Set Relations and Function

SET RELATIONS
AND

FUNCTION

AHMAD SANI BIN AMAN
NUR JANNAH BINTI ZAINOL ABIDIN

MOHD FAIROSS BIN IBRAHIM



SET RELATIONS
AND

FUNCTION

Published by,
POLITEKNIK METrO TASEK GELUGOR
No 25, Jalan Komersial 2
13300 Tasek Gelugor
Pulau Pinang
First Print 2022
All rights reserved. It is not permitted to reproduce any part of contents in SET RELATIONS
AND FUNCTION book, in any form by any means whether electronically, photocopying,
mechanically, recording or otherwise with obtaining written permission from the Director of
Politeknik METrO Tasek Gelugor

PREFACE

This text is a basic introduction to those areas of discrete mathematics. Discrete
mathematics is the common core course for the students in the IT department. In writing
this e-book is to offer fundamental concepts and methods of sets, relations, and functions.
This writing also tries to help those interested in mathematics to keep insight into
mathematical techniques and their importance for application in real life.

The objectives of a study on sets, relations, and functions in discrete mathematics can help
the students to think mathematically. This book emphasizes mathematical reasoning and
problem-solving technique. Each chapter begins with a clear definition, principles, and
theorem, followed by solved examples and exercises.

Our deepest gratitude to Unit e-Pembelajaran, Politeknik METrO Tasek Gelugor, for their
practical guidance, advice, and suggestions during the project and thus inspired us to take
up an exciting and challenging project like this. We want to acknowledge the contribution
of many people who have helped to bring this project successfully.

CONTENTS 1
2
i. Preface 3
ii. Contents 9
10
1. SETS 11
1.1 Representation of A Set 13
1.2 Classification of Sets 14
1.3 Venn Diagram
1.4 Difference of Sets 16
1.5 Complement of A Set 16
1.6 Intersection of Sets 16
1.7 Union of Sets 17
19
2. RELATIONS 21
2.1 Introduction
2.2 Definition Of Relations 23
2.3 Domain And Range 23
2.4 Types Of Relations 26
2.5 Representation Of Relations Using Digraph
28
3. FUNCTIONS
3.1 Definition of A Function
3.2 Classification of Functions

REFERENCE

SYNOPSIS

This book covers the sets, relations, and functions topics in discrete mathematics. The text
explains and guides the students through a discussion on how sets, relations, and functions
were presented. The hands-on exercises help students understand a concept soon after
learning it.

In Sets, the basic definitions and notation of set theory were introduced to show how to
establish properties of sets. The explanation of how to derive its properties was also
discussed.

The mathematics of relations defined on sets, focusing on ways to represent relations and
explore their various properties. A more formal way to refer to the kind of relation is to call
it a binary relation because it is a subset of a Cartesian product of two sets. Binary relations
in this text will mean binary relation when we use the term relation by itself.

Functions are ubiquitous in mathematics and computer science. We consider an additional
wide variety of functions in this part, focusing on those defined on discrete sets. We then
look at properties of functions such as one-to-one and onto, the existence of inverse
functions, and the interaction of composition of functions and the properties of one-to-one
and onto.

CHAPTER 1 : SETS

SOME STANDARD NOTATIONS

For example, = ℎ , , .
Some standard notations to represent sets :
N : the set of natural numbers
W : the set of whole numbers
Z or I : the set of integers
Z+ : the set of positve integers
Z- : the set of negative integers
Q : the set of rational numbers
R : the set of real numbers
C : the set of complex numbers
Other frequently used symbols are :
∈ : 'belongs to'
∉ : 'does not belong to'
∃ : There exists, ∄ : There does not exist.
For example, N is the set of natural numbers, and we know that 2 is a natural number but
−2 is not a natural number. It can be written in the symbolic form as 2 ∈ and −2 ∉ .

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2.1 REPRESENTATION OF A SET

There are two methods to represent a set.

1) Roster method (Tabular form)

In this method, a set is represented by listing elements, separating them by commas and
enclosing them in the curly bracket.

