discrete mathematical structures

Discrete MathematicalStructures

Bernard KolmanRobert C. Busby

Sharon Cutler Ross

离散数学结构

《离散数学》教学组

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中山大学软件学院

2.1 Propositions and Logical Operations

2.2 Conditional Statements

2.3 Methods of Proof

2.4 Mathematical Induction

Chapter 2 Logic

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推理的符号化（形式化）推理规则举例：假言推理（ modus ponens ）如果张三是软院 2011 级学生，那么离散数学是张三的必修课张三是软院 2011 级学生所以，离散数学是张三的必修课

符号化：若 p 则 q

p ------------ q

p => q p ------------ q

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推理规则举例：三段论（ Syllogism ）所有软院 2011 级学生必修离散数学张三是软院 2011 级学生所以，张三必修离散数学

符号化： x(A(x) -> B(x)) A(a) ------------ B(a)

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Reasoning in Mathematics

Gottfried Wilhelm Leibniz’s (July 1, 1646 – November 14, 1716) Symbolic thought: The only way to rectify our reasonings is to make them as tangible as those of the Mathematicians, so that we can find our error at a glance, and when there are disputes among persons, we can simply say: Let us calculate [calculemus], without further ado, to see who is right.

It is obvious that if we could find characters or signs suited for expressing all our thoughts as clearly and as exactly as arithmetic expresses numbers or geometry expresses lines, we could do in all matters insofar as they are subject to reasoning all that we can do in arithmetic and geometry. For all investigations which depend on reasoning would be carried out by transposing these characters and by a species of calculus.

Formal Logic

George Boole (2 November 1815 -8 December 1864 ) invented Boolean Logic.

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1) Statement or Proposition A statement or proposition ( 命题 ) is a declarative

sentence that is either true or false, but not both.

Ex. Which of the following are statements?(a) The earth is round.

(b) 2 + 3 = 5.

(c) Do you speak English?

(d) 3 – x = 5

(e) Take two aspirins.

(f) The temperature on the surface of the planet Venus ( 金星 ) is

800F.

(g) The sun will come out tomorrow.

Yes

Yes

No, it is a question.

No, it is a declaration sentence, not a statement.

No, it is a command.

Yes, it is a statement.

Yes, it is a statement.

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Proposition examples (compound propositions):

2 is an even number and 3 is an odd number.

2 is an even number and 3 is an even number.

2 is an odd number and 3 is an odd number.

2 is an odd number and 3 is an even number.

2 is an odd number and I am a student.

Pattern of the proposition: with connective “and”

Symbol: /\ (conjunction)

Truth of the pattern, or the truth table of “/\”

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Compound proposition examples:

2 is an even number or 3 is an odd number.

2 is an even number or 3 is an even number.

2 is an odd number or 3 is an odd number.

2 is an odd number or 3 is an even number.

2 is an odd number or 2 is an even number.

Pattern of the proposition: with connective “or”

Symbol: \/ (disjunction)

Truth of the pattern, or the truth table of “\/”

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Compound proposition examples:If I am hungry, then I’ll eat.If it is raining, then 2+3 = 5.If it is raining, then 2+3 = 4.If 2 is an even number, then 3+3 = 4.If 2 is an odd number, then 3 +3 = 4.If 2 is an odd number, then the sun rise on the west.Pattern of the proposition: with connective “if … then

…” Symbol: => (implication, antecedent, consequent)Truth of the pattern, or the truth table of “=>”

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Compound proposition examples

2 is not an odd number.

2 is not an even number.

Pattern of the proposition: with connective “not”

Symbol: ~

Truth of the pattern, or the truth table of “~”

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Equivalence of propositions

p => q and ~p \/ q

Bidirectional

p < = > q

Tautologies

Absurdity

Videos:Inventing on Principle

http://worrydream.com/ABriefRantOnTheFutureOfInteractionDesign/

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2) Proposition variable and Compound statementsIn Logic, the letters p, q, r, … denote proposition variable

( 命题变量 ), they are replaced by statements.

p: The sun is shining today.

q: It is cold.

