LINCY JACOB

2015

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PREFACE

The idea of „collection‟ is as old as human civilization. In different context we use

different words and phrases to express this idea. Words like race, class, group,

gang, species etc. and phrases like crowd of people, pack of cards, fleet of ships,

bunch of flowers etc. reflect the idea of collection. Actually they contain different

objects possessing particular characteristics. When we study an object, it is quite

natural to extend our study to other objects having the same characteristics. In

Mathematics we come across different kinds of collection like collection of points,

collection of integers etc. It is to be noted that in each of these collections the

objects are well defined, which means „given an object we can definitely decide

whether it belongs to the collection or not‟. In our day today life we deal with

collection of „known objects‟ and hence they are different, but in theoretical study

we may come across with „unknown objects‟. In such cases, to avoid repetition we

must stress that the objects should be distinct. So we need a well defined collection

of distinct objects and such a collection is called a set. It was George Cantor (1845-

1918) , a German mathematician, who defined set for the first time. He developed

this theory because of his interest on infinite quantities. Kronecker (1810-1893),

Richard Dedekind (1831-1916), Betrand Russel are also mathematicians who

contributed much to the development of the theory of sets. The set theory plays a

very important role in understanding and solving various problems of mathematics.

The purpose of this text is to give the basic ideas of sets, relations and functions.

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CONTENTS

Chapter 1. Sets

(i) Sets and their representations

(ii) Types of sets

(iii) Venn diagrams

(iv) Complement of a set

(v) Operations on sets

(vi) Algebra of sets

(vii) Cartesian product of sets

Chapter 2. Relations

(i) Relation

(ii) Domain and range of a relation

(iii) Inverse relation

(iv) Types of relations

Chapter 3. Functions

(i) Functions

(ii) Graph of a function

(iii) Types of functions

(iv) Linear function

(v) Inverse of an element under a function

(vi) Inverse function

(vii) Composition of functions

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CHAPTER 1.

SET

A set is a well-defined collection of distinct objects. By the term well-defined we mean that

there exists a rule with the help of which we will be able to say whether a particular objects

„belongs to‟ the set or does not belong to the set. The objects in a set are called its members or

elements.

The sets are usually denoted by the capital letters of the English alphabet and the elements are

denoted by small letters.

If x is an element of the set A, we write x ∈ A (read as x belongs to A). If x is not an element of

A then we write x ∉ A (read as x does not belong to A).

Examples:

i) The collection of vowels in the English alphabet is a set. This set contains the

elements a,e,i,o,u.

ii) The collection of the first three odd natural numbers is a set containing the elements

1,3,5.

Representation of a set

A set is often represented in two ways:

i) Roster method (tabular form)

ii) Set-builder method (rule method)

i) Roster Method : In this method, a set is described by listing the elements, separated by commas and

are enclosed within braces.

Examples:

i) The set of first three prime numbers may be represented as {2,3,5} ii) The set of odd natural numbers can be represented as {1,3,5}

ii) Set-builder Method :

In this, the set is represented by specifying the characteristic property of its elements.

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Examples:

i) The set of natural numbers between 1 and 25 can be written as {x∕x ∈ ℕ and

1<x<25}

ii) The set of even numbers can be represented as {x : x=2k, k ∈ ℕ}

Types of sets

1. Null set

A set containing no element is called a null set or void set or empty set. It is denoted by

∅ or { }.

Examples:

i) The set of natural numbers between 4 and 5.

ii) {x : x ∈ ℝ and x2 + 2 = 0}.

2. Singleton set

A set containing only a single element is called a singleton set.

Examples:

i) The set of even prime numbers.

ii) {10}

3. Finite set

A set is said to be finite set if it consist only a finite number of elements. The null set is

regarded as a finite set.

Examples:

i) The set of natural numbers less than 10.

ii) The set of vowels in the English alphabet.

4. Infinite set

A set is said to be an infinite set if it consists of an infinite number of elements.

Examples:

i) Set of natural numbers.

ii) Set of real numbers between 0 and 1.

5. Subsets and Supersets

Let A and B be two sets. If every element of A is an element of B then A is called a

subset of B. Symbolically, we write A⊂ B. If A is a subset of B, we say that B contains A or B is

a superset of A and we write B ⊃ A.

An immediate observation is that the null set ∅ is a subset of every set A. Since, if ∅ ⊄ A, there

is an element of ∅ which is not an element of A, but there is no element in ∅.

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Example:

i) Let A= {2,3} and B = {2,3,4}. Then A is a proper subset of B. That is A⊂ B.

ii) The set of natural numbers is a subset of the set of integers.

6. Equivalent sets

Two sets A and B are said to be equivalent if they are finite and contain the same

number of elements.

