, the physical characteristics of an experiment suggest that the various

outcomes of the sample space be assigned equal probabilities. Such a finite probability space S, where each sample point has the same probability, will be called an equiprobable space. In particular, if S contains n points then the probability of each point is lln.

1 T Furthermore, if an event A contains T points then its probability is r * - = -. In other words, n n number of elements in A number of elements in S =

or number of ways that the event A can occur number of ways that the sample space S can occur P(A) =

We emphasize that the above formula for P ( A ) can only be used with respect to an equiprobable space, and cannot be used in general.

The expression “at random” will be used only with respect to an equiprobable space; formally, the statement “choose a point a t random from a set S” shall mean that S is an equiprobable space, i.e. that each sample point in S has the same probability,


Example 3.1: Let a card be selected a t random from a n ordinary deck of 52 cards. Let A = {the card is a spade}

and B = {the card is a face card, i.e. a jack, queen o r king}

We compute P ( A ) , P ( B ) and P ( A n B ) . Since we have a n equiprobable space,

number of face cards - 12 - 3 P(B) = = number of cards 52 4 number of cards 52 13 - _ - number of spades - 13 = 1

-

number of spade face cards - 5 number of cards 52

- P ( A n B ) =

Example 3.2: Let 2 items be chosen a t random from a lot containing 12 items of which 4 are defective. Let

A = {both items are defective} and B = {both items a r e non-defective)

Find P ( A ) and P(B) . Now

= 66 ways, the number of ways that 2 items can be chosen from 12 items;

= 6 ways, the number of ways that 2 defective items can be chosen from 4 defective items;

= 28 ways, the number of ways that 2 non-defective items can be chosen from 8 non-defective items.

5' can occur in

A can occur in (;)

B can occur in (28)

Accordingly, P ( A ) = = and P(B) = = E. 33

THEOREMS ON FINITE PROBABILITY SPACES

is the sum of the probabilities of its points.

Theorem 17.1:

The following theorem follows directly from the fact that the probability of an event

The probability function P defined on the class of all events in a finite probability space satisfies the following axioms: [P,] For every event A , 0 6 P ( A ) 1.

[P,] [Pal P(S) = 1.

If events A and B are mutually exclusive, then P ( A U B ) = P(A) + P(B)

Using mathematical induction, [PSI can be generalized as follows:

Corollary 17.2: If A, , A,, . . . , AT are pairwise mutually exclusive events, then P(A,UA,U * . U A r ) = P(A,) + P(A,) + . * . + P(AJ

Using only the properties of Theorem 17.1, we can prove (see Problems 4.19-4.22).

Theorem 17.3: If @ is the empty set, and A and B are arbitrary events, then:

(ii) P(Ac) = 1 - P ( A ) ; (iii) P ( A \B) = P ( A ) - P ( A n B) , (iv) A c B implies P ( A ) L P ( B ) .

(i) P(P) = 0;

i.e. P ( A nBc) = P(A) - P ( A n B);

Property (ii) of the preceding theorem is very useful in many applications.

Example 4.1: Refer to Example 3.2, where 2 items a r e chosen at random from 12 items of which 4 are defective. Find the probability p t h a t at least one item is defective.

Now C = {at least one item is defective}

is the complement of B = {both items a r e non-defective}

i.e. C = Be; and we found tha t P(B) = g. Hence 14 19 p = P(C) = P(BC) = 1 - P ( B ) = 1 - 33 = 33

The odds that an event with probability p occurs is de$e$ to be the ratio p : (1 - p ) . Thus the odds that a t least one item is defective is 5 : 33 or 19 : 14 which is read “19 to 14”.

Observe that axiom [P,] gives the probability of a union of events in the case that the events are mutually exclusive, i.e. disjoint.

Theorem 17.4:

Corollary 17.5:

The general formula follows:

For any events A and B, P(AUB) = P ( A ) + P(B) - P ( A n B ) .

For any events A , B and C, P ( A U B U C ) = P ( A ) + P(B) + P(C) - P ( A n B )

- P ( A n C ) - P(BnC) + P ( A n B n C )

Example 4.2: Of 100 students, 30 a re taking mathematics, 20 are taking music and 10 a re taking both mathematics and music. If a student is selected at random, find the probability p tha t he is taking mathematics or music.

Let A = {students taking mathematics} and B = {students taking music}. Then A nB = {students taking mathematics and music} and A U B = {students taking mathematics o r music}.

Since the space is equiprobable, then P ( A ) = - lo, P ( B ) = - s, P ( A n B ) = = k. Thus by Theorem 17.4,

30 - 3 2 0 - 1

p = P ( A u B ) = P ( A ) + P ( B ) - P ( A n B ) = $ + i - f = 5

CLASSICAL BIRTHDAY PROBLEM The classical birthday problem concerns the probability p that n people have distinct

birthdays. In solving this problem, we assume that a person’s birthday can fall on any day with the same probability.

Since there are n people and 365 different days, there are 365” ways in which the n people can have their birthdays. On the other hand, if the n persons are to have distinct birthdays, then the first person can be born on any of the 365 days, the second person can be born on the remaining 364 days, the third person can be born on the remaining 363 days, etc. Thus there are 365 * 364.363 . . (365 - n + 1) ways the n persons can have distinct birthdays. Accordingly,

365-364.363 * e (365 - n + 1) 365 364 363 365 - n + 1 -.-.- . . . - 365“ - 365 365 365 365 P =

It can be shown that for n s 2 3 , p < $; in other words, amongst 23 or more people i t is more likely that a t least two of them have the same birthday than that they all have distinct birthdays.

Solved Problems SAMPLE SPACES AND EVENTS 17.1. Let A and B be events. Find an expression and exhibit the Venn diagram for the

event that: (i) A but not B occurs, i.e. only A occurs; (ii) either A or B, but not both occurs, i.e. exactly one of the two events occurs. (i) Since A but not B occurs, shade the area of A outside of B as in Figure (a) below. Note

tha t Bc, the complement of B , occurs since B does not occur; hence A and Bc occurs. In other words, the event is AnBc.


A but not B occurs. (4

Either A or B, but not both, occurs. ( b )

(ii) Since A or B but not both occurs, shade the area of A and B except where they intersect a s in Figure ( b ) above. The event is equivalent to A but not B occurs or B but not A occurs. Now, as in (i), A but not B is the event A n B c , and B but not A is the event BnAc. Thus the given event is ( A n B c ) u ( B n A c ) .

17.2. Let A, B and C be events. Find an expression and exhibit the Venn diagram for the event that (i) A and B but not C occurs, (ii) only A occurs. (i) Since A and B but not C occurs, shade the intersection of A and B which lies outside of C,

as in Figure ( a ) below. The event is A nBn Cc.

17.3.

I 1

A and B but not C occurs. (4

Only A occurs. ( b )

(ii) Since only A is to occur, shade the area of A which lies outside of B and of C, as in Figure ( b ) above. The event is A nBcn Cc.

Let a coin and a die be tossed; let the sample space S consist of the twelve elements: S = {Hl, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

(i) Express explicitly the following events: A = {heads and an even number appears}, B = {a prime number appears}, C = {tails and an odd number appears}.

(ii) Express explicitly the event that: ( b ) B and C occurs, ( c ) only B occurs.

(iii) Which of the events A, B and C are mutually exclusive?

(i)

(a) A or B occurs,

To obtain A , choose those elements of S consisting of an H and an even number: (H2, H4, H6).

To obtain B, choose those points in S consisting of a prime number: B = (H2, H3, H5, T2, T3, T5). To obtain C, choose those points in S consisting of a T and an odd number: C = {Tl, T3, T5).

A =

(ii) (a) A or B = A u B = (H2, H4, H6, H3, H5, T2, T3, T5) ( b ) B and C = BnC = {T3,T5} (c )

(iii) A and C are mutually exclusive since A n C = @.

Choose those elements of B which do not lie in A or C: BnAcnCc = {H3,H5,T2}.


FINITE PROBABILITY SPACES 17.4. Suppose a sample space S consists of 4 elements:

tion defines a probability space on S?

(i) P(al) = +, P(a,) = $, P(a,) = 4, P(a,) = &. (ii) P(a,) = 4, P(a,) = $, P(aJ = -&, P(a,) = +.

S = {u,, a,, a3, ad}. Which func-

(iii) P(a,) = 3, P(a,) = d, P(aJ = 6 , P(a,) = 6. (iv) P(al) = 4, P(a,) = t , P(a , ) = $, P(aJ = 0.

(i) Since the sum of the values on the sample points is greater than one, + + $ + i + Q = s, the function does not define a probability space on S.

(ii) Since P(a,) = -4, a negative number, the function does not define a probability space on S.

(iii) Since each value is non-negative, and the sum of the values is one, & + & + Q + Q = 1, the function does define a probability space on S.

(iv) The values a re non-negative and add up to one; hence the function does define a probability space on S.

17.5. Let S = {ul, a,, a,, a4}, and let P be a probability function on S.

(i) Find P(a,) if P(a,) = +, P(a,) = &, P(a,) = +. (ii) Find P(ul ) and P(a,) if P(a,) = P(a,) = & and P(a l ) = 2P(a,).

(iii) Find P(a,) if P({a,,a,}) = b, P((a,,a,}) = + and P(a,) = Q.

(i) Let Pia,) = p . Then, for P t o be a probability function, the sum of the probabilities on the sample points must be one:

(ii) Let P(a,) = p , then P ( a l ) = 2 p . Hence 2 p f p - k & + i = 1 o r p = 9. Thus P(a,) = & and P ( a l ) = +.

?I + + + & f 8 = 1 or p = 18

(iii) Let P ( a l ) = p . P(a,) = P({az,u,}) - P(a2) = Q - + = Q

P ( a J = P({a,,u4}) - P(a,) = 4 - Q = Q Then p + 4 + + + Q = 1 or p = +, that is, P(aJ = 4.

17.6. A coin is weighted so that heads is twice as likely to appear as tails. and P(H).

Find P(T)

Let P I T ) = p ; then P(H) = 2 p Now set the sum of the probabilities equal to one: p + 2 p = 1 or p = $. Thus P(T) = p = Q and P (H) = 2 p = 3.

17.7. Two men, 7n1 and m,, and three women, w,, w, and w,, are in a chess tournament. Those of the same sex have equal probabilities of winning, but each man is twice as likely to win as any woman. (i) Find the probability that a woman wins the tournament. (ii) If m, and 20, are married, find the probability that one of them wins the tournament.

probabilities of the five sample points equal to one: p + p + p + 21, + 2 p = 1 or p = +, Set P(zul) = p ; then P(w2) = P(w3) = p and P(ml ) =P(m,) = 2 p . Next set the sum of the

We seek ii) P({wl, w,, wn}) and (ii) P({ml,wl}). Then, by definition,

P({Wl, w,,w,}) = P(Wl) + P(W,) + P(WQ) = 3 + + + 3 = 4 P({rnl, w,}) = P ( m J + P(W1) = 9 + + = $


17.8. Let a die be weighted so that the probability of a number appearing when the die is tossed is proportional to the given number (e.g. 6 has twice the probability of appearing as 3) . Let A = {even number), B = {prime number), C = {odd number).

(i) Describe the probability space, i.e. find the probability of each sample point.

(ii) Find P ( A ) , P(B) and P(C) . (iii) Find the probability that:

prime number occurs; (c) A but not B occurs. ( a ) an even or prime number occurs; ( b ) an odd

(i) Let P(1) = p . Since the sum of the probabilities must be one, we obtain p + 2 p + 3 p + 4 p + 5 p + 613 = 1 or p = 1 / 2 1 . Thus

Then P ( 2 ) = 2 p , P (3 ) = 3 p , P ( 4 ) = 4 p , P ( 5 ) = 5 p and P(6) = 6 p .

1 2 P(1) = 21, P ( 2 ) = E, P ( 3 ) $, P(4) = i, P ( 5 ) = &, P(6) = $

(ii) P ( A ) = P ( ( 2 , 1, 6 ) ) = +, P ( B ) = P ( ( 2 , 3, 5 ) ) = g, P ( C ) = P ( { 1 , 3 , 5 } ) = +. (iii) ( a ) The event that an even o r prime number occurs is A LIB = { 2 , 4 , 6 , 3 , 5 } , o r that 1 does

not occur. Thus P ( A u B ) = 1 - P(1) = g. ( b ) The event tha t a n odd prime number occurs is BnC = ( 3 , 5 } . Thus P ( B n C ) =

8 P ( { 3 , 5 ) ) = 21.

(c) The event that A but not B occurs is A n B c = { 4 , 6 ] . Hence P ( A n B c ) = P ( { 4 , 6 ) ) = 8.

17.9. Find the probability p of an event if the odds that it will occur are a : b, that is, ‘‘a to b”.

The odds that an event with probability p occurs is the ratio p : (1 - p ) . Hence a a - - P - l - p - b or b p = a - a p or a p + b p = a o r p = - a f b

17.10. Find the probability p of an event if the odds that it will occur are “3 to 2”.

2, = f from which p = i. We can also use the formula of the preceding problem t o obtain 1 - P 3 3 the answer directly: p = 5 = = 5.

EQUIPROBABLE SPACES 17.11. Determine the probability p of each event:

(i) an even number appears in the toss of a fa i r die;

(ii) a king appears in drawing a single card from an ordinary deck of 52 cards;

(iii) a t least one tail appears in the toss of three fair coins;

(iv) a white marble appears in drawing a single marble from an urn containing 4 white, 3 red and 5 blue marbles.

The event can occur in three ways (a 2 , 4 o r 6) out of 6 equally likely cases; hence p =

There a re 4 kings among the 52 cards; hence p =

(i)

(ii)

(iii) If we consider the coins distinguished, then there a r e 8 equally likely cases:

= i. = 3.

H H H , H H T , Only the first case is not favorable to the given event; HTH,HTT,THH,THT,TTH,TTT.

hence p = $.

(iv) There are 4 - 3 + 5 = 12 marbles, of which 4 a r e white; hence p = 5 = f .

PROBABILITY [CHAP. 17 192

17.12. Two cards are drawn a t random from an ordinary deck of 52 cards. Find the probability p that (i) both are spades, (ii) one is a spade and one is a heart.

There a re it2) = 1326 ways to draw 2 cards from 52 cards.

There are (i3) = 78 ways to draw 2 spades from 13 spades; hence (i)

number of ways 2 spades can be drawn - __ 78 - - 3 P = number of ways 2 cards can be drawn - 1326 - 51

(ii) Since there a re 13 spades and 13 hearts, there a re 13 * 13 = 169 ways to draw a spade and a heart; hence p = 1326 = =. 102

17.13.Three light bulbs are chosen at random from 15 bulbs of which 5 are defective. Find the probability p that (i) none is defective, (ii) exactly one is defective, (iii) a t least one is defective.

There a re (':) = 455 ways to choose 3 bulbs from the 15 bulbs.

Since there a re 15 - 5 = 10 non-defective bulbs, there a re (f,") = 120 ways to choose 3 non-

defective bulbs. (i)

Thus p = L?!! 455 = %. 91

(ii) There are 5 defective bulbs and (120) = 45 different pairs of non-defective bulbs; hence there

a re 5 * 45 = 225 ways to choose 3 bulbs of which one is defective. 22s 45 Thus p = 45j = 91.

(iii) The event tha t at least one is defective is the complement of the event t h a t none are defective

which has, by (i), probability 91. 24 24 - 67 Hence p = 1-91 - 91.

17.14. Two cards are selected a t random from 10 cards numbered 1 to 10. Find the probability p that the sum is odd if (i) the two cards are drawn together, (ii) the two cards are drawn one after the other without replacement, (iii) the two cards are drawn one after the other with replacement.

(i) There a re (io) = 45 ways t o select 2 cards out of 10. The sum is odd if one number is odd

and the other is even. There are 5 even numbers and 5 odd numbers; hence there a re

5 . 5 5 26 ways of choosing an even and a n odd number. 25 5 Thus p = 45 = 9.

(ii) There are 1 0 . 9 = 90 ways t o draw two cards one af ter the other without replacement. There a re 5 5 = 25 ways t o draw an even number and then a n odd number, and 5 5 = 25 ways to draw an odd number and then an even number; hence p = 7 - - - 5 - g.

(iii) There are 1 0 . 1 0 = 100 ways t o draw two cards one af ter the other with replacement. As in (ii), there a re 5 . 5 = 25 ways t o draw a n even number and then a n odd number, and

5 - 5 = 25 ways to draw an odd number and then an even number; hence p = 100 - - 2. 2 5 - 2 5 - 50 - 1

17.15. An ordinary deck of 52 cards is dealt out a t random to 4 persons (called North, South, East and West) with each person receiving 13 cards. Find the probability p of each event (answers may be left in combinatorial and factorial notation):


(i) (ii) North receives no aces.

(iii) North receives 5 spades, 5 clubs, 2 hearts and 1 diamond.

(iv) North and South receive between them all the spades.

(v) North and South receive between them all the aces and kings.

(vi) Each person receives an ace.

North receives all 4 aces.

(i) North can be dealt 13 cards in (!:) ways. If he receives the 4 aces, then he can be dealt the other 9 cards from the remaining 48 cards in (",") ways. Thus

(ii) North can be dealt 13 cards in (E) ways. If he receives no aces, then he can be deaIt 13 cards

froni the remaining 48 cards in ways. Thus

(iii) North can be dealt 13 cards in (::) ways. He can be dealt 5 spades in ( i3) ways, 5 clubs in 13 13 13 13 ( 5 ) ( 5 ) ( 2 1 )

(3 (F) ways, 2 hearts in (i3) ways, and 1 diamond in (If) = 13 ways. Hence p =

(iv) North and South can be dealt 26 cards between them in (ii) ways. If they receive the 13 spades, then they can be dealt 13 other cards from the remaining 39 cards in (;:) ways.

Hence p = (;!) / (!:).

(v) North and South can be dealt 26 cards between them in (ii) ways. If they receive the 4 aces and the 4 kings, then they can be dealt the other 2 6 - 8 = 18 cards from the remaining 52 - 8 = 44 cards in (18) ways. 44 Thus p = (;:) / (;",.

(vi) The 52 cards can be distributed among the 4 persons in 13! l& ways (an ordered partition). The remaining 48 cards can be distributed Now each person can get an ace in 4 ! ways.

among the 4 persons in 12! 121 12! 12! ways. 481 Thus

48! 133

17 * 25 * 49 - -

4 ! 134 52 51 * 50 * 49

- - 4 ! 1 2 ! 12: l e ! IZ! 521 P =

13! 131 13! 131

17.16. Six married couples are standing in a room.

(i) If 2 people are chosen a t random, find the probability p that (a) they are married, ( b ) one is male and one is female.

(ii) If 4 people are chosen a t random, find the probability p that (a) 2 married couples are chosen, ( b ) no married couple is among the 4, (c) exactly one married couple is among the 4.

(iii) If the 12 people are divided into six pairs, find the probability p that (a) each pair is married, ( b ) each pair contains a male and a female.


(i) There are (7) = 66 ways t o choose 2 people from the 12 people.

( ( 1 ) There are 6 married couples: hence p = 66 = fi.

( b ) There are 6 ways to choose a male and 6 ways to choose a female; hence p = 66 = 11.

6 1

6 6 6

(ii) There are (y) = 495 ways t o choose 4 people from the 12 people.

( a ) There are (:) = 15 ways to choose 2 couples from the 6 couples; hence p 1 a = ,%.

( b ) The 4 persons come from 4 different couples. There are (!) = 15 ways to choose 4 couples from the 6 couples, and there are 2 ways to choose one person from each couple. Hence

15 1

p = - 2 .2 .2 .2 .15 - - 16 495 - 33.

( c ) This event is mutually disjoint from the preceding two events (which are also mutually disjoint) and a t least one of these events must occur. 16 Hence p f $ f = 1 or p = 33.

12! - 12! (iii) There a re 2!2 !2 !2 !2 !2 ! - 5; ways to partition the 12 people into 6 ordered cells with 2 people in each.

((0 The 6 couples can be placed into the 6 ordered cells in 6! ways. Hence p = & = &. ( b ) The six men can be placed one each into the 6 cells in 6 ! ways, and the 6 women can be

placed one each into the 6 cells in 6! ways. 6: 6 ! 16 Hence p = 12!/26 = 231.

17.17. If 5 balls are placed in 5 cells a t random, find the probability p that exactly one cell is empty.

There a re 53 ways to place the 5 balls in the 5 cells. If exactly one cell is empty, then another cell contains 2 balls and each of the remaining 3 cells contains one ball. There a r e 5 ways to select the empty cell, then 4 ways t o select the cell containing 2 balls, and (:) = 10 ways t o select 2 balls to go into this cell. Finally there are 3! ways to distribute the remaining 3 balls among the remaining 3 cells. Thus 125'

p = j.q.103! - - 48 5 5

17.18.A class contains 10 men and 20 women of which half the men and half the women have brown eyes. Find the probability p that a person chosen a t random is a man or has brown eyes.

Let A = {person is a man} and B = {person has brown eyes}. We seek P ( A u B ) .

Then P ( A ) = lo = $, P(Bj = = 1 P ( A n B ) = = 5. Thus, by Theorem 17.4, 30 30 2' 30

p = P ( A u B ) = P ( A ) + P(B) - P ( A n B ) = g + + - Q = 3

PROOFS OF THEOREMS 17.19. Prove Theorem 17.3(i): If $3 is the empty set, then P(@) = 0.

Let A be any event; then A and (3 a r e disjoint and A U $3 = A .

Thus P ( A ) = P ( A U @ ) = P ( A ) + P(@), from which our result follows.

17.20. Prove Theorem 17.3(ii): For any event A, P(AC) = 1 - P ( A ) .

The sample space S can be decomposed into the mutually exclusive events A and Ac: S = A UAC.

Thus 1 = P(S) = P(A WAC) = P ( A ) P(Ac), from which our result follows.

PROBABILITY 195 CHAP. 171

17.21. Prove Theorem 1'7.3(iii):

P(A\B) = P ( A ) - P ( A n B )

A can be decomposed into the mutually exclusive events A \ B and A n B : A = ( A \B) u ( A n B ) . Thus

P ( A ) = P ( A \ B ) + P ( A nB) from which our result follows. A is shaded.

17.22. Prove Theorem 17.3(iv):

If A c B , then P(A) 6 P(B).

If A c B , then B can be decomposed into the mutually exclusive events A and B\A. Thus

The result now follows from the fact that P ( B \A) 2 0.

P ( B ) = P ( A ) + P(B\A)

B is shaded.

17.23. Prove Theorem 17.4: For any events A and B, P ( A U B ) = P ( A ) + P(B) - P ( A n B )

Note tha t A u B can be decomposed into the mutually exclusive events A \B and B : A UB = ( A \B)UB. Thus, using Theorem 1'7.3(iii), we have

P ( A u B ) = P ( A \B) + P ( B ) = P ( A ) - P ( A n B ) f P ( B ) = P ( A ) + P ( B ) - P ( A n B )

~

A U B is shaded.

17.24.Prove Corollary 17.5: For any events A , B and C, P ( A u B u C ) = P ( A ) + P ( B ) + P ( C ) - P ( A n B ) - P ( A n C ) - P ( B n C ) + P ( A n B n C )

Let D = B u C . Then A n D = A n ( B u C ) = ( A n B ) U ( A n C ) and P ( A n D ) = P ( A n B ) + P ( A n C ) - P ( A n B n A n C ) = P ( A n B ) + P ( A n C ) - P ( A n B n C )

Thus P ( A u B u C ) = P ( A U D ) = P ( A ) + P ( D ) - P ( A n D )

= P ( A ) + P(B) + P ( C ) - P ( B n C ) - [ P ( A n B ) + P ( A n C ) - P ( A n B n C ) ] = P ( A ) + P(B) + P ( C ) - P ( B n C ) - P ( A n B ) - P ( A n C ) + P ( A n B n C )

MISCELLANEOUS PROBLEMS 17.25. Let A and B be events with P ( A ) = Q, P ( B ) = & and P ( A n B ) = &. Find (i) P ( A U B ) ,

(ii) &'(Ac) and P ( B c ) , (iii) P(AcnBc) , (iv) P ( A c U B c ) , (v) P ( A n B c ) , (vi) P ( B n A c ) .

(i)

(ii)

(iii) Using De Morgan's Law, (AuB)c = ACnBc, we have

P ( A u B ) = P ( A ) + P ( B ) - P ( A n B ) = Q + & - $ = Q P(Ac) = 1 - P ( A ) = 1 - # = Q and P ( B ) = 1 - P ( B ) = 1 - Q = 1 2

P(AcnBc) = P ( ( A u B ) c ) = 1 - P ( A u B ) = 1 - 8 = 8 (iv) Using De Morgan's Law, ( A nB)c = ACUBC, we have

P(AcuBc) = P ( ( A n B ) c ) = 1 - P ( A n B ) = 1 - & = p

196

Number 1 2 3 4 5 6

Frequency 14 17 20 18 15 16

PROBABILITY [CHAP. 1 7

Equivalently, P(AcUt7c) = P(Ac) + P(Bc) - P(AcnBC) 6 8 2 8 + 1 - = 8 4

(v, P(AnBc) = P(A\B) = P ( A ) - P ( A n B ) = 8 - $ = &

(vi) P(BnAc) = P ( B ) - P ( A n B ) = + - & = 4

17.26. Let A and B be events with P ( A UB) = $, P(Ac) = + and P ( A n B ) = Q. Find (i) P ( A ) , (ii) P(B), (iii) P ( A n B c ) .

(i) P ( A ) = 1 - P ( A c ) = 1 - t = Q

(iii Substitute in P ( A u B ) = P ( A ) + P ( B ) - P ( A n B ) to obtain 4 = + + P ( B ) - & or P ( B ) = 3. (iii, P(AnBc) = P ( A ) - P ( A n B ) = & - & =

number of successes total number of trials ' The relative frequency f =

20 17+18+16 - 17+20+15 - (i) f = = .20 (ii) f = $ = .15 (iii) f = 100 - .51 (iv) f = 100 - .52

17.28. Let S = {ul, a,, . . ., as) be finite probability spaces. Let the number p L j = P(a,) P(b , ) be assigned to the ordered pair (ai, bj) in the product set S x T = ((8, t ) : s E S, t E T } . Show that the p l i define a probability space on S X T, i.e. that the p , are non-negative and add up to one. (This is called the product probabi l i ty space. We emphasize that this is not the only probability function that can be defined on the product set S x T.)

and T = {b,, b,, . . ., b,}

Since P(a,),P(b,) 1 0, for each i and each j , pij = P(uJ P(b,) 2 0. Furthermore,

Pll + P12 + . - . + Plt + P21 + P 2 2 + * * . + Pzt + * ' ' + Psl + Ps2 + . . * + Pst

= P(a1) P(bJ + * * * + P(a1) P(bt) + * * * + P(aJ P(b1) + * * * + P(as) P(b,)

P(al)[P(bl) + . * = + P(b,)] + * * * + P(as)[P(bl) + * * . +P(bt ) ]

= P ( a l ) . l + * " + P(a,) '1 = P(al) + a ' . + P(as) = 1


Supplementary Problems SAMPLE SPACES AND EVENTS 17.29. Let A and B be events. Find a n expression and exhibit the Venn diagram for the event tha t

(i) A or not B occurs, (ii) neither A nor B occurs.

Let A , B and C be events. Find an expression and exhibit the Venn diagram for the event that (i) exactly one of the three events occurs, (ii) a t least two of the events occurs, (iii) none of the events occurs, (iv) A or B, but not C, occurs.

Let a penny, a dime and a die be tossed. (i) Describe a suitable sample space S. (ii) Express explicitly the following events: A = {two heads and an even number appears},

B = {a 2 appears}, C = {exactly one head and a prime number appears}. (iii) Express explicitly the event tha t ( a ) A and B occur, ( b ) only B occurs, (c ) B or C occur.

17.30.

17.31.

FINITE PROBABILITY SPACES 17.32. Which function defines a probability space on S = {a,,a,,a,}?

(i) (iii) P(al ) = Q, &‘(a2) = +, &‘(a,) = 4 (ii) P(al ) = 8, P(a,) = -Q, P(a,) = Q (iv) P(al) = 0, P(az) = 5, P(u3) = $

P(aJ = $, P(ad = g, P b 3 ) = 3

17.33. Let P be a probability function on S = { a l , a ~ , a 3 } . Find P(al ) if (i) P(u2) = Q and P(u3) = 2, (ii) P(al ) = 2 P ( a 2 ) and P(a3) = $, (iii) P({aZ,a3}) = 2 P ( a 1 ) , (iv) P(a3) = 2P(u,) and P(a2) = 3 P ( a 1 ) .

17.34. Let A and B be events with P ( A u B ) = 4, P ( A n B ) = $ and P ( A c ) = 8. P ( A n Bc).

Find P ( A ) , P(B) and

17.35. Let A and B be events with P ( A ) = 4, P ( A U B ) = 3 and P(B) = 3. Find P(AnB) , P(AcnBc), P(AcuBc) and P(BnAc) .

A coin is weighted so tha t heads is three times as likely to appear a s tails. Find P(H) and P(T).

Three students A , B and C a r e in a swimming race. A and B have the same probability of winning and each is twice as likely to win as C. Find the probability t h a t B or C wins.

A die is weighted so tha t the even numbers have the same chance of appearing, the odd numbers have the same chance of appearing, and each even number is twice as likely to appear as any odd number. Find the probability tha t (i) an even number appears, (ii) a prime number appears, (iii) a n odd number appears, (iv) an odd prime number appears.

Find the probability of an event if the odds tha t it will occur a re (i) 2 to 1, (ii) 5 to 11.

In a swimming race, the odds t h a t A will win a re 2 to 3 and the odds tha t B will win a re 1 to 4. Find the probability p and the odds that A or B win the race.

17.36.

17.37.

17.38.

17.39.

17.40.

EQUIPROBABLE SPACES 17.41. A class contains 5 freshmen, 4 sophomores, 8 juniors and 3 seniors. A student is chosen at random

to represent the class. Find the probability that the student is (i) a sophomore, (ii) a senior, (iii) a junior o r senior.

One card is selected at random from 50 cards numbered 1 to 50. number on the card is (i) divisible by 5 , (ii) prime, (iii) ends in the digit 2.

Of 10 girls in a class, 3 have blue eyes. If two of the girls a re chosen at random, what is the probability tha t ( i ) both have blue eyes, (ii) neither has blue eyes, (iii) at least one has blue eyes?

17.44. Three bolts and three nuts a re put in a box. If two par ts a re chosen at random, find the

17.42. Find the probability tha t the

17.43.

probability tha t one is a bolt and one a nut.


17.45. Ten students, A , B, . . . , a r e in a class. If a committee of 3 is chosen at random from the class, find the probability t h a t ( i ) A belongs to the committee, (ii) B belongs to the committee, ( i i i ' d and B belong to the committee, tivi A or B belongs t o the committee.

A class consists of 6 girls and 10 boys. If a committee of 3 is chosen at random from the class, find the probability tha t (i) 3 boys are selected, (ii) exactly 2 boys are selected, (iii) a t least one boy is selected, (iv) exactly 2 girls a re selected.

17.47. A pair of fa i r dice is tossed. Find the probability tha t the maximum of the two numbers is greater than 1.

17.48. Of 120 students, 60 a re studying French, 50 are studying Spanish, and 20 are studying French and Spanish. If a student is chosen a t random, find the probability tha t the student ( i ) is studying French or Spanish, (ii) is studying neither French nor Spanish.

17.49. Three boys and 3 girls sit in a row. Find the probability tha t (i) the 3 girls sit together, (i i l the boys and girls sit in alternate seats.

17.50. A die is tossed 50 times. The following table gives the six numbers and their frequency of

17.46.

occurrence:

17.29.

17.30.

17.31.

17.32.

17.33.

17.34.

17.35.

17.36.

17.37.

17.38.

17.39.

17.40.

17.41.

17.42.

Find the relative frequency of the event ii) a 4 appears, iii) an odd number appears, (iii) a prime number appears.

Answers to Supplementary Problems (i) A u B ~ , (ii) ( A u B ) ~

( i ) (AnBcnCc) u (BnAcnCc) u (CnAcnBc) (iii) ( A u B u C). (ii) ( A n B ) u ( A n C ) U (BnC) (iv) ( A u B ) n CC

(i) S = (HH1, HH2, HH3, HH4, HH5, HH6, HT1, HT2, HT3, HT4, HT5, HT6, TH1, TH2, TH3, TH4, TH5, TH6, TT1, TT2, TT3, TT4, TT5, TT6)

A = (HH2, HH4, HH6}, B = (HH2, HT2, TH2, TT2}, (ii) (iii) ( a ) A n B = (HH2)

C = {HTZ, TH2, HT3, TH3, HT5, TH5)

( b ) B\(AuC) = (TT2) (c) B u C = (HH2, HT2, TH2, TT2, HT3, TH3, HT5, TH5}

(i) no, (ii) no, (iii) yes, (iv) yes

3 5 - 17.44.

17.45. (i) A, (ii) A, (iii) 3 1 (iv) 15 8

27 15 17.46. (i) k, (ii) $, (iii) 28, (iv) E 3 5 17.47. 3 p = 5; the odds a r e 3 to 2.

(i) I, iii) &, (iii) 3 11 17.48. (i) 4, 3 (ii)

1 17.49. (i) i, (ii) 8 17.43. (i) A, (ii) &, (iii) 17.50. (i) &, (ii) 2, (iii) 8

Chapter 18

Conditional Probability. Independence

CONDITIONAL PROBABILITY Let E be an arbitrary event in a sample space S with P(E) > 0. The probability that

an event A occurs once E has occurred or, in other words, the conditional probability of A given E, written P ( A ' E) , is defined as follows:

P ( A n E ) P ( A ' E ) = P(E)

As seen in the adjoining Venn diagram, P(A I E ) in a certain sense measures the relative probability of A with respect to the reduced space E.

in the event A , then In particular, if S is an equiprobable space and we let \A] denote the number of elements

P ( A n E ) - ( A n E ( - P(E) IEl

and so P(A1E) = A n E l (El IS IS

P ( A n E ) = --, P(E) = -

In other words,

Theorem 18.1: If S is an equiprobable space and A and E are events of S, then

number of elements in A n E number of elements in E

number of ways A and E can occur number of ways E can occur

I E,

or P(A1E) =

Example 1.1: Let a pair of fa i r dice be tossed. the dice is a 2.

If the sum is 6, find the probability tha t one of In other words, if

E = {sum is 61 = {(1,5) , (2 ,4) , (3 ,3) , (4 ,2) , (5,111 and A = {a 2 appears on at least one die}

find P ( A I E).

A n E = {(2,4) , (4,2)}. Now E consists of five elements and two of them, (2,4) and (4,2), belong to A :

Then P ( A I E) = i. On the other hand, since A consists of eleven elements,

A = ((2, l), 12,2), (2, 3), (2,4), (2,5), (2 ,6 ) , (1 ,2 ) , (3 ,2 ) , (4,2), (5 ,% (6,211

and S consists of 36 elements, P ( A ) = $.

Example 1.2: A couple has two children; the sample space is S = { b b , b g , g b , g g } with probability & for each point. Find the probability p tha t both children a r e boys if ( i ) i t i s known tha t one of the children is a boy, ( i i ) it is known tha t the older child is a boy.

( i ) Here the reduced space consists of three elements, { b b , b g , g b } ; hence p = &. rii) Here the reduced space consists of only two elements, { b b , b g } ; hence p = 4.

199

200 CONDITIONAL PROBABILITY. INDEPENDENCE [CHAP. 18

MULTIPLICATION THEOREM FOR CONDITIONAL PROBABILITY If we cross multiply the above equation defining conditional probability and use the

fact that A n E = E n A , we obtain the following useful formula.

Theorem 18.2: P ( E f l A ) = P ( E ) P ( A 1 E ) This theorem can be extended by induction as follows:

Corollary 18.3: P ( A l n A z n . . . n A , ) = P ( A 1 ) P ( A 2 1 A 1 ) P ( A 3 1 A l n A z ) P(A4 I A l n A e n A 3 ) . 9 . P(A, I A1 n A p n . * * nA,-l)

We now apply the above theorem which is called, appropriately, the multiplication

Three items are drawn at ran- Find the probability p tha t all three a re

The probability that the first item is non-defective is 5 since 8 of 12 items are non-defective. If the first item is non-defective, then the probability that the next item is non-defective is fi since only 7 of the remaining 11 items are non-defective. If the first two items are non-defective, then the probability tha t the last item is non-defective is & since only 6 of the remaining 10 items are now non-defective. Thus by the multiplication theorem,

theorem. Example 2.1: A lot contains 12 items of which 4 are defective.

dom from the lot one af ter the other. non-defective.

= & . $ . " = 11 10 55

FINITE STOCHASTIC PROCESSES AND TREE DIAGRAMS A (finite) sequence of experiments in which each experiment has a finite number of

outcomes with given probabilities is called a ( f in i te ) stochastic process . A convenient way of describing such a process and computing the probability of any event is by a t ree diagram as illustrated below; the multiplication theorem of the previous section is used to compute the probability that the result represented by any given path of the tree does occur.

Example 3.1: We a r e given three boxes as follows: Box I has 10 light bulbs of which 4 are defective. Box I1 has 6 light bulbs of which 1 is defective. Box I11 has 8 light bulbs of which 3 a re defective.

We select a box a t random and then draw a bulb at random. What is the probability p tha t the bulb is defective?

Here we perform a sequence of two experiments:

(i) select one of the three boxes; (ii) select a bulb which is either defective (D) or non-defective ( N ) .

The following tree diagram describes this process and gives the probability of each branch of the tree:

CHAP. 181 201 CONDITIONAL PROBABILITY. INDEPENDENCE

The probability that any particular path of the tree occurs is, by the multiplication theorem, the product of the probabilities of each branch of the path, e.g.,

Now since there a re three mutually exclusive paths which lead to a defective

the probability of selecting box I and then a defective bulb is 3.5 1 2 = IS. 2

bulb, the sum of the probabilities of these paths is the required probability: = 1 . 2 + ; . ; + 1 . 3 = - 113

3 5 3 8 360

Example 3.2: A coin, weighted so tha t P ( H ) = $ and P(T) = $, is tossed. I f heads appears, then a number is selected at random from the numbers 1 through 9; if tails appears, then a number is selected at random from the numbers 1 through 5. Find the probability p that an even number is selected.

The tree diagram with respective probabilities is

Note t h a t the probability of selecting an even number from the numbers 1 through 9 is 4 since there a re 4 even numbers out of the 9 numbers, whereas the probability of selecting a n even number from the numbers 1 through 5 is 5 since there are 2 even numbers out of the 5 numbers. Two of the paths lead to an even number: H E and TE. Thus

p z P(E) = $ . + + ? . * = - 55 9 3 3 135

Example 3.3: Consider the stochastic process of the preceding example. selected, what is the probability that tails appeared on the coin? we seek the conditional probability of T given E: P(T 1 E).

probability 3.5 = &; tha t is, P ( T n E ) = 5. ample, P ( E ) = %. Thus by the definition of conditional probability,

If a n even number was In other words,

Now tails and an even number can occur only on the bottom path which has Furthermore, by the preceding ex- 1 2 2

9 P ( T n E ) - 2/15 - - P(E) - 581135 - 29 P ( T / E ) = ___ ~

In other words, we divide the probability of the successful path by the probability of the reduced sample space.

INDEPENDENCE

not influenced by whether A has or has not occurred. of B equals the conditional probability of B given A : P(B) for P(B I A ) in the multiplications theorem P ( A nB) = P ( A ) P(B 1 A) , we obtain

We formally use the above equation as our definition of independence.

An event B is said to be independent of an event A if the probability that B occurs is In other words, if the probability

Now substituting P(B) = P(B1A).

P ( A n B ) = P(A)P(B)

1 Definition:] Events A and B are independent if P ( A n B ) = P(A)P(B); otherwise they are dependent.

Example 4.1: Let a f a i r coin be tossed three times; we obtain the equiprobable space



Consider the events A = {first toss is heads}, B = {second toss is heads)

C = {exactly two heads are tossed in a row}

Clearly A and B are independent events; this fact is verified below. hand, the relationship between A and C o r B and C is not obvious. that A and C are independent, but that B and C are dependent. We have

On the other We claim

P ( A ) = P({HHH, HHT, HTH, HTT}) = $ = f P ( B ) = P({HHH, HHT, THH, THT}) = f =

P ( C ) = P({HHT, THH)) = f = Then

P ( A n B ) = P({HHH, HHT}) = 2, P ( A n C ) = P({HHT}) = 8 , P ( B n C ) = P({HHT, THH}) = 2

Accordingly,

P ( A ) P ( B ) = 3.4 = & = P ( A n B ) , P ( A ) P ( C ) = d * $ = 4 = P ( A n C ) , P ( B ) P(C) = 12 * $ = 4 # P ( B n C ) ,

and so A and B are independent;

and so A and C are independent;

and so B and C are dependent.

Frequently, we will postulate that two events are independent, or the experiment itself will imply that two events are independent.

Example 4.2: The probability that A hits a target is 2 and the probability that B hits it is 2. What is the probability that the target will be hit if A and B each shoot a t the target?

Further- more, the probability that A or B hits the target is not influenced by what the other does; that is, the event that A hits the target is independent of the event that B hits the target: P ( A n B ) = P ( A ) P ( B ) . Thus

We are given that P ( A ) = 4 and P ( B ) = 5, and we seek P ( A u B ) .

P ( A uB) = P ( A ) + P ( B ) - P ( A n B ) = P ( A ) + P ( B ) - P ( A ) P ( B ) - t + $ - ' . ? = - 11

4 5 20 -

Three events A, B and C are i ndependen t if: (i) P ( A n B ) = P ( A ) P ( B ) , P ( A n C ) = P ( A ) P ( C ) and P ( B n C ) = P ( B ) P ( C )

i.e. if the events are pairwise independent, and (ii) P ( A n B n C ) = P ( A ) P ( B ) P ( C )

The next example shows that condition (ii) does not follow from condition (i); in other words, three events may be pairwise independent but not independent themselves.