Example 1:
If be the set of vowels of the English alphabet, it can be written in Roster form as :

= { , , , , }

Example 2:
If be the set of natural numbers less than 7, then = {1, 2, 3, 4, 5, 6}, is in the Roster
form.

2) Set-builder form

In this form elements of the set are not listed but these are represented by some common
property.

Example 3:
Let be the set of vowels of the English alphabet, then can be written in the set-builder
form as = { ∶ ℎ ℎ }

Example 4:
Let be the set of natural numbers less than 7, then = { ∶ Î 1£ < 7}

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Example 5:
Write the following in set -builder form :
(a) = {−3, −2, −1,0,1,2,3}
(b) = {3,6,9,12}

Solution:
(a) = { ∶ ∈ − 3 ≤ ≤ 3 }
(b) = { ∶ = 3 ∈ , ≤ 4 }

Example 6:
Write the following in Roster form.
(a) = { ∶ ∈ 50 ≤ ≤ 60 }
(b) = { ∶ ∈ ! − 5 + 6 = 0}

Solution:
(a) = {50, 51, 52,53,54,55,56,57,58,59,60}
(b) ! − 5 + 6 = 0 ; ( − 3) ( − 2) = 0; = 3,2 ; ∴ = {2, 3}

1.2 CLASSIFICATION OF SETS

1) Finite and infinite sets

Let A and B be two sets where = { ∶ } and = { ∶
ℎ }

The number of elements in set is not finite (infinite), while the number of elements in set
is finite.

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is said to be an infinite set, and is said to be a finite set. A set is said to be finite if its
elements can be counted, and it is said to be infinite if it is not possible to count up to its last
element.

2) Empty (Null) Set

Consider the following sets.
= { : ∈ ! + 1 = 0}
= { ∶ ℎ ℎ ℎ 7 ℎ 5}

Set consists of real numbers, but there is no real number whose square is -1. Therefore this
set consists of no element. Similarly, there is no such number that is less than 5 and greater
than 7. Such a set is said to be a null (empty) set. The symbol void denotes it, ∅ or { }. A set
with no element is a null/empty/void set and is denoted by ∅ .

3) Singleton Set :

Consider the following set : = { ∶ }

As there is only one even prime number, namely 2, so set A will have only one element.
Such a set is said to be a singleton. Here = {2}. A set that has only one element is
known as a singleton.

4) Equal And Equivalent Sets

Consider the following examples.
(i) = {1, 2, 3} , = {2,1,3}
(ii) = {1, 2, 3} , = { , , } .

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In example (i) Sets and to have the same elements. Such sets are said to be equal sets,
and it is written as = . In example (ii) Set and have the same number of elements
but different elements. Such sets are said to be equivalent sets and are written as ≈ .

Two sets and are said to be equivalent sets if they have the same number of elements,
but they are said to be equal if they have not only the same number of elements but elements
are also the same.

5) Disjoint Sets

Two sets are said to be disjoint if they do not have any common element. For example, sets
= { 1,3,5} and = { 2,4,6 } are disjoint sets.

Example 7:
Given that = { 2, 4} and = { ∶ ! + 6 + 8 = 0 }
Are A and B disjoint sets ?

Solution:
If we solve ! + 6 + 8 = 0,we get = − 4, − 2.
= {−4, −2} and = {2, 4}. Clearly, and are disjoint sets as they do not have any
common element.

Example 8:
If = { ∶ ℎ ℎ } and = { ∶ ∈ ≤ 5}
Is (i) = , (ii) ≈ ?

Solution:
= { , , , , } and = {1, 2, 3, 4,5}.
Each set has five elements, but elements are different ∴ ≠ but ≈ .

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Example 9:
Which of the following sets are finite or infinite?
= { ∶ }
= { ∶ ∈ ≤ 50}

Solution:
As the number of points on a line is uncountable (cannot be counted) so is an infinite set
while the number of natural numbers up to fifty can be counted, so is a finite set.