Statement or proposition variables can be combined by

logical connectives ( 逻辑联结词 ) to obtain compound

statements( 复合命题 ).

Ex. and

p and q: The sun is shining today and it is cold.

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3) Negation of statement – 否定命题If p is a statement, the negation ( 非 ) of p is the

statement not p, denote by ~p.

~p is the statement “it is not the case that p”.

If p is true, ~p is false. If p is false, ~p is true.

The truth table( 真值表 ) is a table that shows the truth

valuesof a compound statement in terms of its component

parts.

F

T

T

F

~ppThe truth table

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3) Negation of statementnot is not a connective, it does not join two statements.

not p is not compound statement.

not is a unary operation for collection of statements.

~p is statement if p is.

Ex. Give the negation of the following statements:

(1) p: 2 + 3 > 1. (2) q: it is cold.

Solution:

(1) ~p: 2 + 3 1.

(2) ~q: it is not cold.

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4) Conjunction of statements – 合取式If p and q are statements, the conjunction of p and q is

“p and q”, denote by pq.

stands for connective and.

The compound pq is true if both p and q are true,

otherwise, it is false.The truth table of pq

TFFF

T TT FF TF F

pqp q

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4) Conjunction of statements Ex. From the conjunction of p and q for each of the

following.

(a) p: It is snowing, q: I am cold.

(b) p: 2 < 3, q: -5 > -8.

(c) p: It is snowing, q: 3 < 5.

Solution:

(a) pq: it is snowing and I am cold.

(b) pq: 2 < 3 and -5 > -8.

(c) pq: It is snowing and 3 < 5.

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5) Disjunction of statements – 析取式If p and q are statements, the disjunction of p and q is “p

or q”, denote by pq.

denotes the connective or.

The compound pq is true if at least one of p or q is true,

it is false when both p and q are false.The truth table of pq

TTTF

T TT FF TF F

pqp q

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5) Disjunction of statements Ex. From the disjunction of p and q for each of the

following.

(a) p: 2 is a positive integer q: 21/2 is a rational number.

(b) p: 2 + 3 5 q: London is the capital of France.

Solution:

(a) pq: 2 is a positive integer or 21/2 is a rational number.

p is true, q is false, so pq is true.

(b) pq: 2 + 3 5 or London is the capitcal of France.

p and q are false, so pq is false.

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6) Compound statement – 复合命题The compound statement has many component parts,

both of component parts are connected by connective.

pq(pr) is compound statement and has three

compound parts (propositions): p, q and r.

pq(pr) = pqrTTTTTFFF

TTFFTFFF

TTTTTFTF

T T TT T FT F TT F FF T TF T FF F TF F F

pq(pr)q(pr)prp q r

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7) The truth table – 真值表If a compound statement s has n component statements,

there will need to be 2n rows in the truth table for s. its truth

table may be systematically constructed in the following way.

(1). The first n columns of the table are labeled by

statements, further columns are included for all intermediate

expression, the last column is for the full statement.

(2). Under each of the first n headings, we list the 2n

possible n-tuples of truth values for the n component

statements.

(3). For each of the remaining columns, we compute the

remaining truth values in sequence.

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If p and q are statements, the compound statement “if p

then q” denoted by p q, is called a conditional statement

( 条件命题 ), or implication ( 蕴涵 ).

To p q, p is called the antecedent ( 前件 ) or hypothesis

( 假设 ), and q is called the consequent ( 后件 ) or conclusion

( 结论 ).

Ex. Write the implication p q for each of the following.

(a) p: I am hungry. q: I will eat.

(b) p: It is snowing. q: 3 + 5 = 8

Solution:

(a) If I am hungry, then I will eat.

(b) If it is snowing, then 3 + 5 = 8.

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p q: p implies q; q if p; p only if q.

p is a sufficient condition for q, q is necessary condition

for p.

If p q is a implication, the converse ( 逆命题 ) of p q

is the implication q p, and the contrapositive ( 逆否命题 )

of p q is the implication ~q ~p.Ex. Give the converse and contrapositive of the implication “If it is

raining, then I get wet.”