Example 4.3: Let a pair of fair coins be tossed; here S = {HH, HT, TH, TT} is an equiprobable space. Consider the events

A = {heads on the first coin} = {HH, HT}

B = {heads on the second coin} = {HH, TH}

C = {heads on exactly one coin} = {HT, TH}

Then P ( A ) = P ( B ) = P(C) = $ = f and

P ( A n B ) = P({HH}) = $, P ( B n C ) = ({TH}) = $ Thus condition ii) is satisfied, i.e., the events are pairwise independent. However, A n B n C = $3 and so

P ( A n C ) = P({HT}) = t,

P ( A n B n C ) = P(@) = 0 # P ( A ) P ( B ) P ( C )

In other words, condition (ii) is not satisfied and so the three events are not independent.

CHAP. 181 CONDITIONAL PROBABILITY. INDEPENDENCE 203

Solved Problems CONDITIONAL PROBABILITY IN EQUIPROBABLE SPACES 18.1.

18.2.

18.3.

18.4.

18.5.

A pair of fair dice is thrown. Find the probability p that the sum is 10 or greater if (i) a 5 appears on the first die, (ii) a 5 appears on at least one of the dice.

ii) If a 5 appears on the first die, then the reduced sample space is

A = { [5 ,1 ) , 15,2), (5 ,3) , (5,4), (5 ,5 ) , ( 5 , 6 ) } 2 1 The sum is 10 o r greater on two of the six outcomes: (5 ,5 ) , (5 ,6 ) . Hence p = = 3.

If a 5 appears on at least one of the dice, then the reduced sample space has eleven elements: (ii)

B = ((5, I ) , (5,2), ( 5 , 3 ) , 15,4), 15,5), (5 ,6) , (1 ,5) , (2 ,5 ) , 13,5), (4,5), (6,511 3 The sum is 10 or greater on three of the eleven outcomes: (5 ,5 ) , (5 ,6 ) , (6,5). Hence p = a.

Three fair coins are tossed. Find the probability p that they are all heads if (i) the first coin is heads, (ii) one of the coins is heads.

The sample space has eight elements: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

If the first coin is heads, the reduced sample space is A = {HHH, HHT, HTH, HTT}. Since the coins are all heads in 1 of 4 cases, p = $.

If one of the coins is heads, the reduced sample space is B = {HHH, HHT, HTH, HTT, THH. THT, TTH}.

(i)

(ii) Since the coins are all heads in 1 of 7 cases, p = 3.

A pair of fa i r dice is thrown. If the two numbers appearing are different, find the probability p that (i) the sum is six, (ii) an ace appears, (iii) the sum is 4 or less.

. . ., (6 ,6) .

(i)

Of the 36 ways the pair of dice can be thrown, 6 will contain the same numbers: ( l , l ) , (2,2), Thus the reduced sample space will consist of 36 - 6 = 30 elements.

The sum six can appear in 4 ways: (1, 5) , (2 ,4) , (4,2), ( 5 , l ) . the numbers are the same.) Hence p = 30 = 5.

(We cannot include (3 ,3 ) since 4 2

(ii) An ace can appear in 10 ways: (1 ,2 ) , (1,3), . . ., (1 ,6) and (2, l), (3 ,1) , . . ., ( 6 , l ) . Hence 10 1 p = , = , .

(iii) The sum of 4 or less can occur in 4 ways: (3, I), (1,3), (2, l), (1 ,2 ) . Thus p = 6 $.

Two digits are selected at random from the digits 1 through 9. If the sum is even, find the probability p that both numbers are odd.

The sum is even if both numbers a re even or if both numbers a r e odd. There a re 4 even numbers i 2 , 4 , 6 , 8 ) ; hence there are (;) = 6 ways to choose two even numbers. There a re 5 odd numbers i l , 3 , 5 , 7 , 9 ) : hence there are ( 3 ) = 10 ways to choose two odd numbers. Thus there a r e 6 + 10 = 16 ways to choose two numbers such tha t their sum is even: since 10 of these ways occur

10 - 5 when both numbers are odd, p = 16 - 2.

A man is dealt 4 spade cards from an ordinary deck of 52 cards. If he is given three more cards, find the probability p that at least one of the additional cards is also a spade.

2 04 CONDITIONAL PROBABILITY. INDEPENDENCE [CHAP. 18

Since he is dealt 4 spades, there a re 5 2 - 4 = 48 cards remaining of which 1 3 - 4 = 9 are spades. There a re (",") = 17,296 ways in which he can be dealt three more cards. Since there a r e 48 - 9 = 39 cards which a re not spades, there a r e ( y ) = 9139 ways he can be dealt three cards which are not spades. Thus the probability q tha t he is not dealt another spade is q = -; hence p = 1 - q = 8157 17,296 '

18.6. Four people, called North, South, East and West, are each dealt 13 cards from an ordinary deck of 52 cards.

(i) If South has no aces, find the probability p that his partner North has exactly two aces.

(ii) If North and South together have nine hearts, find the probability p that East and West each has two hearts.

(i) There are 39 cards, including 4 aces, divided among North, Eas t and West. There are (::) ways tha t North can be dealt 13 of the 39 cards. There a re (;) ways he can be dealt 2 of the four aces, and (;:) ways he can be dealt 11 cards from the 39 - 4 = 35 cards which a r e not aces. Thus

(;)(::) - 6 . 1 2 . 1 3 . 2 5 . 2 6 - 650 2109

- - p z - - (3 36 * 37 * 38 * 39

(ii) There a re 26 cards, including 4 hearts, divided among Eas t and West. There a re (;!) ways that , say, Eas t can be dealt 13 cards. (We need only analyze East's 13 cards since West must have the remaining cards.) There a re (i) ways Eas t can be dealt 2 hearts from 4 hearts, and (;;) ways he can be dealt 11 non-hearts from the 26 - 4 = 22 non-hearts. Thus

MULTIPLICATION THEOREM 18.7. A class has 12 boys and 4 girls. If three students are selected at random from the

class, what is the probability p that they are all boys? The probability t h a t the first student selected is a boy is 12/16 since there are 12 boys out of

16 students. If the first student is a boy, then the probability tha t the second is a boy is 11/15 since there are 11 boys left out of 15 students. Finally, if the first two students selected were boys, then the probability tha t the third student is a boy is 10114 since there a r e 10 boys left out of 14 students. Thus, by the multiplication theorem, the probability t h a t all three a re boys is

Another Method. There are (i6) = 560 ways to select 3 students of the 16 students, and (y) = 220 ways to select 3 boys out of 12 boys; hence p = 2 = 6.

A T h i i d Method. If the students are selected one af ter the other, then there a re 16 * 15 * 14 ways to select three students, and 12 * 11 * 10 ways to select three boys; hence p = - = $.

18.8. A man is dealt 5 cards one after the other from an ordinary deck of 52 cards. What is the probability p that they are all spades?

The probability tha t the first card is a spade is 13/52, the second is a spade is 12/51, the third is a spade is 11/50, the fourth is a spade is 10/49, and the last is a spade is 9/48. (We assumed in each case tha t the previous cards were spades.) Thus p = 2 . $ 9 $ 0 % * & = -.


18.9. An urn contains 7 red marbles and 3 white marbles. Three marbles are drawn from the urn one after the other. Find the probability p that the first two are red and the third is white.

The probability that the first marble is red is 7/10 since there a re 7 red marbles out of 10 marbles. If the first marble is red, then the probability t h a t the second marble is red is 6/9 since there a re 6 red marbles remaining out of the 9 marbles. If the first two marbles a re red, then the probability tha t the third marble is white is 3/8 since there a re 3 white marbles out of the 8 marbles in the urn. Hence, by the multiplication theorem,

?, & . E . . = 9 8 w

18.10. The students in a class are selected at random, one after the other, for an examination. Find the probability p that the boys and girls in the class alternate if (i) the class consists of 4 boys and 3 girls; (ii) the class consists of 3 boys and 3 girls.

(i) If the boys and girls a re to alternate, then the first student examined must be a boy. The probability tha t the first is a boy is 417. If the first is a boy, then the probability tha t the second is a girl is 3/6 since there a re 3 girls out of 6 students left. Continuing in this manner, we obtain the probability t h a t the third is a boy is.3/5, the fourth is a girl is 2/4, the fifth is a boy is 213, the sixth is a girl is 112, and the las t is a boy is 1/1. Thus

p = *.?.3.2.2.1.1 - r , 6 5 4 3 2 1 - 3 5

(ii) There a re two mutually exclusive cases: the first pupil is a boy, and the first is a girl. If the first student is a boy, then by the multiplication theorem the probability pl tha t the students alternate is

= g.;.;.;.;.;

p 2 = 3 . ~ . $ . $ . 1 . 1 = 2 1 1 YO

If the first student is a girl, then by the multiplication theorem the probability p 2 tha t the students alternate is

6 0

1 Thus p = p , + p z = % + & = $.

CONDITIONAL PROBABILITY 18.11. In a certain college, 25% of the students failed mathematics, 15% of the students

failed chemistry, and 10% of the students failed both mathematics and chemistry. A student is selected at random.

(i) If he failed chemistry, what is the probability that he failed mathematics?

(ii) If he failed mathematics, what is the probability that he failed chemistry? (iii) What is the probability that he failed mathematics o r chemistry?

Let M = {students who failed mathematics} and C = {students who failed chemistry}; then P ( M ) = 25% = 25, P ( C ) = 15% = .15, P ( M n C ) = 10% = . l o

(i) The probability that a student failed mathematics, given tha t he has failed chemistry is P ( M n C ) - . lo - 2 P ( M ) C) = ~ - - - -

P(C) .15 3

(ii) The probability tha t a student failed chemistry, given t h a t he has failed mathematics is

3 (iii) P(MuC) = P ( M ) + P(C) - P ( M n C ) = .25 + .15 - . l o = .30 =


18.12. Let A and B be events with P(A) = +, P(B) = Q and P ( A nB) = $. Find (i) P ( A [ B), iii) P ( B I A ) , (iii) P ( A uB), (iv) P(Ac 1 Bc), (v) P(Bc I A').

1 1 s - (iii) P(A U B ) = P(A) + P(B) - P ( A nB) = 5 + 3 - 4 =

riv) Firs t compute P(Bc) and P(AcnBc). P(Bc) = 1 - P ( B ) = 1 -Q = +. By De Morgan's Law, ( A u B ) ~ = AcnBc; hence P(AcnBc) = P ( ( A d B ) c ) = 1 - P(A U B ) = 1 - 1 1 2 = 2. 1 2

18.13. Let A and B be events with P ( A ) = 8, P(B) = # and P ( A U B ) = 9.

Firs t compute P(A n B ) using the formula P(A UB) = P(A) + P(B) - P(A n B ) :

_ - a - 4 + 8 - P ( A n B ) or P ( A n B ) = 4

Find P ( A 1 B) and P(B I A).

18.14. Let S = { a , b , c , d , e } with P(a) = P(b) = A, P(c ) = and P(d) = P(e ) = &i, and let E = { b , c , d } . Find: (i) P ( A IE) where A = {a, b , c } ; (ii) P ( B I E ) where B = { c , d , e } ; (iii) P(CIE) where C = { a , c , e } ; (iv) P(D IE) where D = { a , e } .

P(* n E ) In each case use the definition of conditional probability

Here P(E) = P ( { b , c , d } ) = P ( b ) + P(c) + P(d) = & J- A P E = { b , c } , P ( A n E ) = P ( b ) + P ( c ) = %+; = &, and P ( A / E ) = ~

P ( x E ) = p.

+ $ = 5. (El

P ( A n E ) - & - 1 S ( i i P(E) - B - 2 '

tii) B n E = {c ,d} , P ( B n E ) = P(c) t

(iii) C n E = {c}, P(CnE) = P ( c ) = +,

(iv) D n E = $3, P ( D n E ) = P ( @ ) = 0,

P ( B n E ) - - - 5 P ( d ) = f + & = i, and P ( B / E ) = ~ - a P ( E ) - % - 6'

P(CnE) 1 1 and P(C 1 E ) = ~ - 5 - -

P ( D n E ) - 0 and P(DI E ) = ___ -

- - P(E) # 3 '

P ( E ) - 3 = O .

18.15. Show that the conditional probability function P(* 1 E ) satisfies the three axioms of Theorem 17.1; that is:

[P,] [P,] [P,]

For any event A, 0 L P ( A 1 E ) f 1. For the certain event S , P ( S I E ) = 1.

If A and B are mutually exclusive, then P ( A uB 1 E ) = P ( A I E ) + P(B ~ E).

Since A n E c E , P ( A n E ) 5 P(E) . Thus P(A I E ) = ~ n E ) 4 1 and is also non-negative;

Since S n E = E ,

If A n B = 8, then A n B n E = ( A n E ) n ( B n E ) = 8, i.e. A n E and B n E are mutually exclusive events; hence P ( ( A n E ) u (BnE)) = P ( A n E ) + P ( B n E ) . But ( A n E ) u ( B n E ) = ( A LB) n E. Thus P((Ac:B) n E ) = P ( A n E ) + P ( B n E ) . Hence

P ( E ) that is, 0 5 P(A 1 E ) f 1. P ( S n E ) - P(E) P(S 1 E ) = ~ - __ P ( E ) P(E) =


Remark: Since P(* I E ) satisfies the aKove three axioms, it also satisfies any consequence of these axioms: for example,

P(AUB I E ) = P(A IE) + P(B1E) - P(Ai7B 1 Ej

18.16. Prove: Let E l , E,, . . .,En form a partition of a sample space S , and let A be any event. Then

P ( A ) = P(Ei)P(A I El) + P(Er) P ( A I Ez) + . . . + P(En) P ( A 1 En) Since the E , forin a partition of S, the E, are pairwise disjoint and S = E I U E z U . ~ ~ U E , , .

Hence

and the E , n A are also pairwise disjoint.

A = S n A = ( E , u E , u ~ ~ ~ u E , , ) n A = ( E l n A ) u (E ,nA) u . . . u (E,nA)

Accordingly, P(A) = P ( E , n A ) P(E,nA) + - . . + P(E , ,nA)

By the multiplication theorem, P ( E , p A ) = P(E, ) P(A I Et); hence

P ( A ) P(E1) P ( A 1 E l ) - P(E,j P(A 1 Ez) + . . - P(E,,) P(A I En)

18.17.Three machines A, B and C produce respectively 50%) 3 0 5 and 20% of the total number of items of a factory. The percentages of defective output of these machines are respectively 3%) 4% and 5 G . If an item is selected at random, what is the probability 71 that the item is defective?

preceding problem,

P ( X ) =

Let X be the event tha t a n item is defective. Then by the

P ( A ) P:X I A ) + P(B) P ( X B ) - P(C) P ( X 1 C ) = 5 O C r . 3 % + 30%*4% + 2 0 5 . 5 % - _ 5 0 , 3 + 3o.L + 2 o . z - 37 = 3.7%

100 100 100 100 100 100 - 1000 -

We can also consider this problem as a stochastic process having the adjoining tree diagram.

18.18. Prove Bayes’ Theorem: Let E l , E,, . . . , E l , form a partition of a sample space S , and let A be any event. Then, for-each i,

P(EJ P(A I E l ) P(E1) P ( A E l ) + P(E2) P ( A 1 Ez) + . . . + P(En) P ( A 1 En) P(Ei A) =

Using the definitlon of conditional probability, we obtain

P(E, n A ) - P(E,) P ( A I EL) P ( A )

- P(Ei) P ( A I El) + P(Ez) P(A I E 2 j + . . - P(En) P(A 1 En) P(E, A ) =

Here we used the multiplication theorem, and Problem 18.16 t o obtain the denominator.

P ( E , n A ) = P(E,) P ( A 1 Ez), t o obtain the numerator,

18.19. In a certain college, 4% of the men and 1% of the women are taller than 6 feet. Furthermore, 6 0 5 of the students are women. Now if a student is selected at random and is taller than 6 feet, what is the probability that the student is a woman?

Let A = {students taller than 6 feet:. We seek P(W 1 A ) , the probability t h a t a student is a woman given that the student is taller than 6 feet. By Bayes’ theorem,


18.20. Find P(B A ) if (i) A is a subset of B, (ii) A and I3 are mutually exclusive. ( i ) If A is a subset of B, then whenever A occurs B must occur: hence P(B 1 A ) = 1. Alternately,

if A is a subset of B then A r B = A ; hence P ( A n B ) - P ( A ) P ( B A ) = ~ ~ P ( A ) - P ( A ) =

(i) (ii)

If A and B are mutually exclusive. i.e. disjoint, then whenever A occurs B cannot occur; hence P ( B I A ) = 0. Alternately, if A and B are mutually exclusive then A n B = 0; hence

(iii

FINITE STOCHASTIC PROCESSES 18.21. A box contains three coins; one coin is fair, one coin is two-headed, and one coin is

A coin is selected a t weighted so that the probability of heads appearing is +. random and tossed. Find the probability p that heads appears.

Construct the tree diagram as shown in Figure (a ) below. Note that I refers t o the fai r coin, Now heads appears along three of the I1 t o the two-headed coin, and I11 t o the weighted coin.

paths: hence 1.1 + 1.1 + 1.1 = - 11 P = 3 2 3 3 3 18

(4 0) 18.22.We are given three urns as follows:

Urn A contains 3 red and 5 white marbles. Urn B contains 2 red and 1 white marble. Urn C contains 2 red and 3 white marbles.

An urn is selected at random and a marble is drawn from the urn. If the marble is red, what is the probability that i t came from urn A?

Construct the tree diagram as shown in Figure ( b ) above. We seek the probability tha t A was selected, given that the marble is red; tha t is, P ( A 1 R).

The probability tha t urn A is selected and a red marble drawn is 3 . 8 = g ; tha t is,

In order t o find P ( A 1 R ) , i t is necessary first to compute P ( A n R ) and P(R) . 1 3 1

P ( A r R ) = f . Since there are three paths leading to a redmarble, P(R) = 1 * 3 + 1 * ~ + 1 . g = 173 Thus

3 8 3 3 3 5 360'

1 P ( A n R ) - s - 45 - P ( A , R ) = ~ - - P(R) - II.? 360 173


18.23.

18.24.

18.25.

Box A contains nine cards numbered 1 through 9, and Box B contains five cards numbered 1 through 5. A box is chosen at random and a card drawn. If the number is even, find the probability that the card came from Box A .

The tree diagram of the process is shown in Figure (a) below. We seek PIA ' E ) , the probability that A was selected, given that the number is even. The

Since 1 4 probability tha t Box A and an even number is drawn is 2 - y = $; that is, P ( A n E ) = $ . there are two paths which lead to an even number, P ( E ) = L*. i -L* 2 9 2 ; = L!? 4 5 . Thus

An urn contains 3 red marbles and 7 white marbles. urn and a marble of the other color is then put into the urn. drawn from the urn.

A marble is drawn from the A second marble is

0) (ii)

6)

(ii)

We

An

Find the probability p that the second marble is red. If both marbles were of the same color, what is the probability p that they were both white?

Construct the tree diagram as shown in Figure ( b ) above.

Two paths of the tree lead to a red marble:

The probability that both marbles were white is & * $ = - ii. The probability that both marbles were of the same color, i.e. the probability of the reduced sample space, is

p = 10 * & 10 + 10 * 10 = !? 50 .

lo.lo+-'- 3 2 7 6 - 1o 1o b g . Hence the conditional probability p = 5 /g = i.

are given two urns as follows: Urn A contains 3 red and 2 white marbles. Urn B contains 2 red and 5 white marbles. urn is selected at random; a marble is drawn and put into the other urn; then

a marble is drawn from the second urn. drawn are of the same color.

Find the probability p that both marbles

Construct the following tree diagram:

210 CONDITIOKAL PROBABILITY. IISDEPENDENCE [CHAP. 18

Note that if urn A is selected and a red marble drawn and put into urn B, then urn C has

Since there a re four paths which lead to two marbles of the same color,

3 red marbles and 5 white marbles.

= $.;.; + $ . $ . 2 + 1.2.2 - L . 5 . L = 901 2 7 3 2 7 2 1680

INDEPENDENCE 18.26. Let A = event that a family has children of both sexes, and let B = event that a

( i ) Show that A and B are independent events if a (ii) Show that A and B are dependent events if a family

family has at most one boy. family has three children. has two children.

f i ) We have the equiprobable space S = { b b b , bbg , b g b , b g g , g b b , g b g , g g b , ggg). Here P ( A ) = 6 = 3 A = W g , bgb, b g g , g b b , g b g , g g b ) and so 8 4

B = Cbgg, g b g , g g b , s w > and so P(B) = + z $ A n B = {bug, gbg, ggb) and so P ( A n B ) =

Since P ( A ) P ( B ) = 2 - 4 = # = P ( A n B ) , A and B are independent.

Iii) We have the equiprobable space 5' = { b b , bg , g b , gg) . Here A = { b g , g b ) and so P ( A ) = 4 B = { b g , gb, gg} and so P ( B ) = 2

A n B = { b g , g b } and so P ( A n B ) = 4 Since P ( A ) P ( B ) # P ( A n E ) , A and B a r e dependent.

18.27. Prove: If A and B are independent events, then Ac and B' are independent events. Let P ( A ) = x and P(B) = y. Then P(Ac) = l - - r and P(Bc) = 1-y. Since A and B are

independent, P ( A nB) = P ( A ) P ( B ) = X U . Furthermore,

P ( A u B ) = P ( A ) + P(B) - P ( A n B ) = x + y - xy

By De Morgan's theorem, ( A u B ) c = AcnBc; hence

P(AcnBc) = P ( ( A u B ) c ) = 1 - P ( A ~ J B ) = 1 - x - y + xy On the other hand,

P(AC) P(Bc) = (1 - x) ( l - y) = 1 - j: - y + Thus P(AcnBc) = P ( A c ) P(Bc) , and so A' and BC are independent.

18.28. The probability that a man will live 10 more years is 4, and the probability that his wife ill live 10 more years is +. Find the probability that (i) both will be alive in 10 years, (ii) at least one will be alive in 10 years, (iii) neither will be alive in 10 years, (iv) only the wife will be alive in 10 years.

Let A = event tha t the man is alive in 10 years, and B = event t h a t his wife is alive in 10 years; then P ( A ) = t and P ( B ) = 4. 11)

( ii) We seek P ( A u B ) . P ( A u B ) = P ( A ) + P ( B ) - P ( A n B ) = &++-A = 4. iiii) We seek P(AcnBc). Now P(Ac) = 1 - P ( A ) = 1 - & = 4 and P(i3c) = 1 - P(B) = 1 - 4 = 3.

Alternately, since ( A U B ) C = ACnBc, P(AcnBc) = P ( ( A u B ) ~ ) = 1 - P ( A L'B) = 1 - 4 = 4. (iv) We seek P ( A c n B ) . Since P(Ac) = 1 - P ( A ) = 2 and Ac and B a r e independent, P(AcnB) =

We seek P ( A n B ) . Since A and B are independent, P ( A n B ) = P ( A ) P(B) = & - + = A.

Furthermore, since Ac and Ec are independent, P(Acn Sc) = P(AC) P(Bc) = 3 ' 4 3 2 = 1. 2

P ( A c ) P ( B ) = f.


18.29. Box A contains 8 items of which 3 are defective, and box B contains 5 items of which 2 are defective. An item is d raxn at random from each box. (i) What is the probability p that both items are non-defective? (ii) What is the probability p that one item is defective and one not? (iii) If one item is defective and one is not, what is the probability p that the defec-

tive item came from box A ?

(i)

(ii)

(iii)

18.30. The

The probability of choosing a non-defective item from A is events a re independent, p = f a ; = 5

Method 1. The probability of choosing two defective items is = &. From fi) the

probability that both a re non-defective is i. Method 2. from B is

defective item froin B is

Consider the events X = {defective item from A } and Y = {one item is defective and one non-defective}. We seek P(X [ Y). By (ii), P (XnY) = p , = and P(Y) = g. Hence

and from B is $. Since the

8 ’

Hence p = 1

The probability p , of choosing a defective item from A and a non-defective item The probability p L of choosing a non-defective item from A and a

= 19. R 20 40

*: = &. 9 1

= $. Hence p = pl + p , = lot- = ?! 4 4 0 .

probabilities that three men hit a target are shoots once a t the target. the target. first man ?

(i) Find the probability (ii) If only one hit the target, what is

Consider the events A = {first man hits the target}, B C = {third man hits the target}; then P ( A ) = +, P ( B ) = independent, and P(Ac) = g, P(Bc) = 9 , P(CC) = t. (i) Let E = [exactly one man hits the target}. Then

E = (AnBcnCc) u (AcnBnCc)

respectively Q, & and +. Each p that exactly one of them hits the probability that i t was the

= {second man hits the target), and and P ( C ) = a. The three events a re

In other words, if only one hit the target, then i t was either only the first man, AnBcrCc , or only the second man, AcnBnCc, o r only the third man, AcnBcnC. Since the three events a re mutually exclusive,

p = P(E) = P(AnBcnCc) + P(AcnBnCc) + P(AcnBcnC) = P(A) P(Bc) P(Cr) + P(Ac) P(B) P(Cc) T P(Ac) P(Bc) P(C) - -.-.---.-.__-.-. 1 3 2 5 1 2 5 3 1 - = L t 5 - 5 = 31

6 4 3 6 4 3 6 4 3 12 36 24 72 -

(ii) We seek P ( A ’ E ) , the probability that the first man hi t the target given t h a t only one man hit the target. Now A n E = A n Bc n Cc is the event t h a t only the first man hi t the target. By (i), P ( A r E ) = P(AnBcnCc) = $ and P ( E ) = g; hence

Supplementary Problems CONDITIONAL PROBhBILITT IN EQUIPROBABLE SP-ACES

18.31.

18.32.

A die is tossed. If the number is odd, what is the probability tha t it is prime?

A letter is chosen a t random from the word “trained”. If i t is a consonant, what is the probability tha t i t is the first letter of the word?

2 12

18.33.

18.34.

18.35.

18.36.

18.37.

18.38.

18.39.

18.40.

CONDITIONAL PROBABILITY. INDEPENDENCE [CHAP. 18

Three fair coins a re tossed. If both heads and tails appear, determine the probability that exactly one head appears.

A pair of dice is tossed. If the numbers appearing a re different, find the probability tha t the sum is even.

A man is dealt 5 red cards from an ordinary deck of 52 cards. What is the probability t h a t they are all of the same suit, i.e. hearts o r diamonds?

A man is dealt 3 spade cards from an ordinary deck of 52 cards. If he is given four more cards, determine the probability tha t a t least two of the additional cards a re also spades.

Two different digits a r e selected a t random from the digits 1 through 9.

( i ) ( i i )

If the sum is odd, what is the probability tha t 2 is one of the numbers selected?

If 2 is one of the digits selected, what is the probability tha t the sum is odd?

Four persons, called North, South, Eas t and West, a re each dealt 13 cards from an ordinary deck of 5 2 cards.

i i ) If South has exactly one ace, what is the probability that his par tner North has the other three aces?

If North and South together have 10 hearts, what is the probability t h a t neither Eas t nor West has the other 3 hearts?

iii)

A class has 10 boys and 5 girls. Three students a re selected from the class at random, one af ter the other. Find the probability tha t ( i ) the first two are boys and the third is a girl, (ii) the first and third a re boys and the second is a girl, (iii) the first and third a r e of the same sex, and the second is of the opposite sex.

In the preceding problem, if the first and third students selected are of the same sex and the second student is of the opposite sex, what is the probability that the second student is a girl?

CONDITIONAL PROBABILITY 18.41. In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both

brown hair and brown eyes.

ti) If he has brown hair, what is the probability t h a t he also has brown eyes?

(ii) If he has brown eyes, what is the probability tha t he does not have brown hair?

(iii) What is the probability tha t he has neither brown hair nor brown eyes?

A person is selected at random from the town.

18.42. Let A and B be events with P ( A ) = a, P(B) = & and P(AUB) = 4. Find (i) P ( A IB), (ii, P(B A ) , (iii) P ( A nBc), (iv) P ( A 1 Bc).

18.43. Let S = (a , b , c , d , e , f} with P(a) = A, P(b) = A, P(c) = 4, P(d) = A, P(e) = & and P ( f ) = A. Let A = { a , c , e } , B = { c , d , e , f } and C = { b , c , , f } . Find (i) P ( A 1 B ) , (ii) P(B I C), (iii) P ( C ' Ac) ,

1 6

(iv) P(Ac 1 C).

18.44. In a certain college, 25% of the boys and 10% of the girls are studying mathematics. The girls constitute 60Cb of the student body. If a student is selected at random and is studying mathematics, determine the probability that the student is a girl.

FINITE STOCHASTIC PROCESSES

18.45. We are given two urns as follows:

Urn A contains 5 red marbles, 3 white marbles and 8 blue marbles.

Urn B contains 3 red marhles and 5 white marbles.

A fa i r die is tossed; if 3 or 6 appears, a marble is chosen from B , otherwise a marble is chosen from A. Find the probability that ( i ) a red marble is chosen, (ii) a white marble is chosen, (iiii a blue marble is chosen.

213 CHAP. 181 CONDITIONAL PROBABILITY. INDEPENDENCE

18.46.

18.47.

18.48.

18.49.

18.50.

18.51.

18.52.

18.53.

18.54.

18.55.

18.56.

Refer to the preceding problem. (i) If a red marble is chosen, what is the probability tha t it came from urn A ? {i i i If a white marble is chosen, what is the probability that a 5 appeared on the die?

An urn contains 5 red marbles and 3 white marbles. A marble is selected a t random from the urn. discarded, and two marbles of the other color are put into the urn. A second marble is then selected from the urn. Find the probability that i i ) the second marble is red, iii) both marbles are of the same color.

Refer to the preceding problem. (i) If the second marble is red, what is the probability that the first marble is red? iii) If both marbles are of the same color, what is the probability tha t they a re both white?

A box contains three coins, two of them fair and one two-headed. A coin is selected at random and tossed twice. If heads appears both times, what is the probability that the coin is two-headed?

We a r e given two urns as follows:

Urn A contains 5 red marbles and 3 white marbles.

Urn B contains 1 red marble and 2 white marbles.

A fa i r die is tossed; if a 3 or 6 appears, a marble is drawn from B and put into A and then a marble is drawn from A : otherwise, a marble is drawn from A and put into B and then a marble is drawn from B.

(i)

(ii)

Box A contains nine cards numbered 1 through 9, and box B contains five cards numbered 1 through 5. A box is chosen at random and a card drawn; if the card shows an even number, another card is drawn from the same box; if the card shows an odd number, a card is drawn from the other box.

(i) What is the probability t h a t both cards show even numbers?

(ii) If both cards show even numbers, what is the probability that they came from box A ? (iii) What is the probability t h a t both cards show odd numbers?

What is the probability that both marbles a re red?

What is the probability tha t both marbles are white?

A box contains a fa i r coin and a two-headed coin. A coin is selected a t random and tossed. heads appears, the other coin is tossed: if tails appears, the same coin is tossed.

(i) Find the probability t h a t heads appears on the second toss.

(ii) If heads appeared on the second toss, find the probability that it also appeared on the first toss.

A box contains three coins, two of them fair and one two-headed. A coin is selected at random and tossed, If heads appears the coin is tossed again; if tails appears, then another coin is selected from the two remaining coins and tossed.

(i) Find the probability that heads appears twice.

(ii) If the same coin is tossed twice, find the probability t h a t it is the two-headed coin.

(iii) Find the probability tha t tails appears twice.

Urn A contains z red marbles and y white marbles, and urn B contains z red marbles and 9 white marbles.

(i) If an urn is selected at random and a marble drawn, what is the probability tha t the marble is red?

(ii) If a marble is drawn from urn A and put into urn B and then a marble is drawn from urn B , what is the probability t h a t the second marble is red?

If

A box contains 5 radio tubes of which 2 are defective. The tubes a re tested one af ter the other until the 2 defective tubes a r e discovered. What is the probability t h a t the process stopped on the (i) second test, (ii) third test?

Refer to the preceding problem. If the process stopped on the third test, what is the probability tha t the first tube is non-defective?


INDEPENDENCE

18.57. Prove: If -4 and B a r e independent, then A and Bc are independent and Ac and B are independent.

18.58. Let A and B be events with P ( A ) = $, P ( A u B ) = + and P ( B ) = p . mutually exclusive. (ii) Find p if A and B are independent.

(i) Find p if A and B are (iii) Find p if A is a subset of B.

18.59. Urn A contains 5 red marbles and 3 white marbles, and urn B contains 2 red marbles and 6 white marbles.

(i) If a marble is drawn from each urn, what is the probability that they a r e both of the same color?

(iii If two marbles a re drawn from each urn, what is the probability tha t all four marbles a re of the same color?

18.60. Let three fair coins be tossed. Let A = {all heads or all tails}, B = {at least two heads) and Of the pairs ( A , B ) , (A , C) and (B , C), which are independent and C = {at most two heads}.

which a re dependent?

18.61. The probability tha t A hits a target is t and the probability tha t B hits a target is Q. (i) If each fires twice, what is the probability t h a t the target will be hi t at least once? (ii) If each fires once and the target is hit only once, what is the probability t h a t A hit the target?

(iii) If A can fire only twice, how many times must B fire so tha t there is at least a 90% probability that the target will be hi t?

18.62. Let A and B be independent events with P ( A ) = 8 and P ( A u B ) = 8. Find (i) P(B) , (ii) P ( A 1 B) , (iii) P(Bc ' A ) .


18.31. $

18.32. 2

18.33. 4

18.34. +

18.37. (i) &, (ii) 5

15 - 18.40. % = $

~

21

18.41. (i) Q , (ii) 3, (iii) fr

(i) Q, (ii) $, (iii) 4, (iv) Q

(i) $, (ii) 3, (iii) 4 , (iv) $

18.42.

18.43.

18.44. Q

18.45. (i) + (ii) & (iii) 4

18.46. (i) a, (ii) &

10 5 9 15 18.47. ( i ) z , (ii) 36

10 9 5 15 18.39. (i) - z * 13 = 91

41 13 (ii) -.-.---

15 14 13 - 91

b

- W 8

15 20 5 273 21 (iii) 91 + - = - 20 3

18.48. (i) 41' (ii) 13

CHAP. 181 CONDITIONAL PROBABILITY. I N D E P E N D E N C E 215

16.49. $

16.50. 5 61 3 8 3il 16 81 1206 (i) 24 + & = z, ( i i ) - + - = ~

<-+: 6 w e ; 9

Tree diagram for Problem 18.50 1

2 15

- 1 1 2 18.51. (i) 5 + = - (ii) 12 = 5, (iii) 4 + Q = Q

- 1 5 '

16.52. (i) 8, (ii) 8

1 1 2 1 16.53. (i) 12 + 12 + 5 = f , (ii) 3, (iii) 12

Tree diagram for Problem 18.53 Tree diagram for Problem 18.55

16.55. (i) h , (ii) &; we must include the case where the three non-defective tubes appear first, since the las t two tubes must then be the defective ones.

18.56. 3 16.60. Only A and B a r e independent.

(i) $, (ii) 9, (iii) 5 18.58.

18.59. (i) 16, (ii) 18.62. (i) 4, (ii) 4, (iii) 3

1 18.61. 1 1

(i) 12, (ii) 9, (iii) 5

7 5 5

Chapter 19

Independent Trials, Random Variables

INDEPENDENT OR REPEATED TRIALS

ment repeated a finite number of times, as the tossing of a coin three times. of repetition is formalized as follows:

We have previously discussed probabiIity spaces which were associated with an experi- This concept

[Definition:I Let S* be a finite probability space. By ?i i n d e p e n d e n t or repea ted trials, we mean the probability space S consisting of ordered n-tuples of elements of S" with the probability of an ?z-tuple defined to be the product of the probabilities of its components:

P((s1, s2, . . . , Sn)) = P(s,) P(s2) . . . P(s,)

Example 1.1: Whenever three horses a, b and c race together, their respective probabilities of winning are +, f and A. and P ( c ) = 8. I f the horses race twice, then the sample space of the 2 repeated trials is

I n other words, S = { u , b , c } with P(a) = +, P(b) =

S = {aa, ab, a c , h a , bb , bc, ca, cb, cc}

For notational convenience, we have written a c f o r the ordered pair ( a , ~ ) . Probability of each point in S is

The

P ( a a ) = P ( a ) P ( a ) = 4.4 = $

P ( a b ) = P ( u ) P ( b ) = + * & = 4 P ( a c ) = P(u) P(c) = i.8 = & P ( b c ) = ilj-

P(ba) = Q P(ca) = & P ( b b ) = & P(cb) = 3

P(cc) = & 1

1

Thus the probability of c winning the first race and a winning the second race is P(ca) = A.

From another point of view, a repeated trials process is a stochastic process whose tree diagram has the following properties: (i) every branch point has the same outcomes; (ii) the probability is the same fo r each branch leading to the same outcome. For example, the tree diagram of the repeated trials process of the preceding experiment is as shown in the adjoining figure. C

Observe that every branch point has the outcomes a, b and c, and each branch leading to outcome a has probability &, each branch leading to b has probability Q, and each leading to c has probability Q.

216

CHAP. 191 INDEPENDENT TRIALS, RANDOM VARIABLES 217

REPEATED TRIALS WITH TWO OUTCOMES We now investigate repeated trials of an experiment with two outcomes; we call one

of the outcomes success and the other outcome failure. Let p be the probability of success, and so q = 1 - p is the probability of failure. Frequently we are interested in the number of successes and not in the order in which they occur. The following theorem applies.

Theorem 19.1: The probability of exactly k. successes in n repeated trials is denoted and given by

The probability of at least one success is 1 - q".

f ( n , k , P) = (3 P k rk

Here (;) is the binomial coefficient (see Chapter 13).

Example 2.1: A fair coin is tossed 6 times or, equivalently, six fair coins are tossed; call heads a success. Then n = 6 and p = q = 4.

(i) The probability that exactly two heads occurs (i.e. k = 2) is

f(6,2,4) = ($)z(+)4 = 6 4

(ii) The probability of getting at least four heads (i.e. k = 4, 5 or 6) is

f(6,4, # f f(6,5,4) + f(6,6, Q) = (3)4 (g)2 + (E) (#5 (4) + (i) ($J6 - 1 5 + 6 + 1 = 11

6 4 6 4 6 4 3 2 -

(iii) The probability of no heads (i.e. all failures) is q6 = (4)6 = A, and so the probability of a t least one heads is 1 - q 6 = 1 - 1 6 4 = a. 6 4

Example 2.2: A fair die is tossed 7 times; call a toss a success if a 5 o r a 6 appears. Then n = 7, p = P({5,6}) = Q and q = 1 - p = 2. 3

(i) The probability that a 5 or a 6 occurs exactly 3 times (i.e. k = 3) is

f ( 7 , 3 , Q ) = (7) (')3 ( " )4 1 2187 3 3 3

(ii) The probability that a 5 or a 6 never occurs (i.e. all failures) is 47 = (Q)' = p. ,187'

hence the probability that a 5 or a 6 occurs a t least once is 1 - q7 = 2187. 2059

If we treat n and p as constants, then the function defined by f (n , k, p ) is called the binomial distribution since for k = 0,1,2, . . . , n it corresponds to the successive terms of the binomial expansion:

( q +p)" = q" + (Y) q n - ' p + (;) q"-Zp2 + * a + p"

= f(n, 0, P) + f(n, 1, P ) + f(n, 2, P) + - . + f (n , n, P)

The use of the word distribution will be explained later in the chapter. This distribution is also called the Bernoulli distribution, and independent trials with two outcomes are called Bernoulli trials.

RANDOM VARIABLES As noted previously, the outcomes of

the experiment, i.e. the sample points of S, need not be numbers. However, we frequently wish to assign a specific number to each outcome of the experiment. Such a n assignment is called a random variable.

Let S be a sample space of some experiment.

In other words:

218 I N D E P E N D E N T TRIALS, RANDOM VARIABLES [CHAP. 19

1 Definition: I A mndom variable X is a function from a sample space S into the real

We shall refer to

numbers.

We shall let R, denote the range of a random variable X : R, = X ( S ) . R, as the range space.

Remark: If the sample space S is nondenumerable, then f o r theoretical reasons lying beyond the scope of this text certain functions defined on S are not random variables. However, the sample spaces here are finite, and every function defined on a finite sample space is a random variable.

Example 3.1: A pair of dice is tossed. numbers between 1 and 6:

The sample space S consists of the 36 ordered pairs of

S = ((1, I), ( 1 , 2 ) , . . . , ( 6 6 ) )

Let X assign to each point in S the sum of the numbers; then X is a random variable with range space

Rx = (2, 3, 4, 5, 6, 7 , 8, 9, 10, 11, 12)

Let Y assign t o each point the maximum of the two numbers; then Y is a random variable with range space

RY = (1, 2 , 3, 4, 5 , 6)

Example 3.2: A sample of 3 items is selected from a box containing, say, 12 items of which 3 a r e defective. The sample space S consists of the (y) = 220 different samples of size 3. Let X denote the number of defective items in the sample; then X is a random variable with range space R, = { 0 , 1 , 2 , 3 } .

PROBABILITY SPACE AND DISTRIBUTION O F A RANDOM VARIABLE Let R, = {x,,x,, . . .,x,} be the range space of a random variable X defined on a

finite sample space S. Then X induces an assignment of probabilities on the range space R, as follows:

p l = P(x,) = sum of probabilities of points in S whose image is x1

The function assigning p i to xi or, in other words, the set of ordered pairs (xl, pl), . . . , usually given by a table

...

is called the distribution of the random variable X . In the case that S is an equiprobable space, we can easily obtain the distribution of a

random variable from the following result.