Example:

Let A = {1, 2, 3} and B = {a, b. c}

A and B are equivalent sets, because n(A) = n(B)

7. Equal sets

Two sets A and B are said to be equal if A ⊆ B and B ⊆ A.

That is, A=B iff every element of A is an element of B and every element of B is an

element of A. If A is a subset of B and A ≠ B then we say A is a proper subset of B and

we write A ⊂ B. If A is a proper subset of B, then there exists an element x ∈ B such that

x ∉ A.

Example:

i) Let A = {1, 2, 3} and B = {2,3,1}. Clearly, A = B

ii) ℕ = ℤ+

Remark: The order in which the elements occur in a set is immaterial.

For example, {1, 2, 3} = {3, 1, 2} = {2, 3, 1} = {2, 1, 3} = {3, 2, 1} = {1, 3, 2}.

8. Power set

Let A be any set. Then the collection of all subsets of A is called the power set of A

and is denoted by P(A). In P(A), every element is a set.

Example: Let A = {1, 2, 3}

Then P(A) = { ∅, {1, 2, 3}, {1}, {2}, {3}, {1,2}, {2,3}, {1,3} }.

Note: If A contains n elements, then P(A) contains 2n elements.

9. Universal set

In set theory if all the sets under consideration are subsets of a fixed set U, then theset

U is called universal set.

Example:

i) For the set of all integers, the universal set can be the set of rational numbers or,

for the matter, the set ℝ of real numbers.

ii) Let A be the set of vowels in the English alphabet. Then the set of all letters of the

English alphabet. Then the set of all letters of the English alphabet may be taken

as the universal set U.

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The easiest way to introduce any concept is its diagrammatic representation. So let us try to give

a diagrammatic representation to the sets. We denote the next section for this.

Venn Diagrams:

The Swiss Mathematician Euler and British Mathematician Venn developed the idea of using

geometrical figures (diagrams) to represent sets. The diagrams drawn to represent sets are called

Venn-Euler diagrams or simply Venn diagrams. In this diagram the universal set U is denoted

by a rectangular region and any subset of U, by a region enclosed by a closed curve lying within

the rectangular region.

The Venn diagram representing the sets A and B such that A ⊂ B is shown is given below

Complement of a set

Let U be the universal set and A ⊆ U. The complement of A with respect U, denoted

as Ac or A′ or 𝐴 , is defined as the set consisting of those elements of U which do not

belong to A.

That is, A′ = {x: x ∈ U: x ∉ A}

The shaded region in the figure represents A′.

U

B

A

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Example: i) Let U = {1, 2, 3, …, 10} and A = {2, 4, 6, 8, 10}, then A′ = {1,3,5,7,9}

Example: ii) Let U = ℕ and A = {x: x ∈ ℕ and x is odd}, then A′ = {x: x is an even natural

number}

Note: (A′)′ = A, U′= ∅ and ∅′ = U

Operation on sets

1. Union of sets:

The union of two sets A and B is the set of all those elements which belong to A or to B or to

both. We use the notation A∪B (read as A union B) to denote the union of A and B.

That is, A∪B = {x: x ∈ A or x ∈ B}

Shaded region in the figure represents A∪B.

The definition can be extended to n sets; n>2.

If A1,A2,A3, …,An are n sets then their union is denoted by ∪i=1nAi and is defined as

∪i=1nAi = {x: x∈ Ai, for some i= 1,2,…n}

Example: i) Let A = {a, b, c} and B = {b, c, d}. Then, A∪B = {a, b, c, d}

ii)Let A = {x: x∈ ℕ and 2< x≤ 6} and B = {x: x∈ ℕ and 5<x≤8}

2. Intersection of sets

The intersection of two sets A and B, denoted by A∩B (read as A intersection B), is

the set consisting of all elements which belong to both A and B.

That is, A∩B = {x: x∈A and x∈B}

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Shaded region in the figure represents A∩B

The definition can be extended to n sets; n>2.

The intersection of n sets A1, A2, …, An is denoted by ∩i=1nAi and is defined as

∩i=1nAi = {x: x∈Ai, for all i=1,2,3,…, n}

Example: Let A = {1,2,3} and B = {2,4,6}.Then , A∩B = {2}.

Disjoint sets

Two sets A and B are said to be disjoint if A∩B = ∅. That is A and B are disjoint sets or

non-overlapping sets if they have no common element(s).

Example: Let A = {x: x∈ ℕ and x is a multiple of 2}, B = {x: x∈ ℕ and x is odd}.

Then A= {2,4,6,…} , B = {1,3,5,…}, therefore A∩B = {∅} or A and B are disjoint sets.

Algebra of sets

The operations on sets such as union and intersection satisfy various laws of algebra.

1. Idempotent Laws: For any set A, we have (i) A∪A = A and (ii) A∩A = A

2. Identity Laws: For any set A, we have (i) A∪ ∅ = A and (ii) A∩U = A

That is, ∅ and U are the identity elements for the union and intersection respectively.