Solution:

p: It is raining; q: I get wet.

the converse is q p: If I get wet, then It is raining.

the contrapositive is ~q ~p: If I don’t get wet, then it is not

raining.

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If p and q are statements, the compound statement p if

and only if q, denoted by p q, is called an equivalence ( 等价 ) or biconditional ( 双条件 ).

The connective if and only if is denoted by the symbol

.

conditional statement ( 条件命题 ), or implication ( 蕴涵 ).

The truth table of p q is following.

The truth table of p q

T

F

F

T

T T

T F

F T

F F

p qp qp q is true only when both

p and q are true or when both p

and q are false.

The equivalence p q can

also be stated as p is a necessary

and sufficient condition for q.

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Ex. Compute the truth table of the statement (p q)

(~q ~p).

Solution:

T

T

T

T

T

F

T

T

F

F

T

T

F

T

F

T

T

F

T

T

T T

T F

F T

F F

(p q) (~q ~p)~q ~p~p~qp qp q

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A statement that is true for all possible values of its

propositional variables is called a tautology( 重言式 / 永真式 ).

A statement that is always false is called a contradiction

( 矛盾 / 永假式 ) or an absurdity ( 谬论 ).

A statement that can be either true or false, depending

on the truth values of its propositional variables, is called a

contingency ( 不定式 ).

Ex.

(a) (p q) (~q ~p) is a tautology.

(b) p ~p is an absurdity.

(c) (p q) (p q) is a contingency.

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If p q is a tautology, p and q are logically equivalent

( 逻辑等价 ), or simply equivalent ( 等价 ).

“p is equivalent to q” is denoted by p q.

Ex. p q q p.

The truth table for (p q) (q p) shows the statement

is a tautology.

T

T

T

T

T

T

TF

T

T

T

F

T T

T F

F T

F F

(p q) (q p)q pp qp q

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Ex. p q is equivalent to (~p q).

Column 1 and 3 in the above table show that for any

truth values of p and q, p q and (~p q) have the same

truth values.

T

F

T

T

F

F

T

T

T

F

T

T

T T

T F

F T

F F

~p q~pp qp q

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Theorem 1 The operations for propositions have the

following properties.(1) Commutative properties

p q q p p q q p(2) Associative properties

p (q r) (p q) r p (q r) (p q) r(3) Distributive properties

p (q r) (p q) (p r) p (q r) (p q) (p r)

(4) Idempotent properties

p p p p p p

(5) Properties of Negation

~(~p) p

~(p q) (~p) (~q) ~(p q) (~p) (~q)

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Theorem 2

(a) (p q) ((~p) q)

(b) (p q) (~q ~p)

(c) (p q) ((p q) (q p))

(d) ~(p q) (p ~q)

(e) ~(p q) ((p ~q) (q ~p))

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Theorem 4 Each of the following is a tautology.

(a) (p q) p (b) (p q) q

(c) p (p q) (d) q (p q)

(e) ~p (p q) (f) ~(p q) p

(g) (p (p q)) q (h) (~p (p q)) q

(i) (~q (p q)) ~p

(j) ((p q) (q r)) (p r)

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Formalization of math statements

Theorem If A and B are finite sets, then |A B|=|A|+|B|-|∪A∩B|.

Implicitly, we are saying that for all sets A and B,

if A and B are finite, then |A B|=|A|+|B|-|A∩B|.∪Here we have a predicate P(x): x is a finite set.

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8) Quantifiers – 量词A = { x | P(x) }

The set A is the collection that an object t for which P(t)

is true.

P(x) is called predicate( 谓词 ) or propositional function

( 命题函数 ).

To each choice of x, P(x) is a proposition that is either

true ot false.

Ex. A = { x | x is an integer less than 8 }.

P(x) is the sentence “x is an integer less than 8”.

P(1) and P(2) are true, P(9) is false.