Theorem 19.2: Let S be an equiprobable space. If R, = {xl, . . . , x,} is the range space of a random variable X defined on S , then

number of points in S whose image is xi number of points in S pi = P(XJ =

Example 4.1: A pair of fa i r dice is tossed. We obtain the equiprobable space 5' consisting of the 36 ordered pairs of numbers between 1 and 6:

S = {(L I), (1 ,2 ) , . . .,

CHAP. 191

xi

''

INDEPENDENT TRIALS, RANDOM VARIABLES 219

2 3 4 ' 5 6 7 1 8 9 10 11 12

3 - - I - - - 3 1 - 5 - 6 - - - 5 2 36 f 6 36 36 36 36 1 36 ,'6 36 36 ,'6

Let X be the random variable which assigns to each point the sum of the two numbers; then X has range space

R, = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

We use Theorem 19.2 t o obtain the distribution of x. There is only one point (1,l) whose image is 2: hence P(2) = &. There a re two points in S, (1 ,Z) and (2, l), having image 3; hence P(3) = z. There a re three points in S , (1,3), (2 ,2) and (3, l), having image 4: hence P ( 4 ) = 36. Similarly, P(5) = x, P(6) = &, etc. The distribution of X consists of the points in R, with their respective probabilities:

2

3 4

Example 4.2: A sample of 3 items is selected a t random from a box containing 12 items of which 3 a re defective. The sample space S consists of the ii') = 220 different equally likely samples of size 3. Let X be the random variable which assigns to each sample the number of defective items in the sample; then Rx = {0,1,2,3). Again we use Theorem 19.2 to obtain the distribution of X.

There a r e (i) = 84 samples of size 3 which contain no defective items; hence There a re 3 * (:) = 108 samples of size 3 containing 1 defective item;

There are (;) * 9 = 27 samples of size 3 containing 2 defective There is only one sample of size 3 containing the 3 defec-

84 P ( 0 ) = m. hence P(1) = g. items; hence P ( 2 ) = &. tive items; hence P(3) =A.

2 -

The distribution of X follows:

Example 4.3: A coin weighted so tha t P iH) = 8 and P(T) = + is tossed three times. The S = {HHH, HHT, HTH, HTT,

a re as follows: probabilities of the points in the sample space THH, THT, TTH, TTT}

p ( H H H ) = g.:F.2 z P ( T H H ) = t . 3 .g = 4 27

P ( H H T ) = 2.2.1 a = - i7 PtTHT) = Q.8.Q = & p ( H T H ) = $.;+*+ = 6 P(TTH) = g.1.2 3 3 = 2 27

p ( H T T ) = +.l.+ = 4 27 P(,TTT) = & * & * & = - 27

3 1 3 27

1

Let X be the random variable which assigns t o each point in S the largest number of successive heads which occurs. Thus,

X(TTT) = 0

X(HTH) = 1, X(HTT) = 1, X(THT) = 1, X(TTH) = 1

X(HHT) = 2, X!THH) = 2

X(HHH) = 3

The range space of X is R, = {0,1,2,3}. follows:

P(0) = P(TTT) = & P(1) = P ( H T H ) + P(HTT) + P(THT) + P(TTH) = & + $ + & + $ P(2) = P(HHT) t P ( T H H ) = &-I $ = 2 P(3) = P(HHH) = &

We obtain P(O), P(1), P(2) and P(3) as

10 =

220 I N D E P E K D E 9 T T I? I A L S , RAND 0 31 V A R I A B L E S ;CHAP. 19

The distribution of X consists of the points in the range space of X with their respective probabilities:

Now consider repeated trials of an experiment with two outcomes: that of success To each sequence of

Then X , 2 is a random variable with with probability p and that of failure with probability Q = 1 - p . n outcomes, let X n assign the number of successes. range space RY,, = {0,1,2, . . ., ti). Using Theorem 19.1, v e obtain

PjO) = f ( t z , O , p ) = Q"

P(1) = f ( i l , 1,p) = ( : ) q r z - 1 ] 1

P ( 2 ) = f( i1,2,p) = (;) q"-'p'

. . . . . . . . . . . . . . . . . . . . . . . . . . P(H) = . f ( i r , n, p ) = p"

In other words, the distribution of X n is the binomial distribution as discussed previously:

EXPECTED VALUE

respective probabilities pl, p 2 , . , . , p , , then the expected value is defined to be If the outcomes of an experiment are the numbers T , , x,, . , ., x r which occur with

E = ~~p~ + ~ , p , + y r p r

In particular, if the xZ and pZ come from the distribution of a random variable X , then the expected value ic, called the expectation of the random variable and is denoted by E ( X ) .

Example 5.1 : Consider the random variable X of Example 4.2. I ts distribution gives the possible numbers of defective items in a sample of size 3 with their respective probabilities of occurrence. Then the expected number of defective items in a sample of size 3, i.e. the expectation of the random variable X, is

E(X) = o.g+ 1.E + 2.sc 3 . L - s 3 220 -

Example 5.2: A f a i r coin is tossed 6 times. with their respective probabilities a re as foilom-s:

The possible numbers of heads which can occur

Note tha t the above table is the binomial distribution with n = 6 The expected value, i.e., the expected number of heads, is

and p = 3.

E = 0 . k + 1-& + 2 - g + 3 * % + 4 . 3 + 5 * & + 6-A = 3

Observe tha t E = np.

CHAP. 191 IKDEPENDENT TRIALS, RANDOM VARIABLES

xi

Pi

22 1

n . . . 0 1 2

qn ( ; ) q " - ' P ( ; ) 9 " - 2 P 2 . . . Pn

The result in the preceding example holds true in general; namely,

xi 2 3 5 -1 -4 -6

Example 5.3: The probability is 0.2 that an item produced by a factory is defective. If 3000 items are made by the factory, then the expected number of defective items is E = np = (3000)(0.2) = 60.

GAMBLING GAMES

to the player. favorable if E is negative. If E = 0, the game is fair.

In a gambling game, the expected value E is considered to be the value of the game The game is said to be favorable to the player if E is positive, and tin-

The negative numbers -1, -4 and -6 correspond t o the fact tha t the player loses if a non-prime number occurs, The expected value of the game is

Thus the game is unfavorable to the player since the expected value is negative.

Example 6.2: A player tosses a fa i r coin. If a head occurs he receives 5 dollars, but if a tail occurs he receives only 1 dollar. His expected winnings E = 5 * + 1 * 4 = 3.

Question: How much money should he pay to play the game so t h a t the game is fa i r? We claim that the answer is 3 dollars. For if he pays 3 dollars t o play the game, then each outcome is decreased by 3 dollars. Thus when a head occurs he wins 5 - 3 = 2 dollars and when a tail occurs he wins 1 - 3 = -2 dollars, i.e., he loses 2 dollars; hence the expected value is now E = 2 * + + (-2) * $ = 0.

The result in the preceding example holds true in general; namely,

Theorem 19.4: Let E be the expected value of a n experiment. If a constant k is added to each outcome of the experiment, then the expected value E* of the new experiment is E + k : E" = E + k.

Example 6.3: A thousand tickets a t $1 each are sold in a lottery in which there is one prize of $500, four prizes of $100 each, and five prizes of $10 each. The expected winnings of one ticket are

500-& + l o o * & + l o * & = 0.95

However, since a ticket costs $1, the expected value of the game is 0.95 - 1.00 = - $0.05, which is unfavorable to the player.

222 INDEPENDEKT TRIALS, RANDOJI VARIABLES [CHAP. 19

Solved Problems

REPEATED TRIALS WITH TWO OUTCOMES 19.1. Find (i) f ( 5 , 2 , + ) , (ii) f ( 6 , 3 , i), (iii) f ( 4 , 3 , t ) . Here

f ( n , k , P ) = (;) Pk (1 - P y k

19.2. A fa i r coin is tossed three times. Find the probability P that there will appear (i) three heads, (ii) two heads, (iii) one head, (iv) no heads.

Method 1. We obtain the following equiprobable space of eight elements:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT]

(i) Three heads (HHH) occurs only once among the eight sample points; hence P = 4. (ii) Two heads occurs 3 times IHHT, HTH, and THH); hence P = # .

(iii) One head occurs 3 times (HTT, THT and TTH); hence P=$.

(iv) No heads, i.e. three tails (TTT), occurs only once; hence P = $.

Method 2. Use Theorem 19.1 with .n = 3 and p = q = 4. (i) Here k = 3 and P = f ( 3 , 3 , 4 ) = (i) ( 3 ) O = 1 . 1 * 1 8 = 1. 8

(ii) Here k = 2 and P = f (3 ,2 , *) = (;) (+)z(4) = 3 1.1 4 2 = a. 8

(iii) Here k = 1 and P = f ( 3 , 1 , 3 ) = (t) (+)l (4)z = 3 9 4 . 4 = #. (iv) Here k = 0 and P = f ( 3 , 0 , 4 ) = (:) ( + ) O ( 4 ) 3 = 1. 1.1 8 8 = 1.

19.3. Suppose that 20% of the items produced by a factory are defective. chosen a t random, what is the probability P that defective, (iii) a t least one is defective?

If 4 items are (i) 2 are defective, (ii) 3 are

Use Theorem 19.1 with n = 4, p = .2 and q = 1 - p = .8.

Here k = 2 and

Here k = 3 and

(i)

(ii)

(iii) P = 1 - 44 = 1 - (.8)4 = 1 - .4096 = .5904.

P = J(4,2, .2) = (i) (.2)2(.8)2 = .1536.

P = f (4 ,3 , .2) = (:) (.2)3(.8) = ,0256.

19.4. Team A has probability ;f of winning whenever it plays. If A plays 4 games, find the probability that A wins (i) exactly 2 games, (ii) a t least 1 game, (iii) more than half of the games.

Here 12 = 4, p = and q = 1 - p = $.

223 CHAP. 191 INDEPENDENT TRIALS, RANDOM VARIABLES

(i)

(ii) Here 44 =

We seek f ( n , k , p ) when k = 2: f ( 4 , 2 , $ ) = (i) ($ )2(&)2 = &. = $ is the probability tha t A loses all four games. Then 1 - 44 = is the

probability of winning at least one game.

(iii) A wins more than half the games if A wins 3 or 4 games. Hence the required probability is

+g = 3 f(4,3,+, + f(4,4,9) = (;) ( 9 ) 3 (g) + ( 9 4 = g 27

19.5. A family has 6 children. Find the probability that there are (i) 3 boys and 3 girls, (ii) fewer boys than girls. Assume that the probability of the birth of either sex i s 3. (i) The required probability is

f ( 6 , 3, 4) = (6) (1)3 (1)s = 20 = -5- 3 2 2 64 1 6

(ii) There a re fewer boys than girls if there a re 0, 1 or 2 boys. Hence the required probability is

.f(6,0,4) + f(6,1, +) + f ( 6 , 2 , 4 ) = (&6 + ($I5 (4) f (g) (4)4 CgP =

19.6. The probability that a college student does not graduate is .3. Of 5 college students chosen at random, find the probability that (i) one will not graduate, (ii) three will not graduate, (iii) a t least one will not graduate.

Here n = 5, p = .3 and q = 1 - p = .7,

f ( 5 , 1,.3) = (;) (.3) (.7)4 = .36015

f ( 5 , 3, .3) = (i) (.3)3 (.7)2 = .1323

(i)

(ii)

(iii) 1 - 45 = 1 - (.7)5 = 1 - .16807 = .83193

Consider n repeated trials with two outcomes and p = q = 3. (i) Find the probability of exactly h: successes.

(ii) Find the probability P of more successes than failures if n is (a) odd, ( b ) even.

(i) The probability of exactly k successes is f (n , k , 3, = (T;) (&)k (&)"+ = (T;) ( 3 ) n

(ii) Recall tha t the distribution of the number of successes appears in the expansion of the binomial ( q + p)" = (4 + 4)":

1 = (1 2 2 + 1)" = (+)n + ( 7 ) (3)" + (;) (4)" + * - ' + (;I ($P + ( Y ) (&). + (4)" Since the binomial coefficients are symmetrical about the center of the expansion and since each term is (4)" times a binomial coefficient, the above terms are also symmetrical about the center of the expansion.

(a) If n is odd, then there is an even number of terms in the above expansion and so there are more successes than failures in exactly the terms in the las t half of the expansion. Thus 1 = P + P or P = Q.

( b ) If n is even, then there is an odd number of terms in the above expansion, with the middle Now there a re more successes than failures in the terms to the term M = (&)(:)".

right of the middle term; hence by the symmetry, 1 = P + M + P or P = A(1- M ) .

224 I N D E P E N D E K T TRIALS, RANDOM VARIABLES 'CHAP. 19

19.8. The probability of team A winning any game is 4. A wins more than half of its games if A plays [ i ) 43 games, (ii) 8 games?

What is the probability P that

(i) By the preceding problem, P = + since n is odd and 71 = q = +. (ii) The middle term M of the expansion (4 + &)8, i.e. the probability t h a t A wins 4 games and

loses 4 games, is (:)(&)8 = g, Then by the preceding problem, P = $(l - M ) = +(l - g) = z. 19.9. How many dice must be thrown so that there is a better than even chance of obtaining

a six? The probability of not obtaining a six on n dice is ( $ ) r l . Hence we seek the smallest 71 for

which (Q)" is less than 4:

Thus 4 dice must be thrown.

19.10.The probability of a man hitting a target is 4. (i) If he fires 7 times, what is the probability P of his hitting the target at least twice? (ii) How many times must he fire so that the probability of his hitting the target at least once is greater than t? (i) We seek the sum of the probabilities for k = 2, 3, 4, 5, 6 and 7 . It is simpler in this case t o

find the sum of the probabilities for k = 0 and 1, i.e. no hits or 1 hit, and then subtract i t from 1. 1187 5103

f ( 7 ,0 ,$ ) = ( $ I i = m, f (7,1,%) = Ci) ($1 ($)6 ~ 16,384

Then p = 1 2187 5101 = 45-17. 16,384 16,384 b lOZ

(ii) The probability of not hitting the target is qn. less than q until q n < Q is obtained:

Thus we seek the smallest n for which qn is Hence compute successive powers of 1 - $ = Q, where q = 1 - p = 1 - $ = $.

(2)l = 3 Q 1. ( $ ) Z = A Q A ; ( 9 ) 3 = !7 Q 1; but ( 9 ) 4 z 81 < I 4 3 ' 1 6 d 6 4 8 256 3

In other words, he must fire 4 times.

19.11.Prove Theorem 19.1: The probability of exactly k successes in n repeated trials is f (n , k , p ) = ([)pk qnPk. The probability of at least one success is 1 - qn.

The sample space of the 7% repeated trials consists of all ordered n-tuples whose components a re either s (success) o r f (failure). The event A o f k successes consists of all ordered n-tuples of which Is components a re s and the other n - k components are f . The number of n-tuples in the event A is equal to the number of ways tha t k letters s can be distributed among the ?I components of an n-tuple; hence A consists of (E) sample points. Since the probability of each point in A is p k qll- k , we have P ( A ) = (:) p k q n - k ,

The event B of no successes consists o f the single n-tuple (f,f, , , .,f), i.e. all failures, whose probability is qn. Accordingly, the probability of at least one success is 1 - qn.

19.12. The mathematics department has 8 graduate assistants who are assigned to the same office. Each assistant is just as likely to study at home as in the office. How many desks must there be in the office so that each assistant has a desk at least 90% of the time?

If there are 8 desks, then 0% of the time an assistant will not have a desk at which t o study.

225 CHAP. 191 INDEPENDENT TRIALS, RANDOM VARIABLES

Suppose theie are 7 desks; then an assistant will not have a desk a t which to study if all 8 assistants are in the office. The probability tha t 8 will study in the office is ( $ ) 8 = & = 0 .45: . Thus there will not be enough desks 0 . d c r of the time.

Suppose there are 6 desks; then an assistant will not have a desk if 8 o r 7 assistants a re in = 3.2%. Thus there the office.

will not be enough desks 3.2% + 0.Ac; = 3.GC; of the time.

Suppose there are 5 desks; then an assistant will not have a desk if 8, 7 o r 6 assistants a re in the office. The probability tha t 6 will study in the office is f ( 8 , 6 , $) = = 10.9%. Thus there will not he enough desks 10.9% + 3.2"r + 0.1G = 14.5% of the time. In other words, there will be enough desks only 85.5% of the time.

Accordingly. 6 desks a re required.

The probability tha t 7 xi11 study in the office is f ( 8 . 7 , i ) =

RANDOM VARIABLES

the

(i)

(ii)

19.13. A pair of fair dice is thrown. Let S be the random variable which assigns to each (i) Find

I distribution of X . point the maximum of the two numbers which occur: X ( ( a , b ) ) = max (a, b).

(ii) Find the expectation of X .

The sample space S is the equiprobable space consisting of the 36 ordered pairs of numbers between 1 and 6: S = {(l, l), (1, Z ) , , . , (6,6)}. Since X assigns to each point in S the maximum of the two numbers in the ordered pair, the range space of X consists of the numbers from 1 through 6: R , = { l , Z , 3 ,4 ,5 ,6} .

Each of three points Each of five points in S ,

P(4) =A, P ( 5 ) = &

There is only one point (1,l) whose maximum is 1; hence P ( 1 ) = & .

Similarly, in S, ( 1 , 2 ) , ( 2 , 2 ) and (2, l), has maximum 2; hence (1 ,3 ) , ( 2 , 3 ) , 13,3), (3 ,2) and (3, l) , has 3; hence P(3) = A . and P ( 6 ) = 1 ~ , .

P(2) = &. 11

The distribution of X consists of the points in Rx with their respective probabilities:

I I I I I

To obtain the expectation of X , i.e. the expected value, multiply each number x, by i ts probability p z and take the sum:

= 4.5 E = 1 . - :i6 + 2 . A - 3 . 2 16 + 4 - $ + 5 - $ + 601 36 - - 36

19.14. Suppose a random variable X takes on the values -3, -1, 2 and 5 with respective

. Determine the distribution and 2 x - 3 x + l x-1 x - 2 and ___ probabilities __ ~ __

expectation of X. 10 10 ' 10 ' 10

Set the sum of the probabilities equal t o 1, and solve to find .2: = 3. Then put x = 3 into the above probabilities and obtain .3, .4, .2 and .1 respectively. The distribution of X is

xi I -3 I -1 2 5 I I I

The expectation is obtained by multiplying each value by its probability

E = (-3)(.3) + (-1)(.4) + 2(.2) + 5(.1) = --.4

and taking the sum:

226 INDEPENDENT TRIALS, RANDOM VARIABLES [CHAP. 19

19.15. Four fair coins are tossed. Let X denote the number of heads occurring. Find the distribution and expectation of the random variable X .

Since this experiment consists of four independent trials, we obtain the binomial distribution nith 7 1 = 4 and p = $:

P(0) = f(4,0,3) = P(3) = 1 ( 4 , 3 , 3 ) = 2L 1 6

P(1) = f ( 4 , 1 , + ) = P ( 4 ) = 1'(4,4,3) = & P(2) = J ( 4 , 2 , 4 ) = &

Thus the distribution is

I I I

and the expectation is Theorem 19.3 which states that the expected value E = ? ~ p = 4 4 = 2.

E = 0 1 + 1 - A + 2 * A + 3 * &. + 4 & = 2. This agrees with 1 6 16 16

19.16. A coin weighted so that P(H) = $ and P ( T ) = & is tossed three times. Let X be the Find the random variable which denotes the longest string of heads which occurs.

distribution and expectation of X . The random variable X is defined on the sample space


The points in S have the following respective probabilities: P(HHH) = 3.8.3 = 27 P(THH) = $ * $ a $ = &

4 4 . 4 6 4

P(THT) = 1.3-1 = 8 4 4 4 6 4

P(HHT) = 3 . 3 . 1 A 4 4 4 6 4

P(HTH) = $ * $ * $ = & P(TTH) = $ * $ a $ = & P(HTT) = $ * $ * $ = & P(TTT) = &.&.& =

Since X denotes the longest string of heads,

X(TTT) = 0; X(HTT) = 1, X(HTH) = 1, X(THT) = 1, X(TTH) 1;

X(HHT) = 2, X(THH) = 2; X(HHH) = 3

Thus the range space of X is R, = {0,1,2,3} . obtained by summing the probabilities of the points in S whose image is xi:

The probability of each number x, in R, is

P ( 0 ) = P(TTT) = & P(1) = P(HTT) + P(HTH) + P(THT) + P(TTH) = P ( 3 ) = P(HHH) =

P(2) = P(HHT) + P(THH) = 8

The distribution of X consists of the numbers in the range space R, with their respective probabilities:

6 4 6 4 6 4 6 4

19.17. Prove Theorem 19.2: Let S be an equiprobable space. range space of a random variable X defined on S , then

If R, = {xl, . . , , xt} is the

number of points in S whose image is xi number of points in S pi = P(x , ) =

Let S have n points and let sl, s2, , . . ,s, be the points in S whose image is x,. We wish to show that P ( x J = r/n. By definition,

CHAP. 191 I N D E P E N D E N T TRIALS, RANDOM VARIABLES

P ( x i ) = sum of the probabilities of the points in S whose image is xi = P(Sl) + P(s2) + . . . + P(s,)

Since S is an equiprobable space, each o l the ?t points in S has probability lln. Hence

r times

19.18. Let X be a random variable defined on a sample space S and with distribution

227

Show that the following are equivalent ways of defining the expectation E ( X ) of the random variable X :

E ( X ) = 2,p1 + * * . + x,p, = c X ( s ) P ( s ) S E S

Let sil, si2, . , . , sini be the points in S with image x i , tha t is, X ( s i j ) = x i .

p , = P ( z J = sum of the probabilities of the points in S whose image is z1

Then,

= P(S1,) + P ( S I * ) + ' . * + P(Slnl) = 3 P(S1j) j = 1

"2

j=1 p , = P ( Z 2 ) = P(S21) + P ( S 2 2 ) + ' ' ' + P(s2n2) = P(s2j)

.............................. . . . ............................ nt

p , = P ( r J = P(S,l) t P ( S r 2 ) + ' a * + P(S,,t) = z P(Stj) j=1

n2 nt

Similarly, Z2Pz = ,z X ( S 2 ; ) P ( S Z , ) , . . ., X t P , = J = l B X(s,,) P(Stj) 3 = 1

Since the slJ range over all the points s E S , we obtain

19.19. Let X and Y be random variables defined on the same sample space S. Let X + Y be the random variable on S defined by

(X + Y ) ( s ) = X ( s ) + Y(s)

Prove that E ( X + Y ) = E ( X ) + E ( Y ) .

Using the result of the preceding problem, we obtain

E(X + Y ) = 2 ( X + Y ) ( s ) P ( s ) = 2 ( X ( s ) + Y ( s ) ) P ( s ) s e s s e S

= 2 ( X ( S ) P ( S ) i Y ( s ) P ( s ) )

= 3 X ( s ) P ( s ) + z Y ( s ) P(s ) = E ( X ) + E ( Y ) s e s

S E S s e S

228 INDEPENDEKT TRIALS, RANDOM VARIABLES [CHAP. 19

EXPECTED VALUES 19.20. A player tosses two fair coins. He wins $2 if 2 heads occur, and $1 if 1 head occurs.

On the other hand, he loses $3 if no heads occur. Determine the expected value of the game and if i t is favorable to the player.

Note that the probability that 2 heads occur is f, that 2 tails occur is f and that 1 head occurs is +. Thus the probability of winning $2 is $, of winning $1 is 4, and of losing $3 is a. Accordingly, E = 2 - & + 1 * 3 - 3 . i = & = .25. In other words, the expected value of the game is 25d in favor of the player.

19.21. A player tosses two fair coins. He wins $3 if 2 heads occur, and $1 if 1 head occurs. If the game is to be fair, how much should he lose if no heads occur?

Let .T denote the amount he should lose if no heads occur. We have t h a t the probability of winning 93 is $, of winning $1 is 3, and of losing $x is a. Since the game is to be fair , set E = 0:

E = 3 . $ + 1 . 1 - - - ~ . 1 = 0 or - - - 5 z - - 0 or x = 5 2 4 4

In other words, the game is fa i r if he loses $5 when no heads occur.

19.22. A player tosses two fair coins. He wins $5 if 2 heads occur, $2 if 1 head occurs and (ii) How much should he pay $1 if no heads occur.

t o play the game if i t is to be f a i r ? (i) Find his expected winnings.

( i ) The probability of winning $ 5 is $, of winning $2 is 6 , and of winning $1 is $; hence E = 5 - t + 2 * 3 + 1 - 4 = 2.50,

If he pays $2.50 to play the game, then the game is fair.

tha t is, the expected winnings a r e $2.50.

(ii)

19.23. A box contains 10 items of which 2 are defective. If four items a re selected from the box, what is the expected number of defective items in the sample of size 4?

There are (io) = 210 different samples of size 4 (without replacement). Since there a re 8 non- defective items in the box, then (:) = 70 different samples contain no defective items. Hence the probability P ( 0 ) of no defective items in the sample is P(0) = = 5 .

(:) = 112 different samples contain 1 defective item: hence P(1) = 2

( B ) = 28 different samples contain the 2 defective items; hence P ( 2 ) =

Thus the expected number of defective items is E = O * ' + 3 l a & + 2m2 15 = 5 *

= &. 28 - 2

- E.

19.24. A fair die is tossed 300 times. Compute the expected number of sixes.

This is an independent trials experiment with n = 300 and p = Q.

Hence by Theorem 19.3, E = n p = 300.Q = 50.

19.25.Determine the expected number of boys in a family with 4 children, assuming the sex distribution to be equally probable. What is the probability that the expected number of boys does occur?

This is an independent (repeated) trials experiment with ?i = 4 and p = 4; hence by Theorem 19.3, E = n p = 4 1 2 = 2. In other words, the family can expect to have 2 boys. The probability t h a t the family has 2 boys is f ( 4 , 2 , 3 ) = 8.

CHAP. 191 INDEPENDENT TRIALS, RANDOM VARIABLES 229

19.26. Prove Theorem 19.3: For the binomial distribution

the expected value is E = np. n 71

E = k ( I ; ) q n - ! i,ib = k=O

n

q " - k p k - 1 = 1 1 p (p + p)!l-l zz n p ( l i - l)!

since q - 11 = 1. 1111 the above summation. n e dropped the term nhen li = 0 since i t is 0 * q" = 0.)

19.27. A fair coin is tossed until a head or 4 tails occurs. Find the expected number of

Two tosses occur if the Three tosses occur if the first t n o are

Four tosses occur if either TTTH or TTTT occurs.

tosses of the coin. Only one toss occurs if heads occurs the first time, i.e. the ererit H.

first is tails and the second is heads, 1.e. the event TH. tails and the third is heads, i.e. the event TTH. Hence P 11 = P I H I = 3, P ( 2 ) = P(TH = $, P 31 = P ( T T H ) = +, and

P(4) = P T T T H I - P T T T T ) = &+A = 8 Thus E = 1 . + + 2 . & + 3 . & + . : . & =

8

19.28.A box contains 8 light bulbs of which 3 are defective. A bulb is selected from the Find the expected number of box and tested until a non-defective bulb is chosen.

bulbs t o be chosen. The probability of choosing a non-defective bulb the first time is P(1) = 8. The probability

of first choosing a defective bulb and then a non-defective one is P ( 2 ) = $.f = $. The probability of first selecting two defective bulbs and then a non-defective one is P ( 3 ) = t.3. f = &. Lastly, the probability of first selecting the three defective bulbs and then a non-defective one is P(4) = 3.2.1.5 , = 1_ 5 6 . Hence

84 E = 1.: L 3 - & + 4.k = 56

-

19.29. Prove Theorem 19.4: Let E be the expected value of an experiment. If a constant h: is added t o each outcome of the experiment, then the expected value E* of the new experiment is E + k.

T ~ , . . . , y r be the outcomes of the original experiment with respective probabilities Let pl,p,, . . , , pr . Note p, + p , + . . . 4 = 1. Thus

E = ( % l + k ) p l - (/l - k ) p , + " ' f ( " t + k ) P t

= xlpl + kp, - . f $ J + kp , + x,p1 + .'.pz + ' ' . + xtp, + k(p1 - 112 - . . ' - Pt )

. . . T a.,P, + kp,

- -

= E + k

230 INDEPENDENT TRIALS, RANDOM VARIABLES [CHAP. 19

Supplementary Problems REPEATED TRIALS WITH TWO OUTCOMES

19.30.

19.31.

19.32.

19.33.

19.34.

19.35.

19.36.

19.37.

A card is drawn and replaced three times from a n ordinary deck of 52 cards. Find the probability that iii two hearts a re drawn, ( i i ) three hearts a re drawn, (iii, a t least one hear t is drawn.

A baseball player’s batting average is .300. that he will get (i) two hits, (ii) a t least one hit?

If he comes to bat 4 times, what is the probability

A box contains 3 red marbles and 2 white marbles. A marble is drawn and replaced three times from the box. Find the probability tha t (i) 1 red marble was drawn, (ii) 2 red marbles were drawn, (iii) a t least one red marble was drawn.

Team A has probability Q of winning whenever i t plays. If A plays 4 games, find the probability tha t A wins ( i ) 2 games, (ii) a t least 1 game, (iii) more than half of the games.

A card is drawn and replaced in an ordinary deck of 52 cards. How many times must a card be drawn so that (i) there is a t least an even chance of drawing a heart, (ii) the probability of drawing a heart is greater than $ ?

The probability of a man hitting a target is 8. hitting the target at least twice? hitting the target at least once is more than 90%.

(i) If he fires 5 times, what is the probability of iii) How many times must he fire so t h a t the probability of

Ten typists are assigned to the same office. Each typist is a s likely to work at home a s in the office. How many typewriters must there be in the office so tha t each typist has a typewriter a t least 90r2 of the time?

RANDOM VARIABLES

19.38.

19.39.

19.40.

19.41.

19.42.

19.43.

19.44.

19.45.

A pair of fa i r dice is thrown. two numbers which occur. Find the distribution and expectation of X .

Let X be the random variable which denotes the minimum of the

Five fa i r coins a r e tossed. Let X be the random variable which denotes the number of heads occurring. Find the distribution and expectation of X.

A fa i r coin is tossed f o u r times. Let X be the random variable which denotes the longest string of heads. Find the distribution and expectation of X .

Solve the preceding problem in the case tha t the coin is weighted so tha t P(H) = 3 and P(T) = +. Two cards a re selected from a box which contains five cards numbered 1 , 1 , 2 , 2 and 3. (i) Let X be the random variable which denotes the sum of the two numbers. Find the distribution and expectation of X . (ii) Let Y be the random variable which denotes the maximum of the two numbers. Find the distribution and expectation of Y .

Refer to the preceding problem. Let X + Y be the random variable defined by ( X + Y ) ( s ) = X ( s ) + Y ( s ) , i.e. which denotes the sum of the two numbers added to the maximum number. For example, ( X + Y)(Z, 3) = 5 A 3 = 8. Find the distribution and expectation of X + Y . Verify that E ( X + Y ) = E ( X ) + E ( Y ) .

A man stands a t the origin. He tosses a coin. If a head occurs, he moves one unit to the right; if a tail occurs he moves one unit t o the left. (i) If X is the random variable which denotes his distance from the origin af ter 4 tosses of the coin, find the distribution and expectation of X . (ii) If Y is the random variable which denotes his position af ter 4 tosses of the coin, find the distribution and expectation of Y .

Solve ( i ) and [ii) of the preceding problem for the case of 5 tosses of the coin.

CHAP. 191

EXPECTED VALUE

INDEPENDENT TRIALS, RANDOM VARIABLES 231

19.46.

19.47.

19.48.

19.49.

19.50.

19.51.

19.52.

19.53.

19.54.

19.55.

19.56.

19.57.

A player tosses three fa i r coins. He wins $5 if 3 heads occur, $3 if 2 heads occur, and $1 if only 1 head occurs. Find the value of the game to the player.

On the other hand, he loses $15 if 3 tails occur.

A player tosses three fa i r coins. He wins $8 if 3 heads occur, $3 if 2 heads occur, and $1 if only 1 head occurs. If the game is to be fair , how much should he lose if no heads occur?

A player tosses three f a i r coins. He wins $8 if 3 heads occur, $3 if 2 heads occur, $1 if 1 head occurs, but nothing if no heads occur. If the game is to be fair , how much should he pay to play the game?

A box contains 8 items of which 2 are defective. A man selects 3 items from the box. Find the expected number of defective items he has drawn.

Of the bolts produced by a factory, 2 5 a r e defective. In a lot of 8000 bolts from the factory, how many are expected t o be defective?

Find the expected number of girls in a family with 6 children, assuming sex distribution t o be equally probable. What is the probability tha t the expected number of girls does occur?

A lottery with 500 tickets gives one prize of $100, 3 prizes of $50 each, and 5 prizes of $25 each. (i) Find the expected winnings of a ticket. fii) If a ticket costs $1, what is the expected value of the game?

A fa i r coin is tossed until a head or five tails occur. Find the expected number of tosses of the coin.

A coin weighted so that P ( H ) = 4 and P ( T ) = 6 is tossed until a head or five tails occurs. the expected number of tosses of the coin.

Find

The probability of team A winning any game is 4. A plays B in a tournament. The first team to win 2 games in a row or a total of three games wins the tournament. Find the expected number of games in the tournament.

A box contains 10 transistors of which 2 are defective. A transistor is selected from the box and tested until a non-defective one is chosen. Find the expected number of transistors to be chosen.

Solve the preceding problem in the case tha t 3 of the 10 items are defective.

Answers to Supplementary Problems 21 27 19.30.

19.31.

19.32. (i) 0.2646, (ii) 0.7599

19.33.

19.34.

19.35. (i) 3, (ii) 5

19.36. (i) g, (ii) 6

19.37. 7

(i) &, (ii) 128, (iii) 1.r~

(i) &, (ii) &, (iii)

54 117 (i) g, (ii) a, (iii)

(i) g, (ii) m, (iii) 544 112

232

19.38.

19.39.

19.40.

19.41.

19.42.

19.43.

19.44.

19.45.

xi c 1 2 3 4

I N D E P E N D E N T TRIALS, RANDOM VARIABLES

5

[CHAP. 19

Pi 5 U l O L - 1 n 32 3 2 3 2 32

xi

Pi

0 1 2 3 4 E ( X ) = $

1 1 3 - 2 - 16 16 1 6 1'6

81 - E(X) = g 8 1 81 81 81

(i) E ( X ) = 3.6 (ii) E ( Y ) = 2.3

x i 3 5 6 7 8 I I I I 1 E ( X + Y ) = 5.9

pi

E ( X ) = 15

.1 .4 .1 .2 .2

19.46. 2%

19.47. $20

19.48. $2.50

19.49.

19.50. 160

19.51. 3,

23

19.52. (i) 75d, ( i i ) - 25@

19.53.

19.54. % 19.55. 23

11 19.56. s

19.57.

Chapter 20

Markov Chains

PROBABILITY VECTORS

non-negative and their sum is 1. A row vector u = (u1,u2, . . .,u,,) is called a probability vector if its components are

Example 1.1 : Consider the following vectors:

Then: 16 = ( 3 0 -1 I) 7 j = (2, 4, 0 1) and zu = (1 1 0 1)

4’ > 4’ ’’ I ’ 4 4 ’ 4 ’ ’ 2

i c is not a probability vector since its third component is negative;

1: is not a probability vector since the sum of its components is greater than 1:

zt‘ is a probability vector.

Example 1.2: The non-zero vector 2’ = (2 ,3 ,5 ,0 ,1) is not a probability vector since the sum of its components is 2 f 3 + 5 - O + 1 = 11. However, since the components of w are non-negative, u has a unique scalar multiple Xu which is a probability vector; it can be obtained from v by dividing each component of v by the sum of the components of v: & G = (z 11’ fi, fl. 0 , E ) . ’ 3 5 1

Remark: Since the sum of the components of a probability vector is one, an arbitrary probability vector with n components can be represented in terms of n - 1 unknowns as follows:

In particular, arbitrary probability vectors with 2 and 3 components can be represented, respectively, in the form

( X l , x2, . . ., x,1-1, 1 - x1 - * . ’ - xn-1)

( x , 1 - x ) and (x, Y, 1 - x - y )

STOCHASTIC AND REGULAR STOCHASTIC MATRICES A square matrix P = ( p i J is called a stochastic matrix if each of its rows is a proba-

bility vector, i.e. if each entry of P is non-negative and the sum of the entries in every row is 1.

Example 2.1 : Consider the following matrices:

(ii) (iii)

( i ) is not a stochastic matrix since the entry in the second row and third column is negative;

is not a stochastic matrix since the sum of the entries in the second row is not 1;

(iii) is a stochastic matrix since each row is a probability vector.

(ii)

233

234 31ARKOV CHAINS [CHAP. 20

We shall prove (Problem 20.7)

Theorem 20.1: If A and B are stochastic matrices, then the product AB is a stochastic matrix. Therefore, in particular, all powers A" are stochastic matrices.

We now define an important class of stochastic matrices whose properties shall be investigated subsequently.

-1 A stochastic matrix P is said to be regular if all the entries of some power Pm are positive.

0 1 Example 2.2: The stochastic matrix A = +) is regular since

is positive in every entry.

1 0

(+ +J Here Example 2.3: Consider the stochastic matrix A =

11 0 1 0

A2 = j4 A3 = ( 3 $1' A4 = (; A) In fact every power A'n will have 1 and 0 in the first row; hence A is not regular.

FIXED POINTS OF SQUARE MATRICES A non-zero row vector u = (ul, u,, . . . , u,) is called a f ixed point of a square matrix A

if u is left fixed, i.e. is not changed,- when multiplied by A :

u A = u We do not include the zero vector 0 as a fixed point of a matrix since it is always left fixed by every matrix A : OA = 0.

Example 3.1: Consider the matrix A = (: :) . Then the vector u = (2, -1) is a fixed point of A . For,

Example 3.2: Suppose u is a fixed point of a matrix A : zcA = u. multiple of u, say hti, is also a fixed point of A .

We claim tha t every scalar For,

(xu)A = k(zcA) = xu

Thus, in particular, the vector 2u = (4, -2) is a fixed point of the matrix A of the example above:

We state the result of the preceding example as a theorem.

Theorem 20.2: If u is a fixed vector of a matrix A , then for any real number h # 0 the scalar multiple XU is also a fixed vector of A .

CHAP. 201 MARKOV CHAINS 235

FIXED POINTS AND REGULAR STOCHASTIC MATRICES The main relationship between regular stochastic matrices and fixed points is contained

in the next theorem

Theorem 20.3: Let

(i)

(ii)

(iii)

whose proof lies beyond the scope of this text.

P be a regular stochastic matrix. Then:

P has a unique fixed probability vector t , and the components of t are all positive;

the sequence P, P2, P 3 , . . . of powers of P approaches the matrix T whose rows are each the fixed point t ; if p is any probability vector, then the sequence of vectors pP,pP2, pP3, . . . approaches the fixed point t.

Note: P" approaches T means that each entry of Prl approaches the corresponding entry of T, and pPn approaches t means that each component of pPn approaches the corresponding component of t.

Example 4.1: Consider the regular stochastic matrix P = ( +). We seek a probability 0 1

vector t = (5, 1 - 2) such that tP = t

Multiplying the left side of the above matrix equation, we obtain

Thus By Theorem 20.3, the sequence P , P 2 , P 3 , . . . approaches the matrix T whose rows are each the vector t:

t = (4, 1 - 4) = (Q, +) is the unique fixed probability vector of P.

T = (i :) = ( *33 *67) .33 .67

We exhibit some of the powers of P to indicate the above result:

Example 4.2: Find the unique fixed probability vector of the regular stochastic matrix

0 1 0

Method 1. We seek a probability vector t = (2, Y, 1 - 2 - y) such that tP = t :

= ( x , y , l - z - y )

Multiplying the left side of the above matrix equation and then setting corresponding components equal to each other, we obtain the system

236 MARKOV CHAINS [CHAP. 20

- 2 1 -lz-l, 2 2 Y = x 3 x + y = 1

Z + ~ - + Z - & = y or z - 3 ~ = -1 or

y = 1 - x - y z + 2 y = 1

2 = 1 5

y = g

Thus t = (&,$, 8) is the unique fixed probability vector of P.

Method 2. We first seek any fixed vector u = (2, y,z) of the matrix P:

$2 = x Z + L Z = y

2

u = z

0 1 0

= (z, Y, 2 ) or

We know tha t the system has a non-zero solution; hence we can arbitrarily assign a value to one of the unknowns. Set z = 2. Then by the first equation x = 1, and by the third equation y = 2. Thus U. = (1 ,2 ,2) is a fixed point of P . But every multiple of u is a fixed point of P ; hence multiply u by 6 to obtain the required fixed probability vector t = QZL = (i,?, $).

MARKOV CHAINS

following two properties:

(i)

We now consider a sequence of trials whose outcomes, say, X , , X,, . . . , satisfy the

Each outcome belongs to a finite set of outcomes {a,, u2, . . . , am} called the state space of the system; if the outcome on the nth trial is al , then we say that the system is in state a? at time n or at the nth step.

(ii) The outcome of any trial depends at most upon the outcome of the immediately preceding trial and not upon any other previous outcome; with each pair of states (a,,aj) there is given the probability p i j that uj occurs immediately after at occurs.

The numbers p, , , called the Such a stochastic process is called a (finite) Alarkov chain. transifion probabilities, can be arranged in a matrix

P12 * *

P = . . . . . . . . . . . . . . . . . .

called the transition matrix.

Thus with each state al there corresponds the ith row (pi,, pi2 , . . . , pl,J of the transition matrix P; if the system is in state ui, then this row vector represents the probabilities of all the possible outcomes of the next trial and so it is a probability vector. Accordingly,

Theorem 20.4: The transition matrix P of a Markov chain is a stochastic matrix.