Example 10:
Which of the following sets are empty?
= { : ! − 1 = 0} .
= { ∶ ∈ − 2 ≤ ≤ 2}

Solution:
Set consists of those irrational numbers which satisfy ! − 1 = 0.
If we solve ! − 1 = 0 we get = ±1. Clearly, ±1 are not irrational numbers.
Therefore is an empty set.
But = {−2, −1, 0, 1, 2}. B is not an empty set as it has five elements.

Example 11:
Which of the following sets is singleton?
= { ∶ ∈ − 2 = 0} = { ∶ ∈ ! − 2 = 0}

Solution:
Set A contains those integers which are the solution of − 2 = 0 or = 2. ∴ = {2}.
is a singleton set.
is a set of those real numbers which are solutions of ! − 2 = 0 or = ± √2
= { −√2, √2} Thus, is not a singleton set.

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6) Subset

Let set be a set containing all students of your school, and be a set containing all students
of class XII of the school. In this example, each element of set is also an element of set .
Such a set is said to be a subset of set . It is written as ⊆

Consider = {1, 2, 3, 4, . . . . . . . . }, = {. . . . . −3 − 2, −1, 0, 1, 2, 3, . . . . . . . }
Clearly, each element of set is an element of set also as ⊆

If and are any two sets such that each element of the set is an element of the set
also, then is said to be a subset of .

Example 12:
If = { ∶ ℎ 5} and
= { ∶ } then is a proper subset of ?

Solution:
It is given that = {2, 3 }, = {2} Clearly ⊆ and ≠
We write ⊂ and say that is a proper subset of .

Example 13:
If = {1, 2, 3, 4}, = {2, 3, 4, 5}. Is A Í B or B Í A?

Solution:
Here 1ÎA but 1ÏB ÞA⊄B.
Also 5ÎB but 1 Ï B ÞA⊄ B.
Hence neither A is a subset of B nor B is a subset of A.

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Example 14:
If = { , , , , }, = { , , , , }. Is A Í B or B Í A or both?

Solution:

Here in the given sets, each element of set A is an element of set B also

\AÍB .......... (i)

and each element of set B is an element of set A also. \ B Í A ......(ii)

From (i) and (ii)

A=B

7) Power Set

Let = { , }. Subset of are f , { }, { } and { , }.
If we consider these subsets as elements of a new set (say) then = {f,{ }, { }, { , }}
is said to be the power set of .

Example 15:
Write the power set of each of the following sets :
(i) = { ∶ Î 2 + 7 = 0}.
(ii) = { ∶ Î 1£ £ 3}.

Solution:
(i) Clearly = f ( )
\ f is the only subset of given set
\ ( ) = { f }
(ii) The set can be written as {1, 2, 3}
Subsets of B are f , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.
\ ( ) = { ∅ , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} } .

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8) UNIVERSAL SET

Consider the following sets.
= { ∶ ℎ }
= { ∶ ℎ }
= { ∶ ℎ }
= { ∶ ℎ }

Clearly, the sets , , are all subsets of .

Example 16:
Which of the following set can be considered as a universal set?
= { ∶ }
= { ∶ }
= { ∶ }

Solution:
As it is clear that both sets and are a subset of .
\ is the universal set for this problem.

1.3 VENN DIAGRAM

British mathematician John Venn (1834-1883 AD) introduced the concept of diagrams to
represent sets. According to him, universal set is represented by the interior of a rectangle
and other sets are represented by the interior of circles.

For example if = {1, 2, 3, 4, 5}, = {2, 4} and = {1,3}, then these sets can be
represented as

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Figure 1.1 : Venn Diagram
Diagrammatic representation of sets is known as a Venn diagram.

1.4 DIFFERENCE OF SETS
Consider the sets = {1, 2, 3, 4, 5} and = {2, 4, 6}.
A new set having those elements in but not is said to be the difference of sets and
and is denoted by − .
\ − = {1, 3, 5}
Similarly, a set of elements in but not in is the difference between and and is
devoted by − .
\ − = {6}
In general, if and are two sets then

− = { ∶ Î and Ï }
− = { ∶ Î and Ï }
The difference of the two sets can be represented using the Venn diagram as :

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Figure 1.2 : Difference of Sets

1.5 COMPLEMENT OF A SET

Let X denote the universal set and Y, Z its sub set where
= { ∶ ℎ }
= { ∶ ℎ }
= { ∶ ℎ }
− is a set having female members of the family
− is a set having male members of the family
− is said to be the complement of and is usally denoted by ′ or "..
− is said to be complement of and denoted by ′ or ".