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9) Universal quantification – 全称量化The universal quantification of a predicate P(x) is the

statement “For all values of x, P(x) is true”, it is denoted by

xP(x).

The symbol is called universal quantifier( 全称量词 ).

Ex. P(x): -(-x) = x, x is a real number.

xP(x) is true because for all real numbers, -(-x) = x.

Q(x): x + 1 = 4.

xQ(x) is false because Q(1) is false.判定 xP(x) 真假的一般方法 :

(1) 当要确定 xP(x) 为“真”时 , 则要确定所有 P(x) 都为“真” ;

(2) 当要确定 xP(x) 为“假”时 , 只要找一个 y, 使得 : P(y)

为“假” .

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9) Universal quantificationA predicate may contain many variables. The universal

quantification may be applied to each of the variables.

if P(x, y) is a predicate, its universal quantification is

xyP(x, y).

For example,

P(x, y): x + y = y + x

xyP(x, y) is true for every x and y.

Q(x, y): x ∪ y = y ∪ x

xyQ(x, y) is true for any set x and y.

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10) Existential quantification – 存在量化The existential quantification of a predicate P(x) is the

statement “There exists a value of x for which P(x) is true”, it

is denoted by xP(x).

The symbol is called existential quantifier( 存在量词 ).

Ex. P(x): x + 1 < 4, x is a real number.

xP(x) is true because P(2) is true.

Q(y): y + 2 = y.

yQ(y) is false because Q(y) is false for any y.判定 xP(x) 真假的一般方法 :

(1) 当要确定 xP(x) 为“真”时 , 只要找到一个 x, 使得 : P(x) 为“真” ;

(2) 当要确定 xP(x) 为“假”时 , 则要确定所有的 x, 都有 : P(x)

为“假” .

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10) Existential quantificationA predicate may contain many variables. The existential

quantification may be applied to each of the variables.

if P(x, y) is a predicate, its existential quantification is

xyP(x, y).

Ex.

P(x, y): x * y = y + x

xyP(x, y) is true because x = 2 and y = 2.

Q(x, y): x∩y = x - y

xyQ(x, y) is false for any set x and y.

x=?

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10) The order of quantifiersThe order of same quantifiers is not important, it does

n’t affect the truth value. The order of different quantifiers is

important, it affects the truth value.

Ex.

P(x, y): x + y = 0

xyP(x, y) is true, xyP(x, y) is false.

Q(x, y): x * y = 0

xyP(x, y) is true, xyP(x, y) is true too.

R(x, y): <x, y> E(G), x, y V(G)

xyR(x, y) is true, xyR(x, y) is false.

v1

v2

图 G

v3

v4

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11) Negation of quantifiersLet p: xP(x), what is the negation of p? (~p)

~p: ~xP(x) = x~P(x).

Let q: xQ(x), what is the negation of q? (~q)

~q: ~xQ(x) = x~Q(x).

3

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Theorem 3

(a) ~(xP(x)) x~P(x)

(b) ~(xP(x)) x(~P(x))

(c) x(P(x) Q(x)) xP(x) xQ(x)

(d) x(P(x) Q(x)) xP(x) xQ(x)

(e) x(P(x) Q(x)) xP(x) xQ(x)

(f) xP(x) xQ(x) x(P(x) Q(x)) is a tautology

(g) x(P(x) Q(x)) xP(x) xQ(x) is a tautology

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If an implication p q is a tautology, where p and q may

be compound statements involving any number of proposition

variables, we say that q logically follows from p.

Suppose that an implication of the form (p1 p2 … pn)

q is a tautology. We say that q logically follows from p1, p2,

…, pn, and writep1

p2

┆

pn

qIf p1, p2, …, pn are true, we know

q is true.

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(p1 p2 … pn) q

The pi’s are called the hypotheses ( 假设 ) or premises

( 前提 ), and q is called the conclusion( 结论 ).

To “prove the theorem” means to show that the implica-

tion is a tautology. If p1, p2,… and pn are true, q will be true.

For this reason, mathematical proofs often begin with

the statement “suppose that p1, p2,… and pn are true” and

conclude with the statement “therefore, q is true.”