Example 5.1: A man either drives his car or takes a train to work each day. Suppose he never takes the train two days in a row; but if he drives to work, then the next day he is just as likely t o drive again as he is t o take the train.

This stochastic process is a Markov chain since the outcome on any day depends only on what happened the preceding day. The transition matrix of the Markov chain is

t d

The state space of the system is { t (train), d (drive)}.

CHAP. 20) SIARKOV CHAINS 237

The first row of the matrix corresponds to the fact tha t he never takes the t ra in two days in a row and so he definitely will drive the day after he takes the train. The second row of the matrix corresponds t o the fact that the day af ter he drives he will drive o r take the train with equal probability.

Three boys A , B and C are throwing a ball t o each other. A always throws the ball to B and B always throws the ball to C; but C is just as likely to throw the ball to B as to A . Let X, denote the n th person to be thrown the ball. The s ta te space of the system is ( A , B , C}. This is a Markov chain since the person throwing the ball is not influenced by those who previously had the ball. The transition matrix of the Markov chain is

Example 5.2:

A B C

B

The first row of the matrix corresponds to the fact that A always throws the ball to B. The second row corresponds t o the fact that B always throws the ball to C. The last row corresponds to the fact that c throws the ball t o A or B with equal probability (and does not throw it to himself).

A school contains 200 boys and 150 girls. One student is selected after another to take a n eye examination. Let X, denote the sex of the n th student who takes the examination. The state space of the stochastic process is {m (male), f (female)}. However, this process is not a Markov chain since, for example, the probability t h a t the third person is a girl depends not only on the outcome of the second trial but on both the first and second trials.

(Random waIk with reflecting barriers.) A man is at an integral point on the x-axis between the origin 0 and, say, the point 5. He takes a unit step to the r ight with probability p o r to the left with probability g = 1 - p , unless he is at the origin where he takes a step to the r ight to 1 o r at the point 5 where he takes a step t o the left to 4. Let X, denote his position af ter n steps. This is a Markov chain with state space {uo, a,, a,, a,, a,, a,) where a, means t h a t the man is a t the point i.

Example 5.3:

Example 5.4:

The transition matrix is a0 a1 a2 a3 a4 a5

Each row of the matrix, except the first and last, corresponds to the fact that the man moves from state ai to s ta te a,+l with probability p or back to state ai-l with probability q = 1 - p . The first row corresponds to the fact tha t the man must move from state a. to s ta te al, and the las t row that the nian must move from state a5 to s ta te a,.

HIGHER TRANSITION PROBABILITIES The entry p l j in the transition matrix P of a Markov chain is the probability that the

system changes from the state a[ to the state a, in one step: a, -+ acl. Question: What is the probability, denoted by pi;), that the system changes from the state at t o the state al in exactly n steps:

The next theorem answers this question; here the pi;) are arranged in a matrix the n-step transition matrix:

U, + a k , + * * * + ak,-l + a,

called

Theorem 20.5: Let P be the transition matrix of a Markov chain process. n-step transition matrix is equal to the nth power of P: P(") = P".

Then the

Now suppose that, a t some arbitrary time, the probability that the system is in state ai is 11, ; we denote these probabilities by the probability vector p = (p , , p2, . . . , p,,,) which is called the probability distribution of the system a t that time. In particular, we shall let

p'0' = (@', p y , . . .) p:;))

P ( l l ) = ( P y , P?), * * . > P,, )

denote the initial probability distribution, i.e. the distribution when the process begins, and we shall let

denote the nth s t ep probability distribution, i.e. the distribution after the first n steps. The following theorem applies.

( n )

Theorem 20.6: Let P be the transition matrix of a Markov chain process. If p = ( p J is the probability distribution of the system a t some arbitrary time, then pP is the probability distribution of the system one step later and pPn is the probability distribution of the system n steps later. In particular,

p ( n ) = p ( 0 ) p n * ( I ) = p<o'p p) = p"'p @ 3 ) = p(2'p, . . . ) f 9

Example 6.1 : Consider the Markov chain of Example 5.1 whose transition matrix is

t d

Here t is the s ta te of taking a t ra in to work and d of driving t o work. ample 4.1,

By Ex-

Thus the probability tha t the system changes from, say, state t t o state d in exactly 4 steps is 8, i.e. p $ ) = #.

Now suppose that on the first day of work, the man tossed a f a i r die and drove to work if and only if a 6 appeared. In other words, p(O) = (&&) is the initial arobabilitv distribution. Then

Similarly, pj : ) = #, = and p(4) = 11. dd 1 6

is the probability distribution af ter 4 days, i.e. piq) = and p ( 4 ) = d 96'

Example 6.2: Consider the Markov chain of Example 5.2 whose transition matrix is

A B C

Suppose C was the first person with the ball, i.e. suppose p ( 0 ) = ( O , O , 1) is the initial probability distribution. Then

0 1 0 p"' = p'0'P = (0,0,1)


Thus, af ter three throws, the probability tha t A has the ball is &, t h a t B has the ball is $ and tha t C has the ball is $: pa31 = $, p:) = 1 and p y ) = fr. Consider the random walk problem of Example 5.4. Suppose the man began a t the point 2; find the probability distribution af ter 3 steps and af ter 4 steps, i.e. ~ ( 3 )

and pC4).

Example 6.3:

Now p ( 0 ) = ( O , O , 1, 0, 0,O) is the initial probability distribution. Then

p“’ = p(0’P = (0 , q , 0, p , 0 , 0 ) p‘2’ = p“’P = (q! , 0 , 2pq, 0, p2, 0 ) p‘3’ = p(”P = (0 , q L + 2p2q, 0, 3p2q, 0, p3)

p(4) = p ( 3 ) ~ = ( a 3 - 2 p q 2 , 0 , pq2 + 2p3q + 3 ~ 2 q 2 , 0 , 3 ~ 3 q + 9 , o ) Thus af ter 4 steps he is a t , say, the origin with probability q3 + 2p2q2.

STATIONARY DISTRIBUTION OF REGULAR MARKOV CHAINS Suppose that a Markov chain is regular, i.e. that i ts transition matrix P is regular.

By Theorem 20.3, the sequence of n-step transition matrices Pn approaches the matrix T whose rows are each the unique fixed probability vector t of P ; hence the probability p:;) that u3 occurs for sufficiently large n is independent of the original state a, and it approaches the component t j of t . In other words,

Theorem 20.7: Let the transition matrix P of a Markov chain be regular. Then, in the long run, the probability that any state u3 occurs is approximately equal to the component t3 of the unique fixed probability vector t of P .

Thus we see that the effect of the initial state or the initial probability distribution of the process wears off as the number of steps of the process increase. Furthermore, every sequence of probability distributions approaches the fixed probability vector t of P , called the stationary distribution of the Markov chain.

Example 7.1: Consider the Markov chain process of Example 5.1 whose transition matrix is t d

t 0 1

P = d (& 4) By Example 4.1, the unique fixed probability vector of the above matrix is (4, g ) . Thus, in the long run, the man will take the train to work 4 of the time, and drive to work the other 8 of the time.

Consider the Markov chain process of Example 5.2 whose transition matrix is Example 7.2: A B C

P = B

By Example 4.2, the unique fixed probability vector of the above matrix is (&, 3, g). Thus, in the long run, A will be thrown the ball 20% of the time, and B and C 40% of the time.

240 I IARKOV CHAINS [CHAP. 20

ABSORBING STATES A state ut of a Markov chain is called absorbing if the system remains in the state ai

once it enters there. Thus a state az is absorbing if and only if the ith row of the transition matrix P has a 1 on the main diagonal and zeros everywhere else.

Example 8.1:

Example 8.2:

Example 8.3:

Example 8.4:

Suppose the following matrix is the transition matrix of a Markov chain:

a1 a2 a3 a4 a5

The states a? and a5 are each absorbing, since each of the second and fifth rows has a 1 on the main diagonal.

(Random walk with absorbing barriers.) Consider the random walk problem of Example 5.4, except now we assume that the man remains at either endpoint whenever he reaches there. This is also a RIarkov chain and the transition matrix is given by

an ai a, a3 a4 a5

a1 q o p o o o ““I‘ O O O O O\

” I a5 a4 \ 0 0 0 0 0 1

o o o q o

We call this process a random walk with absorbing barriers, since the a. and In this case, phn) denotes the probability that the man

Similarly, p p ) denotes the proba- a5 a r e absorbing states. reaches the s ta te a,, on o r before the izth step. bility tha t he reaches the a5 on o r before the nth step.

A player has, say, z dollars. H e bets one dollar a t a time and wins with probability p and loses with probability q = 1 - p . The game ends when he loses all his money, i.e. has 0 dollars, or when he wins 3 - x dollars, i.e. has N dollars. This game is identical to the random walk of the preceding example except tha t here the absorbing barriers are at 0 and N .

A man tosses a fa i r coin until 3 heads occur in a row. Let X , = k if, a t the itth trial, the las t tail occurred at the (n - k)-th trial, i.e. X , denotes the longest string of heads ending a t the nth trial. This is a RIarkov chain process with state space {ao, al, a2, as}, where a, means the string of heads has length i. The transition matrix is

a0 a1 a2 a3

a, + + o o

u1 a2 ;i f ;) a3

Each row, except the last, corresponds to the fact t h a t a string of heads is either broken if a tail occurs or is extended by one if a head occurs. The las t line corresponds t o the fact tha t the game ends if three heads a r e tossed in a row. Note tha t a3 is an absorbing state.


Let ui be an absorbing state of a hlarkov chain with transition matrix P. Then, for i # j , the n-step transition probability p:;) = 0 for every n. Accordingly, every power of P has a zero entry and so P is not regular.

Theorem 20.8: If a stochastic matrix P has a 1 on the main diagonal, then P is not regular (unless P is a 1 x 1 matrix).

Thus:

Solved Problems PROBABILITY VECTORS AND STOCHASTIC MATRICES 20.1. Which vectors are probability vectors ?

(i) u = (&, 0, -&, 4, Q), (ii) v = (4, 0, f t , +, h), (iii) w = (4, 0, 0 , +, 4). A vector is a probability vector if its components a r e non-negative and their sum is 1.

(i) ZL is not a probability vector since its third component is negative.

(ii) v is not a probability vector since the sum of the components is greater than 1. (iii) PL' is a probability vector since its components a r e non-negative and their sum is 1.

20.2. MuItiply each vector by the appropriate scalar to form a probability vector: (i) (2, 1, 0, 2, 3), (i) The sum of the components is 2 - 1 + 0 + 3 + 2 = 8; hence multiply the vector, i.e. each

component, by & to obtain the probability vector (t, Q, 0, $, 8). (ii) The sum of the components is 4 + 0 + 1 + 2 + 0 + 5 = 12; hence multiply the vector, i.e. each

component, by to obtain the probability vector (g, 0, &, Q, 0 , A). (iii) The first component is positive and the third is negative; hence i t is impossible to multiply the

vector by a scalar to form a vector with non-negative components. Thus no scalar multiple of the vector is a probability vector.

(iv) Every scalar multiple of the zero vector is the zero vector whose components add to 0. Thus no multiple of the zero vector is a probability vector.

(ii) (4, 0, 1, 2, 0, 5), (iii) (3, 0, -2, l), (iv) (0, 0, 0, 0, 0).

20.3. Find a multiple of each vector which is a probability vector: (i) (g, $, 0, 2, g), (ii) (0, 3, 1, Q , $).

In each case, first multiply each vector by a scalar so that the fractions are eliminated.

Firs t multiply the vector by 6 to obtain (3 ,4 ,0 ,12 ,5) . Then multiply by 1/(3 + 4 + 0 + 12 + 5) = & t o obtain (&,&,O,$,&) which is a probability vector.

Firs t multiply the vector by 30 t o obtain (0,20,30,18,25). Then multiply by 1/(0 -t 20 + 30 + 18 + 25) = & t o obtain (0,~' 3. 93'93) which is a probability vector.

(i)

(ii) 20 30 18 25

20.4. Which of the following matrices are stochastic matrices?

(i) A is not a stochastic matrix since it is not a square matrix.

(ii) B is not a stochastic matrix since the sum of the components in the last row is greater than 1. (iii) C is a stochastic matrix.

(iv) D is not a stochastic matrix since the entry in the first row, second column is negative.


20.5. Let A = az bz c, be a stochastic matrix and let u = ( U I , U P , U ~ ) be a proba- (1: 1: 1:) bility vector. Show that uA is also a probability vector.

b , c1

ZLA = (u1, u g , 243) (nt i' :I\ = (ulal + u2a2 + u3a3, u l b , + u2bZ + ~ $ 3 , ulcl+ u z c ~ + ~ b 3 ~ 3 )

Since the ui, a,, bi and cl a re non-negative and since the products and sums of non-negative numbers a re non-negative, the components of wl are non-negative a s required. Thus we only need t o show tha t the sum of the components of uA is 1. Here we use the fac t tha t z i 1 + u 2 + u 3 ,

a, + b , + cl , u2 + b, + c2 and u3 + b, + c3 are each 1:

~ 1 ~ 1 + ~ 2 ~ 2 + 2 ~ 3 ~ 3 + ~ ~ l b l + uzb, + ~ 3 b 3 + ~ 1 ~ 1 + ~ 2 ~ 2 + ~ 3 ~ 3

= z l l ( ~ 1 + b , + ~ 1 ) + + b, + c,) + Z L ~ ( U ~ + b 3 + ~ 3 )

= Z l l * l + u 2 . l + u , * l = u1 + u 2 + UQ = 1

20.6. Prove: If A = (aij) is a stochastic matrix of order n and u = (ul,u,, , . . , uJ is a probability vector, then uA is also a probability vector.

The proof is similar to t h a t of the preceding problem for the case n = 3 :

a12 . . ' [::: :1 . . * ::I . . . . . . . . . . . . . . . . . ZLA = ( ~ 1 , ~ 2 , . . ., u,)

ann . . . an1

= (ulall + uza21 + . - * + unanl, ula12 + u2a2, + * + * + unaR2, . . . , ulaln + uZa2,, + * - 9 + unarm)

Since the ui and aZj are non-negative, the components of uA are also non-negative. Thus we only need to show t h a t the sum of the components of uA is 1:

~~u~~ + z ~ ~ u ~ ~ + . + ununl + u1a12 + u2a22 + * - + u,an2 + * + + ulaln + uZazn + * * * + u,a,, - - u,(all+ a12 + * * * + Ul,) + u,(azl + a22 + * * * + a,,) + . ' ' + un(anl + a,, + * * * + a,,)

2 L 1 ' 1 + u2. l + * ' . + U n ' l = u,+u,+ ' . . +u, = 1 - -

20.7. Prove Theorem 20.1: If A and B are stochastic matrices, then the product A B is a stochastic matrix. Therefore, in particular, all powers An are stochastic matrices.

The ith-row si of the product matrix AB is obtained by multiplying the ith-row ri of A by Since each ri is a probability vector and B is a stochastic matrix, by the matrix B:

the preceding problem, si is also a probability vector. Hence A B is a stochastic matrix. si = riB.

20.8. Prove: Let p = (p l , p2 , . . .,p,) be a probability vector, and let T be a matrix whose rows are each the same vector t = ( t l , t , , . . . , tRJ. Then pT = t .

Using the fact tha t p1 + p2 + . . . + pm = 1, we have

tm\ i t l t , * . *


REGULAR STOCHASTIC MATRICES AND FIXED PROBABILITY VECTORS

20.9. Find the unique fixed probability vector of the regular stochastic matrix A = What matrix does An approach? 2 2

We seek a probability vector f = (Y, 1 - 2) such tha t tA = f:

Multiply the left side of the above matrix equation and then set corresponding components equal to each other to obtain the two equations

$ x + + - + z = 2, $ Z + & - f r x = 1 - 2

Solve either equation to obtain x = Q . Thus t = (3 , +) is the required probability vector.

Check the answer by computing the product tA:

The answer checks since tA = t . The matrix An approaches the matrix T whose rows a r e each the fixed point t : T = ( f i) .

z

20.10. (i) Show that the vector u = ( b , a ) is a fixed point of the general 2 x 2 stochastic

matrix P = ('I" l a b ) .

(ii) Use the result of (i) to find the unique fixed probability vector of each of the following matrices:

1--a (i) UP = ( " a ) ( a ) = ( b - a b + a b , a b f a - a b ) = ( b , a ) = u.

1 - b

(ii) By (i), z t = (1,i) is a fixed point of A . Multiply u by 3 to obtain the fixed point (3,2) of A which has no fractions. Then multiply (3 ,2 ) by 1/(3 + 2) = 6 to obtain the required unique fixed probability vector (#, Q) .

By (i), u = ($, 4) is a fixed point of B. Multiply u by 6 to obtain the fixed point (4 ,3 ) , and then multiply by 1 / ( 4 + 3 ) = 3 t o obtain the required unique fixed probability vector

Hence (8,3) and the probability vector ( A , 6) cq, $). By (i), ZL = (.8, .3) is a fixed point of C.

a re also fixed points of C.

20.11. Find the unique fixed probability vector of the regular stochastic matrix

P = 0 1 0

Method 1. We seek a probability vector t = (2, y, 1 - x - y) such tha t tP = t:

= (Z,Y,1- - -Y)

0 1 0


R L

R .8 .2

L ( . 6 A ) P =

The probability distribution for the first t r ia l is p = ( . 5 , - 5 ) . To compute the probability distribution for the next step, i.e. the second trial, multiply p by the transition matrix P:

( .5 , .5) (*8 *2) = (.7, .3) .6 .4

Thus on the second trial he predicts that 70% of the mice will go right and 30% will go left. To compute the probability distribution for the third trial, multiply that of the second trial by P :

(.7, .3) ( *8 ‘2) = (.74, .26) .6 .4

Thus on the third trial he predicts that 74% of the mice will go right and 26% will go left.

We assume t h a t the probability distribution for the thousandth t r ia l is essentially the stationary probability distribution of the Markov chain, i.e. the unique fixed probability vector t of the transition matrix P . By Problem 20.10, u = (.6, .2) is a fixed point of P and SO t = ($,$) = (.75,.25). Thus he predicts that, on the thousandth trial, 75% of the mice will go to the right and 25% will go to the left.

1 0 20.17. Given the transition matrix P = (~ *) with initial probability distribution

p(O) = (i,+). Define and find: (i) pi:), (ii) pC3’ , (iii) p i 3 ) .

(i) pa:) is the probability of moving from state u2 to state a l in 3 steps. It can be obtained from the 3-step transition matrix P3; hence first compute P3:

p 2 =(; l ) J p 3 =(; ;) Then p;) is the entry in the second row first column of P3: p:) = 2. 8

(ii) p ( 3 ) is the probability distribution of the system af ter three steps. It can be obtained by successively computing p ( 1 ) , p ( 2 ) and then ~ ( 3 ) :

However, since the 3-step transition matrix P3 has already been computed in (i), p ( 3 ) can also be obtained as follows:

(iii) p r ) is the probability tha t the process is in the s ta te a2 af ter 3 steps; it is the second component of the 3-step probability distribution ~ ( 3 ) : p?) = L 12‘


20.18.

20.19.

0 ” 2

Given the transition matrix P = (4 4 i) and the initial probability distribution

p(O) = (3 , O,&) . Find: (i) and p t i ) , (ii) p(4) and p?), (iii) the vector that p(O) P” approaches, (iv) the matrix that Pi* approaches.

0 1 0

(i) Firs t compute the 2-step transition matrix P2:

0 1 1 0 1 1 - - 1 3

pa = ‘. ; ;) (: f ”) = (i i - - 0 1 0 0 1 0 a 2

Then p::) = 4 and pi:) = 0, since these numbers refer to the entries in P2.

To compute ~ ( ~ 1 , use the 2-step transition matrix P2 and the initial probability distribution (ii) p(0J:

p ( 2 ) p(0)p2 = (1 2 0 ) and p(4) = p(2)Pz = (1 1 - 3 ’ 3 ’ 4 ’ 12’ A)

Since pi41 is the third component of p(4), p;) = Q.

(iii) By Theorem 20.3, p(o)Pn approaches the unique fixed probability vector t of P. To obtain t , first find any fixed vector u = (x, y, z ) :

3y = x + x + + y + z = y (%2/* 4 (O 4 ’ 3 ”\ = (x, y, z ) or

0 1 0

Find any non-zero solution of the above system of equations. Set z = 1; then by the third equation x = 2, and by the first equation y = 4. Thus u = (2,4,1) is a fixed point of P and so t = (+,+,+).

(iv) Pn approaches the matrix T whose rows are each the fixed probability vector of P ; hence

Pn approaches (!

In other words, p(O)Pn approaches ($,$,$).

2 4 r I 1 7

$ . - - _ I 7

A salesman’s territory consists of three cities, A , B and C. He never sells in the same city on successive days. If he sells in city A , then the next day he sells in city B. However, if he sells in either B or C, then the next day he is twice as likely to sell in city A as in the other city. In the long run, how often does he sell in each of the cities?

The transition matrix of the problem is as follows: A B C

A 0 1 0

p = * V ; )

(2, v , 4 (E 1 $ = (% Y, 4 01

c z 3

We seek the unique fixed probability vector t of the matrix P. Firs t find any fixed vector u = (x ,y , z ) :

0 1 0 $y + 3 2 = 2 s + $ z = y

ly = z

Set, say, z = 1. u = ( ~ , 3 , 1 ) . obtain the required fixed probability vector run he sells 407;. of the time in city A , 45% of the time in B and 15% of the time in C.

Then by the third equation y=3, and by the first equation x =+. Thus Also 3u = (8,9,3) is a fixed vector of P . Multiply 32~ by 1/(8+ 9 + 3) = $ to

Thus in the long

8

t = (z 5’ 20’ ”) 20 = (.40, .45, J5) .


20.20. There are 2 white marbles in urn A and 3 red marbles in urn B . At each step of the process a marble is selected from each urn and the two marbles selected are interchanged. Let the state a, of the system be the number i of red marbles in urn A . (i) Find the transition matrix P. (ii) What is the probability that there are 2 red marbles in urn A after 3 steps? (iii) In the long run, what is the probability that there are 2 red marbles in urn A ? ( i ) There a re three states, ao, a, and a2 described by the following diagrams:

A B A B A B a0 a1 a2

If the system is in state a,, then a white marble must be selected from urn A and a red marble from urn B , so the system must move to state a,. Accordingly, the first row of the transition matrix is (0,1,0).

Suppose the system is in state a,. It can move to state a. if and only if a red marble is selected from urn A and a white marble from urn B; the probability of tha t happening is 1. 1 = 1. The system can move from state al to a, if and only if a white marble is selected from urn A and a red marble from urn B ; the probability of tha t happening is 1 . 2 2 3 = 1. a Accordingly, the probability t h a t the system remains in s ta te a , is p , , = 1 - Q - Q = 4. Thus the second row of the transition matrix is (Q, 4,g). (Note tha t p , , can also be obtained from the fac t t h a t the system remains in the state a , if either a white marble is drawn from each urn, probability 4 - i = Q, o r a red marble is drawn from each urn, probability 4 * -g = 4; thus p11 = & f 4 = 4.)

Now suppose the system is in state a2. A red marble must be drawn from urn A . If a red marble is selected from urn B , probability Q, then the system remains in state a,; and if a white marble is selected from urn B , probability $, then the system moves to state al. Note that the system can never move from state a2 t o the state uo. Thus the third row of the transition matrix is (0, +, i).

Thus p l o = Q. 2 3 6

Thus p12 = 4.

That is, a, a1 a2

a0 1

a1 a2 (; 3 3 ;) P = - -

(ii) The system began in state ao, i.e. p(O) = (l,O,O). Thus:

p"' = pco'p = (0 , 1, O), p ( 2 ) = pcup = (1 6 , ~, 1 g), p'3' = p(2)P = (L 12' 36' ") 18

5 Accordingly, the probability tha t there a re 2 red marbles in urn A af te r 3 steps is g.

(iii, We seek the unique fixed probability vector t of the transition matrix P . Firs t find any fixed vector u = (x,y,z):

Qy = x x + & y + Q z = y

& y + Q z = z

0 1 0

(5, Y, 2) Q Q 4 = (x, Y, 4 or (o g J

Set, say, x = l . Then by the first equation y = 6 , and by the third equation x = 3 . Hence zi = (1,6,3). Multiply u by 1/(1+ 6 + 3) = & to obtain the required unique fixed probability vector t = (.1, .6, .3). Thus, in the long run, 30% of the time there will be 2 red marbles in urn A .

Note that the long run probability distribution is the same as if the five marbles were placed in a n urn and 2 were selected at random to put into urn A .

20.21. A player has $2. He bets $1 at a time and wins with probability 8. He stops playing (i) m'hat is the probability that he has lost his money

(ii) What is the probability that the game lasts if he loses the $2 or wins $4. at the end of, a t most, 5 plays? more than 7 plays?


This is a random walk with absorbing barriers at 0 and 6 (see Examples 8.2 and 8.3). The transition matrix is

a2 a3 a4 u5 a6

1 0 0 0 0 0 0 \ :: / 4 0 * 0 0 0 0 ‘ a , o g o + o o o ’ ~ 3 0 0 3 0 + o o i

as o o o o + o + ,

I

a 4 0 0 0 + O + 0

a 6 0 0 0 0 0 0 1 /

p‘4’ = p‘3’P = (#, 0 , A, 0, *, 0, A) p‘5’ = p‘4’P = (#, A, 0, &, 0 1 L)

I P =

with initial probability distribution p ( 0 ) = (0, 0 , 1, O , O , 0,O) since he began with $2.

(i) We seek p;5), the probability that the system is in s ta te a. af ter five steps. 5th step probability distribution ~ ( 5 ) :

Compute the

p“’ = p‘0’P = (0, +, 0, 4, 0, 0, 0 )

p ‘ 2 ) = p“’P = (&O, *, 0, *, 0, 0 )

p‘3’ = p‘2’P = (*, 4, 0, #, 0, &, 0) ’ 8’ 16

Thus p i s ) , the probability that he has no money af te r 5 plays, is 8. (ii) Compute p ( 7 ) : p ( 6 ) = p ( 5 ) p = (29, 0, A, 0 13 0 1) p ( 7 ) p ( 6 ) P = (2 2 0 27 0 13 i)

6 4 ’ 6 4 ’ ’ 8 ’ 6 4 ’ 6 4 ’ ’ 128’ ’ 128’ 6

The probability that the game lasts more than 7 plays, i.e. tha t the system i s not in state a, or a6 af ter 7 steps, i s & + + = g.

20.22. Consider repeated tosses of a fair die. Let Xn be the maximum of the numbers occurring in the first n trials.

(i) Find the transition matrix P of the Markov chain. Is the matrix regular? (ii) Find p C 1 ) , the probability distribution after the first toss. (iii) Find p ( z ) and p C 3 ) .

(i) The state space of the Markov chain is { l , 2,3,4,6,6}. The transition matrix is 1 2 3 4 5 6

P = o o o + g g

0 0 0 0 g Q

We obtain, for example, the third row of the matrix as follows. Suppose the system is in state 3, i.e. the maximum of the numbers occurring on the first n t r ia ls is 3. Then the system remains in state 3 if a 1, 2, o r 3 occurs on the (n + 1)-st trial; hence p33 = 6. On the other hand, the system moves to s ta te 4 , 5 or 6, respectively, if a 4 , 5 o r 6 occurs on the (n 4- 1)-st trial; hence p34 = p,, = pS6 = Q. The system can never move to state 1 or 2 since a 3 h a s occurred on one of the trials; hence p31 = p32 = 0. Thus the third row of the transition matrix is (0, O,:, i, i, i). The other rows are obtained similarly.

in row 6.

On the first toss of the die, the state of the system XI is the number occurring; hence

3

The matrix is not regular since s ta te 6 is absorbing, i.e. there is a 1 on the main diagonal

(ii) p“’ = (Q, Q, Q, Q, Q, 9).

(iii) #2’ = p“’ p = (2- 3 s I 2 “)* p ‘ 3 ) = p(2’ p = (L 1 19 37 61 91). 369 361 36’ 36, 3 6 ’ 36 216’ 216’ 216’ 216’ 216’ 216


20.23. Two boys b , and b, and two girls gl and g, are throwing a ball from one to the other. Each boy throws the ball t o the other boy with probability 8 and to each girl with probability 4. On the other hand, each girl throws the ball to each boy with probability 4 and never to the other girl. In the long run, how often does each receive the ball?

This is a Markov chain with state space {b l , b,, g,, g 2 } and transition matrix

bl b2 91 9 2

b, 0 8 ' 4 h

P = g1 b2 [; 1 ;) gz t z

We seek a fixed vector u = (2, u, z , w) of P: (2, y, x , w)P = (2, u, x , w). Set the corresponding components of UP equal to u t o obtain the system

+ y + + z + a w 2 = 5

+ 5 + + z + $ w = u h x + & y = x

$ x + & y = w

We seek any non-zero solution. Set, say, x = 1; then w = 1, 5 = 2 and y = 2. and so the unique fixed probability of P is t = ($,Q,&,&). receives the ball Q of the time and each girl Q of the time.

Thus u = (2 ,2 ,1 ,1) Thus, in the long run, each boy

20.24. Prove Theorem 20.6: Let P = (p, ,) be the transition matrix of a Markov chain. If p = ( p J is the probability distribution of the system at some arbitrary time k , then pP is the probability distribution of the system one step later, i.e. at time k + 1; hence pPn is the probability distribution of the system n steps later, i.e. at time k + n. In particular, pc1) = p(O)P, p C 2 ) = p ( l ) P , . . . and also pen) = p(0 )PR.

Suppose the state space is {a l , u2, . . .,am}. The probability that the system is in state a, a t time k and then in state a, a t time k + 1 is the product pjp, , . Thus the probability that the system is in state at a t time k f 1 is the sum

m

P l P l t + ~ 2 ~ 2 1 + + Pmpmi = jg, P ~ P , ,

Thus the probabiilty distribution at time k + 1 is m m \

P*

However, this vector is precisely the product o f the vector p = (pi) by the matrix P = (p i j ) : p* = pP.

20.25. Prove Theorem 20.5: Let P be the transition matrix of a Markov chain. Then the

Suppose the system is in state ai at, say, time k. We seek the probability 13:;) that the system is in state aj a t time k + n . Now the probability distribution of the system at time k , since the system is in state ai, is the vector ei = (0, . . ., 0,1,0, . . ., 0) which has a 1 at the ith position and zeros everywhere else. By the preceding problem, the probability distribution a t time k + n . is the product eiPn. But ei Pn is the ith row of the matrix Pll. Thus P!?) i s the jth component of the ith row of Pn, and so P(n) = Pn.

n-step transition matrix is equal to the nth power of P: Pn) = Pn.

MISCELLANEOUS PROBLEMS 20.26. The transition probabilities of a Markov chain can be represented by a diagram,

called a transi t ion diagram, where a positive probability p i j is denoted by an arrow from the state ui to the state a j . Find the transition matrix of each of the following transition diagrams:


(i) Note first t h a t the s ta te space is {al, az, u3$ and so the transition matrix is of the form a1 a2 a3

P = a3

The ith row of the matrix is obtained by finding those arrows which emanate f rom a, in the diagram; the number attached to the arrow from a, to aJ is the j t h component of the ith row. Thus the transition matrix is

a, a2 a3

P =

(ii) The state space is {a1,a2,a3,a4}. The transition matrix is

a1 a2 a3 a4

P = a3

20.27. Suppose the transition matrix of a Markov chain is as follows: a1 a2 a3 a4

0 0

a4

Is the Markov chain regular? Note tha t once the system enters the state a , or the state a2, then i t can never move to

s ta te a3 or state a4, i.e. the system remains in the s ta te subspace {a l ,uZ} . Thus, in particular, pi;) = 0 for every n and so every power Pn will contain a zero entry. Hence P is not regular.

20.28. Prove: An rn x m matrix A = (a lJ ) has a non-zero fixed point if and only if the determinant of the matrix A - I is zero,

u = (q, xz, . . . , x m ) such t h a t

Here I is the identity matrix. The matrix A = (aij) has a non-zero fixed point if and only if there exists a non-zero vector

all a12 . . .

. . . . . . . . . . . . . . . . . . . a m 1 am2 * "


or (allxl + a21x2 + ... + umlzm, u12x1 + a,,%, + ... + um2xm, . . . . almxl + a 2 m ~ 2 + ... + ammxm) = (x1, x2, . . . . xm)

or, equivalently, if there exists a non-zero solution to the homogeneous system

allxl + u21x2 + ... + arnlzm = z1 (all-l)xl + a21x2 + ... + amlxm = 0

a12x1 + uX2x2 + + amzxm = z2

aln,sl + aXmz2 + * . * + ammxm = x,

... a1221 + (u*2- l)s, + * * . + am2xm = 0

almXl + a 2 m ~ 2 + ... + (amm- l ) xnL = 0

.................................. .............................................

Now m homogeneous equations in ?n unknowns have a non-zero solution if and only if the determinant of the matrix of coefficients is zero. But the matrix of coefficients of the above homogeneous system of equations is the transpose of the matrix

1 0 * * * a l l - 1 a12 . * * all a12 * * .

. . . ... ... A - I = ................. .......... ..........................

am1 am2 ’ * ‘ amm 0 0 * . . aml am, . . . amm-l

Since the determinant of a matrix is equal to the determinant of its transpose, A has a non-zero fixed point if and only if det(A - I ) = 0.

20.29. Prove: Every stochastic matrix P = ( p J has a non-zero fixed point, Since the rows of the stochastic matrix P a r e probability vectors, the sum of its column

vectors is

Accordingly, the sum of the columns of the matrix P - Z is

Thus the columns of the matrix P - I are dependent (see Page 114). But the determinant of a matrix is zero iff its columns are dependent; hence det ( P - I ) = 0 and so, by the preceding problem, P - Z has a non-zero fixed point.

20.30. Suppose m points on a circle are numbered respectively 1, 2, , . .,m in a counterclockwise direction. A particle performs a “random walk” on the circle; it moves one step counterclockwise with probability p or one step clockwise with probability g = 1 - p . Find the transition matrix of this Markov chain.

transition matrix which appears to the left below. The state space is {1,2, . . . . m}. The diagram to the right below can be used to obtain the

1 2 3 4 ... r n - 2 m - 1 m

...

...

... ...................................................

m-2 ... ni- 1 ...

1 P 0 0 0 0

0 P 0 0 0

Q 0 P 0 0

m - 1 0 0 0 P 0

0 0 0 0 Q

CHAP. 201 MARKOV CHAINS


253

PROBABILITY VECTORS AND STOCHASTIC MATRICES 20.31. Which vectors are probability vectors?

(i) (5, 4, -d., 4) (ii) (4, 0, Q, 4.9) (iii) (A, 4, Q, 0, $1.

20.33. Which matrices are stochastic?

REGULAR STOCHASTIC MATRICES AND FIXED PROBABILITY VECTORS 20.34. Find the unique fixed probability vector of each matrix:

20.35. (i) Find the unique fixed probability vector t of P = (; : :I. 0 1 0

(ii) What matrix does Pn approach? (iii) What vector does (a, $, +)Pn approach?

20.36. Find the unique fixed probability vector t of each matrix:

20.37. (i) Find the unique fixed probability vector t of P = 1; ; ,” ; ) * \ 0 + 0 1 I (ii)

(iii) What vector does (t, 0, +, &)Pn approach? (iv) What vector does (+, O,O, &)Pn approach?

What matrix does Pn approach? 2

20.38. (i) Given that t = ( I , O , &) is a fixed point of a stochastic matrix P , is P regular?

t = (t,f,f,&) is a fixed point of a stochastic matrix P , is P regular? (ii) Given that

20.39. Which of the stochastic matrices are regular?

20.40. Show that (cf + ce + de , uf + b f + ae , ad + bd + bc) is a fixed point of the matrix

1 - a - b a b P = ( c 1 - c - d d )

e f 1 - e - f

254

MARKOV CHAINS

MARKOV CHAINS [CHAP. 20

20.41.

20.42.

20.43.

20.44.

20.45.

20.46.

20.47.

20.48.

20.49.

20.50.

A man’s smoking habits a r e as follows. If he smokes filter cigarettes one week, he switches to non-filter cigarettes the next week with probability .2. On the other hand, the probability that he smokes non-filter cigarettes two weeks in succession is .7. In the long run, how often does he smoke filter cigarettes?

A gambler’s luck follows a pattern. is .6. However, if he loses a game, the probability of losing the next game is .7. even chance tha t the gambler wins the first game.

(i) What is the probability tha t he wins the second game?

(ii) What is the probability tha t he wins the third game?

(iii) In the long run, how often will he win?

For a Markov chain, the transition matrix is P = (i i) with initial probability distribution

If he wins a game, the probability of winning the next game There is a n

1 1

p(0 ) = (&,$). Find: (i) $:I; (ii) p i ; ) ; (iii) ~ ( 2 ) ; (iv) p i 2 ) ; (v) the vector p(O)Pn approaches; (vi) the matrix Pn approaches.

For a Markov chain, the transition matrix is P = and the initial probability dis- Z T Z

tribution is p(O) = (T,z,O). 1 1 Find (i) p i t ) , (ii) p;:), (iii) P ( ~ ) , (iv) p y ) .

Each year a man trades his car for a new car. If he has a Buick, he trades it for a Plymouth. If he has a Plymouth, he trades i t for a Ford. However, if he has a Ford, he is just a s likely to trade i t for a new Ford as to trade i t for a Buick or a Plymouth. In 1955 he bought his first car which was a Ford. (i) Find the probability tha t he has a (a) 1957 Ford, ( b ) 1957 Buick, (c) 1958 Plymouth,

( d ) 1958 Ford.

(ii) In the long run, how often will he have a Ford?

There a re 2 white marbles in urn A and 4 red marbles in urn B. At each step of the process a marble is selected from each urn, and the two marbles selected a re interchanged. Let X , be the number of red marbles in urn A af ter n interchanges. (i) Find the transition matrix P. (ii) What is the probability tha t there a re 2 red marbles in urn A af ter 3 steps? (iii) In the long run, what is the probability t h a t there a r e 2 red marbles in urn A ?

Solve the preceding problem in the case tha t there a r e 3 white marbles in urn A and 3 red marbles in urn B.

A fair coin is tossed until 3 heads occur in a row. Let X, be the length of the sequence of heads ending at the nth trial. (See Example 8.4.) What is the probability that there a re a t least 8 tosses of the coin?

A player has 3 dollars. A t each play of a game, he loses one dollar with probability 3 but wins two dollars with probability &. He stops playing if he has lost his 3 dollars or he has won at least 3 dollars.

(i) (ii)

Find the transition matrix of the Markov chain.

What is the probability tha t there a r e at least 4 plays to the game?

The diagram on the r ight shows four compartments with doors leading from one to another. A youse in any compartment is equally likely to pass through each of the doors of the compartment. Find the transition matrix of the Markov chain.


20.51. Find the transition matrix corresponding to each transition diagram:

20.52. Draw a transition diagram for each transition matrix:

a1 a2

Answers to Supplementary Problems 20.31. Only (iii).

20.32. (i) (3/13, 0, 2/13, 5/13, 3/13)

(ii) (8/18, 2/18, 0, 1/18, 3/18, 0, 4/18) (iii) (4/45, 24/45, 6/45, 0, 3/45, 8/45)

20.33. Only (ii) and (iv).

20.34. (i) (6/11, 5/11), (ii) (10/19, 9/19), (iii) (5/13, 8/13), (iv) (z , Q) 20.35. (i) t = (4/13, 8/13, 1/13), (iii) t = (4/13, 8/13, 1/13)

20.36. (i) t = (2/9, 6/9, 1/9), (ii) t = (5/15, 6/15, 4/15)

20.37. (i) t = (2/11, 4/11, 1/11, 4/11), (iii) t , (iV) t

20.39. Only (iii).

20.41. 60% of the time.

20.42. (i) 9/20, (ii) 87/200, (iii) 3/7 of the time.

(:: ::) 20.43. (i) 9/16, (ii) 3/8, (iii) (37/64, 27/64j, (iv) 27/64, (v) (.6, .4), (vi)

20.44. (i) 3/8, (ii) l/2, (iii) (7/16, 2/16, 7/16), (iv) 7/16

20.45. (i) (a) 4/9, ( b ) 1/9, (c ) 7/27, (d) 16/27. (ii) 50% of the time

0 1 0

(ii) 3/8 (iii) 215

256 MARKOV CHAINS

1 0 1 0 o\

20.47. (i) P = 1 1) (ii) 32/81 (iii) 9/20 o a a +

\ o 0 1 o/

1 0 0 0 0 0 0 ~ 0 0 ~ 0 0 0

o % o o ~ a o

[CHAP. 20

20.52. (i) 3 (ii) - t

Chapter 21

Inequalities

THE REAL LINE We

assume the reader is familiar with the geometric representation of R by means of the points on a straight line. As shown below, a point, called the origin, is chosen to represent 0 and another point, usually to the right of 0, to represent 1. Then there is a natural way to pair off the points on the line and the real numbers, i.e. each point will represent a unique real number and each real number will be represented by a unique point. For this reason we refer to R as the real line and use the words point and number inter-

The set of real numbers, denoted by R, plays a dominant role in mathematics.

changeably. 4 --

I I l 1 , I I I I I I

-4 -3 -2 -1 0 1 2 3 4

The real line R

POSITIVE NUMBERS

positive numbers; those numbers to the left of 0 are the negative numbers. positive numbers can be completely described by the following axioms:

[P,] rPP]

Those numbers to the right of 0 on the real line R, i.e. on the same side as 1, are the The set of

If a E R, then exactly one of the following is true: a is positive; a = 0; -a is positive. If a, b E R are positive, then their sum a + b and their product a - b are also positive.

It follows that a is positive if and only if -a is negative.