If is the universal set and is its subset then the complement of is a set of those elements
which are in which are not in . It is denoted by ′ or "

. . ′ = − = { ∶ Î and Ï }
The complement of a set can be represented using Venn diagram as :

Figure 1.3 : Complement of Set

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Example 17:
Given that

= { ∶ ℎ 10} and
= { ∶ ℎ 10}
Find (i) − (ii) − (iii) is − = − ?

Solution:
It is given that = {2, 4, 6, 8, 10}, = {1, 3, 5, 7, 9}
Therefore,
(i) − = {2, 4, 6, 8, 10}
(ii) − = {1, 3, 5, 7, 9}
(iii) Clearly from (i) and (ii) − ≠ − .

Example 18:
Let be the universal set and its subset where
= { ∶ ∈ ≤ 10 }
= { ∶ ℎ 10}
Find (i) " (ii) Represent " in Venn diagram.

Solution:
It is given = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. , = { 2, 3, 5, 7}
(i) " = − = {1, 4, 6, 8, 9, 10}
(ii)

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1.6 INTERSECTION OF SETS
Consider the sets = {1, 2, 3, 4} and = { 2, 4, 6}
It is clear that some elements are common to both sets and . A set of these common
elements is the intersection of A and B and is denoted by ∩ .
Here ∩ = {2, 4 }
If and are two sets, then the set of those elements which belong to both sets is said to
be the intersection of and . It is devoted by ∩ .

∩ = { ∶ ∈ ∈ }
∩ can be represented using the Venn diagram as :

Figure 1.4 : Intersection of Sets
Example 19:
Given that
= { ∶ 52 } and
= { ∶ 52 }
Find (i) ∩ (ii) Represent ∩ by using the Venn diagram.

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Solution:
(i) As there are only four kings out of 52 playing cards, set has only four elements.

Set has 13 elements as there are 13 spade cards, but out of these 13 spade
cards, there is one king also. Therefore there is one common element in and .
\ ∩ = { }.
(ii)

1.7 UNION OF SETS

Consider the following examples :
(i) is a set having all players of the Indian men cricket team, and is a set having all

players of Indian women cricket team. Clearly, and are disjoint sets. Union of
these two sets is a set having all players of both teams, and it is denoted by ∪ .
(ii) is a set having all cricket team players, and has all players of your school's
Hockey team. Suppose three players are common to both the teams then the union
of and is a set of all players of both the teams but three common players to be
written once only.

If and are only two sets, then union of and is the set of those elements which belong
to or .

In set builder form :
∪ = { ∶ ∈ ∈ }
OR
∪ = { ∶ ∈ − ∈ − ∈ ∩ }

∪ can be represented using the Venn diagram as :
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Figure 1.5 : Union of Sets
( ∪ ) = ( − ) + ( − ) + ( ∩ ) .
or ( ∪ ) = ( ) + ( ) − ( ∩ )
where ( ∪ ) stands for the number of elements in ∪ so on.
Example 20:

= { ∶ ÎZ+ and £ 5}
= { ∶ ℎ 10}
Find (1) ∪ (ii) represent ∪ using the Venn diagram.
Solution:
(i) We have, = {1, 2, 3, 4, 5} and = {2, 3, 5, 7}.
\ ∪ = {1, 2, 3, 4, 5, 7}.
(ii)

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CHAPTER 2 : RELATIONS

2.1 INTRODUCTION

Most of us are familiar with the term “less than,” “is parallel to,” “is a subset of,” and so on.
Relations can be referred to as a generalization of functions that represent arbitrary subsets
of × . It can be considered as the existence or nonexistence of a certain connection
between pairs of objects taken in a definite order. Formally, we can define a relation in terms
of these “ordered pairs”. An ordered pair of elements and , where is designated as the
first element and as the second element, is denoted by ( , ).