The proof does not show that that q is true, but simply

shows if the pi are all true, then q has to be true.

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Methods of Proofs

Arguments based on tautologies represent universally correct methods of reasoning. The validity only depends on the form of the statements involved and not on the truth values of the variables they contain. Such arguments are called rules of inference.

A mathematical proof of a theorem must start with the hypotheses, proceed through various steps, each justified by some rule of reference, and arrive at the conclusion.

(see page 63).

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Ex. ((p q) (q r)) (p r)

The argument

is valid, and so is a rule of inference.

Ex. Is the following argument valid?

p q

q r

p r

If you invest in the stock market, then you will get rich.

If you get rich, then you are happy.

If you invest in the stock market, then you are happy.The argument is valid, although the conclusion may be

false.

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A important rule of inference is

That is, p is true and p q is true, so q is true.

The rule is called modus ponens ( 假言推理 ), or the

method of asserting ( 论断方法 ).

p

p q

q

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Ex. Is the following argument valid?


Smoke is healthy.

If Smoke is healthy, then cigarettes are prescribed by physicians.

Cigarettes are prescribed by physicians.The argument is valid. However, the conclusion is false.

Ex. Is the following argument valid?If taxes are lowered, then income rises.

Income rises.

Taxes are lowered.Solution:

p: taxes are lowered, q: income rises.

p q may be true with p being false. Then the

conclusion p is false. The argument is not valid.

p q

q

p

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Ex. Let n be an integer. Prove that if n2 is odd, then n is

odd.

Proof:

p: n2 is odd. q: n is odd.

We have to prove that p q is true.

Suppose that ~q is true, i.e. n is not odd.

Then n = 2k, where k is an integer.

n2 = (2k)2 = 4k2 = 2(2k2), so n2 is even.

So we have that if n is even, then n2 is even, which is the

contrapositive of the given statement.

Hence the given statement has been proved.

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An important indirect proof technique is proof by

contradiction( 反证法 ). This method is based on the

tautology

((p q) (~q)) ~p

Thus the rule of inference is valid.

p q

~q

~pSuppose we prove that a statement q logically follows from

statements p1, p2, …, pn.

If p1 p2 … pn (~q) implies a contradiction, then at least

one of the statements p1, p2, …, pn, (~q) must be false.

If all the pi’s are true, then ~q must be false, so q is true.

This is proof by contradiction.

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Ex. Prove there is no rational number p/q whose square

is 2. In other words, show 21/2 is irrational.

Proof:Assume (p/q)2 = 2 for some integers p and q, which have no common

factors.Then p2 = 2q2, so p2 is even.Since the square of an odd number is odd, so p is even.Suppose that p = 2k for some integer k.

2q2 = (2k)2 = 4k2

q2 = 2k2

Thus, q2 is even, and q is even.Now p and q are even, and have a common factor 2.This is a contradiction to the assumption. Thus the assumption is

false.

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In order to prove a theorem of the form (p1 p2 … pn)

q, we begin with the hypothesis p1, p2, …, pn and show that

some result r1 logically follows. Then, using p1, p2, …, pn, r1,

we show that some other statement r2 logically follows.

We continue this process, producing intermediate

statements r1, r2, …, rk, called steps in the proof, until we can

finally show that the conclusion q logically follows from p1, p2,

…, pn, r1, r2, …, rk.

Each logical step must be justified by some valid proof

technique, based on the rules of inference, or on some other

rules that come from tautological implications. 4

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Ex. Prove or disprove the statement that if x and y are

real number, (x2 = y2) (x = y).

Proof:The statement is restated in the following form

xy((x2 = y2) (x = y))To prove it, we would need to prove that for all x and y

it is true.To disprove it, we need only find one example for which

the implication is false.Let x = -3, y = 3.Since (-3)2 = 32, but -3 3, i.e. x y.Thus the statement is false.The example of x and y is called a counterexample( 反

例 ).

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We have the statement nn0P(n), where n0 is some

fixed integer.