3. Commutative Laws: If A and B are any two sets then

(i)A∪B = B∪A and (ii) A∩B = B∩A

That is, Union and intersection are commutative.

4. Associate Laws: If A,B,C are any three sets then

(i) A∪(B∪C) = (A∪B)∪C and (ii) A∩(B∩C) = (A∩B)∩C

5. Distributive Laws: If A,B,C are any three sets then,

(i)A∪(B∩C) = (A∪B)∩(A∪C) and (ii) A∩(B∪C) = (A∩B)∪(A∩C)

That is, Union and intersection are distributive over intersection and union

respectively.

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6. De-Morgan’s Laws: If A and B are any two subsets of U, then

(i) (A∪B)′ = A′∩B′. That is, complement of union of two sets is equal to the

intersection of their complements.

(ii) (A∩B)′ = A′∪B′. That is, complement of intersection of two sets is equal to

the union of their complements.

7. Difference of two sets interms of intersection and complement: If A and B are any

two sets, then A-B = A∩B′.

We have A-B = {x: x∈ A and x∉B} = {x: x∈A} and {x: x∈B′} = A∩B′.

Ordered pair Let A and B be two sets. If a ∈ A, b ∈ B then an ordered pair whose first

component is „a‟ and the second component is „b‟ is denoted by (a,b) and is defined

as follows:

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

Cartesian Product of Sets

Let A and B be two sets. The set of all ordered pairs (a, b) of elements a∈ A, b∈B

is called the Cartesian product of sets A and B and is denoted by A×B.

Thus A×B = {(a,b) : a∈A and b∈B}.

Example: Let A = {1,2,3} and B = {3,4}.

A×B = {(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)}

Remark:

i) If A = ∅ or B = ∅, then A×B is defined as ∅.

ii) A×B ≠ B×A, unless A = B

iii) If A and B are two disjoint sets, A×B and B×A are also disjoint.

iv) If A and B are two finite sets, then n(A×B) = n(A).n(B)

v) If either A or B is an infinite set and the other is a non-empty set then A×B is

an infinite set.

Illustrative Examples

Example 1. Find x and y if (x-3, 2) = (5, 2x-y)

Solution From the equality of the ordered pairs we have, (x-3) = 5 and (2x-y) =2.

Therefore x = 8 and y = (2x-2) = 16-2 = 14

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Example 2. If A = {1,2,3} and B = {2,4}, find

(i) (A×B), (B×A)

(ii) (A×B) ∩ (B×A)

Solution (i) A×B = {(1,2), (1,4), (2,2), (2,4), (3,2), (3,4)}

B×A = {(2,1), (2,2), (2,3), (4,1), (4,2), (4,3)}

(iii) (A×B) ∩ (B×A) = {(2,2)}

Example 3. If A = {1,2,3}, B = {2,4}, C = {2,5};

(i) Find (B-C), (A×B, (A×C)

(ii) Verify that A×(B-C) = (A×B) - (A×C)

Solution (i) B-C = {4}

A×B = {(1,2), (2,2), (3,2), (1,4), (2,4), (3,4)}

A×C = {(1,2), (2,2), (3,2), (1,5), (2,5), (3,5)}

(iii) A× (B-C) = {(1,4), (2,4), (3,4)}………………(a)

(A×B) - (A×C) = {(1,4), (2,4), (3,4)}………(b)

From (a) and (b), we have A× (B-C) = A×B - A×C

Example4. If A×B = {(1,3), (1,5), (2,3), (2,5), (3,3), (3,5)} ; find A and B.

Solution We know that A is the set of all first co-ordinates of elements of A×B and B is the set

of all second coordinates of elements of A×B.

Therefore, A = {1,2,3} and B = {3,5}.

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CHAPTER 2.

RELATION

Let A and B be two sets. A relation from A into B is a subset of A×B.

Thus, R is a relation from A to B if R ⊆ A×B.

If (a,b) ∈ R, then we write it as aRb and read it as a is R-related to b or simply, a is related to b, if

there is no ambiguity arises over R.

If A = B, we say that R is a relation in the set A or R is a relation on A. That is, A relation in A is

a subset of A×A.

Remark: If A contains „m‟ elements and B contains „n‟ elements then A×B contains mn

elements and hence there are 2mn

subsets of A×B. So the total number of relations from A to B is

2mn

. These include the trivial relation ∅ and the universal relation A×B.

Domain and range of a relation

Let R be a relation from a set A to a set B. Then the domain of R, denoted as dom R, is the set of

all first components of the ordered pairs of R and the range of R, denoted as range R, is the set of

all second components of the ordered pairs of R. B is called the co-domain of R.

Thus, dom R = {a ∈ A : (a, b) ∈ R for some b∈B} and

range R = {b ∈ B: (a, b) ∈ R for some a ∈ A}

Clearly, dom R is a subset of A and range R is a subset of B.