Example 1.1: We show, using only [P,] and [PJ, that the real number 1 is positive.

(-l)(-1) = 1 is positive. cannot both be positive. so 1 is positive.

The real number -2 is negative. For, by the preceding example, 1 is positive and so, by [P2], the sum 1 + 1 = 2 is positive; hence -2 is not positive, i.e. -2 is negative.

By [PI], either 1 or -1 is positive. If -1 is positive then, by [P2], the product But this contradicts [P,] which states tha t 1 and -1

Hence the assumption that -1 is positive is false and

Example 1.2:

ORDER An order relation in R is defined using the concept of positiveness.

1-1 The real number a is less than the real number b, written a < b, if the difference b - a is positive.

257

258 INEQUALITIES [CHAP. 21

The following notation is also used: a > b , read a is greater than b, u 6 b , read a is less than o r equal to b, a 2 b, read a is greater than or equal to b,

means b < a means a b means a is to the right of b on the real line R

Example 2.1: 2 < 5, -6 f -3, 4 6 4, 5 > -8

Example 2.2: A real number 3: is positive iff x > 0, and z is negative iff z < 0. For z f 0, x2 is always greater than 0: x2 > 0.

Example 2.3: The notation 2 < x < 5 means 2 < x and also z < 5; hence x will lie between 2 and 5 on the real line.

We refer to the relations <, >, and 1 as inequalities in order to distinguish them We from the equality relation = .

now state basic properties about inequalities which shall be used throughout. We also shall refer to < and > as strict inequalities.

Theorem 21.1: Let a, b and c be real numbers.

(i) The sense o f an inequality is not changed if the same real number is added to both sides:

If a 0, then ac < be. If a 4 b and c > 0, then ac 4 be.

(iii) The sense of an inequality is reversed if both sides are multiplied by the same negative number:

If a be. I f a 6 b and c < 0. then ac be.

(FINITE) INTERVALS

satisfying Let a and b be real numbers such that a<b . Then the set of all real numbers x

a < x < b is called the open interval from a to b a 6 x 4 b is called the closed interval from a to b a < x 6 b is called the open-closed interval from a to b a L x < b is called the closed-open interval from a to b

The points a and b are called the endpoints of the interval. Observe that a closed interval contains both its endpoints, an open interval contains neither endpoint, and an open-closed and a closed-open interval contains exactly one of its endpoints:

1 " " c -

a b a b

open interval: a < x < b closed interval: (L 5 x 6 b

CHAP. 211 INEQUALITIES 259

- " - - - d

a b U b

open-closed interval: a < x f b closed-open interval: a 6 2 a or x A a is called an infinite i n t e m n l . Let a be any real number. Then the set of all real numbers x satisfying x < a, x 6 a,

LINEAR INEQUALITIES IN ONE UNKNOWN Every linear inequality in one unknown x can be reduced to the form

a x b or a x k b If a Z 0 , then both sides of the inequality can be multiplied by l l a with the sense of the inequality reversed if a is negative. Thus the inequality can be further reduced to the fo rm

x < c , x g c , n . > c or x z c whose solution set is an infinite interval.

Example 3.1 : Consider the inequality 5~ + 7 posing), we obtain

Multiplying both sides by + (or: dividing both sides by 3) we finally obtain

22 + 1. Adding -22 - 7 to both sides (or: trans-

5 2 - 2 2 5 1 - 7 or 3 x 5 -6

4 : - 2 0 2 x 6 -2

Example 3.2: Consider the inequality 22 + 3 < 42 + 9. Transposing, we obtain

2 2 - 42 < 9 - 3 o r -22 < 6

AIultiplying both sides of the inequality by -3 (or: dividing both sides by - 2 ) and reversing the inequality since -+ is negative, we finally obtain -

u r -3 0 3 x > -3

ABSOLUTE VALUE The absolute value of a real number x, written 1x1, is defined by

= { x if X A O

-n: if x < O 1x1

that is, if x is non-negative then = x, and if x is negative then 1x1 = -x. Thus the absolute value of every real number is non-negative: 1x1 2 0 f o r every x E R.

Geometrically speaking, the absolute value of x is the distance between the point x on the real line and the origin, i.e. the point 0. Furthermore, the distance between any two points a, b E R is la- b( = Ib - a ' .

Example 4.1:

Example 4.2:

,-21 = 2, 171 = 7, 1-7, = 7 , , - f i l = fi ' 3 - 81 = 1-51 = 5 and I8 - 3' l - - 151 = 5

Example 4.3: The statement 1x1 < 5 can be interpreted to mean tha t the distance between x and the origin is less than 5; hence z must lie between -5 and 5 on the real line. In other words, - - -

-5 n 5 Ix' < 5 and -5 < x < 5

have identical meaning and, similarly, - - 5 n 5 1x1 f 5 and - 5 5 2 6 5

have identical meaning.

The graph of the function f(x) = 1x/, i.e. the absolute value function, lies entirely in the upper half plane since f ( x ) 2 0 for every x E R:

2 4

- -4 - 2 t t

Graph of f(z) = 1x1

The central facts about the absolute value function are the following:

Theorem 21.2: Let a and b be any real numbers. Then:

(i) la1 (ii) -la1 a la1 (iii) lab1 = /a1 * lbl

(iv) la + bl 6 la1 + lbl

0, and la1 = 0 iff a = 0

(v) la + bl 2 la1 - Ibl

Solved Problems INEQUALITIES 21.1. Write each statement in notational form:

(i) a is less than b. (iv) a is greater than b.

(ii) a is not greater than b. (v) a is not less than b. (iii) a is less than or equal to b. (vi) a is not greater than or equal to b.

A vertical or slant line through a symbol designates the opposite meaning of the symbol,

( i ) a < b, (ii) a 4 b, (iii) a 5 b, (iv) a > b, (v) a 4 b, (vi) a b

21.2. Rewrite the following geometric relationships between the given real numbers using the inequality notation:

(i) y lies to the right of 8. (ii) x lies to the left of -3.

(iii) x lies between -3 and 7. (iv) w lies between 5 and 1.

Recall tha t a 8 or 8 < g. (iii) -3 < z and x < 7 or, more concisely, -3 < 2 < 7 .

(ii) z < -3. (iv) 1 < w < 5.

21.3. Describe and diagram each of the following intervals: (i) 2 < x < 4, (ii) -1 6 x A 2, (iii) -3 < x 4 1, (iv) -4 6 x < -1, (v) x > -1, (vi) x 2.

(i) All numbers greater than 2 and less than 4: l2 0 2 4 r b - -

(ii) All numbers greater than or equal to -1 and less than or equal to 2: -1, ' '

(iii) All numbers greater than -3 and less than or equal to 1:

(iv) All numbers greater than or equal to -4 and less than -1: ' ' I 1 ' '

(v)

(vi) All numbers less than or equal to 2: 4,

- - , , , _ , , -2 0 2

- 4 -2 0 2

I I - *

- 2 0 2 All numbers greater than -1:

- -2 0 2

21.4. Solve and diagram the solution set of each inequality: (i) 4x - 3 L 2x + 3,

(i) Transpose to obtain: 4x - 2x 4 3 + 3 or: 22 4 6

(ii) 3x + 2 < 6n: - 7.

4, - - -3 0 3

Divide by 2: x'3

(ii) Transpose t o obtain: 32 - 6x < -7 - 2 or: -3x < -9 Divide by -3: r x > 3 - L

-3 0 3

Note tha t the inequality is reversed since -3 is negative.

21.5. Solve each inequality and diagram its solution set: (i) 3%-1 4n:+2, (ii) x - 3 > 1 + 3 x , (iii) 2%-3 4 5x-9. (i) Transpose to obtain: 3x - 42 4 2 + 1

or: -x 2 3 4, I

Divide by -1: x ' -3 - 3 0 3

(ii) Transpose to obtain: x - 32 > 1 + 3 or: -2x > 4

4 , - Divide by -2 : 2 < -2 - 2 0 2

(iii) Transpose t o obtain: 22 - 5% 6 -9 + 3 or : -32: f -6

Divide by -3: 2 2 2 - - *

- 2 0 2

In each case the inequality was reversed since division was by a negative number.

x - 2 3 21.6. Solve: 4% + - < 2 x - &

Multiply both sides by 12: 62 + 4 ( ~ - 2 ) < 2 4 ~ - 1 or: 6% + 42 - 8 < 2 4 ~ - 1

Transpose: 6% + 4x - 2 4 ~ < 8 - 1

or: - 1 4 ~ < 7 Divide by -14: 5 > -Q

21.7. Solve each inequality, i.e. rewrite the inequality so that x is alone between the inequality signs: (i) 3 L x - 4 5 8, (ii) -1 6 x + 3 6 2, (iii) -9 6 3x

(i)

(ii) (iii) Divide each side by 3 (or: multiply by 5) to obtain -3 x 4.

(iv) Divide each side by -2 (or: multiply by -+) and reverse the inequalities to obtain -2 f x

12, (iv) -6 A -2s 4 4.

Add 4 to each side to obtain 7 f z 6 12. Add -3 to each side to obtain -4 5 x 4 -1.

3.

21.8. Solve each inequality: (i) 3 < 2x - 5 < 7, (ii) -7 - 2% + 3 A 5. (i) Add 5 to each side to obtain:

Divide each side by 2:

Add -3 t o each side to obtain:

Divide each side by -2 and reverse inequalities:

8 < 2 2 < 12

4 < 2 < 6

(ii) -10 f -2x 6 2. -1 6 x 5 5.

ABSOLUTE VALUE 21.9. Evaluate: (i) 13 - 51, (ii) 1-3 + 51, (iii) 1-3 - 51, (iv) 13 - 71 - 1-51.

(i) 13 - 51 = '-21 = 2 (ii) 1-3 + 5 = 121 = 2

(iii)

(iv)

1-3 - 51 = 1-81 = 8 13 - 71 - 1-51 = 1-41 - 1-51 = 4 - 5 = -1

21.10. Evaluate: (i) 12 - 81 + '3 - 11 (ii) 12 - 51 - 14 - 71

(iii) 4 + 1-1 - 51 - 1-81 (iv) 13 - 61 - 1-2 + 41 - 1-2 - 31

(i) 1 2 - 8 1 + ' 3 - 1 1 = 1-61+l21 = 6 + 2 = 8 (ij) 12-51 - 14-71 = 1-31 - 1-31 = 3 - 3 = 0

(iii) 4 + 1-1 - 51 - 1-81 = 4 + 1-61 - 1-81 = 4 + 6 - 8 = 2

(iv) 13- 61 - 1-2+4/ - 1-2 - 31 = 3 - 2 - 5 = -4

21.11. Rewrite without the absolute value sign:

(i) 1x1 6 3, (ii) (x -21 < 5 , (iii) (2x -31 5 7.

(i) -3 5 x 5 3

(ii) - 5 < 2 - 2 < 5 or - 3 < x < 7

(iii) -7 f 22 - 3 5 7 or -4 5 22 5 10 or -2 f x 5 5

21.12. Rewrite using an absolute value sign: (i) -2 L x 6 , (ii) 4 < x < 10. Firs t rewrite the inequality so tha t a number and its negative appear at the ends of the

Add -2 to each side to obtain -4 6 x -2 4 4 which is equivalent to I X -21 6 4.

Add -7 to each side to obtain -3 < x- 7 < 3 which is equivalent to

inequality.

(i)

(ii) I$-- 71 < 3.

PROOFS 21.13. Prove: If a < b and b < c, then a < c.

By definition, a < b and b < c means b - a and c - b are positive. By [P2], the sum of two positive numbers ( b - a) + (c - b ) = c - a is positive. Thus, by definition, a< c.

21.14. Prove Theorem 21.1: (i) If a < b, then a + c be.

(i) By definition, a < b means b - a is positive. But

( b + ~ ) - ( a + ~ ) = b - a -

Hence ( b + c) - (a + c ) is positive and so a + c < b + c.

By definition, a < b means b - a is positive. But c is also positive; hence by [P2] the product c ( b - a ) = bc- ac is positive. Accordingly, ac< bc.

(iii) By definition, a bc.

(ii)

21.15. Prove Theorem 21.2(iv): la + bl L la/ + J b J .

Method 1. Since la[ = ?a, --/a1 f a 5 la\; also, -161 f b f l b [ . Then, adding,

- ( la l + bl) f a + b 6 la1 + Ib/

Therefore la+ bl 6 ' a : + Ibl I = la1 + 161 Method 2.

Now ab f lab1 = la1 Ibl implies 2ab 6 2 (a1 lbl, and so

( a + b ) i = a? + 2ab + b2 5 a2 + 2 la( lb( + b2 = la12 + 2 lb( + [bl2 = (la1 + lW But d m = la + bl and so, by the square root of the above, 1 0 + bl 6 la1 4- lbl.

21.16. Prove: (a - b / la1 + Ibl. Using the result of the preceding problem, we have la- bl = ( a + (--b)\ 6 la1 + 1-41 = la/ + \ b / .

21.17. Prove: la - c J 6 ]a - bJ + ] b - el. l ~ - ~ c / = ! ( a - b ) + (b -C)[ f la -b l + / b - c /

21.18. Prove that 3 is a positive number. By [P,], either -4 or 8 is positive.

(-4) + (-4) = -1 is also positive. Thus we have a contradiction, and so & is positive.

Suppose -3 is positive and so, by [P2], the sum But by Example 1.1, the number 1 is positive and not -1.

21.19. Prove that the product a . b of a positive number a and a negative number b is negative.

But a ( -b) = - (a - b). Thus -(a * b) is positive and so, by [P,], a * b is negative. If b is negative then, by [PI], -b is positive; hence by [P2] the product a (-b) is also positive.

21.20. Prove that the product a - b of negative numbers a and b is positive. If a and b are negative then, by [PI], --a and -b are positive. Hence by [P2], the product

But a . b = (-a) - (-b), and so a* b is positive. (-a) * (-b) is positive.

21.21. Prove: If a and b are positive, then a < b iff a2 < b2. Suppose a < b. On the other hand, suppose a2 < b2.

Since a and b are positive, a2 < ab and ab < bz; hence a2 < bz,

Then b 2 - a2 = ( b + a ) ( b - a ) is positive. Since a and b are positive, the sum b + a is positive; hence b - a is positive or else the product ( b + a ) ( b - a) would be negative. Thus, by definition, a < b.

21.22. Prove that the sum of a positive number a and its reciprocal l l a is greater than or equal to 2:

a - 1 # 0 and so

if a > 0, then a + l l a 2 2. If a = 1, then lla = 1 and so a + lla = 1 + l = 2. On the other hand, if a # 1, then

(u-1)* > 0 or a2- 2 a + 1 > 0 or az+ 1 > 2a Since a is positive, we can divide both sides of the inequality by a to obtain a+ lla > 2.

Supplementary Problems INEQUALITIES

21.23. Write each statement in notational form:

( i ) x is not greater than y. (iii) m is not less than n.

(ii) Y is greater than or equal to t. (iv) u is less than w.

21.24. Using inequality notation, rewrite each geometric relationship between the given real numbers:

( i ) x lies to the right of y. (iii) x lies between 4 and - 6 .

(ii) r lies to the left of t . (iv) y lies between -2 and -5.

21.25. Describe and diagram the following intervals:

(i) -1 < x 6 3, (ii) 1 f x 6 4, (iii) -2 f x < 0, (iv) x f 1.

21.26. Solve: (i) 6% + 5 6 2x -7 , (ii) 2x - 3 > 4x+ 5.

21.27. Solve: (i) 5 - 2 < 2x -6, (ii) 3% + 2 2 62+ 11.

2 - 1 21.28. Solve: - + 1 5 - 1 3 2 2 '

21.29. Solve, i.e. rewrite so tha t x is alone between the inequality signs: (i) -2 f x - 3 f 4, (iii) -12 < 42 < -8. (ii) -5 < x + 2 < 1,

21.30. Solve: (i) 4 6 -2x f 10, (ii) -1 < 2 s - 3 < 5, (iii) -3 5 - 2x 6 7.

ABSOLUTE VALUES

21.31. Evaluate: (i, 14 - 71, (ii) 17 - 41, (iii) 14 f 71, (iv) 1-4 - 71.

21.33. Evaluate: (i) I 1-31 - 1-91 I, (ii) ! 12 - 61 - 11 - 91 I.

21.34. Rewrite without the absolute value sign: (i) 1x1 5 8, (ii) 1 % - 31 < 8, (iii) 12x + 41 8, (iv) 12-2x1 < 8.

Rewrite using the absolute value sign:

(i) -3 < x < 9, (ii) 2 x 8, (iii) -7 < x < -1, (iv) -1 f 2 x + 3 f 5 .

21.35.

THEOREMS ON INEQUALITIES

21.36. Prove: If a c / b .

21.37. Prove: If a < b and c < d, then a + c < b +d.

21.38. Prove: If a and b are positive, then 6 f ( a + b) /2 .

21.39. Prove: If a < where Q, 6 , c and d are positive, then ~ a + c < c* b d b S d d

21.40. Prove: For any real numbers a, b and c, (i) 2ab f a2 + b2 and (ii) ab + ac + be f u2 + b2 + c2.

THEOREMS ON ABSOLUTE VALUES 21.41. Prove: (i) (--a1 = 1a1, (ii) u2 = Ia12, (iii) 'a] = @, (iv) 1x1 < a iff -a < x < a. 21.42. Prove Theorem 21,2(iii): lab/ = /a] lbl.

21.43.

21.44. Prove: la' - lbl 6 ja- bl.

Prove Theorem 21.2(v): ]a1 - /bl 5 la + bl.

Answers to Supplementary Problems 21.23. (i) x 4 y, (ii) r 2 t , (iii) m 4: n, (iv) u < w.

21.24. (i) x > y or y < x, (ii) T < t, (iii) -6 < x < 4, (iv) -5 < y < -2

r - c - 21.25. (i) - , (iii) I 1 I

- 4 -2 0 2 4 -4 - 2 0 2 4

21.26. (i) x f -3, (ii) .T < -4

21.27. (i) x > 4, (ii) x 6 -3

21.28. x 2 7

21.29. (i) 1 f x 6 7, (ii) -7 < x < -1, (5) -3 < x < -2

21.30.

21.31. (i) 3, (ii) 3, (iii) 11, (iv) 11

21.32. (i) -2, (ii) 7, (iii) 6, (iv) 3

21.33. (i) 6, (ii) 4

21.34.

21.35. (i) 1%-31 < 6, (ii) 'x-51 f 3, (iii) '~$41 < 3, (iv) 12%+1l f 3

(i) -5 f x f - 2 , (ii) 1 < x < 4, (iii) -1 f x f 4

(i) -8 f x f 8 , (ii) -5 < x < 11, (iii) -6 L x 6 2, (iv) -3 < x < 5

Chapter 22

We assume the reader is familiar with the Cartesian plane, denoted by R2, as shown on the right. Here each point P represents an ordered pair of real numbers (a, b ) and vice versa; the vertical line through the point P meets the horizontal axis (x axis) a t a, and the horizontal line through P meets the vertical axis (y axis) a t b. Thus in discussing the Car- tesian plane R2 we use the words point and

b

Points, lines and Hyperplanes

* P

a x axis -

DISTANCE BETWEEN POINTS The distance d between two points 11 = (x,,y,) and q = (x2,y2) is given by

d = d(., - x,)2 + (y, - y,)2

For example, the distance between p = (3, -1) and q = (7,2) is

d = d(7 -3)' 3. (2 -(-l))' = d16 + 9 = 5

We shall denote the distance between p and q by d ( p , q ) .

INCLINATION AND SLOPE OF A LINE The inclination B of a line I is the angle that I makes with the horizontal; that is, 6 = 0"

if 1 is horizontal or B is the smallest positive angle measured counterclockwise from the positive end of the x axis to the line 1.

The slope 172 of a line is defined as the tan- gent of its angle t? of inclination: 7n = tan8. In particular, the slope of the line passing through two points p = (x,, y,) and q = (x2, y,) is given by

!I* - 1-11 tv = tan0 = ~ x, - x,

The slope of a vertical line, i.e. when x2=x1, is not defined.

Two distinct lines I, and I , are parallel if and only if their slopes, say nz, and m2 respectively, are equal: On the other hand, the lines are perpendicular if and only if the slope of one is the negative reciprocal of the other: m, = -Urn2 or 7n,?n, = -1.

The range of 0 is given by 0" 4 8 < 180".

m, = nz,.

266

CHAP. 221 POINTS, LINES A N D HYPERPLANES 267

LINES AND LIKEAR EQUATIONS

form

i.e. each point on 1 is a solution to ( 1 ) and each solution to (1) is a point on 1. Conversely, every equation of the form ( 1 ) represents a line in the plane. Thus, in discussing the Cartesian plane, we sometimes use the terms line and linear equation synonymously. (See Chapter 11.)

Every line 1 in the Cartesian plane R' can be represented by a linear equation of the

( 1 ) a.15 + by = c , where u and b are not both zero

HORIZONTAL AND VERTICAL LINES The equation of a horizontal line. i.e. a line parallel to the n: axis, is of the form

y = 1;

where k is the point a t which the line intersects the y axis. In particular, the equation of the :I' axis is y = 0.

The equation of a certical line, i.e. a line parallel to the y axis, is of the form

r = k

where lc is the point a t which the line intersects the 3' axis. In particular, the equation of the y axis is z' = 0.

x = a

y = b

a -

POINT-SLOPE FORRl A line is completely determined once its direction and a point on the line are known.

The equation of the line having slope 02 and passing through the point (xl ,yl) is

y - y1 = m( .u - X I )

Example 1.1: Find the equation of the line having slope 3 and passing through t h e point ( 5 , -21. Substi tute in the above formula to obtain y + 2 = 31, - 5) or y = 3% - 17.

Example 1.2: Find the equation of the line passing through the two points (3,2) and 15, -6).

The slope m of the line is 112 = ____ - - -4. Thus the equation

of the line is

Y z - Y I - G - 2 x q - x l 5 - 3

y - 2 = -4( 7. - 3 ) or 4 2 + y = 1-1.

SLOPE-INTERCEPT FORM The equation of the line having slope vz and intersecting the y axis at k , i.e. having

?/-intercept k , is

Thus if the equation of a line 1 is a? - by = c, where b f 0, then

y = 0 2 3 ' + k

a C 2 / = --.r + - 11 13

and so m = - a / b is the slope of 1.

Example 2.1: The slope of t he line I! = -2.). + 8 is -2.

Example 2.2: Find the slope 7n of the line 2.x - 3y = 5. Rewrite the equation in the f o r m 1/ = 2 X - - 3 . ,~ 3 , hence ) ) I = 3.

268 POINTS, LINES AND H Y P E R P L A N E S [CHAP. 22

PARALLEL LINES. DISTANCE BETWEEN A POINT AND A LINE Let I be the line ax + by = c. The set of all

lines parallel to I is represented by the equation

ax + by = h. (2)

i.e. every value for k determines a line parallel to 1 and every line parallel to l is of the above form f o r some k . We call k a p a m m e t e r , and the set (2) a one parameter family of lines.

The distance d between a point 11 = ( x ~ , y ~ ) and the line 1 is given by

lax, + byl - c1

V G T i 7 d =

In particular, the distance between the origin (0,O) and the line Z is

Example 3.1: Find the line I parallel to the line 2r - 51/ = 7 (3 ,4) .

and passing through the point The line 1 belongs to the family

2x - 5y = Iz Since the point 1 3 , l ) belongs t o 1 it must satisfy the equation of I :

2 . 3 - 5 . 4 = k o r k = -14

Thus the equation of I is 2%-55y = -14.

Example 3.2: Let 1 be the line 3.c - 4y = 2.

d =

The distance d between the point (2, - G ) and 1 is

1 3 . 2 + 4 * ( - - 6 ) - 2 1 - - 1-201 = 5 JIm

2 - 121 - l/m 5 '

The distance p between the origin and 1 is p = ___ -

EUCLIDEAN m-SPACE Let ic and v be vectors with 112 components of real numbers:

u = (ul, u, , . . . , uIn) and

Recall (see Chapter 9) the following operations of vector addition, scalar multiplication, and dot product :

v = (vl, v2, . . ., v,)

u + v = ( 2 1 , 4 G,, 26, + v*, . . . , un, + v,J

2 1 , 2 ' , + 1 L 2 V 2 + . - * + Ui,,7jli,

ku = (k'u,, ku2, . . . , ku,), f o r any scalar k

U ' v =

The set of all such vectors with the above operations is called Euclidean m-space o r m dimensional Euclidean space o r , simply, m-space, and will be denoted by R".

The distance between the above vectors u and v, written d(u,v), is defined by

POINTS, LINES AND HYPERPLANES 269 CHAP. 221

The norm (or: l e n g t h of the vector ZL, written ]lull, is defined by

ilull = -\/... = dzq + u; + . * * + Uf,

Observe that d(u, v) = Iju - ~ 1 1 1 .

is zero: u-v = 0.

the zero vector, also called the origin, belongs to every axis.

The vectors u and 1: are said to be orthogonal (or: perpendicular) if their dot product

The set of vectors of the form 10, . . . , 0, Z L ~ , 0, . . ., 0) is called the xi axis. Note that

Example 4.1 : Consider the following vectors in R4: u = (1, - 2 , 4 , 1 ) and 2) = (3,1, -5,O). Then:

Z L + v = (1 + 3 , -2 - 1, -1 - 5 , 1 + 0 ) = (4, -1, -1, 1)

7u = (7, -14, 28, 7 )

U.W = 1 . 3 + ( - 2 ) * 1 + 4 . ( -5 ) + 1 . 0 = 3 - 2 - 20 + 0 = -19

diu,w) = d(3 - 1 ) 2 + (1 + 2 ) 2 + (-5 - 4)2 + (0 - 1 ) 2 = a

Example 4.2: We frequently identify a point in, say, R2 with the arrow from the origin to the given point. Then the arrow corresponding to the sum p + q is the diagonal of the parallelogram determined by the arrows corresponding to the points p and q as shown on the right. Furthermore, llpll, the norm of p, is precisely the length of the arrow to p . The sum p + q.

BOUNDED SETS A subset A of Rm is bounded iff there exists a

real number m > 0 such that llull < nz for every point u E A . In the case of the plane R2, this says that the points of A lie inside of the circle of radius rn centered a t the origin, as indicated on the right. A is bounded.

+$+ -m

HYPERPLANES Consider a linear equation in the m unknowns x,, . . ., xm:

b CIXl + C 2 X , + * * * + C m X m = ( 3 )

Using the notation of the dot product of RI'~, the above equation can be written in the form

C ' X = b

where c = (c1,c2, . . .,cJ and x = (x1,z2, . . .,x,). that c # 0, i.e. that the ci in (3) are not all zero.

Definition:

Unless otherwise stated, we assume

The set of all points in Rm which satisfy the equation c x = b, i.e. the set of - all solutions of the linear equation (3), is called a hyperplane.

270 POINTS, LINES AND H Y P E R P L A K E S [CHAP. 22

Example 5.1: The hyperplanes i n the Car- tesian plane R’ a re the s t ra ight lines a s discussed a t the beginning of this chapter.

The hyperplanes in 3-space RJ, i.e. in solid geometry, cor-

lustrated on the right. In particular, the equation of the x y p l a n e is z = 0, the ?z-plane is y=O, and the ?/:-plane is x = 0.

Example 5.2:

respond to the planes a s il- Y

Hyperplane in R 3 .

Observe that the “dimension” of a hyperplane in R’ is one, and in R3 is two; that is, we can identify the points of a hyperplane in R2 with the points on the real line R and we can identify the points on a hyperplane of R3 with the points in R2 with the further property that distances between corresponding points are equal. In each case the dimension of the hyperplane is one less than the dimension of the Ivhole space. In fact, this property completely characterizes hyperplanes in general; namely,

Theorem 22.1: A subset A of R”‘ is a hyperplane if and only if there exists a one-to-one correspondence between A and Ril’-i such that distances between corresponding points are equal.

The proof of this theorem is beyond the scope of this text.

PARALLEL HYPERPLANES. DISTANCE BETWEEN A POINT AND A HYPERPLANE

Let N be the hyperplane c , ~ , + CJ, t . - . + cliplII = b . Then

el,?-, + C J , + . * . + C l n X n l = k _ _ is the equation of the family of all hyperplanes parallel to a, i.e. which do not have a point in common with a. The distance d between the point 71. = (ul, u2, . . ., un,) and N is given by

a =

In other words, the distance between u and the hyperplane c * R: = b is

1 c , l ~ l t c,u, + . . . + e,,LzL;,, - 2, 1 dcf + c; + * * * + c;l

1c-u - bl

llcll d =

In particular, the distance p between the origin, i.e. the zero vector, and ~r is

- P I IC I l

- - Ibl dcf + c; + - - - + Ct P =

Example 6.1: I n solid geometry, i.e. R3, the family of hyperplanes parallel to the r p p l a n e , i.e. to z = O , is represented by x = k where k is the point at which the hyperplane intersects the i axis. (See diagram be1ow.r

Y

Two members of the family z = k.

CHAP. 221 POINTS, LINES A N D HYPERPLANES 271

Example 6.2: Let iy be the hyperplane 3 2 - 4y + x - 2w 1 7 in R4. point (1,-4,-2,3) and a is

The distance d between the

4 i 3 . 1 - 4.(-4) + 1*(-2) - 2 ' 3 - 7 d3' + (-4)2 + 12 + (-2)Z

- - ~ d = fi

- 7 - - 171

d32 + (-4)' + 1 2 + (-2)2 The distance p between a and the origin is p =

fi'

Solved Problems DISTANCE I N THE PLANE 22.1. Find the distance d between ( i ) ( - 3 , 5 ) and (6,4), (ii) ( 5 , -2) and (-1, -4).

In each case use the formula d = ~ ( X ' - X ~ ) ~ + (yz-yJ'.

(i) d = d(6+ 3)? + (4-5)2 = f ? = fi (ii) d = d(-l-5)* + (-4+2)2 = d m = fi

22.2. Show that the points p = (--1,4), q = ( 2 , l ) and r = (3,5) are the vertices of an isosceles triangle.

Compute the distance between each pair of points:

d ( p , q ) = d(2 + 1)' + (1 -4)2 = &m = fi

d(q , r ) = d ( 3 - 2)' + ( 5 - 1)2 = d r n = m Since d ( p , T ) = d ( q , T ) , the triangle is isosceles.

22.3. Show that the points p = (4, -2), q = (-1,3) and r = (1 ,4) are the vertices of a right triangle, and find its area.

Compute the distance between each pair of points:

d ( p , q ) = l/-l- 4)' + (3+2)2 = d r n = rn d(p,r) = d(1- 4)2 + (4 + 2)2 = d w - 5 =

d(q,r) = d(1+1)"(4-3)2 = m = 6 Since d(q, T ) Z + d ( p , ~ ) 2 = d ( p , q)2 , the triangle is a r ight triangle.

The area of the right triangle is A = +(a)(fi) = 2.

22.4. Find the point equidistant from the points a = (-4,3), b = ( 5 , 6 ) and c = (4, -1). Since d ( p , a) = d ( p , b ) ,

3z + y = 6

Let p = (x, y) be the required point. Then d ( p , a ) = d ( p , 6 ) = d ( p , c).

d(z + 4)2 + (y-3)Z = d ( z - 5)2 + (y- 6)2 or

Since d ( p , a ) = d ( p , c ) ,

d ( z +4)2 + (y-3)2 = d ( x - 4)2 + (y + 1)2 or 2% - y = -1

Solve the two linear equations simultaneously to obtain x = 1 and y = 3. required point.

Thus p = (1,3) is the

[CHAP. 22 2'72 POINTS, LINES AND HYPERPLANES

Show that the points p = (1, -2), q = (3,2) and 1' = (6,8) are collinear, i.e. lie on the same straight line. Method 1.

Compute the distance between each p'air of points:

d ( p , q ) = d ( 3 - 1)Z + ( 2 + 2 ) 2 = lLi-TT6 = &G = 2 6

d ( p , r ) = d(6-1)2 + ( 8 + 2 ) 2 = d m = = 5 6

Since d ( p , T ) = d ( p , q) + d(y , T ) , the points lie on the same straight line.

Method 2.

through p and r: Compute the slope m, of the line passing through p and y, and the slope m2 of the line passing

8 - (-2) - lo = 2 - 5 m2 = ~

2 - (-2) = 4 = 2, 2 6 - 1

ml = ~

Since m, = m2, the three points lie on the same line.

3 - 1

LINES 22.6. Plot the following lines:

(i) x = -4, x = -1, x = 3 and x = 5;

( i ) Each line is vertical and meets the x axis at its respective value of x. See diagram below.

(ii) Each line is horizontal and meets the ~j axis at its respective value of y. See diagram below.

(ii) y = 4, y = 1, y = -2, and y = -3.

y = 4 1 5

y = l

, I I t *

- 5 5

y = - 2

y = - 3 r t - 5

22.7. Find the equation of the line (i) parallel to the x axis and passing through the point (-2, - 5 ) , (ii) parallel to the y axis and passing through the point (3,4).

( i ) If the line is parallel t o the z axis, then i t is horizontal and its equation is of the form y = k. Since i t passes through the point (-2,-5), k must equal the y-component of the point. Thus the required equation is y = - 5 .

If the line is parallel to the u axis, then i t is vertical and its equation is of the form 2 = k. Since it passes through the point (3,4), k must equal the 2-component of the point. Thus the required equation is r = 3 .

fii)

22.8. Plot the following lines: (i j 2% - 3y = 6, (ii) 4x t 3y = 6, (iii) 3x + 2y = 0. In each case find at least two points on the line, preferably the intercepts, and as a check

a third point. The line drawn through these points is the required line.

CHAP. 221 POINTS, LINES AND HYPERPLANES 273

-- 5

22.9. Find the equation of the line (i) passing through (3,2) with slope -1, through (-4,1) with slope 4, (iii) passing through (-2, -5) with slope 2.

point. (i) y - 2 = -l(.z-3) o r x + y = 5

(ii) y - 1 = i ( x t 4 ) or x - 2y = -6

(iii) y + 5 = 2(2 + 2) or 22 - y = 1

(ii) passing

In each case use the formula y - y1 = m(.r - xl) where m is the slope and (xl, yl) the given

22.10. Find the equation of the line passing through each pair of points: (i) (1,3) and (4, -5); (ii) (-2, -6) and ( 3 , l ) .

Y2 - Y1 2 2 - $1

In each case, first find the slope nt of the line using the formula m = ___ and then find the equation of the line using the point-slope formula.

8 3

The equation of the line is y - 1 = - (x-3) o r 72 -5y = 16.

The equation of the line is Y - 3 = --(X - 1) or 8~ + 376 = 17. - 5 - 3 -

4 - 1 3 ' 1 1 - 6 - 7 7 3 f 2 5 ' 5

(i) m = - - - -

(ii) m = - - -

22.11. Find the equation of the line passing through each point and parallel to the given line: (i) (3, -2) and 5% + 4y = -5; (ii) (1,6) and 3x - 7y = 4.

(i) The line belongs t o the family 6n: + 4y = k. 5 . 3 i 4 * (-2) = k Thus the required line is 52 + 41/ = 7.

The line belongs t o the family 32 - 71/ = k . 3 * 1 - 7 * 6 = k or k = -39.

Substitute (3,-2) into the equation to obtain

Substitute (1 ,6) into the equation to obtain

or k = 7.

(ii) Thus the required line is 32 - 7y = -39.

22.12. Let ax + by = c be the equation of a line 1. Show that

is the equation of the family of lines perpendicular to 1.

the family of lines perpendicular to I .

bx - ay = k (4

If b = 0, then 1 is vertical and ( 1 ) is the equation of the family of horizontal lines; hence ( 1 ) is

274 POINTS, LINES AND HYPERPLANES [CHAP. 22

On the other hand, if a = 0 then 1 is horizontal and (1) is the equation of the family of vertical

Lastly, suppose b # 0 and a # 0. Then the slope of 2 is - a / b and ( 1 ) is the equation of the But ( - a l b ) ( b / a ) = -1; hence, again, ( 1 ) is the family of lines

lines and so is the family of lines perpendicular to 1.

family of lines with slope b l a . perpendicular to 1.

22.13. Find the equation of the line passing through the given point and perpendicular to the given line: (i) (2,5) and 32 + 4y = 9; (5) (3, -1) and 2x - 49 = 7.

i i ) The line belongs to the family 4r -3y = k. Substitute ( 2 , 5 ) into the equation t o obtain 8 - 15 = k or k = -7. Thus the required line is 45. - 3y = -7.

The line belongs to the family 4x - 2y = k. 12 - 2 = k or k = 10. Thus the required line is 4r + 2y = 10 or 2 2 + y = 5 .

(ii) Substitute (3, -1) into the equation t o obtain

22.14. Determine the distance d of each line from the origin: (i) 3% - 4 ~ = 8, (ii) 5 x + 12y = -4.

4 - - _ - (ii) d = 1-41 - 181 - (i) d = lei

(rn 5 ' @Ti5 13' Use the formula d = d r n '

22.15. Determine the distance d between each point and line: (i) (5,3) and 5x-12y = 6; (ii) (-3,-2) and 4 x - 5y = -8.

lax1 + by1 - C I Use the formula d = & F i z .

EUCLIDEAN m-SPACE 22.16. Find the distance d(u, v) and the dot product u * v for each pair of vectors:

(i) u = (3,-2,0,5) and 2) = (6,2,-3,-4); (ii) u = (2 ,3 , -5 , -7 , l ) and v = (5,3,-2,-4,-1); (iii) ZL = (4,-1,6,5) and v = (1,-4,2,0,7).

Use the formulas d ( u , v) = d ( v l - u1)2 + * + (wnL - u,,,)2 and U * w = UlW, + * * . + umvm.

(iii) The distance d ( u , w ) and the dot product U " U are not defined since the vectors belong to different spaces: u E R4 and v E R5.

22.17. Find the norm of each vector: ZL = (2,-1,3,5); v = (3,-2,-1).

In each case use the formula ~ l u ~ = dut + ui + . . . + u$

22.18. Let u = (1, -3, 2, 4) and v = (3, 5, -1, -2). Find: (i) u + v; (ii) 5u; (iii) 2u - 371; (iv) 'urn v; (v) ~ lu l~ and 11v11.

CHAP. 221 POINTS, LINES AND HYPERPLAKES 273

(i) u + v = ( l + 3 , -3+5, 2 - 1 , 4 - 2 ) = ( 4 , 2 , 1 , 2 ) (ii) 5u = (5 - 1, 5 * (-3), 5 - 2, 5 . 4 ) = (5 , -15, 10, 20) (iii) 2 u - 3v = (2, -6, 4, 8) + (-9, -15, 3 , 6 ) = (-7, -21, 7 , 14) (iv) U . L ' = 3 - 15 - 2 - 8 = -22 (v) 1 1 ~ 1 = d 1 + 9 + 4 + 1 6 = m. , J v l l = d 9 + 2 5 + 1 + 4 = a.

22.19. For each pair of vectors, determine li so that the vectors are orthogonal: (i) (3 , -2, k , 5 ) and (2k, 5, -3, -1); (ii) (2 ,3k , -4,1,5) and (6, -1,3,7,2k).

In each case, set the product of the vectors equal to zero and solve for k. (i) 3 2k + (-2) * 5 -- k (-3) + 5 (-1) = 0 or 6k - 10 - 3k - 5 = 0 or k = 5

(ii) 2 . 6 A 3 k * ( - 1 ) - ( - 4 ) . 3 + 7 + 5 * 2 k = 0 o r 12 - 3k - 12 + 7 + 10k = 0 o r k = -1

22.20. Determine which of the following subsets of R" are bounded: probability vectors; (iv) D consists of all points with positive components.

(i) A consists of all (iii) C is finite;

Recall that X C R ~ is bounded if there exists a real number T ~ Z > 0 such t h a t IIu(, f m for

(ii) B consists of the points on the x, axis;

every u E X .

(i) ( i i )

(iii)

(iv)

A is bounded since IIuI/ f 1 for every probability vector u .

B is not bounded; for if rn is any positive real number, then (m + 1, 0 , 0 , . . ., 0) belongs t o B and has norm greater than rn.

C is bounded. Choose rn to be the maximum norm of the vectors in C; then rn satisfies the required property.

D is not bounded.

HYPERPLANES 22.21. Find the distance d of each hyperplane from the origin:

(i) 2 ~ - 3 y + 5 x = 8; (ii) 2 ~ + 7 y - z + f z u = -4. bi

I e l l In each case use the formula d = - where the hyperplane is c x = b .

- 4 ~ -~ 1-41 iii) d = 8 - - 181 (i) d =

d 2 2 + (-3)Z + 52 6 . d 4 + 4 9 + 1 + 4 6'

22.22. Determine the distance d between each point and hyperplane: (i) (4,0, -2, -3) and x - 6.y + 5 x + 22.9 = 3; (ii) (2 , -1 ,3 ,7 , -2, -4) and 3x, - 5 x , - 2x, + 4x, + X, - 3x, = -9.

where u is the point and c * x = b the hyperplane. ~ * l t - b ' In each case use the formula d = 1~ i

' 4 + 0 - 10 - 6 - 31 - 15 i i i) cl = / 6 + 5 - 6 + 2 8 - 2 + 1 2 + 9 1 - 52 - 13 -~ (i) d =

d l + 3 6 + 2 5 + 4 m' ~ / 9 + 2 5 + 4 + 1 6 + 1 + 9 a 2 '

22.23. Find the equation of the hyperplane passing through the given point and parallel to the given hyperplane: (i) (3,-2,1, -4) and 2 x + 5 y - 6 2 - 2 ~ : = 8; (ii) (1 , -2 ,3 ,5) and 4 ~ - 5 y + 2 x + w = 11.