2.2 DEFINITION OF RELATIONS

Let us begin with a definition of relations.

Suppose and are sets. Binary relation or relation, between the sets and is the
subset of the direct product × . Relation, is a set of ordered pairs where each first
element comes from and each second element comes from . That is, for each pair ∈
and ∈ , exactly one of the following is true:

(i) ( , ) ∈ ; we then say “ is in relation, to ”, written .
(ii) ( , ) ∉ ; we then say “ is not in relation to ”, written .

We hold that ⟺ ( , ) ∈

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If is a relation from a set to itself, that is, if is a subset of ! = × , then we say that
is a relation to . The domain of a relation is the set of all first elements of the ordered
pairs which belong to , and the Range is the set of second elements.
Example 1:
Let = { ℎ }, = { ℎ } and = {( , ) ∶
ℎ }. Thus denotes ‘ ℎ ’.

Solution:
Examples are ( ) ( ), ( ) ( ), ( ) ( ), etc.

Example 2:
Let = = {1, 2, 3, 4, 5, 6} and = {( , ) ∶ }. Since A is a small finite set
we can list the elements of the relation:

Solution:
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)}.

2.3 DOMAIN AND RANGE

The domain is the set of all first elements of the ordered pairs. On the other hand, the Range
is the set of all second elements of the ordered pairs. However, Range includes only the
elements used by the function
If there are two sets, A and B, and relation R have order pair (x, y), then:

• The domain of , ( ), is the set { | ( , ) ∈ }
• The Range of , ( ), is the set { | ( , ) ∈ }

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Example 3:
Let, = {1, 4, 6} and = {1, 4, 7}

• Case 1: If relation is ‘ ’ then = {(1, 1), (4, 4)}
( ) = { 1, 4}, ( ) = { 1, 4}

• Case 2: If relation is ‘ ℎ ’ then = {(1, 4), (1, 7), (4, 4), (4, 7)}
( ) = { 1, 4}, ( ) = { 4, 7}

• Case 3: If relation is ‘ ℎ ’ then = {(4, 1), (6, 1), (6, 4)}
( ) = { 4, 6}, ( ) = { 1, 4}

Example 4:
State the domain and range of the following relation:
{(4,3), (−1,7), (2, −3), (7,5), (6, −2)}

Solution:
= {4, −1, 2, 7, 6}, = {3, 7, −3, 5, −2}

Example 5:
From the following Arrow Diagram, find the Domain and Range and depict the relation
between them?

Solution:
= {3, 4, 5}, = {3, 4, 5, 6}
Relations, = {(3, 4), (4, 6), (5, 3), (5, 5)}

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2.4 TYPES OF RELATIONS

The types of relations are nothing but their properties. There are different types of relations,
namely reflexive, symmetric, and transitive, which are defined and explained through real-
life examples.

1) Reflexive Relations

A relation on a set is reflexive if for every ∈ , that is, if ( , ) ∈ for every
∈ . Thus is not reflexive if there exists ∈ such that ( , ) ∉ .

Example 6:
The relation = {(1,1)(2,2)(3,3)} is reflexive over the set = {1,2,3}.

Example 7:
Consider the following five relations on the set = {1, 2, 3, 4}:

1 = {(1, 1), (1, 2), (2, 3), (1, 3), (4, 4)}
2 = {(1, 1)(1, 2), (2, 1), (2, 2), (3, 3), (4, 4)}
3 = {(1, 3), (2, 1)}
4 = ∅, ℎ
5 = × , ℎ
Determine which of the relations are reflexive.

Solution:
Since contains the four elements 1, 2, 3, and 4, a relation on is reflexive if it includes
the four pairs (1, 1), (2, 2), (3, 3), and (4, 4).
Thus only 2 and the universal relation 5 = × are reflexive.
Note that 1, 3, and 4 are not reflexive since, for example, (2, 2) does not belong to any
of them.

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Example 8:
Let = {1, 2, 3} and = {(1, 2), (2, 2), (3, 1), (1, 3)}. Is the relation reflexive?