Example: Let A = {1,2,3,4} and B = {2,5,7,9}

Then, R = {(1,2), (1,5), (3,9), (4,2)} is a relation from A to B because R is a subset of A×B.

Therefore, dom R = {1,3,4} range R = {2,5,9} and co-domain = {2,5,7,9}

Inverse relation

Let A and B be two non empty sets and R be a relation from A to B. Then the inverse of R,

denoted by R¯1 , is a relation from B to A and is defined by

R¯1 = {(b,a) : (a,b) ∈ R}, Thus (a,b) ∈ R ⇔ (b,a) ∈ R¯1

Clearly, the range of R is the domain of R¯1 and conversely.

In the example given above, R¯1 = {(2,1), (5,1), (9,3), (2,4)}

dom R¯1 = {2,5,9} = range R and range R¯1 = {1,3,4} = dom R.

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Types of Relations

(a) Reflexive Relation: A relation R in a set A is said to be reflexive if aRa for all a ∈ A.

That is , if (a,a) ∈ R for every a ∈ A

In other words, every element is related to itself.

Examples:

(i) The relation “is equal to” in the set of real numbers is a reflexive relation because

each number is equal to itself.

(ii) The relation “is parallel to” in the set of all lines in a plane is reflexive because each

line is parallel to itself.

(b) Symmetric Relation: A relation R in a set A is said to symmetric if aRb ⇒ bRa; a,b∈A.

That is, if (a,b) ∈ R ⇒ (b,a) ∈ R.

Examples:

(i) The relation „is parallel to‟ in the set of all lines in a plane is a symmetric relation

because if line L1 is parallel to line L2 then line L2 is parallel to line L1.

(ii) The relation „is less than‟ in the set of real numbers is not a symmetric relation

Note: If R is a symmetric relation then R¯1 = R then R is a symmetric relation.

(c) Transitive Relation: A relation R in a set A is said to be transitive if aRb and

bRc ⇒ aRc for a,b,c ∈ A. That is (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R.

Examples:

(i) The relation „is parallel to‟ in the set of all lines is a transitive relation because if line

L1 is parallel to line L2 and line L2 is parallel to line L3 then line L1 is parallel to line

L3.

(ii) Let A = {1,2,3}. Then R1 = {(1,1) , (2,2) , (3,3) , (1,2) , (2,1)} is a transitive relation

while R2 = {(1,1) , (2,2) , (3,2) , (2,3)} is not a transitive relation, because (3,2) ∈ R

and (2,3) ∈ R, but (3,3) ∉ R.

(iii) The relation „is less than‟ in the set of real numbers is a transitive relation.

(d) Anti Symmetric Relation: A relation R in a set A is said to be an anti symmetric if aRb

and bRa ⇒ a = b. That is, (a,b) ∈ R and (b,a) ∈ R ⇒ a = b.

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Example: The relation R in the set of natural numbers defined by aRb if „a is a divisor of b‟ is

an anti symmetric relation because a/b and b/a implies a = b.

(e) Equivalence Relation: A relation R in a set A is said to be an equivalence relation if R is

reflexive, symmetric and transitive. Thus, R is an equivalence relation on A (in A) if :

(i) aRa, ∀ a∈R (ii) aRb ⇒ bRa , a,b∈ A and (iii) aRb and bRc ⇒ aRc , a,b,c∈ A.

Examples:

(i) The relation „is parallel to‟ in the set of all lines in a plane is an equivalence

relation.

(ii) Let A = {1,2,3}, then R1={(1,1), (2,2), (3,3), (1,3), (3,1)} is an equivalence

relation, but it is not reflexive.

Note: The smallest equivalence relation in a non-empty set A is the relation „is equal to‟ in A

and the largest equivalence relation is A×A.

Working Rule: If R is a relation on a set A, then to prove:

(i) Reflexivity: Take an arbitrary element a∈A and prove that aRa

(ii) Symmetry: Take any two elements a,b∈A and assume that aRb, prove that bRa

(iii) Transitivity: Take any three elements a,b,c ∈A and assume that aRb and bRc,

prove that aRc.

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Illustrative Examples

Example 1 If R = { (n,n3) : n is a prime number less than 10}

(i) Write R explicitly

(ii) Find the domain and range of the relation

Solution (i) We have, R = {(n,n3) : n is a prime number less than 10}

= {(2,8), (3,27), (5,125), (7,343)}

(ii)Domain of R = {2,3,5,7}

Range of R = {8, 27, 125, 343}

Example 2 If R is the relation „less than‟ from A = {1,2,3,4,5} to B = {1,4,5},

(i) Write down the set of ordered pairs of R.

(ii) Find R¯1 .