(i) The hyperplane belongs to the family 2 x f 5 y - 6 2 - 2 ~ = k . Substitute (3 , -2 ,1 , -4) into the equation to obtain 6 - 10 - 6 - 8 = k or k = -2. The required hyperplane is

(ii) The hyperplane belongs to the family 4 x - 5 y + 2 x + w = k . Substitute (1 , -2 ,3 ,5 ) into the equation to obtain 4 + 1 0 + 6+ 5 = k or k = 25. The required hyperplane is

2~ + 5~ - 62 - 2 % ~ = -2.

4 ~ - 5 5 ~ 1 2 z + u : = 25.

276 POINTS, LINES AND HYPERPLAKES [CHAP. 22

22.24. Show that the equation of the hyperplane (Y in Rm which intersects the xi axis at aj f 0 is

2 , 2 , X - + - + . . . $2 1 1 a1 a2 am

Let clxl i czxz+ + c,x, = b be the equation of a. Substituting each x,-intercept (0, . . ., 0 , a,, 0 , . . ., 0) into the equation, we obtain the homogeneous system

ale1 - b = 0

a2cz - b = 0

a,c, - b = 0

where the ci and b are the unknowns. The system is in echelon form and we can arbitrarily assign a value to b . Set b = 1; then c i = l / n i . Thus the equation of a is as claimed.

22.25. Find the equation of the hyperplane in R4 which intersects the axis at -2, 4, 3 and -6, respectively.

By the preceding problem, the equation is

x1 2 2 5 xq - + - + 3 + - = 1 or -2 4 3 -6

6x1 - 35, - 4~~ 1 2x4 = -12

22.26. Find the equation of the hyperplane 01 in R4 which passes through the points (1,3, -4, a), (@,I, 6, -3), ( O , O , 2 , 5 ) and (O,O, 3,7).

Let a x + b y + c z + d w = e be the equation of 01. Substituting each point into the equation, we obtain the following homogeneous system of linear equation where a , b , c , d and e are the unknowns:

a + 3 b - 4 c + 2 d - e = 0 a + 3 b - 4 c + 2 d - e = 0 b + G c - 3 d - e = 0

2 c + 5 d - e = 0 I 3 c + 7 d - e = 0 Or I d - e = 0

b + 6 c - 3 d - e = 0

2 c + 5 d - e = 0

The system has been reduced to echelon form on the right. There is only one free variable, i.e. we can arbitrarily assign a value to e . Set e = 1. Then d = 1 by equation four, c = -2 by equation three, b = 1 6 by equation two, and a = - 4 1 by equation one. Thus the required equation is

The hyperplane a is uniquely determined since there is only one free variable in the In other words, every value of e determines a different equation but all

- 4 l ~ + l G y - 2 ~ + ~ = 1. Remark:

homogeneous system. these equations a r e multiples of each other and so they determine the same hyperplane a.

22.27. Show that there is more than one hyperplane in R4 which passes through the points (1,6, -1,4), (@,I, 2,2), (0,1,1,6) and (0,1,3, -2)-

Let ax + b y + c z + d w = e be the general equation of the hyperplanes passing through the given points. Substitute each point into the general equation and then reduce the system of equations to echelon form as follows:

~ + 6 b - c + 4 d - e 1 0 a + G b - c f 4 d - e = 0 a + 6 b - c + 4 d - e = 0 b + 2 c + 2 d - e = 0

or or C - 4 d = O

b + 2 c + 2 d - e = 0 b + 2 c + 2 d - e = 0

b S c + 6 d - e = 0 C - 4 d 1 0

b + 3 c - 2 d - e = 0 C - 4 d = 0

The system has two free variables, e and d ; t h a t is, we can arbitrarily assign values to both e and d and obtain a n equation of a hyperplane passing through the given points. Accordingly, all the equations will not be multiples of each other and so there is more than one hyperplane passing through the given points.

277 CHAP. 221 POINTS, L I N E S A N D H Y P E R P L A N E S

1 2 3

D 2 -1 -1

3 2 2

22.28. Show that there is a t least one hyperplane in Rm passing through any m arbitrary points.

clxl + c2x2 + * * * + c,z, = b , we obtain a system of n z homogeneous linear equations in the m + 1 unknowns, el, . . . , c , and b . Thus the system has a t least one free variable and so has a non-zero solution. The non-zero solution determines a hyperplane passing through the given points.

Substituting each of the points into the equation

= 1( -2+2) - 2 ( 4 + 3 ) + 3 ( 4 + 3 ) = 0 - 14 + 21 = 7

22.29. Show that any r < m arbitrary points in Rttz do not uniquely determine a hyperplane, i.e. that there is more than one hyperplane passing through the r points.

Substituting each of the r points into the equation clxl + c2x2 + * a * + c,x, = b, we obtain a system of r homogeneous equations in the m + 1 unknowns, el, . . ., c, and b. Since r < m, the system has a t least two free variables. Accordingly, all the solutions of the system will not be multiples of each other and so they determine more than one hyperplane.

1 3 -2

(i) D = 2 -1 1

4 5 -3

= l ( 3 - 5 ) - 3(-6-4) - 2 ( 1 0 + 4 ) = -2 + 30 - 28 = 0

Substitute in the second equation to obtain z = 2, and then substitute in the first equation to find x = 1. Thus the hyperplanes intersect in the unique point (1, -3,2).

22.31. Determine whether the given hyperplanes in R3 intersect in a unique point:

(i) x + 3 ~ - 2 x = 1, 2 x - y + x = 2 and 4x+5y-3x = 4;

(ii) x + y + x = 1, 2 x + y - x = 6 and 3 x - 3 y + x = -5.

In each case, compute the determinant D of the matrix of coefficients:

Since D = @ , the hyperplanes do not intersect in a unique point.

1 3 -3 11

Since D#@, the hyperplanes intersect in a unique point.


Show that iii arbitrary hyperplanes u1 x = b,, ti, * x = b,, . . . , urn* x = bm in Rrn intersect i ; ~ a unique point if and only if the ??z rectors P L ~ , U ~ , . . .,urn are (linearly) independent.

The system of linear equations in m unknowns has a unique solution if and only if the detemiinarit of the matrix A of coefficients is not zero. However, the determinant of A is not zero if and only if the rows of A are independent vectors. The result now follows from the fact that the rows of A are the vectors z i l .

MISCELLANEOUS PROBLEMS 22.33. Prove the Cauchy-Schwarz Inequality:

and v = ( t i l , . . .,vJ in Rm, For any pair of vectors u = (ul, . . .,urn)

m

11/ * III 6 c Iulvll 6 1 ull 110 1 2 = 1

If 7~ = 0 o r w = 0, then the inequality reduces to 0 6 0 f 0 and is therefore true. Thus we need only consider the case in which z( # 0 and w # 0, i.e. in which I J Z L I I # 0 and I I w I J # 0. Furthermore,

m

a = 1 IU. w1 = IuIz.I + . . . 1 ZtlnlJnll 6 Iulwl( + . ' . + ~ z ~ l , l w m / = (u,7J,l

Thus we need only prove the second inequality.

Now for any real numbers x , y E R. 0 f (z - Y ) ~ = r2 - 2ey + y2 or, equivalently,

2xy f 2 2 + y2 (1 )

Set Y = ? ( , I / z ~ l l and y = 1~~ / t in ( 1 ) to obtain, for any 1,

But, by definition of the norm of a vector, Thus summing ( 2 ) with respect t o i and using

((all = 2 a; = 2 ( z L i i 2 and Iztiwil = izti ' l w i l ,

(IwI( = 2 v? = 2 iv[l2. we have

tha t is,

hIultiplyinp both sides by I I L ' 1 w'l, we obtain the required inequality.

22.34. Prove Minkowski's Inequality: For any pair of vectors u = (ul, . . .,urn) and ZI = (vl, . . . ,vrn) in R", I ' Z L + vll 6 I(u(I + ~ ~ v ~ ~ .

If 1 + % I / = 0, the inequality clearly holds. Thus we need only consider the case in which 7i + 2) # 0 .

Now 7 ( 1 + v,I 6 I U , ~ + I w l ~ fo r any real numbers wul E R. Hence

IIu + w / p = I: ( Z L , + w , ) 2 = B IU, + w,/2 = 2 Ik + 4 1% + w,1

f 2 I ? ( , + w11 (IZL,i + Iv,l) = B I Z f , + vtl ILL,/ + B 1 % + wtl Iw,l

But by the Cauchy-Schwarz Inequality (see preceding problem),

~ ~ u , + w , 1 I ~ , l f I Z L + V I I llz~ll and ~ l z ~ , f w l ~ p,I ( l ~ + v l \ l I ~ 1 1

CHAP. 221 POINTS, LINES AND HYPERPLANES 279

SuppIementary Problems DISTANCE BETWEEN POINTS 22.35. Compute the distance between each pair of points:

(iii) (4,Oi and (-2, -71; (iv) (-3, -51 and 14, -11.

Determine k so that the distance between (1,3) and (2k, 7) is 5.

Find the point equidistant from ( 1 , 2 ) , i - l , 4 ) and (5,3).

Determine li so that the points (2, l), (-2, k ) and 16,4) a r e collinear, i.e. belong to the same straight line.

Show tha t the points 16,5), (2, -5) and (-1,2) are the vertices of a right triangle, and find its area.

(i) (5,-1) and ( 2 , G ) ; (ii) ( 2 , 2 ) and ( 5 , -5);

22.36.

22.37.

22.38.

22.39.

LINES 22.40.

22.41.

Plot the following lines:

Find the equation of the line: (i) parallel to the J- axis and passing through the point (3, -4); (ii) parallel to the u axis and passing through the point (-2,l) ; (iii) horizontal and passing through the point 15, -7); (iv) vertical and passing through the point (6,-3).

(i) x = -5, .t = -2, x = 1, x = 4; (ii) y = - 4 , y = -1, y = 2, ZJ = 6.

22.42. Plot the following lines: (i) 3 2 - 5y = 15 and 2x + 3y = 12; (ii) 42 - 3y = -6 and 5% + 2 ~ 1 = 8 ; (iii) 2 x + 5 y = 0 and 4x-3y = 0.

22.43. Find the slope of each line: (i) y = -2x i 8; iii) 22-5y = 14; [iii) z = 3y-7; (iv) x = -2; (v) ?J = 7 .

22.44. Find the equation of the line: (i) passing through (3 , -2) with slope &; (ii) passing through (-5,O) and (3 ,5 ) ; (iii) passing through (1, -3) and parallel to 2 s - 3y = 7; (iv) passing through (2 , -1) and perpendicular to 3 s + 4y = 6.

Determine the distance of each line from the origin: (i) 4x + 31/ = -8; (ii) 2x + 5y = 3;

Determine the distance between each point and line: (i) (2, -1) and 6s - 8y = 15; (ii) ( 3 , l ) and 3: - 3y = -6; (iii) (-2,4) and 12% + 5y = 3.

22.45. (iii) 12s - 5y = 26.

22.46.

EUCLIDEAN m-SPACE 22.47. Let u = (1, - 2 , 5 ) and v = (3,1, -2). Find:

(i) u + v; (ii) 2 2 ~ - 5v; (iii) 2 6 . v; (iv) I'dI; (v) d(u, w).

22.48. Let u = ( 2 , 1 , - 3 , 0,4) and v = (5, -3, -1,2,7). Find: (i) u+v; (ii) 32~-21;; (iii) u-v; (iv) I;v:l; (v) d(u,v), i.e. IIu-dI.

For each pair of vectors determine the value of k so t h a t the vectors a re orthogonal: (i) (5,k,-4,2) and (1, -3 ,2 ,2k);

Determine the value of k so t h a t the distance between ( 2 , k , 1 , - 4 ) and ( 3 , - 1 , 6 , - 3 ) is 6.

Show that a hyperplane is not bounded.

Show tha t a set A is bounded if and only if there exists a real number m* > 0 such tha t -m* 6 u, 5 177

22.49. (ii) (1 ,7 , k + 2 , -2) and ( 3 , k , - 3 , k ) .

22.50.

22.51.

22.52. for every component of any vector u E A .

HYPERPLANES 22.53. Calculate the distance of each hyperplane from the origin, i.e. the zero vector:

(i) 2x - y + 32 (iii) 3% + 2y + 22 - w = -6. 4u: = 7; (ii) 4x - 12y - 32 = -10;

22.54. Calculate the distance between the point and the hyperplane: ( i ) 12, -3 , l ) and 3 x + 4 y - l 2 x = 8; (ii) (5,-2,1,-3) and r + 72/-5x+5w = 1; (iii) (4,1,-2) and x + 8 y - 4 x = -7.

Find the equation o f the hyperplane in R3 which: ( i ) contains (2,-3,-5) and is parallel t o the xx plane; (ii) intersects the x, y and x axes at 2, -3 and 4 respectively; (iii) contains (1, -2,2), (0, 1 ,3) and (0,2, -1); (iv) contains (1,-5,4) and is parallel to 3%-7y-6x = 3.

Find the equation o f the hyperplane in R4 which: (i) contains the xl, x2 and x4 axes; (ii) contains (1,5,-4,3) and is parallel to xl-3x2-4x3-2x1 = -1; (iii) contains (1,1,1, l), ( O , l , -3, l), (O,O, 1,4) and (O,O, 2, -1); (iv) intersects the axes at 3, -1,2, and G , respectively.

Determine whether the given hyperplanes in R3 intersect in a unique point:

22.55.

22.56.

22.57. 2 - 4 y - 2 x = 1

\3x- y - x = o 1 I 5 2 + 2 y + x = 4

2 $ 2 y + x = 1 . v + 3 y + 5 x = 1

(i) 52 + y + x = 2 ; (ii) 32 - 2y - x = 2 ; (iii) 22 + 2y + x = 2 . 1 12"- y - 2 2 = 7

22.58. Show tha t the intersection of r<?n hyperplanes in Rm is either empty or contains an infinite number of points.


22.36. k = 2 o r k = -1

22.37. (2.3, 5.3)

(i) m, (ii) m, (iii) m, (iv) a

22.38. k = -2

22.39. 29

22.41. (i) y = -4, [ii) x = -2, (iii) = -7, (iv) x = 6

22.43. (i) -2, (ii) 2/5, (iii) 1/3, (iv) not defined, (v) 0

22.44. (i) 2 - 4 y = 11, (ii) 5x-8y = -25, (iii) 2%-33y = 11, (iv) 42-3y = 11

22.45. (i) 8/5, (ii) 3 / 6 9 , (iii) 2

22.46. (i) 4, (ii) G / n , (iii) 7/13

22.47. (i) u + v = (4, -1,3); (ii) 2u - 5w = (-13, -9,20); (iii) u v = - 9; (iv) I I U / I = 6; (v) d(u, w) = &i 22.48. (i) u + w = (7, -2, -4,2,11); (ii) 3u - 2w = (-4,9, -7, -4, -2); (iii) u w = 38; (iv) ;1w11 = fi =

2 a ; (v) d(2(,2)) = a 22.49. (i) k = 3; (ii) k = 3/2

22.50. k = 2 or k = -4

22.53. (i) 7/m, (ii) 10/13, (iii) 6/m = 2 / f i

22.54. (i) 2, (ii) 3, (iii) 3

22.55.

22.56.

(i) y = -3;

(i) x3 = 0; (ii) x1 - 3x2 - 433 - 22, = -4; (iii) 20x1 - 23x2 - 52, - x4 = -9; (iv) 2x1 - 62, f 3x3 + (ii) 6%- 4y f 3 x = 12; (iii) 13xl-4y + x = 7; (iv) 32 - 7y- 6x = 14

xq = 6

22.57. (i) No. (ii) No. (iii) Yes.

Chapter 23

Convex Sets and linear Inequalities

LINE SEGMENTS AND CONVEX SETS The line segment joining two points P = (ul, a2, . . .,am) and Q = (bl, b,, . . ., b,,) in

Rm, denoted by P&, is the set of points

where 0 L t f 1 tP + (1 - t)Q = ( ta , + (1 - t ) h , ta2 + (1 - t ) h , , . ., ta,,, + (1 - t )b j , l ) ,

A subset X of Rm is convex iff the line segment joining any two points P and Q in X is contained in X : P,Q E X implies PQ c X . Example 1.1.

The rectangular and elliptical areas below are each convex since the line segment joining any two points P and Q, in either figure, is contained in the figure. On the other hand, the U-shaped area to the right below is not convex, since the line segment joining the points P and Q also has points lying outside of the figure.

Example 1.2. Let X and Y be any two convex sets in R7n. We show that

the intersection X n Y is also convex. For let P and Q be any two points in X n Y ; then P, Q E X and P, Q E Y. Since X and Y a r e each convex, the line segment P& is contained in both X and Y: P&cX and P&cY. Hence P& is contained in the intersection: P& c X n Y . Thus X n Y is convex.

In fact, Theorem 23.1: The intersection of any number of convex sets is convex.

We next state a theorem about line segments which will be used throughout.

Theorem 23.2: Let f be a linear function defined over a line segment PQ. Then for every point A E P&, either

(i) f ( P ) 6 f (A) 6 f ( Q ) or (ii) f(Q) f ( A ) f ( p )

In particular, if f (P ) = f ( Q ) then f has the same value at every point in P&.

In other words, the value of a linear function f at any point of a line segment lies between the values of f a t the end points.

Remark: A function f defined over Rm is said to be linear if it is of the form

f = f(x1, X2) . . . , Xm) = C l X l + czxz + ' . . + C n 1 X m

281

282 CONVEX S E T S AND L I N E A R INEQUALITIES ;CHAP. 23

LINEAR INEQUALITIES The linear equation a,x, + a,x, + * * . + amxm = b partitions R" into 3 subsets:

(i) those points which satisfy the given equation; (ii) those points which satisfy the inequality (iii) and those points which satisfy the inequality

a,x , + a,zz + - - + amxnl b.

The first set, as defined previously, is a hyperplane; the other two sets are called open hal f spaces.

Accordingly, the so lu t ion se t or, simply, solution (or: graph) of the linear inequality a lx l + a2xz + - + a,,,~,,, 6 Z, (or: a l x l - ULT? - . * + + amxnl 2 b )

consists of the points of the hyperplane a,x, + a,x, + 9 - . + a n l ~ m = b, called the bound- i n g hyperplane , and of one of the open half spaces; such a set is termed a closed ha l f space or, simply, a ha l f space.

An important property of half spaces follows.

Theorem 23.3: Half spaces are convex.

inequality In particular, the solution of the linear

ax + b y c (or: ax+ by 2 c )

is called a ha l f plane. Geometrically speaking, the half plane consists of all of the points lying on the bounding line ax + by = c and on one side of the line.

Example 2.1. To draw the graph of the inequality 2 r - 3 y f -6,

first plot the line 2 x - 3 3 ~ = -6. Then choose an arbi- t r a r y point that doesn't belong to the line, say ( O , O ) , and test it to see if i t belongs to the solution set. Since (0,O) does not satisfy the given inequality, the points on the other side of the line belong to the solution set as shown on the right. Graph of 2% - 3y f -6

5

POLYHEDRAL CONVEX SETS AND EXTREME POINTS The intersection X of a finite number of (closed) half spaces in R" is called a polyhedral

convex se t . A point P E X is called an e x t r e m e poin t or c o m e r po in t of X if P is the intersection of 7n of the bounding hyperplanes determining X . Observe that Theorem 23.1 and Theorem 23.3 justify the use of the term convex; that is, half spaces are convex and so their intersection is also convex.

Theorem 23.4: The solution of a system of linear inequalities

a11n.1 + a12xz + * . . -a1,,>:c,,1 6 bl a21r1 + a 2 2 5 2 + * ' ' -a2ii,:c,n 6 bz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

is a polyhedral convex set.

The theorem follows from the fact that the solution to each inequality is a half space and that the solution of the system is the intersection of the individual solutions.

CHAP. 231 CONVEX S E T S A N D LINEAR INEQUALITIES 283

x - y = 4 -i y = -1 - B = (3, -l), the solution of

Remark: Since the sense of an inequality can be changed by multipIying che inequality by, say, -1, there is no loss in generality in assuming the inequalities in Theorem 23.4 are of the same given sense.

- 5

Example 3.1. The bounded polyhedral convex set in the figure

on the r ight is the solution of the following system of five linear inequalities:

x + u + x f 2 2 - - y 2 0

x'l x'0 2 ' 0

The points A , B , C, D, E and 0 in the diagram

solution; each point is the intersection of three of the bounding planes. For example, D is the intersection of the planes x + y + x = 2, x - y = 0 and x = 1; solving the three equations simultaneously. we obtain D = (a,+, 1). Similarly we can obtain A = (2 ,0 ,0) , B = (1, l , O ) , C = ( l , O , l ) , E = ( O , O , 1) and 0 = ( O , O , O ) .

are the corner points, i.e. extreme points of the Y

Remark: Although there are (i) = 10 different ways to select 3 of the 5 given bounding planes in the above example, four of the ten sets of equations will not determine an extreme point. In general, a set of m bounding hyperplanes of a polyhedral convex set X in R" does not determine an extreme point of X if and only if one of the following holds:

(i) its solution is empty, i.e. the equations are inconsistent;

(ii) its solution is infinite, i.e. the equations are dependent; (iii) its solution is unique but does not belong to X .

POLYGONAL CONVEX SETS AND CONVEX POLYGONS

convex set and, if it is bounded, a convex polygon. extreme point of X iff it is the intersection of two of the bounding lines of X.

In the special case of the plane R*, a polyhedral convex set X is called a polugonal Furthermore, a point P E X is an

z = 1

E

284 CONVEX SETS A N D LINEAR INEQUALITIES [CHAP. 23

Although there a re (i) = 6 different ways to select 2 of the 4 bounding lines,

p + 2 y = 4 E = (1, -3), the solution of , does not satisfy z -y L 4

lx = 1

J x - y = 4

\ y = -1 F = (6, -1), the solution of , does not satisfy y 5 -1

In other words, E and F do not belong t o the convex polygon.

Example 4.2. Consider the following four systems of inequalities:

x + 2 y f 4 x t 2 y f 4 x f 2 y 5 4 z + 2 y 4

x - y 5 0 x - y L 0 2 - - y =' 0 x - y g 0

y 22 -1 y -1 y 6 -1 y f -1

( i ) (ii) (iii) (iv)

In each case the bounding lines of the given half planes are x + 2y = 4, x - y = 0 and y = -1. a r e (i) = 3 different ways to select 2 of the 3 bounding lines. Furthermore,

There

x - 7 J = 0

y = -1 A = (-1, -l), is the solution of

x+21/ = 4

y = -1 B = (6,-l), is the solution of

x + 2 y = 4 -i x - y = 0 4 4 C = (3,s)' is the solution of

-5 1

CHAP. 231 CONVEX SETS AND LINEAR INEQUALITIES 285

I /

(iii) (iv)

We now discuss the solution of each system which is sketched above.

Here the solutions a re bounded and hence called a convex polygon. All three points A , B and C a r e i ts extreme points.

Here the solutions a re not bounded. Only the points A and C a r e i ts extreme points.

, tha t is, without the inequality Here the solutions a re the same as the solutions to

x + 2y 6 4. In this case we say tha t the inequality x + 2y f 4 is redztndant (or: superfluous). The point A is the only extreme point.

Here the solution is empty, and we say tha t the system is inconsistent. There a re no extreme points.

x - y 6 0

--1 1

LINEAR FUNCTIONS ON POLYHEDRAL CONVEX SETS We seek the maximum or minimum value of a linear function

f = ~ 1 x 1 + ~ 2 x 2 + * * + C m X m

where the unknowns are subject to the following constraints:

U l l X l + a12x2 + ’ ’ ‘ + U l n i X r n 6 61 U Z l X l + az2x2 + * * ’ + UZrnXrn I ba . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

U , l X l + ar2x2 + * * . + amxm 6 b,

That is, we seek the maximum or minimum value of a linear function over a polyhedral convex set. The following fundamental theorem applies:

Theorem 23.5: Let f be a linear function defined over a bounded polyhedral set X in R”. Then f takes on its maximum (and minimum) value a t an extreme point of x.

Thus the solution of the above problem consists of two steps:

(i) Find the extreme points of X . (ii) Evaluate the function f at each extreme point.

The maximum of the values of f in step (ii) is the maximum value of f over the entire set X, and the minimum value of f in step (ii) is the minimum value of f over X .

286 CONVEX SETS AND LINEAR INEQUALITIES [CHAP. 23

Example 5.1.

given by Find the maximum and minimum value of the function f = 2 r i 5y over the convex polygon X

2 + 2 y L 4 2 - y 5 4

x t l y 2 -1

The extreme points of X a r e as follows (see Example 4.1):

A = (1,-1), B = (3, -l), C = (4,O) and D = (1,;)

Evaluate f a t each of these points: f ( A ) = f(1,-1) = 2 . 1 - 5 . 1 = -3 f (C) = f ( 4 , O ) = 2 . 4 + 5.0 = 8 j ( B ) = f(3,-1) = 2 . 3 - 5 . 1 = 1 f ( D ) = f'(l,i) = 2 . 1 + 5 . ; = 9.5

Thus the maximum value of f over the entire convex polygon X is 9.5 and occurs at D, and the minimum value is -3 and occurs at A .

We shall prove Theorem 23.5 f o r the case m = 2 as a solved problem (see Problem 23.17). However, it is instructive to show here how this result follows from geometric considera- tions in the case of the preceding example.

We can consider the equation f = 2% + 5y as a one parameter family of lines in the plane R2 (see adjoining diagram). We seek the maximum (or: minimum) value of f with the condition that the line f = 2x+5y intersects the given convex polygon X . As indicated by the diagram, starting with lines of the family with large values of f , the family will first intersect X at the point D. Similarly, starting with lines having small values of f , the family will first intersect X at the point A . In other words, the linear function f = 2 x f 5 y will take on its maximum and minimum values over X at the points D and A, respectively.

Example 5.2.

The family f = 2x + 5y and X a r e plotted.

Find the maximum value of the function f = x + 2y 4- 82 over the polyhedral convex set X given by

x + y + z 6 2 x - y 2 0

z " 1 2 2 0 2 ' 0

The extreme points of X are (see Example 3.1, Page 283).

A = (2,0,0), B = (1,1,0), C = (1,0,1), D = (+ ,+ , I ) , E = ( O , O , l ) , 0 = ( O , O , O )

Evaluate f a t each of the six extreme points:

f ( A ) = 2, f ( B ) = 3, f(C) = 9, f ( D ) = 9.5, f ( E ) = 8, f(0) = 0

Thus, by Theorem 23.5, the maximum value of f over the entire polyhedral convex set X occurs a t D and is 9.5.

Consider now a function f of the form

f = ~ 1 x 1 + ~ 2 x 2 + * * * + CmXm + d

over a polyhedral convex set X in Rm. The function f differs from the linear function

CHAP. 231 CONVEX S E T S AND LINEAR INEQUALITIES 287

g = c1x1 + c2x2 + * * . + CmX,n

a t every point of X by the fixed amount d ; hence f takes on its maximum (and minimum) value a t the same point as g. In other words,

Theorem 23.6: A function of the form f = C l X l + CZXZ + * * * + C,nXm + d

defined over a bounded polyhedral set X in Rm takes on its maximum (and minimum) value a t an extreme point of X.

Example 5.3. Find the minimum value of the function

f = 2 z - 4 4 1 1 + 3 z + 2

over the bounded polyhedral convex set X of the preceding example.

Evaluate f a t each of the six extreme points of X :

f ( A ) = f ( 2 , 0 , 0 ) = 4 + 0 + 0 + 2 = 6 f ( D ) = f(+, +,1) = 1 - 2 + 3 + 2 = 4

f ( B ) = f ( l , l , O ) = 2 - 4 + 0 + 2 = 0 f ( E ) = f ( O , O , 1) = 0 4 0 + 3 + 2 = 5

f(C) = f ( l , O , 1) = 2 + 0 + 3 T 2 = 7 f(0) = f ( O , O , 0 ) = 0 + 0 + 0 + 2 = 2

Thus the minimum value of f over X occurs a t the point B and is 0.

Remark:

However, the following theorem always applies (see Problem 23.9):

If a polyhedral convex set X is not bounded, then a linear function f may or may not take on a maximum (and minimum) value.

Theorem 23.7: Let f be a linear function defined over a polyhedral convex set X. Then (i) f will take on a maximum (minimum) value over X if and only if the values of f are bounded above (below) and (ii) the maximum (minimum) value will always occur a t an extreme point of X.

Solved Problems LINEAR INEQUALITIES IN TWO UNKNOWNS 23.1. Graph each inequality:

(i) 2x + y 4 -4, (ii) 4% - 3 y -6, (iii) 2x - 3y 6 0.

In each case, first plot the line bounding the half plane by finding at least t w o points on the line; then choose a point which doesn't belong t o the line and test i t in the given inequality.

(i) On the line 22 + y = -4: if R: = 0, I / = -4: if y=O, z = - 2 . Draw the line through (0,-4) and ( - 2 , O ) . Test, say, the origin (0,O). Since (0,O) does not satisfy 21 + y 5 -4, i.e. 0 =f -4, shade the side of the line which does not contain the origin. ( i ) Graph of 2 x + y 6 -4.

288 CONVEX SETS AND LINEAR INEQUALITIES [CHAP. 23

(ii) On the line 42 - 3y = -6: if x = 0, y = 2; if y = 0, z = -1.5. Draw the line through (0,2) and (-1.5,O). Test the origin (0,O). Since (0,O) satisfies the given inequality 4x - 3y 2 -6, shade the side of the line containing the origin.

(iii) On the line 22 - 3y = 0: if z = 0, y = 0; if z = 3, y = 2. Draw the line through (0,O) and ( 3 , 2 ) . Since the origin belongs to the line, test the point (2,O). Since (2,O) does not satisfy the given inequality 22 - 3y 6 0, shade the other side of the line.

-5 1 (ii) Graph of 42-3y 5 -6. (iii) Graph of 22 - 3y 4 0.

23.2. Graph each inequality: (i) x 2, (ii) x 1, (iii) y -1.

(i)

(ii)

(iii)

First plot the line x = 2 , the vertical line meeting the x axis a t 2. the left of the line since the x-coordinates of these points are less than 2.

First plot the line z = 1, the vertical line meeting the x axis a t 1. Then shade the area to the right of this line since the x-coordinates of these points are greater than 1.

First plot the line y = -1, the horizontal line meeting the y axis a t -1. Then shade the area above the line since the y-coordinate of these points are greater than -1.

Then shade the area to

(i) Graph of 2 4 2. (ii) Graph of x 1. (iii) Graph of y 2 -1.

23.3. Graph each inequality: (i) 1 x 4, (ii) -2 4 y 4 1.

(i) Recall that 1 5 x 6 4 means 1 fz and 3 5 4 . Plot the vertical lines s = l and x = 4 ; then shade the area between the two lines as shown in Fig. (i) below.

Plot the horizontal lines y = -2 and y = 1; then shade the area between the lines as shown in Fig. 6.i) below.

(ii)


y = -2

- 4

(i) Graph of 1 f z 5 4. (ii) Graph of -2 f y 5 1.

23.4. Plot the solution of each pair of inequalities:

(i) 2% - 39 6 6 (ii) 2x + 5y 10 (iii) -1 x 6 3 2 x + y 6 4 X L l x - y 6 2

In each case, plot each half-plane with different slanted lines; the cross-hatched area is the required solution.

Graphsof 22-3y f 6 and 2 z + y f 4.

(ii)

22 -3376 5 6

2 z + y f 4 Solution of

Graphs of 2 2 + 5y 10 and z 2 1. 22 + 5y f 10 { 2’1

Solution of

( I J 1 . 1 1 ~ ~ piub cllci L ~ L G u - Y, Y I I G v r l V l l Y l ...vvl---D ---- .. I____ -. -. ~~

the left of the line since the 2-coordinates of these points a re less than 2.

(ii) Firs t plot the line x = 1, the vertical line meeting the s axis at 1. Then shade the area to the

(iii) F i r s t plot the line y=-1, the horizontal line meeting the y axis at -1. Then shade the area

right of this line since the s-coordinates of these points a re greater than 1.

above the line since the y-coordinate of these points a re greater than -1.

(i) Graph of x 6 2. (ii) Graph of z 5 1. (iii) Graph of y 2 -1.

23.3. Graph each inequality: (i) 14 x 4 4, Recall tha t 1 6 x 5 4 means 1 5 x and x r 4 . Plot the vertical lines x = l and x = 4 ; then shade the area between the two lines as shown in Fig. (i) below.

(ii) -2 4 y 4 1.

(i)

(ii) Plot the horizontal lines y = -2 and y = 1; then shade the area between the lines a s shown in Fig. (ii) below.

Each half plane is plotted.

Observe that the solution of the system is not bounded.

(ii)

/ / l / i l

- 5

The required solution.


Here the solution of the system is bounded.


CHAP. 231 CONVEX SETS A N D LINEAR INEQUALITIES

LINEAR FUNCTIONS OVER POLYHEDRAL CONVEX SETS 23.6.

23.7.

23.8.

291

Find the maximum and minimum of each function over the convex polygon X shown in the diagram on the right:

8 -

A = (0 ,2 )

c = ( 5 , l ) g = x-22J D = (3,5)

f = 2x +52J B = (4, -1)

- 2 h = - 3 x f 2 J

-2 - Evaluate each function at each extreme, i.e. corner point: X is shaded.

f ( A ) = t ( 0 , 2 ) = 0 + 10 = 10 y ( A ) = 0 - 4 = -4 h(A) = 0 + 2 = 2

f ( B ) = t(4,-1) = 8 - 5 = 3 g ( B ) = 4 + 2 = 6 h ( B ) = -12 - 1 = -13

f (C ) = t ( 5 , l i = 10 + 5 = 15 y(C) = 5 - 2 = 3 h ( C ) = -15 + 1 = -14

f ( D ) = t (3 ,5) = 6 + 25 = 31 ~ ( 0 ) = 3 - 10 = -7 h ( 0 ) = -9 + 5 = -4

Thus: maximum of f over X is 31 and occurs at the point D maximum of g over X is 6 and occurs a t the point B maximum of h over X is 2 and occurs a t the point A

minimum of i over X is 3 and occurs a t the point B minimum of g over X is -7 and occurs a t the point D minimum of lz over X is -14 and occurs a t the point C.

and

Find the maximum of the function f = 2 x - y over the convex polygon X of the preceding example.

Evaluate t' a t each of the four extreme points of X :

f ( A ) = i ( 0 , 2 ) = -2 f (C ) = f ( 5 , l ) = 9

f ( B ) = f(4, -1) = 9 f ( D ) = f (3 ,5) = 1

Thus the maximum of f over X is 9 and occurs at two of the corner points, B = (4,-1) and C = ( 5 , l ) : hence by Theorem 23.2 it will occur at every point on the line segment joining B and C.

The above can be interpreted geometrically: the member of the family of parallel lines determined by the equation f = 2% - y which has the maximum value of f and intersects X , passes through both B and C and so contains every point on the line segment joining B and C, as shown in the diagram.

The diagram also indicates tha t f takes on its minimum value at the point A .

Find the extreme points of the polygonal convex set X determined by the system 3, r+2y G 6 x-y 0

x G 4 Method 1 (Algebraic).

There a r e (i) = 3 ways to select 2 of the 3 bounding lines:

23.5. Plot the solution of each system of inequalities: (i) 3x +2y 6 (ii) ,T + 4y L 4

x-y 4 n: ‘1 -1 $ 5 2 y 1 0

(i)


Observe tha t the solution of the system is not bounded.

(ii)



Here the solution of the system is bounded.

The required solution,

_ _ . ~ . ~ iiiinimum value of It over X.

X and the family g X f 76. X and the family h = % - 276.


23.10. Find the maximum and minimum of the function f = x - 2y + 4 over the convex polygon X given by

x s y 5 4

Ic + 2y + -2

x-y 2 -2

x g 3

To obtain the extreme points of X, plot X as shown in the diagram. The points A , B . C and D a r e the corner points of X . The coordinates of each point can be obtained by finding the intersection of the appropriate pair of bounding lines:

x + 2 y = -2

x - y = -2

x + 2 y = -2

A = (-2,O) is the solution of

{ 2 = 3

{x+: 1 ," i x - y = -2

B = (3, -%) is the solution of

C = ( 3 , l ) is the solution of

x + y = 4 D = (1,3) is the solution of

Next evaluate the function f at each extreme point:

f(A) = --2'0+4 = 2 f (C) = 3 - 2 1 - 4 = 5

f ( B ) = 3 + 5 + 4 = 12 f ( D ) = 1 - 6 + 4 = -1

Thus the maximum of f over X is 12 and occurs at B, and the minimum is -1 and occurs a t D.

23.11. Find the extreme points of the polyhedral convex set X given by

2 + y + x g 3

2 x + 2 y + 2 4 x-y 0

X k O

2 ' 0

A point p E R3 is an extreme point of X iff it is the intersection of three of the bounding planes of X. There a re (g) = 10 ways to choose 3 of the 5 bounding planes:

x + y + x = 3

2 x + 2 y + x = 4

z = o (ii)

x + y + z = 3

2 x + 2 y + z = 4

x - y = 0 i i i)

x + y + z = 3

x = o

x = o

(vi) 1 x + y + z = 3

2 - y = 0 x = o

(v)

x - t y + z = 3

x - - y = 0

x = o

(iv)

x + y + z = 3

2 x + 2 y + x = 4

z = o (iii)

2 x + 2 y + x = 4 2 - y = 0

2 = 0

(vii)

294 CONVEX SETS AND L I N E A R INEQUALITIES CHAP. 23

2.'. 2y - z = 4

.t - g = 0

z -7 0

I I I I ' -y x = o = O I 2 - 0

Each system except (iii) yields a unique solution:

(vi) A , = (0, 3, 0)

(vii) A , = (0, 0, 4)

! i i i ) No solution, i.e. system is inconsistent. (viii) A s = (1, 1, 0 )

I I V ) A , = (0, 0 , 3 ) ( I X ) A , = (0, 2, 0)

(x) A i o = (0, 0, 0) rv) 3 3 A , = (2 - 0 ) ' 2 '

To find the extreme points of X, test each of the above points in the given inequalities. The following are not extreme points of X:

A = ($, f , 0) does not satisfy 2.x + 2u - z =' 4

A , = (0, 3,O) does not satisfy 2a i 2u x 4

A ; = ( O , O , 4) does not s a t ~ s f y .L y - z L 3

The other six points do satisfy the given inequalities and are the extreme points of X. The adjacent diagram sho\\s the position of these points. Note that we obtained the extreme points purely algebraically.

23.12. Find the maximum and minimum values of the function f = 4x-2y-x over the bounded polyhedral convex set X of the preceding problem.

Evaluate t' a t each of the six extreme points of X: t'(A1) 1 f(+,+,2) = 2 - 1 - 2 -1 !(A,) = f ( l , l , O ) = 4 - 2 + 0 = 2

f (AJ =- f(0,1,2) = 0 - 2 - 2 = -4 J(A, ) = f ( 0 , 2 , 0 ) = 0 - 4 + 0 -4

t(A4) 1 f (O,O,3) 0 - 0 - 3 -3 f (A1,) = f(0, 0 , O ) =z 0 + 0 + 0 z= 0

The maximum value of f over X is 2 and occurs at the point A s . The minimum value of f over X is -4 and occurs at the extreme points A , and A , and so also occurs at each point on the line segment joining A , to A,.

MISCELLANEOLIS PROBLEMS 23.13. Prove: Let f be a linear function defined over R":

f(X1, x 2 , . . . , X!!!) = C l X l + C 2 X r + . * . + C n l X r n

Then f o r any vectors P,Q E R"' and any real number k ,

(i) f ( p + Q) = f(P) + f(Q), (ii) f ( k P ) = k f(0

23.14. Prove: Let f be a linear function over R”’ with f ( P ) 6 f ( Q ) . Then for 0 6 t 6 1, f(&). f (p) L f ( t P + (1 - t )&)

Since t and ( 1 ~ f ) a r e each non-negative, t ( P ) f f ( Q ) implies

t j ( P ) 5 t f ( Q i and (1- t ) f ( P ) (1 - t ) f ( Q j

Hence, using the iesult of the preceding problem, we have

f ( P ) t f ( P ) + (1 - t ) f ( P ) t t ( P ) f (1 ~ tj f ( Q ) = t ( t P -- (1 - t ) Q )

and t ( f P - (1 - t )Q) = t f ( P ) + (1 - t ) fCQ) t f ( Q ) + (1 - t ) t ( Q ) = f ( Q )

23.15. Prove Theorem 23.2: Let f be a linear function defined over a line segment p& in R”‘. f ( P ) . Then for every point A E P&, either (i) f ( P ) A f ( A ) 6 f ( Q ) or (ii) f(Q) A f ( A )

In particular, if f ( P ) = f (&) then f’ has the same value a t every point in P&.

t between 0 and 1. The proof folloivs from the preceding problem and the fact tha t A = f P f (1 - t )Q for some

23.16. Prove Theorem 23.3: Let X be a half space a,x, + * . . + unIsm 4 b. Then X is convex: P , Q E X implies P& C X .

Now X consists of all the points A E R”’ such tha t f ( A ) 5 b where f = alxl + ~ 2 % ~ f . . . + a,,,r,,,

Hence if P , Q E X , then f(Pj 6 b and f ( & ) f b. Thus, by Theorem 23.2, if A E P& then either

t ( P ) 6 f ( A ) f f ( Q ) aE b or f ( Q ) 6 f ( A ) 5 f ( P ) f b

In either case T”A) 5 b and so W c X .

23.17. Prove Theorem 23.5 (for the case w i = 2): Let f be a linear function defined over a convex polygon X . Then f takes on its maximum (and minimum) value at a corner point of X .

Suppose that t’ takes on its largest corner value at the corner point P . Let A be any other point in X. We want to prove that f ( A ) 5 j ( P ) .

Since X is bounded, the line passing through the points P and A will intersect a bounding line of X a t a point Q such tha t A lies between P and Q: A E PQ. Furthermore, Q must P

\ / lie on the line segment joining two corners, say, U and V : Q E U T .