Solution:
The relation R is not reflexive as for every ∈ , ( , ) ∉ , i.e., (1, 1) and (3, 3) ∉ .

2) Symmetric Relation

In discrete mathematics, symmetric relation between two or more elements of a set is such
that if the first element is related to the second element, then the second element is also
related to the first element as defined by the relation. As the symmetric name relations
suggests, the relation between any two set elements is symmetric. A relation R on a set A is
called symmetric if ( , ) ∈ holds when ( , ) ∈ .

Example 9:
The relation = {(4,5), (5,4), (6,5), (5,6)} on set = {4,5,6} is symmetric

Example 10:
The relation = {(1, 2), (2, 1), (3, 2), (2, 3)} on set = {1, 2, 3} is symmetric.

3) Transitive Relation

Transitive relations are binary relations defined on a set. If the first element is related to the
second element, and the second element is related to the third element of the set, then the
first element must be related to the third element. For example, if for three elements , ,
in set , if = and = , then = . A relation R on a set is called transitive if
whenever ( , ) ∈ and ( , ) ∈ , then ( , ) ∈ , for all , , ∈

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Example 11:
The relation = {(1, 2), (2, 3), (1, 3)} on set = {1, 2, 3} is transitive.

Example 12:
Define a relation R on a set = { , , } as = {( , ), ( , ), ( , )}. Determine if is
a transitive relation.

Solution:
As we can see that ( , ) ∈ and ( , ) ∈ , and for to be transitive ( , ) ∈ must
hold, but ( , ) ∉ .

So, R is not a transitive relation.

Exercise

1. For each of the following relations on the set {1, 2, 3, 4}, decide whether it is reflexive,
symmetric and/or transitive.
a) {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (4, 4)}
b) {(1, 3), (1, 4), (2, 3), (3, 4)}
c) {(1, 1), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 3), (3, 4)}

2. Consider the following relations on {1, 2, 3, 4}:
1 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 4), (4, 1), (4, 4)},
2 = {(1, 1), (1, 2), (2, 1)},
3 = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1), (4, 4)},
4 = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)},
5 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)},
6 = {(3, 4)}.
Which of these relations are reflexive, symmetric and/or transitive?

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2.5 REPRESENTATION OF RELATIONS USING DIGRAPH
A relation can be represented using a directed graph. The number of vertices in the graph
equals the number of elements in the set from which the relation has been defined. For each
ordered pair ( , ) in the relation , there will be a directed edge from the vertex to vertex
. If there is an ordered pair ( , ), there will be a self-loop on vertex .
Example
Let = {0, 1, 2, 3} and define relations , , and on A as follows:
= {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}
= {(0, 0), (0, 2), (0, 3), (2, 3)}
= {(0, 1), (2, 3)}.
Solution :

Digraph R

Digraph S

Digraph T

Exercise
Draw the directed graph that represents the relation
{( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}.
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CHAPTER 3 : FUNCTION

3.1 DEFINITION OF A FUNCTION

Consider the relation

∶ {( , 1), ( , 2), ( , 3), ( , 5)}

In this relation, we see that each element of A has a unique image in B. This relation f from
set to where every element of A has a unique image in is defined as a function from
to B. So we observe that in a function, no two ordered pairs have the same first element. We
also see that ∃ an element ∈ , i.e., 4 which does not have its preimage in . Thus here:
(i) the set will be termed as co-domain and
(ii) the set {1, 2, 3, 5} is called the range.

From the above, we can conclude that range is a subset of co-domain. Symbolically, this
function can be written as

∶ →

Example 1:
Which of the following relations are functions from A to B. Write their
domain and range. If it is not a function, give a reason?
(a) { (1, −2), (3, 7), (4, −6), (8, 1) }, = {1,3,4,8}, = {−2, 7, −6, 1, 2}
(b) { (1,0), (1, − 1), (2, 3), (4, 10) }, = {1,2,4} , = {0, −1,3,10}
(c) { ( , ), ( , ), ( , ), ( , )} , = { , , , , } = { , }