Solution

(i) Given R = {(a,b) : a∈A , b∈B and a<b}.

Therefore, R = {(1,4), (1,5),(2,4), (2,5), (3,4), (3,5), (4,5)}

(ii) R¯1 = {(4,1), (5,1), (4,2), (5,2), (4,3), (5,3),(5,4)}

Example 3 Let A = {1,2}. Find:

(i) A×A

(ii) The number of relations on A.

Solution

(i) A×A = {(1,1), (1,2), (2,1), (2,2)}

(ii) Number of subsets of A×A = 24 = 16

Therefore, Number of relations on A = 24 = 16

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CHAPTER 3.

FUNCTIONS

In this section we consider a special type of relations known as functions. The word function first

appears in a Latin manuscript written by G.W.Leibnitz (1646-1716)

Definition: Let A and B be two non-empty sets. A function f from A to B is a relation from A

to B such that domain of f is A and no two ordered pairs in f have the same first component.

That is, A function or map f from A to B is a rule which associates to each element x ∈ A, a

unique element y ∈ B. The unique element y of B is called image of x under f and is written as

y = f(x). Also , x is called the pre-image of y under the function f.

If f is a function from A to B, then we write f: A → B

Each function f from A to B gives rise to a set {(x, f(x)) / x ∈ A} of ordered pairs. It is customary

to refer this set itself as the function f. So f ⊆ A×B. Hence ,

(i) Every function from A to B is a relation from A to B; but a relation from A to B need

not be a function.

(ii) A relation from A to B is a function if and only if every element in A has a unique

image in B.

The set A is called the domain of f and B is called the co-domain of f. The subset of B

formed by the image of the elements of A under f is called the range of f and is denoted by

ran f.

That is, ran f = {f(x): x ∈ A}

In our day today life we can see many situations where the concept of function works. For

example, it is an observed fact that if the length of the foot of a person is x inches, then the

size of the most suitable shoe is about (3x – 22) inches. So if we take y as the size if his shoe,

then y = 3x – 22. Clearly, y is a function of x. If a man‟s shoe size is 9 inches we can find

the length of his foot and conversely, if the length of the foot is known we can find the size

of the shoe.

Example 1. Let A = {1,2,3} and B = {a,b}. Let f = {(1,a), (2,a), (3,b)}, f is a function from A

into B because domain of f = A and no two ordered pairs have the same first component.

Example 2. Let A = {1,2,3} and B = {a,b}. Let f = {(1,a), (1,b), (2,a), (3,b)}. Here domain of

f = A. But f is not a function from A to B because the ordered pairs (1,a) and (1,b) have the same

first component.

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Graph of a function

If f : A → B is a function, by the graph of f, we mean the subset {(x,f(x)) : x ∈ A} of A×B.

The concept of the graph of a function is illustrated by the following examples.

Example 1. f : R → R is the function defined by f(x) = 2x-1.

(i) Write the graph of f in the roster form

(ii) Sketch the graph of f

Solution: (i) Graph of f = {(x, f(x)) : x ∈ R} = {(x,2x-1) : x ∈ R}

(ii) We may prepare the following table

Plotting the points (0,-1), (1,1), (2,3) and drawing the graph of f is a straight line.

Note: If the graph of a function is a straight line then it is called a linear function. So the

function given in the above example is a linear function.

Example 2. Draw the graph of the function f : R → R defined by f(x) = |x|

Solution: We have |x| = {x,x≥0 , and f = {(x,f(x)), x ∈ R}}

{-x, x≤ 0 , and f = {(x,f(x)), x ∈ R}}

The points (x,f(x)) lie on the line y = x for x ≥ 0 and on the line y = -x for x < 0. The graph of

f(x) is shown below:

X 0 1 2

f(x) -1 1 3

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Note : The function f is called the modulus function

Equal functions: Two functions f, g: A→B are said to be equal if and only if

f(x) = g(x) ∀ x ∈ A.

Example: Let A = {a,b,c} and B = {p,q,r}

Let g = {(a,p), (b,q), (c,r)}; h = {(b,q), (c,r),(a,p)}

Clearly g and h are functions from A to B. Also g(a) = p = h(a), g(b) = q = h(b) and

g(c) = h(c). Therefore, g and h are equal functions.

Types of functions

(i) Into function: A function f : A → B is said to be an into function if there exists at least

one element in B having no pre-image in A. In other words, a function f: A→B is an into

function if ran f ≠ B.

Example. Let A = {a1, a2, a3,a4, a5}, B = {b1,b2, b3, b4, b5}. Consider the function f : A → B

represented by the diagram given below.

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f: A → B is an into function, because the element b2 in B have no pre-images in A.

Note that ran f ≠ B.

(ii) Onto function: A function f : A → B is said to be an onto function or surjective

function or surjection if each element in B has at least one pre-image in A.

i.e., if f : A → B is an onto function, then corresponding to each y ∈ B there exists at

least one x ∈ A such that y = f(x).