) V

Suppose f ( P ) < f ( A ) ; then by Theorem 23.2, f ( P ) < f ( A ) f f ( Q ) . That is, f ( A ) lies between f ( P ) and f ( Q ) . On the other hand, by Theorem 23.2, either i (Q! f f ( U ) or f(&) f f(V’). Combining the various inequalities, we finally have

f (p ) f ( A ) f(Q) f ( l 7 o r f ( P ) < f ( A ) 4 f ( Q ) f W ) However, both cases contradict the assumption tha t f ( P ) was the largest corner value. Accordingly, f(P) < f ( A ! IS impossible and so f ( A ) f t ( P ) .

23.18. Prove Theorem 23.5: convex set X in R”. extreme point of X .

Let f be a linear function defined over a bounded polyhedral Then f takes on its maximum (and minimum) value a t a n

The proof is by induction on m. If m = 1, then X is a line segment and by Theorem 23.2 f takes on its nlasimum (and minimunii value a t an extreme point, i.e. an end point.

296 CONVEX SETS AND LINEAR INEQUALITIES 'CHAP. 23

We assume the theorem holds fo r a polygonal convex set in Rm-1. Let A E X. Since X is bounded, the point A belongs to some line segment p& where P and Q each belongs to a bounding half space of X . By Theorem 23.2, f ( A ) must lie between f ( P ) and f(Q) and so f must take on its inaximum land minimum) value a t a point on a bounding half space.

Kow every point B E X on a bounding half space of X belongs to the intersection Y of X and a hyperplane H . However, Y is a bounded polyhedral convex set of H and H is isomorphic to R1"-1. Thus by induction f takes on its maximum value a t an extreme point of Y which is also an extreme point of X.

Supplementary Problems LINEAR INEQUALITIES IN TWO UNKNON'NS

23.19. Graph each inequality: (i) x - 2y f 4, (ii) 3x + y f 6, (iii) 2x + 5y 2 0, (iv) 3x - 5y -15.

23.20. Graph each inequality: (i) x f 3, iii) x 2 -4, (iii) y 6 2, iiv) y 1- -3.

23.21. Graph each inequality: (i) -2 f x f 2, (ii) 0 5 y 5 3.

23.22. Plot the solution to each system: ii) 2.c - y 6 4 (ii) x + y f 5 (iii) 22 4- 3y 6 (iv) -1 5 x 6 4

x 5 1 x - 2 y 2 2 - 1 f y 6 2 - 2 f y 5 1

23.23. Plot the solution to each system: i i ) x - y e 2 (ii, 3 r 1 2y 6 6 (iii) x - 2y 2 -4

2 x f y 2 -2 2 - y ' l - 2 5 x 5 1

u s 1 x 2 -1 y 2 -3

LINEAR FUNCTIONS OVER POLYHEDRAL CONVEX SETS

23.24. Find the maximum and minimum values of each function over the convex polygon X shown in the diagram on the right: ( i ) f = 3 x f 2 y ; f i i ) .g = --z + 2y - 4; (iii) h = x - 2y.

A = (-1,l) B (1,-2) D = (5,2)

C = (4, -1)

23.25. Find the maximum and minimum values of , f = 3 r - y over the convex polygon

x + 2 y f 4 x - y f 1

x % -1

23.26. Find the maximum and minimum values of the function f = 22 + y + 3 over the convex polygon given by

2 - y 2 0 3 x + y 2 3

x r 4

23.27. Draw the convex polygon X given by x + y 6 3 x - y 2 -3

y ' 0 x 2 -1 2 4 2

Find the corner points of X and the maximum and minimum values of the function f = 22 - y over X .


23.28. Draw the convex polygon X given by s + 2 y 6

x - y 2 0

2 ' 4

y r o

Find the corner points of X and the maximum and minimum values of the function f = 2 x - 3 y over X.

23.29. Find the extreme points of the following bounded polyhedral convex set X in R3: 3 x + 3 y + x 5 6

x - y 2 0 2'0

y ' 0 2 ' 0

2 5 3 Find the maximum and minimum values of the function f = x + 2y - x over X .

23.30. Draw the polygonal convex set X given by z f 2 y ' 4 2 - y 2 -2

2 ' 2

Find the maximum and minimum values (if they exist) of each function over X

f = 2 x + + , g = 2 2 - y y , h = x - 3 y


(ii) (iii)

23.20.

(ii) (iii)

298

23.21.

23.22.

23.23.

CONVEX SETS AND LINEAR INEQCALITIES

4 1

;CHAP. 23

-4

(ii)

\

-4

(ii) (iii)

\ 4i

i i'i (ii) (iii)

23.24. (i) max f = 19, min f = -1; i,ii) max g = -1, min g = -10; (iii) rnax h = 6, min h = -3

23.25. max f = 5, min f = -5.5

23.26. max f = 15, niin f = 2

23.27. A = (-1, 0) B = (2, 0 ) c = (2, 1) D = (0 , 3) E = (-1, 2)

Maxi = 4, a t B. Minf = -4, a t E.

CHAP. 231 CONVEX S E T S A N D L I N E A R INEQUALITIES 299

23.28. A = (0, 0) B = (4, 0) c = (4, 1) D = (2, 2)

Maxf = 8, a t B. Min f = -2, a t D .

23.29. The corner points are: (O,O,O), (2,0,0), ( 1 , 1 , 0 ) , (0,0,3), (1,0,3), (4, $,3) . Max f = 4, min i = -3.

23.30. Maxf = 5 a t A . niin t’ does not exist. Max g and niin g do not exist. Min h = -6 a t 8. max h does not exist.

Chapter 24

linear Programming

LINEAR PROGRAMMING PROBLEMS

of a linear function

called the objec t ive function, over a polyhedral convex set X which includes the condition that the unknowns (variables) are non-negative:

Linear programming problems are concerned with finding the maximum or minimum

f = C I X I + C Z X Z + * * . + CntXrn

21 2 0, X Z ~ 0 , . . ., X , n 2 0

Each point in X is called a feasible solution of the problem, and a point in X a t which f takes on its maximum (or minimum) value is called an o p t i m m solution.

Example 1.1. (Rlaximum Problem.) 16 sq. yd. cotton, 11 sq. yd. silk, 15 sq. yd. wool.

A suit requires the following: 2 sq. yd. cotton, 1 sq. yd. silk, 1 sq. yd. wool. A gown requires the following: 1 sq. yd. cotton, 2 sq. yd. silk, 3 sq. yd. wool. I f a suit sells for $30 and a gown for $50, how many of each garment should the tailor make t o obtain the maximum amount of money?

A tailor has the following materials available:

The following table summarizes the above facts:

Let 5 be the number of suits the tailor makes and y the number of gowns. We seek to maximize

f = 30% + 50y subject to 2 x + y L_ 16, x + 2y f 11, x i - 3y 15, r 2 0, y % 0

Observe that the inequality 2 x + y 6 16 corresponds to the fact tha t 2 sq. yd. of cotton is required for each suit and 1 sq. yd. for each gown, but only 16 sq. yd. is available. The inequalities x + 2 y f 11 and x + 3 y f 15 correspond to the analogous facts about the silk and wool,

the above inequalities. Fig. 24-1 is the graph of the convex polygon X given by

The corner points of X a r e

0 = ( O , O ) , A = (8,0), B = (7,2), C = (3 ,4) , D = (0,5) The values of the objective function f = 302 + 50y at the corner points a re

f(0) = 0 , J ( A ) = 240, f ( B ) = 310, f(C) = 290, f ( D ) = 250

Thus the maximum of f over X is 310 and occurs at the point B = ( 7 , 2 ) . In other words, the maximum amount of money the tailor can get is $310, and this will happen if he makes 7 suits and 2 gowns. Fig. 24-1

300

CHAP. 241 LINEAR PROGRAMMING 301

Example 1.2. (Minimum Problem.) A person requires 10, 1 2 and 12 units of chemicals A, B and C, respectively, for his garden. A liquid

product contains 5 , 2 and 1 units of A, B and C, respectively, per j a r ; and a dry product contains 1, 2 and 4 units of A , B and C, respectively, per carton. If the liquid product sells for $3 per j a r and the dry product sells for $2 per carton, how many of each should he purchase to minimize the cost and meet the

The requirements? following table summarizes the above facts:

Units per j a r Units per carton Units needed

Chemical A 5 1 10 Chemical B 2 2 12 Chemical C 1 4 12

cost $3 $2

Let x denote the number of j a r s bought and y the number of cartons. We seek to minimize the function g = 32 + 2y

subject to the conditions 1 2

5x + y 2 10, 22 + 2y 2 12, z + 4 y 2 12 P and the fact that z and ?j a r e non-negative, z 0, u 2 0.

this minimum linear programming problem. The corner points of T are

P = (0, lo), Q = (1,5), R = (4,2), S = (12,O) Nthough T is not bounded, the objective function y = 32 + 2y is never negative over T and so takes on a minimum value a t a corner point of T .

the minimum value of g over T is 13 and occurs a t Q = (1,5). In other words, he should buy 1 j a r and 5 cartons.

Fig. 24-2 is the graph of the polygonal convex set T of 8

4

Since g ( P ) = 20, g(&) = 13, g(R) = 16, g(S) = 36 4 8 12 16

Fig. 24-2

DUAL PROBLEMS Every maximum linear programming problem has associated with it a corresponding

minimum problem called its dual and, conversely, every minimum linear programming problem has associated with i t a corresponding maximum problem called its dual. The original problem is termed the prinzal problem. To be more specific, we define each of the following problems to be the dual of the other: (i) Maximum problem:

Maximize f = C I X I + ~ 2 x 2 + . * + C m X m

0, x z 0 , .... X m 10.

Minimize g = b1zu1 + b 2 ~ 2 + - * * + b k W k

and

(ii) Minimum problem:

subject to allwl + alzc2 + * * * + aklwii 2 c1

a12wl + a 2 m 2 + . . . + ak~wk 2 cz

almzuI + ~ 2 ~ ~ 2 + * * + akmwk C m

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

302 LINEAR PROGRAMMING [CHAP. 24

Let us write the coefficients of the above inequalities in the form of a matrix as follows with the coefficients of the objective function as the last row:

(i) Maximum problem: (ii) Minimum problem:

In each case the matrix of coefficients of the dual problem can be obtained by taking the transpose of the matrix of coefficients of the primal problem. (Note that there is no last entry in the last row and last column of either matrix.)

Example 2.1.

(primal) problem and then take its transpose: Find the dual of the maximum problem of Example 1.1. Write the matrix of coefficients of the given

Primal problem Dual problem

3 15 16 11 15

30 50 *

Thus the dual minimum problem is as follows:

Minimize g = 16u + l l v + 15w

subject to 2u + v f w 2 30 u + 2v -I- 3w 2 50

and ZL 2 0, v 0, w 0.

Example 2.2.

given (primal) problem and then take i ts transpose: Find the dual of the minimum problem of Example 1.2. Write the matrix of coefficients of the

Primal problem Dual problem

1 4 12 10 12 1 2 3 2 *

Thus the dual maximum problem is as follows:

Maximize f = 10% + 12v + 12w

subject t o 5u + 2v + w f 3

ZL+2V+4W 6 2

and 21 2 0, v 2 0, w 5 0.

In general, a linear programming problem need not have an optimum solution. How- ever, the following theorem, which is the main result in the theory of linear programming, applies:


Duality Theorem: The objective function f of a maximum linear programming problem takes on a maximum value if and only if the objective function g of the corresponding dual problem takes on a minimum value and, in this case, m a x f = min g. Furthermore, if P and Q are feasible solutions such that f ( P ) = g(&), then P and Q are optimum solutions of their corresponding problems.

The proof of this theorem lies beyond the scope of this text.

Example 2.3. Consider the maximum problem of Example 2.2. The

graph of the polyhedral convex set X given by the inequalities of t h a t problem appears on the right. The corner points a r e

P = ( lOil9, 0, 7/19)

Q = (3 /5 , 0, 0)

R = (114, 7/8, 0)

s = 10, 1, 0 )

T = (0, 0 , 3) 0 = (0, 0, 0 )

2)

The objective function t' = lox + 12u + 12w takes on its maximum value a t the point R and max f = 13. On the other hand, min g = 13 for the corresponding miniinum problem appearing in Example 1.2, as predicted by the Duality Theorem. U

MATRIX NOTATION

as follows: Let the matrix A , the column vectors b and x, and the row vectors c and w be defined

c = ( C l , c2, . . ., e m ) b = x = A = u": . . . . . . . . * . . . . . . . 1:: : : : I:), p, [;:I ' 20 = (201, '202, . . ., W k )

a k l a h 2 ' ' ' a k m b k / X m

Then the maximum and minimum linear programming problems (1) in the preceding section can be rewritten in the following short compact form:

(i) Maximum problem: Maximize f = cx subject to A x b subject to zuA c ( 1 ) and X 0. and 20 0.

(ii) Minimum problem: Minimize g = wb

Remark: If ZL and v are vectors, then zc than or equal to the corresponding component of v.

v shall mean that each component of u is less

INTRODUCTION TO THE SIMPLEX METHOD Let X be the polyhedral convex set, i.e. set of feasible solutions, of the above maximum

linear programming problem. An optimum solution can be found by evaluating the objective function f a t each corner point of X. However, if the number of unknowns and inequalities is large, then i t can be a long and tedious task to find all the corners of X . An alternate way of finding an optimum solution is the following:


(1) Select a corner of X as a starting point. (2) Choose an edge through this corner such that f increases in value along the edge. (3) Proceed along the chosen edge to the next corner. (4) Repeat steps (2) and (3) until an optimum solution is obtained, i.e. a corner is

reached such that f does not increase in value along any of its edges.

The above process is accomplished mathematically by the so-called simplex method. Observe, first of all, that the maximum linear programming problem (I), Page 301, is equivalent to the following problem which contains only linear equalities and not linear inequalities; the variables xm 1, . . , , rFl1 t lc are called "slack" variables:

Maximize f = clx1+cLx3+ . . . + c , x m + O ~ x , l t ~ + O ~ ~ m + z + . . . + O * x m t h

subject to ~ ~ 1 1 ~ 1 + U U X ~ + . . . + a l m X m + ~ ~ - 1 = bl a 2 1 X 1 + a 2 2 2 2 + . + + azmxm + X n i t 2 = bz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2) f Z m t k = bk a k l X l + a h a x 2 c ' ' ' + a k m X m

and x1 A 0, X L 0, . . ., X m 0, X m t l 0, . . . . X m + k % 0.

The algorithm of the simplex method consists of four steps (see Page 307) which correspond, respectively, to the above steps. For the sake of clarity we shall explain each step separately.

Remark: In order to avoid special cases, we shall assume the following additional conditions on our linear programming problems: (i) Positive condi t ion: Each component of b is non-negative ( b % 0) and each

row of A contains a t least one positive entry. (ii) Non-degeneracy condi t ion: If r is the rank of A , i.e. the number of inde-

pendent rows (or columns) of A, then c is not a linear combination of less than T rows of A and Z, is not a linear combination of less than r columns of A .

INITIAL SIMPLEX TABLEAU

equivalent problem (2) is as follows: The initial simplex tableau of the maximum linear programming problem (1) or the

P1 Pa * ' * Pm

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a k l a k z " ' akm 1 0 0 " ' 1 I

I I I I

J indicators

Observe that the rows of the tableau, excluding the last, are the coefficients of the linear equalities in ( 2 ) . The columns of the coefficients of the non-slack variables are labeled Pi, respectively; their significance will appear later. The entries in the last row, excluding the last entry, are called i nd ica tor s and are the negatives of the coefficients of the objective function f in ( 2 ) . The entry in the lower right hand corner is zero; i t corresponds to the value of the objective function f at the origin.


subject to 5 u + 2w + w 5 3

Remark: Although the simplex method is stated in terms of the maximum linear programming problem, the method will give the optimum solution to both the maximum problem and to its dual minimum problem.

5 2 1 1 0 3

Example 3.1. Maximize j = 30.1. - 50y

subject to 22: + y 6 16 2: + 22/ 5 11 3: + 3y 6 15

and n 0 , u ’ O

0 -1 0

Initial simplex tableau:

Pl pi?

15

-30 -50 0 0 0 0

0 2 0 - 1 1 1 5

7 1 0 1 6

Example 3.2. Maximize f = 1 0 ~ ~ + 12v + 12w

-30 -50 , 0 0 0

Initial simplex tableau:

Pl pz p3

0

ti + 2v T 4w ‘ 2

and Z L ~ O , w 2 0 , W k O

Example 3.3. Initial simplex tableau:

Maximize f = x 2y + 8x PI pz p3

subject to z r 1

. r . + y + x f 2 - z + y ‘ 0

-1 1 l o

and 3 - 5 0 , y z o , 75’0

PIVOT ENTRY O F A SIMPLEX TABLEAU A pivot entry or simply a pivot of a simplex tableau is obtained as follows:

(i) Arbitrarily select a column which contains a negative indicator.

(ii) Divide each positive entry in this column into the corresponding element in the last column.

The entry which yields the smallest quotient in step (ii) is the pivot; its column is called the pivotal column and its row the pivotal row. If the tableau has no negative indicator, then it is a terminal tableau and has no pivot. Properly interpreted, the terminal tableau yields a solution to our problem. The procedure which follows represents a method of passing to such a terminal configuration.

Example 4.1.

Consider the following simplex tableaux:

Pl p2

I 1 2 1 1 0 0 1 1 2 1

P I pz p3

0 0 1 I 1 0 0 1 1

1 @ 0 1 - 1 1 0 1 2 1

(ii)


0 1 0 - 1 1 1 4

0 0 0 40 -10 290

(iii) f iv)

(i) Either the first column o r the second column can be chosen a s the pivotal column. In the first case 2 is the pivot since 312 is less than 12/1 and 5/3. In the second case 1 is the pivot since 5/1 is less than 1212. We need not consider the quotient 3/(-1) since -1 is not positive.

Here the second column must be chosen as the pivotal column since i t is the only one with a negative indicator. Furthermore, 1 is the pivot since 2/1 is less than S/2.

(iii) The fifth column is the pivotal column and 3 is the pivot since 6/3 is less than 4/1.

(iv) There is no pivotal column and no pivot since there is no negative indicator. This tableau is a

(ii)

terminal tableau.

CALCULATING THE NEW SIMPLEX TABLEAU Let D be a simplex tableau with rows R, whose pivot appears in the rth row and

sth column; say D is the matrix (&) and so d,, is the pivot. A new tableau D* with rows RF is calculated from D (using appropriate “elementary row operations”) so that 1 appears in the original pivot position and zeros elsewhere in that column. This is accomplished as follows:

(i) First divide each entry of the pivotal row R, of D by the pivot d,, to obtain the corresponding row R: of D*::

1 drs

R: = -I?,

If the pivotal column of D is labeled P,, then this row of D* is labeled (or relabeled) Pi.

(ii) Every other row €2: of D* is obtained by adding the appropriate multiple of the above row R: to the row Ri of D:

If one of these rows of D is labeled Pi, then the corresponding row of D* is given the same label; otherwise the row is not labeled.

R: = -disRT -k R,

Example 5.1.

Consider the following tableau with a pivot entry circled:

R,: R,:

R$

R,: I I I

1 - 8 - 6 1 0 0 0 1 0 1

We remark tha t the circled 2 is a pivot entry since i t appears in a column with a negative indicator and 412 is less than both 22/1 and 914. We now calculate the new tableau step by step:

(i) Divide each entry of the pivotal row R, by the pivot 2 t o obtain the new row RZ; label Rl with the label of the pivotal column, i.e. PI:

CHAP. 241 L I N E A R PROGRAMMING 307

2

R4 (ii) RF is obtained by multiplying the above row R: by -1 and adding i t t o R,: RT = -R,* + R,;

R: is obtained by multiplying R,* by -4 and adding i t to R3: R; = -4R* J- R3; RZ is obtained by multiplying RZ by 8 and adding i t to R,: RT = SR; + R4. These rows do not change labels. Thus the completed new tableau is

PI p ,

The intermediate tableau appearing in the preceding example was illustrated separately from the completed tableau only for the sake of clarity; in general, we calculate the new tableau directly from the given tableau. We remark that if we choose the second column as the pivotal column and so obtain a different pivot, then we would calculate a different new tableau. This corresponds to the fact that in moving from corner to corner there may be more than one edge on which the objective function f increases in value; hence there is more than one way to finally reach the optimum solution.

The simplex method begins with the initial tableau and then calculates new tableaux one after the other until a terminal tableau is obtained. We remark that in each step a t most one row, the pivotal row, changes its label; the columns keep the same label throughout.

Example 5.2. Consider the following tableau with a pivot entry circled:

Pl p2

Pl

0 0 0 -10 0 4 0 16

Divide each entry of the pivotal row by the pivot 2 to obtain the row o l o l g a

The new tableau is Pl p2

(*I

(**)

1 0 0 ~ 0 1 4 5 1 2 1 1

It is calculated as follows:

(i) (ii)

(iii) the third row is (**); it i s labeled P,, the label of the pivotal column of (*); (iv) the fourth row is obtained by multiplying ( * c ) by 10 and adding the result t o the fourth row of (*).

Note t h a t the new tableau is a terminal tableau, i.e. it has no negative indicators.

the first row is obtained by multiplying (**) by -3 and adding the result t o the first row of (*); the second row is obtained by multiplying (**) by fr and adding the result t o the second row of (*); it retains the label PI;

308 L I N E A R PROGRAMMING [CHAP. 24

INTERPRETING THE TERMINAL TABLEAU

ming problem ( I ) , Page 301, we obtain the following terminal tableau: Suppose after performing the algorithm of the simplex method on the linear program-

. . . . . . some labeled with Pi . . . . . . . . . . . . . . . . . . . . . . . . . .

a * . . . * * a . . . I I I

. . . * I 4 , (I, . . . q k I v I * 4:

Then: (i)

(ii)

(iii)

v, the entry in the lower right hand corner of the tableau, is the maximum value of f (and so by the duality theorem is the minimum value of g: m a x f = m i n g = v); the point Q = (q , , qL, . . . . (I,) is the optimum solution of the minimum problem; its components are the last entries in the columns corresponding to the slack variables; the point P = (p l , p l , . . . . p,,,), where

' I 1 = ), 0 if no row is labeled P, 1 the last entry of a row labeled Pt

is the optimum solution of the maximum problem.

Thus a necessary condition (and so a check) that the calculations are correct is that f ( P ) = g ( Q ) = v. Example 6.1.

Find the optimum solution Q = ( q l , . . . . q k ) of the minimum problem, the optimum solution P = (p , , . . ~,p,?,) of the maximum problem, and v = max f = min g according to each of the following terminal tableaux:

PI p2 p, PI P , P ,

::: *

* :* * ::: * *

(i) (ii)

(i)

(ii)

Q = ( 3 , 4 , 5 ) , P = (0 ,2 ,6) , w = max f = min g = 19. no row is labeled P,.

Q = ( 2 , 5 ) , P = (0,7,0), w = max f = min g = 11. a r e each 0 since no row is labeled P , o r P,.

Note that the first component of P is 0 since

Note that the first and third components of P

ALGORITHM OF T H E SIMPLEX METHOD

tively, to the steps on Page 304: The algorithm of the simplex method consists of four steps which correspond, respec-

(1) Construct the initial tableau. (2) Find and circle a pivot entry in the tableau. (3) Calculate the new tableau from the given tableau and circled pivot. (4) Repeat steps (2) and (3) until a terminal tableau is obtained.

The following examples illustrate the use of this algorithm.

CHAP. 241 L I N E A R PROGRAMMING 309

-30 -50

Example 7.1. Applying the simplex method to Example 1.1, Page 300, we obtain the following sequence of tableaux:

0 0 0 0

Pl p2

0 0 9 1 1 5 13

Pl

p2 0 -1 , 0 0 0 40 -10 290

(iii)

Thus, by the terminal tableau (iv), the optimum solution is P = (7,2) since 7 and 2 are the last entries in the rows labeled Pl and P , respectively, and the maximum value of f is 310. This agrees with the previous solution of the problem. Observe that the numbers 0, 250, 290 and 310 which appear respectively in the lower right hand corners of the tableaux correspond precisely to the values of the objective function f at the points 0, D , G and B in Fig. 24-1.

Example 7.2.

initial tableau of the dual maximum problem: We solve the problem in Example 1.2, Page 301, by the simplex method; the first tableau is the

I ; : I : : I : ) 0 -10 -12 -12 0 0 0

Pl p2 p3

PZ

-4 0 12 12

(ii)

Remark: The rows and columns of the tableaux in the simplex method are labeled so that we can obtain the optimum solution of the maximum problem from the terminal tableau. However, if only the optimum solution of the minimum problem is required, then we can dispense with the labels. For example, the labels in the preceding example were not needed.

310 LI N E A R PRO GRAM PVI I N G [CHAP. 24

Solved Problems DUALITY 24.1. Find the dual of each problem.

i i ) Maximize f = 3n: t 5 y + 22 (ii) Minimize g = 22 + 6y + 72

x + 2y + 52 2 4 2 x - y + 2 x A 3

subject to 2x - y + 32 4 6 subject to x + 2 y + 4 x 4 8

and X k O , 2 / 2 0 , 2 2 0 . 3 x + 5 y + x 2 1

and X k O , y”0, 2 2 0 .

In each case form the matrix of coefficients of the given problem with the objective function as the last row, and then form the transpose matrix.

(i) Coefficient matrix: Transpose:

2 - 1 3 6 2 1 3

6 8 *

(ii) Coefficient matrix:

6 7 *

Transpose:

1 2 3

Dual problem: Minimize g = 621 + 8v subject to 2u + v 2 3

- u + 2 v ?= 5 3u + 4v 2 2

and u k 0 , v k 0 .

Dual problem: Maximize f = 4 2 ~ + 3v + w subject to u + 2v + 3w f 2

2u - v + 5tu f 6 5 u + 2 v + w f 7

and u s o , v s o , w s o .

24.2. Find the dual of each problem.

(i) Maximize f = 5x - y + 22 (ii) Minimize y = 2r + 5s + t subject to 7x + 2y - x L 8

x - 3 9 - 2 8 A 4 subject to 9r - 3s + 5t 2 7

6 r - 4 s - 3t L 2 ~ x - Y + ~ z 6 5

and r Z 0 , s k 0 , t h o . and x ” 0 , 21’0, X L O .

ii) Firs t multiply the second inequality by -1 to change its sense to 6.

Coefficient matrix: Transpose: Dual problem: Minimize g = 8zc - 4v + 5w subject to 7u - v -t 3w 2 5

7 2 - 1 8 7 -1 3

2 u + 3 v - w 1 -1

and u 5 0 , v t o , w t o . [-: 5 -: -1 -j [-: 8 -4 : -; -j - ~ t 2 ~ t Gw 2 2

iii) Firs t multiply the second inequality by -1 to change its sense to 5 .

Coefficient matrix: Transpose:

9 - 3 5 7 9 -6 2

7 -2 *

Dual problem: Maximize f = 72 - 2y

subject to 9x - Gy f 2 -3x+ 4y 5 5

5x t 3y f 1

and z 2 0 , y t o .

CHAP. 241 L I N E A R PROGRAMMING

2 1 1 1 0 0

311

7

SIMPLEX METHOD 24.3. Find the initial tableau of each problem.

(i) Maximize f = 5% f 2 y - x (ii) Minimize y = 2r + 5s + t subject to x - 3y f 4 x 6 8

3x + 2y + 62 L 5 2 ~ + 7 y - ~ f 9

and X L O , y’0, , 2 2 0 .

subject to 9 r t 2s + 4t 2 6 6 7 ’ + 3 ~ - 2 t 8

and

(i) The initial tableau is of the form w[ with the columns of A labeled Pi.

Pl pz p3 pi 2 7 - 1 0 0 1 1 9

- 5 - 2 1 1 0 0 0 1 0 1

(ii) Find the dual maximum problem and then construct the initial tableau as in (i).

PI pz

24.4. Find a pivot entry of each of the initial tableaux in the preceding problem. In each case, (1) select any column with a negative indicator as the pivotal column,

(2) divide each positive entry in this column into the corresponding entry in the las t column, and (3) choose as the pivot t h a t entry which yields the smallest quotient in 12).

(i) We can choose either the first o r second column as the pivotal column. ( a ) If the first column is chosen, f o r m the quotients 8/1, 513 and 912. Since 5/3 is the smallest

number of the three, the entry 3 in the second row is the pivot.

( b ) If the second column is chosen, form the quotients 512 and 917. Since 9/7 is the smaller of the two numbers, the entry 7 in the third row is the pivot. Note tha t the quotient 8/ ( -3) is not considered since -3 is not a positive number.

We can choose either the first o r the second column as the pivotal column. In the first column, 9 is the pivot since 2/9 is the smallest of the numbers 2/9, 5/2 and 114. In the second column, 6 is the pivot since 2/6 is the smaller of the numbers 216 and 513.

(ii)

1111 Coefficient matrix:

\ 2 6 7 *

Transpose: Dual problem: 1 2 3 2 Maximize f = 4u + 3v + w

subject to ZL + 2v + 32u 6 2 Z u - v + f w 6 6 5 u + 2 v + w f 7

and U h O , ? . ‘ s o , w “ 0 .


(i) Maximize f = 5% - 3 + 22 (ii) Minimize y = 2 r + 5s + t subject to ‘ix + 2y - x _L 8

x - 3 y - 2 2 1 4 3 x - y + 6 ~ 6 5

and x”0, y”0, x - - ” o .

subject to 9 r - 3s + 5t 1 7 6r - 4s - 3t f 2

and T ~ O , s”0, t ” 0 .

(i) First multiply the second inequality by -1 to change its Sense to 6 ,

Coefficient matrix: Transpose: Dual problem: 2 -1 8 7 - 1 3 5 Minimize g = 8u - 4v 3. 5w

subject to 7u- v + 3w 5

- ~ + 2 v + 6 w 2 2 2 u + 3 v - w 2 -1

and u ” 0 , v s o , W S O . 5 -1 8 -4

(ii) First multiply the second inequality by -1 to change its Sense to k .

Transpose: Dual problem: Coefficient matrix:

Maximize f = 72 - 276 subject to

- 3 2 + 4 y 5 5 5 x + 3 y f 1

9 - 3 5 7 9 -6 2 9% - 6y f 2

7 -2 * and z ” 0 , y 1 0 .

Multiply ( 1 ) by t and add the result to the second row to obtain 3 1 1 1 1 0 0 z 1 6 1 f j

Multiply ( 1 ) by 7 and add the result to the last row to obtain o o o $ z 4 1 s

Thus the calculated tableau is

Pl

p2


0 0

24.7. Using the pivot circled in the following tableau, calculate the new tableau and then find a pivot entry of it.

pi p ,

0 3 - 7 12

2 3 ' 1 0 0

I 3 8 ' 0 1 o I 1 : l

Divide each entry in the pivotal row, i.e. the third row, by the pivot 1 to obtain the (same) row

1 2 0 0 1 3 ( 1 )

Multiply ( 1 ) by -2 and add the result t o the first row to obtain

0 - 1 1 0 - 2 1

Multiply (1) by -3 and add the result to the second row to obtain

0 2 0 1 - 3 2

Multiply ( 1 ) by 2 and add the result t o the last row to obtain

0 - 6 0 0 2 6

Thus the new tableau is Pl pz

where the third row is labeled P I , the label of the original pivotal column.

the pivotal column. The second column of this tableau has the only negative indicator and so must be chosen as

Since 2/2 is less than 3 / 2 , the 2 in the second row is the pivot.

24.8. Using the tableau and the pivot obtained in the preceding problem, calculate the new tableau.

Divide each entry of the pivotal row, i.e. the second row, by the piTiot 2 to obtain the row

(1 ) 0 1 0 1 - 3 2 2 1

Then the new tableau is Pl p2

PZ 2 2

Pl 0 -1 4


24.9. In each of the following terminal tableaux, find the optimum solution Q = (ql, . . ., q,) of the minimum problem, the optimum solution P = (pl, . . .,p,) of the maximum problem, and v = m a x f = ming.

P l p2 p3 Pl p , p3

p3 I:; * * * * *

p3 * * 8: a: * 3 4 Pl * * * * *

* * * 7 1 2 12

(i) (ii) Recall that: (1) the components of Q a r e the las t entries, respectively, in the columns cor-

responding t o the slack variables; (2) the component pi of P is 0 if no row is labeled P, or the last entry in a row labeled P i ; ( 3 ) v is the entry in the lower right hand corner of the tableau.

(i) Note t h a t the second component of P is 0 since no row

(ii) Q = (3 ,4 ) , P = (0,0:2), v = 5 . Note t h a t the first two components of P are 0 since no row

Q = (7,1, 2), P = (5,0,3), v = 12. is labeled P2.

is labeled P , or P2.

24.10. Solve the following linear programming problem by the simplex method. Maximize f = 4 x - 2 y - x

subjectto x + y + x f 3 2 x + 2 y + x A 4

x - y 0 and X k O , y’0, x ” 0 .

We obtain the following sequence of tableaux:

Pl p , p3 Pl p2 p3

-4 2 0

(i) (ii)

Pl p2 p3

(iii) Thus, by the terminal tableau, the maximum of f is 2 and occurs at the point P = (1 ,1 ,0) .

(See Problem 23.11, Page 293.)

24.11. Solve the following linear programming problem by the simplex method. Maximize f = 2x-32j + x

subject to 3x + 6 y + x 6 4 x + 2 y + x A 4

x - y + x 3

and x s o , y’0, 2’0.

CHAP. 241 LIiYEAR PROGRAMMING 315

3 6 1 1 0 0 6

We obtain the following sequence of tabIeaux:

Pl p , p3 Pl p , p3

1 - 1 0

-2 3 - 1 0 0 1 3

0 0 0 0

1 4 2 1 1 0 1 0 1 4 1

0 3 0 0 1 2 ' 0 3 3 5

41 P 3 1 -1

- 1 2 0 0 0 1

Pl 2 3 3 :i

0 - 1 4 s 3 3 3 P3 0 - 2

(iii) Thus, by the terminal tableau, the maximum value of t' is 1013 and occurs at the

P = (1/3,0,8/3). point

24.12. Solve the following linear programming problem by the simplex method. Minimize g = 6x +5y + 2 x

subject to .P + 3y + 2x -I 5 2 x + 2 y + x 3 2

4,2:-2y+3x '1 -1

and c r o , y'0, z r o .

We obtain the following sequence of tableaux.

Pl pz p3 P l p , p3

0 0

-6 -2 0 0 0

(i) (ii)

Pl p2 p3

- . . 1 2 1 Pt 2 . 2 2

2 2

(iii)

Note first tha t ( I ) IS the tableau of the dual maximum problem. Note also tha t the third I'OW of (iii) is relabeled P, , the label of the pivotal column of the precedlng tabIeau (ii). By the terminal tableau, the minimum value of g IS 5 and occurs at the point Q = (0 ,0 ,5/2) .

LINEAR PROGRAMMING WORD PROBLEMS 24.13. A tailor has 80 sq. yd. of cotton material and 120 sq. yd. of wool material. A suit

requires 1 sq. yd. of cotton and 3 sq. yd. of wool, and a dress requires 2 sq. yd. of each. How many of each garment should the tailor make to maximize his income if


Cotton Wool

(i) a suit and dress each sells for $30, (ii) a suit sells for $40 and a dress for $20, (iii) a suit sells for $30 and a dress for $20?

1 2 80

3 2 120

Construct the following table which summarizes the given facts:

Suit I Dress ~ Available I

(ii) Here the objective function is g = 40z+20y . Evaluate g at the corner points:

g(0 ) = 0, g ( A ) = 800, g(B) = 1400, g(C) = 1600

The maximum value of g occurs a t C = (40,O). Thus he should make 40 suits (and no dresses), for which he will receive $1600.

(iii) Here the objective function is It = 3@x+20y. Evaluate h at the corner points:

h(0) = 0, h(A) = 800, h(B) = 1200, h(C) = 1200

The maximum value of h occurs a t both B = (20,30) and C = (40,O). Thus the tailor should make either 20 suits and 30 dresses o r only 40 suits; in either case he will receive $1200. (Actually, any point on the line segment joining B and C is an optimum solution.)

24.14. A truck rental company has two types of trucks; type A has 20 ft3 of refrigerated space and 40 f t3 of non-refrigerated, and type B has 30 ft3 each of refrigerated and non-refrigerated space. A food plant must ship 900 f t3 of refrigerated produce and 1200 f t3 of non-refrigerated produce. How many trucks of each type should the plant rent so as to minimize cost if (i) truck A rents for 30e per mile and B for 40@ per mile, (iii) truck A rents for 2OC per mile and B for 30c per mile?

(ii) truck A rents for 30c per mile and B for 50C per mile,


Type A Type B Requirements

Non-ref rigerated 1200

Let x denote the number of trucks of type A and y the number of type B t h a t the plant rents. Then x and y are subject to the following conditions:

20x + 30y 2 900, 40% + 30y 5 1200, x 2 0, 0


The set of feasible solutions is drawn on the right; its corner points a r e

P = (0,40), Q = (15, 20), R = (45,O)

Evaluate f at the corner points: (i) Here the objective function to minimize is f =

30s + 4011.

f ( P ) = f ( 0 , 4 0 ) = 1600, f ( Q ) = f(15,ZO) = 1250, f ( R ) = f(45,O) = 1350

The minimum value of f occurs a t Q = (15,ZO). Thus the plant should rent 15 trucks of type A and 20 of B at a total cost of $12.50 per mile.

40

30

20

10

10 20 30 40 50

(ii) Here the objective function is g = 3 0 x f 5 0 y . Evaluate g a t the corner points:

g(P) = 2000, g(&) = 1450, g(R) = 1350

The minimum value of g occurs at R = (45,O). Thus the plant should rent 45 trucks of type A (and none of type B ) at a total cost of $13.50 per mile.

(iii) Here the objective function is h = 202 + 30y. Evaluate h at the corner points: h(P) = 1200, h(Q) = 900, h(R) = 900

The minimum value of h occurs at both Q = (15,ZO) and R = (45,O). Thus the plant can rent either 15 trucks of type A and 20 of B or only 45 of type A a t a total cost of $9.00 per mile.

24.15. A company owns two mines: mine A produces 1 ton of high grade ore, 3 tons of medium grade ore and 5 tons of low grade ore each day; and mine B produces 2 tons of each of the three grades of ore each day. The company needs 80 tons of high grade ore, 160 tons of medium grade ore and 200 tons of low grade ore. How many days should each mine be operated if it costs $200 per day to work each mine?


Mine B Requirements 1 High Medium grade grade

Low grade 200

Let x denote the number of days mine A is open and y the number of days B is open. Then x and y a re subject to the following conditions:

x + 2y 80, 3% + 2g 5 160, 52 + 2y 2 200, 2 5 0 , 2( 2 0

We wish t o minimize the function g = 2002+200y subject to the above conditions.

Method 1. Draw the set of feasible solutions as on the’right.

The corner points are: p = (0, loo), Q = (20,50), R = (40,20), S = (80,O).

Evaluate the objective function g = 200% 4- 2 0 0 ~ at the corner points: g ( P ) = 20,000, g(Q) = 14,000, g(R) = 12,000, g(S) = 16,000. The minimum value of g occurs at R = (40,ZO). Thus the company should keep mine A open 40 days and mine B open 20 days for a total cost of $12,000.

Method 2. Consider the dual problem:

Maximize f = 80% + 160v + 2OOw

subject to u + 3w + 5w 200 2u + 2w + 2w 6 200

and u 2 0 , w + o , W’O.

318 L I N E A R PROGRAMMING [CHAP. 24

Using the simplex method we obtain the following sequence of tableaux:

1 1 1 0 ; 100

0 -80 -120 0 40 8,000

r-a 40 1 40 20 I 12,000 I (iii)

Thus the minimum value of g is 12,000 and occurs at the point (40, 20). not necessary since we were only solving the minimum problem.)

(Note tha t labels were

24.16. Suppose, in the preceding problem, the production cost at mine A is $250 per day and at mine B is $150 per day. How should the production schedule be changed?

In this case, the objective function to minimize is g = 2502 + 150y subject to the same set of feasible solutions, i.e. conditions. Evaluate g at the corner points of the set of feasible solutions:

g ( P ) = g(0,lOO) = 15,000, g ( Q ) = g(20, 50) = 12,500, g(R) = g(40,20) = 13,000, g(S) = g(80,O) = 20,000

The minimum value of g occurs at Q = (20,50). 20 days and mine B open 50 days f o r a total cost of $12,500.

Thus the company should now keep mine A open

Supplementary Problems DUALITY 24.17. Find the dual of each problem.

(i) hlaximize f = 52 - 2y + 32 (ii) Minimize g = 122 + 3y + 72

subject to 32 - 4y + 22 f 9

2% + 7y + 52 f 11

and 2’0, y A 0 , 2 2 0 .

subject to 6% - 2y + 6x 3

2 2 + 3y - 4x ’ -2

3 2 + 9 y + x ’ 8

and 2 2 0 , 2 / 2 0 , x ” 0 .


(i) Maximize f = 32 + 216 - 2x (ii) Minimize g = T + 7s + 2t

subject to 42 + 3y - 22 f 11

5% - 2y + 72 2 2

x + y + 3 z f 15

and 2’0, y’0, 2 ’ 0 .

subject to 7r + 3s + 5t 9

4r -i- 8s - t 10

SIMPLEX METHOD 24.19. Find the initial tableau and all possible pivot entries for each of the problems given in Problem 24.17.