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(d) { (2,4), (3,9), (4,16), (5,25), (6,36 }, = { 2,3,4,5,6} , = {4,9,16,25,36 }
(e) { (1, −1), (2, −2), (3, −3), (4, −4), (5, −5)} , = { 0,1,2,3,4,5} ,
= {−1, −2, −3, −4, −5}
Solution :
(a) It is a function. = {1,3,4,8} , = {−2, 7, −6,1}
(b) It is not a function. Because the first two ordered pairs have the same first elements.
(c) It is not a function. = { , , , } ¹ A, = { , }
(d) It is a function. = {2,3,4,5,6} , = {4,9,16,25,36}
(e) It is not a function . = {1,2,3,4,5} ≠ , = {−1, −2, −3, −4, −5}
Example 2:
State whether each of the following relations represents a function or not.

Solution:
(a) f is not a function because the element b of does not have an image in .
(b) f is not a function because the element c of does not have a unique image in .
(c) f is a function because every element of has a unique image in .
(d) f is a function because every element in has a unique image in .

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Example 3:

Which of the following relations from → are functions?

(a) = 3 + 2 (b) < + 3 (c) = 2 ! + 1

Solution :
(a) = 3 + 2

Here corresponding to every element ∈ , ∃ a unique element ∈ .
\ It is a function.
(b) < + 3.
For any real value of we get more than one real value of .
\ It is not a function.
(c) = 2 ! + 1
For any real value of x, we will get a unique real value of y.
\ It is a function.

Example 4:
For the function ( ) = = 2 + 1, find the range when =
{−3, −2, −1,0,1,2,3} .

Solution :
For the given values of , we have
( −3 ) = 2 (−3 ) + 1 = −5
( −2 ) = 2 ( −2) + 1 = −3
( −1) = 2 ( −1 ) + 1 = −1
( 0 ) = 2 ( 0 ) + 1 = 1
(1 ) = 2 (1) + 1 = 3
( 2 ) = 2 ( 2 ) + 1 = 5
( 3 ) = 2 ( 3 ) + 1 = 7
The given function can also be written as a set of ordered pairs.
i.e., {( −3, −5 ) , ( −2, −3 ) , ( −1, −1 ) , ( 0,1 ) (1, 3 ) , ( 2,5 ) ( 3,7 )}
∴ = {−5, −3, −1, 1, 3, 5, 7}

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3.2 CLASSIFICATION OF FUNCTIONS

Let f be a function from to . If every element of set is the image of at least one element
of the set , i.e. if there is no unpaired element in the set , we say that the function f maps
the set onto the set . Otherwise, we say that the function maps set into set . Functions
for which each element of the set is mapped to a different element of the set are said to
be one-to-one.

Figure 3.1 : One-To-One Function
The Domain is { , , }
The co-domain {1, 2, 3, 4}
The range is {1, 2, 3}

A function can map more than one element of set to the same element of set . Such a
type of function is said to be many-to-one.

Figure 3.2 : Many-To-One Function

The domain is { , , }
The co-domain is {1,2,3,4}
The range is {1, 4 }

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A function that is both one-to-one and onto is said to be a bijective function.

Figure 3.3 : Bijective Function
Relation, which is one-to-many, can occur, but they are not function. The following figure
illustrates this fact.

Figure 3.4 : Relation But Not Function
Example 5:
Without using graph, prove that the function
∶ → defiend by ( ) = 4 + 3 is one-to-one.
Solution :
For a function to be one-one function
( ") = ( !) Þ " = ! " ", ! ∈ domain
\ Now f ( x1 ) = f ( x2 ) gives
4 + 3 " = 4 + 3 ! or " = !
\ is a one-one function.

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REFERENCE
Epp, Susanna S. (2010). Discrete Mathematics With Applications, Fourth Edition. Boston:

Brooks/Cole
Kwang, H (2015). A Spiral Workbook for Discrete Mathematics. New York: Open SUNY

Textbooks
Rosen, Kenneth H. (2012). Discrete Mathematics And Its Applications, Seventh Edition. New

York: McGraw-Hill
Seymour, L., Lipson, Lars M. (2007). Discrete Mathematics, Third Edition. New York: McGraw-

Hill

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