Remark: (i) f : A → B is an onto function if ran f = B. A function which is not „onto‟ is „into‟

and vice versa.

(ii) If f is onto then n(B) ≤ n(A)

Example. (i) Let A = {l,m,n,o} and B ={5,6,7}. Consider the function f : A → B defined by

f(l) = 5, f(m) = 6, f(n) = 6, f(o) = 7

Then ran f = {5,6,7} = B. Therefore f is an onto function. The diagrammatic

representation of f is given in the figure below:

A B

The following figure f : X→Y is not an onto function, because there exist 4 in Y, having no

pre-image in X

X Y

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(iii) One-to-one function

A function f : A → B is said to be one-to-one if different elements in A have different

images in B. One-one function is otherwise known as injective function or injection

or one-to-one function.

Thus, f : A → B is one-to-one ⇔ x1, x2 ∈ A, x1 ≠ x2 ⇒ f(x1) ≠ f(x2)

In other words, f : A → B is one-to-one ⇔ f(x1) = f(x2) ⇒ x1 = x2

Example: Consider the function f: A → B and g: A → B represented by the following diagrams.

s

Clearly, f : A → B is a one-one function but g : A → B is not one-to-one because two distinct

element -2 and 2 have the same image 4 under g.

Note: If f: A → B is one-to-one then n(A) = n(B).

(iv) Many-one function:

A function f: A → B is said to be a many-one function if two or more elements of A

have the same image in B. In other words a function which is not one-one is called a many-one

function.

Example: Consider the function f: A → B.

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f: A → B is many-one because two distinct elements a1 and a5 in A are having the same image b2

in B.

(v) Bijective function

A function f: A → B is a said to be a bijective function or simply a bijection if it is

both one-to-one (injective) and onto (surjective).

Example: Consider the function f from A to B defined by the diagram given below

Clearly , f is one-to-one and onto and hence is a bijective map.

Note: If f is a bijection then n(A) = n(B)

(vi) Constant function

A function f: A → B is called a constant function if the same element in B is assigned

as the image of every element in A.

i.e., f: A → B is a constant function if f(a) = b ∀ a ∈ A and for a fixed b ∈ B.

Example: Consider the function f: A → B defined by the given diagram

22

Since f(A) = 3, therefore f: A → B is a constant function.

Remark: f : A → B is a constant function if ran f is a singleton set.

(vii) Identity function

Let A be any set. The function Ix : X → X defined by Ix(y) = y, ∀ y ∈ X is called the

identity function on X.

Example: Let X = {1,2,3,4}. The Ix = {(1,1), (2,2), (3,3), (4,4)}

The diagrammatic representation is given below

X X

Linear function

The function f : R→R defined by f(x) = ax + b, a and b are constants and a ≠ 0 is called a

linear function. Every linear function is a bijection. The graph of a linear function is a straight

line.

Inverse of an element under a function

Let f be a function from A to B and y ∈ B. The inverse of the element y under the function f,

denoted by f-1

(y), is defined as: f-1

(y) = {x: x∈ A and f(x) = y}. Clearly f-1

(y) ⊆ A

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Example. Consider the function f: A→B given in the following diagram.

Here, f(1) = {a}, hence f¯1(a) = {1} an so on.

Inverse Function

Let f be a function from A to B and let b ∈ B. We have seen that f-1

(b) may consists of a single

element, more than one element or no element at all. But if f : A → B is a bijection then for each

b ∈ B, f-1

(b) will consist of a single element.

i.e., for each b ∈ B, there exists a unique element a ∈ A such that f(a) = b.

So, we may define a function, f-1

: B→A by

f-1

(b) = a ⇔ f(a) = b

The function f-1

is called the inverse function of f.

Illustrative Examples

Example 1. Examine each of the following relations given below and state in each case, giving

reasons whether it is a function or not?

(i) R = {(2,1),(3,1).(4,2)}

(ii) R = {(2,2),(2,4),(3,3),(4,4)}

(iii) R = {(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)}

Solution. (i) Since 2,3,4 are the elements of domain of R having unique images, this relation R

is a function.

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(ii)Since the same first element 2 corresponds to two different images 2 and 4, this

relation is not a function.

(iii)Since every element has one and only one image, this relation is a function.

Example 2. Let f: N → N be defined by f(n) = 5n.

(i) If f(n) = 1 find n. Is n an element of N?

(ii) Use (i) to show that f is not „on to‟

(iii) If m and n are two element of N such that f(m) = f(n), prove that m = n

(iv) Is f a one to one function?