If we prove P(n) is true for all integers n n0, the

following steps shall be done.

Step 1. P(n0) is true;

Step 2. If P(k) is true for some k n0, then P(k+1) must

be true.

So P(n) is true for n n0.

The method of proof is called the principle of

mathematical induction ( 数学归纳法 ).

Step 1 is the basis step ( 基础步 ), step 2 is called the

induction step ( 归纳步 ).

53


Ex. Show, by mathematical induction, that for all n 1,1 + 2 + 3 + … + n = n(n+1)/2

Proof:Let P(n) be the statement 1+2+3+…+n = n(n+1)/2.(1). Basis Step

P(1): 1 = 1(1+1)/2, so P(1) is true.(2). Induction Step

Assume that for some fixed k 1, then P(k) is true. P(k): 1+2+3+…+k = k(k+1)/2

Then P(k+1) is the following statement.P(k+1): 1+2+…+k+(k+1)= (k+1)((k+1)+1)/21+2+…+k + (k+1) = k(k+1)/2 + (k+1)

= (k+1)(k/2 + 1) = (k+1)(k+2)/2 = (k+1)((k+1)+1)/2

So P(k+1) is true.Thus by the mathematical induction, P(n) is true for all n 1.

54


Ex. Let A1, A2, …, An be any n stes, we show by mathematical

induction that ∪i=1..nAi = ∩i=1..nAi.

(it is an extended version of one of De Morgan’s laws)Proof:Let P(n) be the predicate that the equality holds for any n sets. We

prove by mathematical induction that for all n 1, P(n) is true.(1). Basis Step

P(1): A1 = A1, so P(1) is true.

(2). Induction Step Assume that for some fixed k 1, then P(k) is true.

P(k+1): ∪i=1..(k+1)Ai = ∩i=1..(k+1)Ai

The left-hand sideof P(k+1) is∪i=1..(k+1)Ai = A1 A∪ 2 … A∪ k A∪ k+1

= (A1 A∪ 2 … A∪ k) A∪ k+1

= (A1 A∪ 2 … A∪ k)∩Ak+1

= (∩i=1..kAi)∩Ak+1

= ∩i=1..(k+1)Ai

So P(k+1) is true.Thus by the mathematical

induction, P(n) is true for all n 1.

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Strong Induction - 强归纳法In strong form of mathematical induction, or strong

induction, the induction step is to show that

P(n0)P(n0+1)…P(k-1)P(k) P(k+1)

is a tautology for k n0.

We must prove: if each P(j) is true (j=n0..k), then P(k+1)

is true.

56


Ex. Prove that every positive integer n > 1 can be written uniquely as p1

a1p2a2…ps

as, where the pi are primes and p1<p2<…<ps.

Proof:(1). Basis Step

Here n0 is 2, P(2) is clearly true, since 2 = 21.

(2). Induction Step Assume that P(2), P(3), …, P(k) are true. Now we need to prove that P(k+1) is true.

(2.1). (k+1) is a prime P(k+1) = (k+1)1, so P(k+1) is true.

(2.2). (k+1) is not prime Let (k+1) = mn, where 2 m, n k. P(m) = q1

b1q2b2…qu

bu, P(n) = r1c1r2

c2…cvcv.

k+1 = mn = q1b1q2

b2…qubu r1

c1r2c2…cv

cv = p1a1p2

a2…psas

if pi = qj = rl, then ai = bj+cl, otherwise pi = qj and ai = bj, or pi = rl

and ai = cl.

Since the factorization of m and n are unique, so is the factorization of (k+1).

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Inductive Definitions, Induction and Recursion

Definition of naturals

Definition of well formed formulas

58

Summary

Important concepts: propositions, connectives, quantifiers, truth value, truth table, tautology, absurdity, contingency, logically equivalent, reference rules, induction

To be able to construct the truth table of a proposition, decide if a proposition is a tautology and if two propositions are equivalent.

To be to do mathematical induction proofs. Understand mathematical proofs formally. Understand inductive definitions and recursion.

discrete mathematical structures

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