L I N E A R PROGRAMMING 319 CHAP. 241

5 - 5 2 0 1 0 6 0 0 1 4

24.20. Calculate the new tableau from each tableau and circled pivot.

0) pi p, p, (ii) P I p, p,

3 Pl

- 3 - 4 2 1 0 0 0 1 0

24.21. In each of the following terminal tableaux, find the optimum solution Q = (q,, . . . , qk) of the minimum problem, the optimum solution P = (pl, . . ., p,) of the maximum problem, and ZI = max f = rnin g.

P z I ; :I I T p , $ * * * * -

2

3 1 5 9 $ * *

21.22. Solve by the simplex method: Maximize f = 9z + 2y + 52

subject t o 2 s + 3y - 5 x f 1 2 2.1: - y + 3x 6 3 3x + y - 2x f 2

and X’O, y’0, 2’0.

24.23. Solve by the simplex method: Minimize g = 4x + 5y + x

subject t o .?: + y + x 1 3 2 2 + 2y - 2 ’ 2

x + 2 y ’ 4

and 2’0, y’0, x 1 0 .

LINEAR PROGRAMMING WORD PROBLEMS 24.24.

24.25.

24.26.

24.27.

A carpenter has 90, 80 and 50 running feet of plywood, pine and birch, respectively. Product A requires 2, 1 and 1 running feet of plywood, pine and birch respectively: and product B requires 1, 2 and 1 running feet of plywood, pine and birch respectively. ti) If A sells for $12 and B sells for $10, how many of each should he make to obtain the maximum gross income? (ii) If B sells for $12 and A sells for $10, how many of each should he make?

Suppose tha t 8, 1 2 and 9 units of proteins, carbohydrates and fa t s respectively a re the minimum weekly requirements for a person. Food A contains 2, 6 and 1 units of proteins, carbohydrates and fa t s respectively per pound; and food B contains 1, 1 and 3 units respectively per pound. (i) If A costs 85c per pound and B costs 40r per pound, how many pounds of each should he buy per week to minimize his cost and still meet his minimum requirements? (ii) If the price of A drops to 75( per pound while B still sells a t do6 per pound, how many pounds of each should he then buy?

A baker has 150. 90 and 150 units of ingredients A , B and C respectively. A loaf of bread requires 1, 1 and 2 units of A , B and C respectively; and a cake requires 5 , 2 and 1 units of A , B and C respectively. ( i ) If a loaf of bread sells for 35C an$ a cake for 80<, how many of each should he bake and what would his gross income be? ( i i ) If the price of bread is increased to 45C per loaf and a cake still sells for 80$, how many of each should he then bake and what would be the gross income?

A manufacturer has 240, 370 and 180 pounds of wood, plastic and steel respectively. Product A requires 1, 3 and 2 pounds of wood, plastic and steel respectively; and product B requires 3, 4 and 1 pounds of wood, plastic and steel respectively. (i) If A sells for $4 and B for $6, how many of each should be made to obtain the maximum gross income? (ii) If the price of B drops t o $5 and A remains a t $4, how many of each should then be made t o obtain the maximum gross income?

320 L I K E A R PROGRAMMIKG [CHAP. 24

24.28. An oil company requires 9,000, 12,000 and 26,000 barrels of high, medium, and low grade oil, respectively. It owns two oil refineries, say A and B. A produces 100, 300 and 400 barrels of high, medium and low grade oil respectively per day; and B produces 200, 100 and 300 barrels of high, medium and low grade oil respectively per day. (i) If each refinery costs $200 per day t o operate, how many days should each be run to minimize the cost and meet the requirements? (ii) If the cost a t A rises to $300 per day and B remains at $200 per day, how many days should each be operated?

MISCELLANEOUS PROBLEMS 24.29.

24.30.

Solve Problem 24.16 by the simplex method.

Show that the maximum linear programming problem (I), Page 301, is equivalent t o the problem (Z), Page 304.

24.17.

24.18.

24.19.

24.20.

24.21.

24.22.

24.23.

24.24.

24.25.

24.26.

24.27.

24.28.

Answers to Supplementary Problems (i) Minimize g = 9u + llv (ii) Maximize .f = 3u - 2v + 8w

subject to 3 u + 2v 2 5 - 4 u + 7 v 22 -2 2 u + 5v 2 3

subject t o G Z L + 2v + 3w f 12 -221 + 3v + 9w f 3

57L - 4v -t w f 7

and u s o , V’O. and 24. 2 0, v 2 0, w 5 0.

(i) Minimize g = l l u - 2a + 15w (ii) Maximize f = 9% - 1Oy

subject to 4u - 5v + to 2 3 subject t o 7.7: - 4y L_ 1 3u + 2v + to 2 2 32 - 8y f 7

-2U - 7V + 3ZC+ 5 -2 5 x + y 6 2

and u 5 0, v ’ 0, zc 2 0. and s 2 0 , y’0.

(i) P , P, P3

-5 2 - 3 0 0 0

(ii) P I P , P ,

0

P,

(i) Q = (3,1,5), P = ( & , & , O ) , v = 9. (ii) Q = (3,1), P = ( 0 , 0 , 7 , 0 ) , w = 11.

The rnaximuin value of f is 49 and occurs at P = (0 ,12,5) .

The minimum value of g is 11 and occurs at the point Q = (0 ,2 ,1) .

(i) 40 of A and 10 of B for $580.

(i) 1 of A and 6 of B for $3.26.

(i) 50 loaves of bread and 20 cakes for $33.50.

(i) 30 of A and 70 of B for $540. (ii) 70 of A and 40 of B for $480.

(i) A run 50 days and B run 20 days.

iii) 20 of A and 30 of B for $560.

(ii) 3 of A and 2 of B for $3.05.

(ii) 70 loaves of bread and 10 cakes for $39.50.

(ii) A run 20 days and B run 60 days.

Chapter 25

Theory of Games

INTRODUCTION TO MATRIX GAMES Every matrix A defines a game as follows:

(1) There are two players, one called R and the other C. (2) A p l a y of the game consists of R’s choosing a row of A and, simultaneously, C’s

choosing a column of A . (3) After each play of the game, R receives from C an amount equal to the entry

in the chosen row and C O ~ L I I I ~ I I , a negative entry denoting a payment from R to C.

Example 1.1. Consider the following matrix games:

I:[ (i) (ii)

In the first game, if R consistently plays the first row, hoping to win an amount 3 or 4, then C can play the second column and YO win an amount 1. However, if C consistently plays the second column then R can play the second row and so win a n amount 1. Thus we see tha t if either player should consistently play a particular row or column, the other player can take advantage of tha t fact.

Now in the second game, player R can guarantee winning 1 or more by consistently playing row 2. Player C can then minimize his loss by playing column 1. Thus in this game i t is “best” if R plays a given row and C plays a given column.

On the other hand, suppose a competitive situation is as follows:

(i) there are two people (or: organizations, companies, countries, . . .); (ii) each person has a finite number of courses of action; (iii) the two people choose a course of action simultaneously; (iv) for each pair (i, j ) of choices, there is a payment b , , from one given player to

Then the above competitive situation is equivalent to the matrix game determined by the matrix B = (b l J ) .

the other.

Example 1.2. If the sum of the fingers

shown is even, R wins the sum from C ; if the sum is odd, R loses the sum t o C. The matrix of this game is a s follows:

Each of the two players R and C sin~ultaneously shows 1 or 2 fingers.

Player C 1 2

Player R 2

We shall show later (Example 3.1) t h a t this game is “favorable” to player C.

321

322 THEORY O F GAMES [CHAP. 25

We remark that matrix games and the above competitive games which can be represented as matrix games are termed two-person zero sum games. The zero sum corresponds to the fact that the sum of the winnings and losses of the players after each play is zero. There also exists a theory for n-person games and non-zero sum games but that lies beyond the scope of this text.

STRATEGIES

By a strategll for R in a matrix game, we mean a decision by R to play the various rows with a given probability distribution, say to play row 1 with probability pl, to play row 2 with probability p2, . . . . This strategy for R is formally denoted by the probability vector p = (pl, p,, . . . , pk). For example, if the matrix game has two rows and R tosses a coin to decide which row to play, then his strategy is the probability vector p = (a, 3).

Similarly, by a s t ra t egy for C we mean a decision by C t o play the various columns with a given probability distribution, say to play column 1 with probability q,, to play column 2 with probability qs, . . . . This strategy for C is formally denoted by the probability vector (I = (q , ,qs , . . .,q,).

A strategy which contains a 1 as a component and 0 everywhere else, i.e. where R decides to play consistently a given row or C decides to play consistently a given column, is called a pure s t ra t egy ; otherwise it is called a m i x e d s t ra t egy .

OPTIMUM STRATEGIES AND THE VALUE OF A GAME

For simplicity, we shall first consider a 2 X 2 matrix game, say

A = m

Suppose player R adopts the strategy 11 = (p l , p2 ) and player C adopts the strategy q = (q,,q,). Then R plays row 1 with probability pl and C plays column 1 with probability ql, and so the entry a will occur with probability p,q,. Similarly, b will occur with probability p,qz, c with probability pLql and d with probability p 2 q 2 . Hence the expected winning (see Page 220) of R is

where qt, At and p t denote the transpose of g, A and p , respectively

Similarly, if A is any kxnz matrix game and player R adopts the strategy p = (p,, . . . , p k ) and C adopts the strategy Q = (ql, . . . , qJ, then the expected winnings of R is given by

E(p, q ) = p A qt = q A t p t

Now suppose that v1 is the maximum number for which there is a strategy pa for R such that the expected winnings of II is at least v1 fo r any strategy adopted by C:

E(po,q) 2 v1, for every strategy q of C (1 1

CHAP. 251 THEORY O F GAMES 323

-4 , 2

Since V I is the maximum of such numbers. the strategy p" is in a certain sense a "best" strategy that R can adopt; we call p o an optinzunz strategy for R. We show in Problem 25.9, Page 233, that ( I ) is equivalent to the condition

p"A 2? ( V l , 2'1, . . . , Vl)

On the other hand, suppose v2 is the minimum number for which there is a strategy qo for player C such that the expected loss of C is not greater than l j 2 for any strategy adopted by R:

Then the strategy q0 is in a certain sense a "best" strategy that C can adopt; we call qo an optimum s t m t e y y for C. Similarly. ( 2 ) can be shown to be equivalent to the condition

E(p, q") 4 vL, for every strategy 11 f o r R ( 2 )

qo A' 6 ( ~ r , 2'2, . . ., 71.)

The fundamental theorem on game theory by von Neumann states that V I and v z exist f o r any matrix game and are equal to each other: v = V I = vz. This number u is called the value of the game, and the game is said to be f a i r if v=O. We observe that (1) and ( 2 ) imply that v1 6 E(po, qo) 6 vz ; hence by von Xeumann's theorem we have the equality relation

-1 3

We summarize and formally define the terms introduced above:

FDefinition:] In a matrix game A , a strategy p o for player R is an optimum strategy, a strategy qo f o r player C is an optii?zum strategy, and a number v is the calzie of the game if P O , qo and 21 are related as follows:

(i) p o A 1 ( v , ~ , . . .,v); (ii) @ A t L ( v , ~ ! , . . .,v) If v = 0, the game is said to be fa i r .

We point out that there may be more than one pair of strategies for which (i) and (ii) hold; that is, a player may have more than one optimum strategy.

We remark that the solution of a matrix game means finding optimum strategies for the players and the value of the game. Before we give the solution to an arbitrary matrix game by means of the simplex method developed in the preceding chapter, we shall discuss two special cases which occur frequently and which offer simple solutions: strictly determined games and 2 x 2 games.

STRICTLY DETERMINED GAMES A matrix game is called strictly de t e iw i i l ed if the matrix has an entry which is a

minimum in its row and a maximum in its column; such an entry is called a "saddle point". The following theorem applies.

Theorem 25.1: Let 2' be a saddle point of a strictly determined game. Then an optimum strategy for R is always t o play the row containing v, an optimum strategy for C is always to play the column containing o, and v is the value of the game.

Thus in a strictly determined game a pure strategy for each player is an optimum strategy.

Example 2.1. Consider the following game: I 3 I 0 1 - 4 1 - 3 1

I


If we circle the minimum entry in each row and put a square about the maximum entry in each column, we obtain

A =

The point 1 is both circled and “boxed” and so is a saddle point of A . Thus w = 1 is the value of the game, and optimum strategies po for R and qo for C are as follows:

pa = (O,l,Oj and qo = ( O , O , 1 , O j

Tha t is, R consistently plays row 2, and C consistently plays column 3.

value of the game: We now show that po, q 0 and w do satisfy the required properties to be optimum strategies and the

3 0 -4 -3

(i) pOA = ( O , l , O ) ( f 3 1 2!) = (2 ,3 ,1 ,2j (1,1,1,1) = (w,v,v,wj

2 -1

We remark that the proof of Theorem 25.1 is similar to the proof above fo r the special case.

2 x 2 MATRIX GAMES

Consider the matrix game A = l+l

If A is strictly determined, then the solution is indicated in the preceding section. Thus we need only consider the case in which A is non-strictly determined. We first state a useful criterion to determine whether or not a 2 X 2 matrix game is strictly determined.

Lemma 25.2: The above matrix game A is non-strictly determined if and only if each of the entries on one of the diagonals is greater than each of the entries on the other diagonal: (i) a,d > b and a,d > c ; or (ii) b, c > a and b,c > d.

The following theorem applies.

Theorem 25.3: Suppose the above matrix game A is non-strictly determined. Then p o = (xl,xz) is an optimum strategy for player R, qo = (y,,y2) is an optimum strategy for player C, and v is the value of the game where

d - c a - b

d - b a - c

x, = II‘, = a + d - b - c ’ a + d - b - c

y 1 = a + d - b - c ’ zJ2 = a + d - b - c

and ad - bc ~ + d - b - ~ 2 , =

CHAP. 251 THEORY O F GAMES 325

Note that the five quotients above have the same denominator, and that the numerator of the formula f o r v is the determinant of the matrix A .

Example 3.1. Consider the following matr ix game (see Example 1.2):

Using the above formula, we can obtain the value 2: of the game:

2 . 4 - ( -3) . ( -3) - - -1 2 f 4 - 3 f 3 12 v =

Thus the game is not fa i r and is in favor of the column player C. Optimum strategies p o for R and qo for C a r e as follows: po = (L 12'12 A) and qo = (&,A).

RECESSIVE ROWS AND COLUMNS Let A be a matrix game. Suppose A contains a row ra such that r l L r , for some other

row r,. Recall that r l L r l means that every entry of rz is less than or equal to the corresponding entry of T,. Then rz is called a yecessive row and r, is said to dominate it. Clearly, player R would always rather play row r, than row rl since he is guaranteed to win the same o r a greater amount in every possible play of the game. Accordingly, the recessive row r? can be omitted from the game.

On the other hand, suppose A contains a column Ca such that c , h c 3 for some other column cj. Then cl is called a recessive coLumri and c, is said to dominate it. For analogous reasons player C would always rather play column c, than column el. Hence the recessive column c, can be omitted from the game.

We emphasize that a recessive row contains numbers smaller than those of another row, whereas a recessive column contains numbers larger than those of another column.

. Note that ( - 5 , - 3 , l ) 6 (2,-1,2) , i.e. every entry

Example 4.1.

Consider the game A =

-2

in the first row is f the corresponding entry in the second row. Thus the first row is recessive and can be omitted from the game and the game may be reduced to

1 2 1 - 1 I 2

Now observe tha t the third column is recessive since each entry is 2 the corresponding entry in the second column. Thus the game may be reduced to the 2 X 2 game

A* = Fl The solution to A* can be found by using the formulas in Theorem 25.3 and is

Thus the solution to the original game A is

326 THEORY O F GAMES JCHAP. 25

a11 a12 . . . al,,, a21 a.2 * * * U2r,1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . a k l ak2 ' ' ' a h n l

SOLUTIOS OF A MATRIX GAME BY THE SIMPLEX METHOD Suppose we are given a matrix game A. We assume that A is non-strictly determined

and does not contain any recessive rows or columns. We obtain the solution to A as follows.

(1) Add a sufficiently large number h. to every entry of A to form, say, the following matrix game which has only positive entries:

ult'! \ / a l l a12 . . *

I A' x 1

a2,!( 1 nil a22 . . . . . . . . . . . . . . . . . . . 1 1 \ ah1 a k 2 ah!,, 1

(The purpose of this step is to guarantee that the value of the new matrix game is positive.)

(2) Form the following initial tableau: PI Pr * . ' P,,,

1 0 . ' . 0 1 0 1 . . . 0 1

0 0 - . . 1 1

. . . -1 -1 -1 I 0 0 * ' . ~~ ~ ~

This tableau corresponds to the linear programming problem:

Maximize f = cc'i - xz + - . + T , ( ~

subject t o al i?-l + ~ ~ 1 2 x 2 + * - . + ui,,,Tl,l f 1

ar13'1 t a22xz + * * . + Cl l r , !5m 6 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . ahi.1'1 + n k 2 x 2 + * ' ' + a h n i s ) ' i t r 1

and s.1'0, X S S O , . . . , X , , , ' O

(3) Solve the above linear programming problem by the simplex method. Let P be the optimum solution to the maximum problem, Q the optimum solution to the dual minimum problem, and 2' ' = f ( P ) = g ( Q ) the entry in the lower right hand corner of the terminal tableau. Set

0 0

1 2: = - - k 1 1

v* V * (iii) (i) p o = -Q (ii) qo = --+ P

Then p o is an optimum strategy f o r player R in game A, qo is an optimum strategy f o r player C, and c is the value of the game.

We remark that the games ,4 and A have the same optimum strategies for their respective players and that their values differ by the added constant k ; that is, p o is an optimum strategy fo r player R in game A * , 4" is an optimum strategy for player C in game A*, and 2' 4 h. = 1/v* is the value of the matrix game A 4 .

Example 5.1. Consider the mati I X game

CHAP. 25! THEORY O F GAMES

4 3 1 1 0 0 1

327

1 4 1

Add 3 to each entry of A t o form the matrix game

0 1 0 1 1

The initial simplex tableau and successive tableaux are as follows:

-1 -1 -1

(iii)

Thus P = C O , & , i ) , Q = ( i , d ) and z- = 4; hence l /v* = 81'3. Then for the original game A ,

p o = (8/3)& = (2/3, 1/3) is an optimum strategy for player R, qo = (8/3)P = (0, 1/3, 2/3) is a n optimum strategy fo r player C, v = 8/3 - 3 = -1/3 is the value of the game.

Observe that the game is favorable t o the column player C.

ExampIe 5.2. Consider the matrix game A shown below:

0 0 0 0

A = ;

-1 -1

Adding 2 t o each entry of A , we obtain the matrix game A' on the right.

The initial tableau and the successive tableaux are as follows:

P , P1 P, P , P , P ,

~~

(iii)


Thus P = (1/22,3/22,9/22), Q = (3/22,1/22,9/22) and v* = 13/22; hence l/w* = 22/13. Accordingly, for the original game A ,

po = (22/13)Q = (3/13, 1/13, 9/13) is a n optimum strategy for R, qo = (22/13)P = (1/13, 3/13, 9/13) is an optimum strategy for C, w = 22/13 - 2 = -4/13 is the value of the game.

2 x m AND m x 2 MATRIX GAMES Consider the 2 x m matrix game

Suppose we apply the simplex method t o the above game. Note that the optimum solution of the corresponding maximum linear programming problem will contain at most two non-zero components; hence so will an optimum strategy for the column player C. In other words, the 2 x m matrix game A can be reduced to a 2 x 2 game A*.

In the next example we show a way of solving such a 2 x nz game A and, in particular, how to reduce the game to a 2 x 2 game A X . A similar method holds for any m X 2 matrix game. Example 6.1.

Consider the following matrix game (see Example 5.1):

3 -1 A = (-2 1 -:)

Let p0 = (2, 1-2) denote an optimum strategy for R and let v be the value of the game. Then

or V L 5 x - 2

v f -22 + 1 V L 2 - 1

32 - 2 ( 1 - 2 ) 5 w

- x + ( l - x ) 5 v or

O - ( l - x ) f v

Let us plot the three linear inequalities in w and x on a coordinate plane where 0 f x f 1, as shown in the diagram. The shaded region depotes all possible values of v; however, R chooses his optimum strategy so tha t v is a maximum. This occurs at the point W which is the intersection of v = -2x+ 1 and u = x - 1; tha t is, where x = Q and

strategy for R is ($,Q). To find an optimum strategy for C, we can set qo =

(y, z, 1-y -x ) and use the condition qoAt 5 (w, v, v) = (-4, -Q, -Q). Ob- serve that the two lines intersecting at W are determined by only the second and third columns of A . Omit the other

4

v = -&. Thus the value of the game is -$ and the optimum 2

1

However, a simpler way is as follows. -1

column and reduce the game to the 2 X 2 niatrix game -2

- 3

A* = El - 4

By Theorem 25.3, the solution t o A* is

V' = -9, po' = (3 , Q) and q0' = (I 2 ) 3 ' 3

CHAP. 251 THEO.RY O F GAMES 329

An optimum strategy for the column player C in the matrix game A is then q 0 = ( O , & , $ ) . (Note that v = v’ and po = PO‘, as expected.)

In the diagram above, the three hounding lines intersect the line E = 0 at -2, 1 and -1 respectively, and the line r = l at 3, -1 and 0, respectively. Observe that these numbers correspond t o the components of the r o w of the original matrix A .

Remark:

SUMMARY In finding a solution to a matrix game, the reader should follow the following steps:

(1) Test to see if the game is strictly determined.

(2) Eliminate all recessive rows and columns.

(3) In the case of a 2 x 2 game, use the formulas in Theorem 25.3.

(4) In the case of a 2 X m game or an ni X 2 game, reduce the game to a 2 X 2 game as in

(5 ) Use the simplex method in all other cases.

Example 6.1.

Solved Problems MATRIX GAMES 25.1. Which of the following games are strictly determined? For the strictly determined

games, find the value v of the game and find an optimum strategy po for the row player R and an optimum strategy qo for the column player C.

1 I -1 0 1 3 1 - 1

-8 -3 7 3 4

( i ) (ii) (iii)

In each case, circle the minimum entries in each row and put a box around the maximum entries in each column as follows:

@ 1 -3

(i)

mi01 1 I (ii) (iii)

(i)

(ii)

The -1 in row 2 and column 2 is a saddle point; hence z1 = -1, p0 = (O,l,O), q0 = (0 ,1 ,0) .

There is no saddle point; hence the game is non-strictly determined. Note tha t both 3’s in the first column are “boxed”, since both a re maximum entries in the column.

(iii) The 2 in row 2 and column 3 is a saddle point; hence v = 2, p o = (0,1, O), QO = (0 ,0 ,1,0) .

330 [CHAP. 25

25.2. Find the solution to

THEORY O F GAMES

each of the following 2 x 2 games.

(ii)

If the game is non-strictly determined, p o ( y l , x2) and 40 = (u l , ur) .

(ii

lii)

(iii)

iivj

d - c XI = x2 = 0 ‘ d - b - c ’

( i i i ) (iv)

use the following formulas of Theorem 25.3, where

a - b d - b a + c l - b - c ’ a + d - b - c ’

ad - bc . u = a - e a f d - b - ~ ’ a + d - - b - ~ Llq =

The game is non-strictly determined. F i r s t compute (L t d - b - c = 2 + 2 - 0 + 1 = 5. Then

That is, (g, +) is an optimum strategy for R and [g , Q I is a n optimum strategy for C. value of the game is ti = favor of the TOW player R.

This game is strictly determined with saddle point -1. Thus w = -1; and p @ = ( 1 , O ) and qo = (0 , l), i.e. R should always play row 1 and C should always play column 2.

The game is non-strictly determined, F i r s t compute a - d - b - c = 1 - 5 4 3 ’ 1 = 10. .yl = 8/10 = .8, x2 = . 2 , y1 = .6, yz = .4.

The game is not fa i r and, since v is positive, the game is in favor of the row player R.

The game is non-strictly determined. Then x1 = 5/13, y2 = 8/13, y 1 = 8/13, y, = 5/13. and (A,&) is a n optimum strategy for C. The value of the game is L’ = The game is not fa i r and is in favor of the row player R.

The Observe tha t the game I S not Pair and is in 2 . 2 - 0 . ( - 1 ) - L

- 5 . 5

Then That is, (A, .2) is a n optimum strategy for R and

- .2. 1 . 5 - ( - 3 ) * ( - 1 ) - i.6, .4) is an optimum strategy for C. The value of the game is t = 10

Firs t compute a + d - b - c = -3 - 3 - 2 - 5 = -13. Thus (&,&) is an optimum strategy for R

( - 3 ) . ( - 3 ) - 2 . 5 - - - 1 - 13 13 *

25.3. Find a solution to the matrix game

First circle each row minimum and box each column maximum:

Observe that there is no saddle point and so the game is non-strictly determined. We now test for recessive rows and columns. Note tha t the third column is recessive since each entry is larger than the corresponding entry in the second column; hence the third column can be omitted to obtain the game

-5

CHAP. 251 THEORY OF GAMES 331

-1 5

1 -3 25.4. Find a solution t o the matrix game A =

Note tha t the second row is recessive since each entry is smaller than the corresponding entry in the first row. Thus the second row can be omitted to obtain the game

1 - 2

-2 5

I 1 1 - 3 1

Using the formulas in Theorem 25.3, the solution to the above 2 2 game is - f i

11' 11 t 11' 11 v' = -13/11,

Thus a solution to the original game is

,,o' = (2 5 ) (Io' = ( 2 -)

- 4 v = -13/11, pa = (fi. O,,), (I" = (5, h. 0 )

Set p o = (:r , 1 - x ) . Then p o A 5 iil, I ' , c , i ' j is equivalent to

(x, 1-2:) (-; -; -; -;) ( V , V , C , " )

or -2% + 1 2 v

8 ~ - 3 2

3 x - 2 % v

- 7 x + 5 5 v

Plot the above four linear inequalities as shown in the diagram. Note tha t the maximum L' satisfying the inequalities occurs a t the intersection of the lines -2x+ 1 = 2; and 3 ~ - 2 = v, tha t is, a t the point x = and L' = -L. Thus v = -8 is the value of the game, and p o = (2, 8) is a n optimum strategy fo r player R.

Since the two lines determining L* come from the first and third columns of the matrix game A , omit the other columns t o reduce A to the 2 X 2 matrix game

A :1 - - Fl. A solution of A " , by Theorem 1 1 - 2

Thus q 0 = (g, 0 , b, 0 ) is an optimum strategy for player C in game A . I Note that v = v' and po 1 P O ' , as expected.)

25.5. Find a solution t o the matrix game A = El. or 4 7 - 1 4 v

-3.r - 1 f 21

23. f v


Plot the above three inequalities (where O‘x‘l) as shown in the diagram. We seek the miximum v satisfying the given inequalities; note that i t occurs at the intersection of the lines -3% + 1 = v and 2x = v, i.e. at the point r = 1 and v = %. Thus v = + is the value of the game and qo = (&,Q) is an optimum strategy for player C.

Since the two lines determining v come from the second and third rows of A , omit the other row to obtain

the 2 X 2 matrix game A:: = Fl. A solution

of A’?, by Theorem 23.5, is

5

3’ = 2, = (2 5) qo’ = (I, +) 5 5 ’ 5 ’ 5

Thus po = (0, $, 8) is a n optimum strategy for player R . (Note that u = v’ and qo =qo’ , as expected.)

25.6. Find a solution to the matrix game mi. -1

The reader should first verify tha t the game is non-strictly determined and that there are no recessive rows or columns. We find a solution by the simplex method. F i r s t add 4 to each t o obtain the game

3 4

The initial and subsequent tableaux follow:

Pl p2 p3 Pl pz p3

4

(i) (ii)

entry

(iii)

(Renzayk: Since the las t row in (iii) has no negative indicators, the tableau is a terminal one

!?, Accordingly, po = ($)Q = and so all of its entries need not be computed.) Thus P = (L l i ’ 5 17’ O ) , Q = (0 ’ 1 7 ’ ”) 17 and ~ 4 : =&; hence 1/~:,:

9 3 (0 , i,~), 4 0 = ( y ) P = (i 5 0 ) and 2: = 17 - 4 = -5 5 ’ 5 ’ 5 5 ’

25.7. Two players R and C simultaneously show 2 or 3 fingers. If the sum of the fingers shown is even, then R wins the sum from C; if the sum is odd, then R loses the sum to C. Find optimum strategies f o r the players and to whom the game is favorable.

CHAP. 251 THEORY OF GAMES 333

The matrix of the game is 2 3

The game is non-strictly determined; hence the formulas in Theorem 25.3, Page 234, apply. Firs t compute a + d - b - c = 4 + 6 + 5 + 5 = 20. Then

That is, (g, &,) is an optimum strategy fo r R and (&,%) is a n optimum strategy f o r C. The value 20 - 2 0 , and, since i t is negative, the game is favorable t o C . of the game is = 4 . 6 - - ( - 5 ) ' ( - 5 ) - -L.

25.8. Two players R and C match pennies. If the coins match, then R wins; if the coins do not match then C wins. Determine optimum strategies for the players and the value of the game.

The matrix of the game is H T

The game is non-strictly determined. x1 = x, = y1 = y2 = 4. the game is

Using the formulas in Theorem 25.3, Page 234, we obtain The value of Thus (g, &) is an optimum strategy for both R and C.

4 = 0: hence the game is fair. = 1.1 - ( - l ) . ( - i )

THEOREMS 25.9. Let p o be a given strategy for player R in a matrix game A . Show that the following

conditions are equivalent:

(i) p o A qt v, for every strategy q for C; (ii) p o A 2 (v, v, . . . , v). Assume that iii) holds. Then, for any strategy q for C,

pOA qt 2 (v, v, . . ., v) qt = v

On the other hand, assume tha t (i) holds and tha t pOA = (a l , a2 , . . ., a k ) . Choosing the pure strategy q = (1,0, . . . ,O), we have

pOA qt = (al, aL, . . ., ak) qt = a , s v

Similarly, a2 2 w, . . ., ak 5 v. In other words, @ A 2 (v, v, . . ., w).

25.10. Let p = ( a l , a r , . . ., ak) and p* = (b l , bz, . . ., bk) be probability vectors. Show that each point on the line segment joining p and p*, i.e. t p -t (1 - t)pa where 0 g t 4 1, is also a probability vector.

Observe that t p + (1 - t)p" = (tal + (1 - t ) b l , ta2 + (1 - t ) b z , . . ., t a L + (1 - t ) b k )

Since t , 1 - f , the a, and the b, are all non-negative, every i. Furthermore,

ta , + (1 - t ) b , is also non-negative for

fa , + (1 - t ) b , + ta2 + (1 - t ) b , + * ' ' + tak + (1 - t)bh = ta, + ta, + . * . + tah + (1 - t )b , + (1 - t ) b z

= t ( a , + a 2 + ~ ~ ~ + a ~ ) + ( 1 - t ) ( b l + b , + ~ ~ ~ + b ~ ) = t + ( l - t ) = 1

* . . + (1 - t ) b k

334 THEORY O F GAMES 'CHAP. 2 6

25.11. Let p o and p be optimum strategies for player R in a matrix game A with value c . t 6 1, is also an optimum strategy for R.

By the pieceding problem, p is a probability vector, hence we need only show that

Show that p' = t po + (1 - t ) p , where 0

p A ( L , I , ,v). We are given that

pOA ( L , I!, .v) and p A 4 ( L , L , ,v)

Thus p'A == ( tpo - i l - t ) p ) A tpOA + (1 - t ) ~ A

2 t (v ,v , . . , l ) (1 - t ) ( C , w, . . ., w) = (v,w, . . , L )

25.12. Show that if a and b are saddle points of a matrix A , then a = b. If n and b are in the sanie row, then each is a minimum of the row and so a = t i . Similarly

Now assume that a and b appear in different columns and different rows as illustrated below:

if u and b are in the same column, then n and b are maxima of the column and so a = b.

*: :,: ::. .,. .'. Q * :I * ::i r;;l ::: f @ :,: * \

::: * :,: * * :: :.: .::

\ : , : ::L p-J Q ;.k pJ :: :i i' \ * * :.: Q * :.k i:: * I

Since a is a saddle point, a G d and c f a ; and since b is a saddle point, b s c and d E b. Thus

a ' d s b ' c 5 a

Accordingly, the equality relation holds above: a = d = b = c.

25.13. Show that the game A = is strictly determined.

Observe that each a is a minimum of the first row. If a b then the first a is a saddle point, and if u 1 c then the second a is a saddle point.

Thus n e are left with the case that b > a and c > a, i.e. where b and c are maxima of their b then c is a saddle point.

Accordingly, in all cases A has a saddle point and so A is strictly determined.

respective columns. Now if b f (3 then b is a saddle point, and if c

25.14. Prove Lemma 25.2: The matrix game A = is non-strictly determined

if and only if each of the entries on one of the diagonals is greater than each of the entries on the other diagonal: (i) a,d > b and a,d > c; (ii) b,c > a and b,c > d. or

If ( i ) or l i i ) holds, then there is no saddle point for A and so A is non-strictly determined.

Now suppose A is non-strictly determined.

Case (1): a < b. c then a is a saddle point; hence IL < c. d c o r else d is a saddle point. c is a saddle point. Accordingly, d % b leads to a contradiction, and so d < b.

Case (2 ) : n > b.

By the preceding problem, a# b . We consider

We need only show that But a < c and d > c implies

Similarly, d < c.

two cases.

If a

The proof is similar to the proof in Case (1).

335 CHAP. 251 THEORY O F GAMES

-1

25.15. Prove Theorem 25.3: Suppose the matrix game A = l++l is not strictly determined. Then

4 -2

a - c a - b a - b a - c ?Jo = ( a + d - b - c ' a + d - b - c

ad - bc a + d - b - c u =

are respectively an optimum strategy fo r R, an optimum strategy for C and the value of the game.

Set p0 = (x, 1-x) and q0 = (y, 1-y), Then pOA 2 (w, w) and qoAf 6 (w, w) is equivalent to

or ( a - c ) z + c 2 v ( a - b ) y + b f t? and

Solving the above inequalities as equations f o r the unknowns 2, y and v, we obtain the required result.

We remark tha t a + b - c - d # 0 by the preceding problem, i.e. by Lemma 25.2. We also note that in the case of a 2 X 2 non-strictly determined matrix game A , we have the following stronger result: poA = (w, v) and qoAt = ( t i , w),

( b - d ) ~ + d 5 v ( c - d ) y 4- d f v

25.16. Let A be a matrix game with optimum strategy po for R, optimum strategy qo f o r C, and value v; and let k be a positive number. Show that f o r the game kA, p o is an optimum strategy for R, qo is an optimum strategy for C, and kv is the value of the game.

We are given that pOA (w, w, , . ., c) and qOA f (w, W , . . ., c). Then pO(kA) = k(p0A) 2 k(w, v, . . ., w) = (kv, kv, . . . , kv)

qO(kA)t = qOkAt = k(q0At) f k ( ~ , W , . . ., W ) = (lit], kv, . . ., kw) which was to be shown.

Supplementary Problems MATRIX GAMES 25.17. Find a solution t o each game, i.e. an optimum strategy po for R, an optimum strategy qo f o r C,

and the value of the game.

(9

25.18. Find a solution to each game.

-3 (4

-4 -4

25.19. Find a solution t o each game. - (ii) (iii)

25.20. Find a solution to each game.

(i)

-1

25.21. Consider the game . Find a solution to the game if i i i a < 1, (ii) 1 < a < 3, (iii) a > 3.

25.22. Each of two players R and C has a dime and a quarter. They each show a coin simultaneously. If the coins are the same, R wins C's coin; if the coins a re different, C wins R's coin. Represent the game as a matrix game and find a solution.

25.23. Each of two players R and C has a penny, nickel and dime. They each show a coin simultaneously. If the total amount of money shown is even, R wins C's coin; if i t is odd, C wins R's coin. Represent the game as a matrix and find a solution.

THEOREMS 25.24. Suppose every entry in a matrix game A is increased by an amount k. Show that the value of

the game also increases by k , but tha t the optimum strategies remain the same.

25.25. Show that if every entry in a matrix game is positive, then the value of the game is positive.

Answers to Supplementary Problems 25.17. (i) p o = (0, l), q0 = (0, l), w = 1; (ii) po = (l,O), q0 = (0, I ) , v = 2; (iii) p0 can be any strategy,

yo = ( l , O ) , v = 1.

25.18. (i) po = (,7,.3), qo = (.4, .6), w = .2; (ii) po = (3/4,1/4), qo = (5/12,7/12), v = 1/4; (iiij p o = (7/8,1/8), qo = (5/8,3/8), v = 3/8.

25.19. (i) p o = (5/8,3/8), qo = (3/8,0,5/8), w = 1/8; (ii) PO = (5/7,2/7), q0 = (5/7,2/7,0), v = 3/7; (iii) P O = (2/7,0,5/7), q0 = (4/7,3/7), w = 13/7.

25.21. (i) p o = (l,O), q 0 = (O,l), w = 1; (ii) p o = (0,1), q0 = (0,1), w = a ;

25.22. =I; p o = (5/7, 2 / 7 ) , q' = (i, g) , w = 0.

1 5 10

25.23. 5 ; po = ( l O / l l , 0, l / i l ) , yo = (1/2, 0, 1/2), w = 0.

10 -10 j -10 I 10

I N D E X

Absolute value, 259 Absorbing states, 240 Algebra of propositions, 11 Algebra of sets, 38 Algorithm of the simplex method, 308 Angle of inclination, 266 Argument, 26

Bayes’ theorem, 207 Biconditional statement, 19 Binomial coefficients, 141

Binomial distribution, 217 Birthday problem, 188 Bounded sets, 269 Bounding hyperplane, 282

C player, 321 Cartesian product, 50

plane, 266 Cauchy-Schwarz inequality, 278 Cell, 60 Class, 37

equivalence, 60 Closed half space, 282 Closed interval, 258 Co-domain of a function, 66 Coefficients,

theorem, 142

binomial, 141 of equations, 106

Collection, 37 Column vector, 81 Combinations, 161 Complement of a set, 37 Components of vectors, 81, 83 Composition function, 68 Compound statements, 1 Conclusion of an argument, 26 Conditional probability, 199 Conditional statement, 18 Conjunction, 1 Consistent equations, 109 Constant, 106 Contradiction, 10 Contrapositive, 19 Converse statement, 19 Convex sets, 281

polygonal, 283 polyhedral, 282

Corner point, 282 Counting, fundamental principle of, 152 Cross-partitions, 162

De Morgan’s laws, 11 Degenerate equations, 107 Dependent vectors, 114 Detachment, law of, 26 Determinants, 127 Disjoint sets, 37 Disjunction, 2

exclusive, 3, 15 Distance, 266 Distribution, 218

binomial, 217 Domain of a function, 66 Dot product, 84 Dual problem, 301 Duality theorem, 303

Echelon form, 109 Element, 35 Elementary event, 184 Empty set, 37 Equations, linear, 104

Equiprobable space, 186 Equivalence class, 60 Equivalence relation, 59 Equivalent, logically, 10 Euclidean space, 268 Event, 184 Exclusive disjunction, 3, 15 Expected value, 220

of a game, 221 Extreme point, 282

system of, 107

Factorial, 141 Fair game, 221, 323 Fallacy, 26 False statement, 1 Family, 37

Feasible solution, 300 Finite set, 36 Fixed point, 234 Frequency, relative, 184 Function, 66

of lines, 268

composition of, 68 identity, 70 inverse, 70 linear, 67, 281 objective, 300 polynomial, 68

337

I N D E X

Games, gambling, 221

Games, matrix, 321 fair , 221

fair , 323 strictly determined, 323 value of, 323

Graph of a function, 67

Half space, 282 plane, 282

Homogeneous equations, 111 Hyperplane, 269

Identity function, 70 Image, 66 Implication, 28 Inclination of a line, 266 Inconsistent equations, 109

Independent events, 201

Independent vectors, 114 Inequalities, 258

linear, 259 Infinite set, 36 Initial simplex tableau, 304 Inner product, 84 Intersection of sets, 37 Intervals, 258 Inverse,

function, 70 matrix, 132 relation, 59 statement, 19

inequalities, 285

trials, 216

Invertible matrix, 131

Joint denial, 16

Line, in the plane, 267 real, 257 segment, 281

equation, 104 function, 67, 281 inequality, 259 programming, 300 space, 83

Logical implication, 28 Logically equivalent, 10

Linear,

Mapping, 66 Markov chain, 236 Matrix, 90, 112

addition of, 90 game, 321 multiplication of, 92 non-singular, 133 regular, 234 scalar multiplication of, 91 singular, 133 square, 93 stochastic, 233

Matrix (cont.) transition, 236 transpose, 94

Member, 35 Minkowski’s inequality, 278 Mixed strategy, 322 Multinoniial coefficients, 144 Mutually exclusive events, 185

Negation, 3 Negative numbers, 257 New simplex tableau, 306 Non-singular matrix, 133 Norm of a vector, 269 Null set, 36 Numbers, real, 256

Objective function, 300 Odds, 188 One-to-one function, 69 Onto function, 69 Open half space, 282 Optimum solution, 300 Optimum strategy, 322 Order, 257 Ordered pair, 50 Ordered partition, 163 Origin, 257

Pair, ordered, 50 Parallel lines, 266 Parameter, 268 Partitions, 60, 162

ordered, 163 Pascal’s triangle, 143 Permutation, 152 Perpendicular lines, 266 Pivot, 305 Pivotal column, 305

row, 305 Plane, Cartesian, 266 Plane, half, 282 Play of a game, 321 Players, R and C, 321 Point, 266

corner, 282 extreme, 282 fixed, 234 saddle, 323

Polygonal convex set, 283 Polyhedral convex set, 282 Polynomial, 68 Premises, 26 Primal problem, 301 Probability, 184

conditional, 199 distribution, 238 vector, 233

Product set, 50 Proper subset, 36 Proposition, 3 Pure strategy, 322

Quotient set, 60

finite mathematics schaum

Documents