Solution

(i) f(n) = 1 ⇒ 5n = 1 ⇒ n = 1/5 ∉ N

(ii) (i) shows that 1 has no pre-image, hence f is not „on to‟

(iii) Let m, n ∈ N such that f(m) = f(n) Since, 5m = 5n ⇒ m = n

(iv) (iii) shows that f is one-to-one

Example 3. Given an example of a map which is (i) one-to-one but not onto (ii) neither

one-to-one nor onto (iii) not one-to-one but onto.

Solution. (i) Consider the function f :N → N defined by f(x) = 3x ∀ 𝑥 ∈ N

Let x1, x2 ∈ N such that x1 ≠ x2.

Therefore,3x1 ≠ 3x2 ⇒ f( x1) ≠ f(x2) (since f is one-to-one)

ran f = { 3,6,9…} ≠ N, the codomain (since f is not onto)

(ii) Consider the map f : I → I defined by f(x) = |x|

f(3) = |3| = 3 and f(-3) = |-3| (since f(3) = f(-3) = 3

⇒ distinct elements having the same image

⇒ f is not one-to-one.

ran f = {0,1,2,…} ≠ I, the codomain of f (since f is not onto).

Hence, f is neither one-one nor onto.

(iii) Let A = {-2,0,2} and B = {0,2}.

Define f : A→B by f(x) = |x|.

Therefore f(-2) = |-2| = 2, f(0) = 0, f(2) = 2.

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Since f(-2) = f(2) = 2, f is not one-to-one.

range f = {0,2} = B, codomain of f (since f is onto).

Hence f is not one-to-one but onto

Example 4. Let f : Q→Q be defined by f(x) = 2x + 3.

(i) If f(x) = f(y), prove that x = y

(ii) If f(x) is a rational number, prove that x is also a rational number.

(iii) Hence prove that f is invertible. Find f¯1.

Solution:

(i) Let x, y ∈ Q such that f(x) = f(y)

⇒ 2x + 3 = 2y + 3

⇒ 2x = 2y

⇒ x = y

(ii) Suppose f(x) = y, a rational number

i.e., 2x + 3 = y, Therefore, x = y-3/2, which is also a rational number

(iii) (i) shows that f is „one-one‟ and (ii) shows that f is onto‟. Therefore f is invertible.

To find f¯1, let y = f(x) ⇒ y = 2x + 3 ⇒ x = y-3/2 .Since f¯1 (y) = y-3/2. Thus, we define :

f¯1: Q→Q by f¯1(x) = x-3/2 ∀ x ∈ Q.

Example 5.

f: N→N is the function defined by f(n) = {(n+1)/2, if n is odd and n/2, if n is even} (i) Find f(1)

and f(2)

(ii) Hence, show that f is not a bijection

Solution

(i) f(1) = 1+1/2 = 1 and f(2) = 2/2 = 1

(ii) Here, 1 ≠ 2, but f(1) = f(2) , since f is not one-to-one.

Hence, f is not a bijection.

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Composition of functions

Let A,B,C be three non-empty sets and let f : A→B , g :B→C be two functions. Let a∈A. Since f

is a function from A to B, corresponding to a∈A, there exists a unique f(a) ∈ B. Since g is a

function from B to C, corresponding to f(a) ∈ B there exists a unique element g(f(a)) in C. Thus

corresponding to each element a∈A, there exists a unique element g(f(a)) in C. So, from f and g

we can define a new function from A to C. This new function is called the product or

composition or composite function of f and g and is denoted by g◦f

Definition: Let f: A → B and g: B → C be two functions. Then the function g◦f : A → C, defined

by (g◦f)(a) = g(f(a)), ∀ a ∈ A, is called the composition of f and g.

Example (i) Consider the function f :R → R and g : R → R defined by f(x) = x2 and g(x) = cos x.

(f◦g)(x) = f(g(x)) = f(cos x) = cos2x

(g◦f)(x) = g(f(x)) = g(x2) = cos (x

2)

f◦g and g◦f are both defined; but f◦g ≠ g◦f

Example (ii) Let A = {p,q,r}, B = {a,b,c,d} and C = {x,y,z}.

Let f : A → B and g : B → C be defined by f = {(p,a), (q,c),(r,d)} and g = {(a,x),(b,x), (c,y),(d,z)}

(g◦f)(p) = g(f(p)) = g(a) = x, (g◦f)(q) = g(f(q)) = g(c) = y,

(g◦f)(r) = g(f(r)) = g(d) = z. Then g◦f = {(p,x), (q,y), (r,z)}. In this case f◦g is not defined.

Remark: g◦f is defined only when ran f ⊆ dom g.

f◦g is defined only when ran g ⊆ dom f

Note: Even if f◦g and g◦f are defined , they may not be equal

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Illustrative Examples

Example1. Let A = {1,2,3,4}. Let f = {(1,4), (2,1), (3,3), (4,2)} and g = {(1,3), (2,1), (3,2),

(4,4)}. Find (i) f◦g (ii) g◦g

Solution (i) (f◦g)(1) = f(g(1)) = f(3) = 3

(f◦g)(2) = f(g(2)) = f(1) = 4

(f◦g)(3) = f(g(3)) = f(2) = 1

(f◦g)(4) = f(g(4)) = f(4) = 2

Therefore, (f◦g) = {(1,3), (2,4), (3,1), (4,2)}

(ii) (g◦g)(1) = g(g(1)) = g(3) = 2

(g◦g)(2) = g(g(2)) = g(1) = 3

(g◦g)(3) = g(g(3)) = g(2) = 1

(g◦g)(4) = g(g(4)) = g(4) = 4

Therefore, (g◦g) = {(1,2), (2,3), (3,1), (4,4)}

Example2. If f, g : R → R are defined by f(x) = x2+3x+1, g(x) = 2x-3 find:

(i) f◦g (ii) f◦f (iii) (g◦f)(2) (iv) (g◦g)(-5) (v) Prove that f◦g ≠ g◦f

Solution (i) (f◦g)(x) = f(g(x)) = f(2x-3) = (2x-3)2 + 3(2x-3) + 1 = 4x

2-6x+1

Therefore, (f◦g)(x) = 4x2-6x+1, ∀ x ∈ R

(ii) (f◦f )(x) = f(f(x)) = f(x2+ 3x + 1 = (x

2+3x+1)

2 + 3(x

2+3x+1) + 1

= x4 + 6x

3+14x

2+15x+5, ∀ x ∈ R

(iii) (g◦f)(2) = g(f(2)) = g(22+3(2) +1) = g(11) = 22-3 = 19

(iv) (g◦g)(-5) = g(g(-5)) = g(2(-5)-3) = g(-13) = -29

(v) g◦f(x) = g(f(x)) = g(x2+3x+1) = 2(x

2+3x+1) = 2x

2+6x-1

Therefore, from (i) f◦g ≠ g◦f.

Example3. If f :R → R is defined by f(x) = x2+1, find f(f(x))

Solution f(f(x)) = f(x2+1) = (x

2+1)

2+1 = x

4+ 2x

2+2.

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Exercises

1. (i) Show that for any sets A and B; A = (A∩B) ∪ (A-B) and A∪(B-A) = A∪B

(ii) Let R = {(x,y): x,y ∈ N and x+3y=12}, N being the set of all natural numbers. Write R

and R¯1 as sets of ordered pairs. Also find the domain and range of R.

2. (i) Let f = {(1,1), (2,3), (0,-1), (-1,-3)} be a function from Z to Z defined by f(x) = ax+b,

for some integers a,b. Determine a,b.

(ii) If f : A→B and g :B→C are onto functions, show that g◦f is also an onto function.

3. (i) If the functions f and g are given by f = {(1,2), (3,5), (4,1)} and

g = {(2,3), (5,1), (1,3)} find f◦g and g◦f

(ii) A survey shows that 76% of the Indians like oranges, whereas 62% like bananas.

What percentage of the Indians like both oranges and bananas?

(iii) Is it true for any sets A and B, P(A)∪P(B)? Justify your answer

4. (i) If A×B = {(a,x), (a,y), (b,x), (b,y)}. Find A and B.

(ii) Show that „similarity‟ is an equivalence relation in the set of all triangles in a plane.

(iii) Find the domain of the function f(x) = [(x2+2x+1) / (x

2-8x+12)]

5. (i) Let A = {1,2,3,4} and R= {(a,b): a∈A, b∈A, a divides b}. Write R explicitly

(ii) Show that the function f :R→R defined by f(x) = 14x3+15 ; x∈ R, is bijective.

6. f :R→R is a linear function such that f(0) = 3 and f(-1) = 1.

(i) Find f(x)

(ii) f◦f(2)

(iii) f¯1(x)

7. (i) Let U be the set of all triangles in a plane. If A is the set of all triangles with at least

one angle different from 600, what is A′?

(ii) If P(A) = 1. Find the set A

Answers

1. (ii) R = {(3,3), (6,2), (9,1)}, R¯1 = {(3,3), (2,6), (1,9)}; dom R = {3,6,9},

Range R = {1,2,3}

2. (i) a = 2, b = -1

3. (i) f◦g = {(2,5), (5,2), (1,5)}; g◦f = {(1,3), (3,1), (4,3)} (ii) 38% (iii) False

4. (i) A = {a,b} B = {x,y} (iii) Domain = R-{2,6}

5. (i) {(1,1), (1,2), (1,3), (1,4), (2,2), (2,4), (3,3), (4,4)}

6. (i) 2x+3 (ii) 4x+9 (iii) (x-3)/2

7. (i) Set of all equilateral triangles (ii) ∅

REFERENCE

Dr. Joseph Mathew-Systematic Approach to Mathematics (Standard XI)