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3. Cottages, a Moderate Canteen, Girls Hostel and a few cottages providing limited
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end examinations.
BT0063-Unit-01-Set Theory
Unit 1 Set Theory
Structure
1.1 Introduction
Objectives
1.2 Sets and Their Representations
1.3 The Empty Set
1.4 Finite and Infinite Sets
1.5 Equal and Equivalent Sets
1.6 Subsets
1.7 Power Set
1.8 Universal Set
1.9 Venn Diagrams
1.10 Complement of a Set
1.11 Operations on Sets
1.12 Applications of Sets
1.13 Cartesian Product of Sets
1.14 Summary
1.15 Terminal Questions
1.16 Answers

4. 1.1 Introduction
The concept of set is basic in all branches of mathematics. It has proved to be of particular
importance in the foundations of relations and functions, sequences, geometry, probability theory
etc. The study of sets has many applications in logic philosophy, etc.
The theory of sets was developed by German mathematician Georg Cantor (1845 – 1918
A.D.). He first encountered sets while working on problems on trigonometric series. In this unit,
we discuss some basic definitions and operations involving sets.
Objectives:
At the end of the unit you would be able to
• understand the concepts of sets
• perform the different operations on sets
• write the Power set of a given set
1.2 Sets and their Representations
In every day life, we often speak of collection of objects of a particular kind such as pack of
cards, a herd of cattle, a crowd of people, cricket team, etc. In mathematics also, we come across
various collections, for example, collection of natural numbers, points in plane, prime numbers.
More specially, we examine the collections:
1. Odd natural numbers less than 10, i.e., 1, 3, 5, 7, 9
2. The rivers of India
3. The vowels in the English alphabet, namely a, e, I, o, u
4. Prime factors of 210, namely 2, 3, 5 and 7
5. The solutions of a equation x2 – 5x + 6 = 0 viz, 2 and 3
We note that each of the above collections is a well defined collection of objects in the sense that
we can definitely decide whether a given object belongs to a given collection or not. For
example, we can say that the river Nile does not belong to collection of rivers of India. On the
other hand, the river Ganga does belong to this collection. However, the following collections
are not well defined:
1. The collection of bright students in Class XI of a school
2. The collection of renowned mathematicians of the world
3. The collection of beautiful girls of the world
4. The collection of fat people
For example, in (ii) above, the criterion for determining a mathematician as most renowned may
vary from person to person. Thus, it is not a well defined collection.
We shall say that a set is a well defined collection of objects. The following points may be noted:

5. 1. Objects, elements and members of a set are synonymous terms. These are undefined
2. Sets are usually denoted by capital letters A, B, C, X, Y, Z etc.
3. The elements of a set are represented by small letters a, b, c, x, y, z etc.
If a is an element of a set A, we say that ‘a belongs to A’. The Greek symbol is used to denote
the phrase ‘belongs to’. Thus, we write . If b is not an element of a set A, we write
and read ‘b does not belong to A’. Thus, in the set V of vowels in the English alphabet, but
. In the set P of prime factors of but .
There are two methods of representing a set:
i) Roster or tabular form
ii) Set builder form.
i) In roster form, all the elements of a set are listed, the elements being separated by
commas and are enclosed within braces { }. For example, the set of all even positive integers
less than 7 is described in roster form as {2, 4, 6}. Some more examples of representing a set
in roster form are given below:
a) The set of all natural numbers which divide 42 is {1, 2, 3, 6, 7, 14, 21, 42}. Note that
in roster form, the order in which the elements are listed is immaterial. Thus, the above
set can also be represented as
{l, 3, 7, 21, 2, 6, 14, 42}.
b) The set of all vowels in the English alphabets is {a, e, i, o, u}.
c) The set of odd natural numbers is represented by {1, 3, 5,. . .}. The three dots tell us
that the list is endless.
It may be noted that while writing the set in roster form an element is not generally
repeated, i.e., all the elements are taken as distinct. For example, the set of letters
forming the word “SCHOOL” is
{S, C, H, O, L}.
ii) In set builder form, all the elements of a set possess a single common property which is
not possessed by any element outside the set. For example, in the set “{a, e, i, o, u}” all the
elements possess a common property, each of them is a vowel in the English alphabet and no
other letter possesses this property. Denoting this set by V, we write
V = {x : x is a vowel in the English alphabet}.
It may be observed that we describe the set by using a symbol x for elements of the set (any other
symbol like the letters y, z etc. could also be used in place of x). After the sign of ‘colon’ write

6. the characteristic property possessed by the elements of the set and then enclose the description
within braces. The above description of the set V is read as ‘The set of all
x such that x is a vowel of the English alphabet’. In this description the braces stand for ‘the set
of all’, the colon stands for ’such that’.
For example, the following description of a set
A = {x : x is a natural number and 3 < x < 10)
is read as “the set of all x such that x is a natural number and 3 < x < 10″. Hence, the numbers 4,
5, 6, 7, 8 and 9 are the elements of set A.
If we denote the sets described above in (a), (b) and (c) in roster form by A, B and C,
respectively, then A, B and C can also be represented in set builder form as follows
A = {x : x is a natural number which divides 42}
B = {y : y is a vowel in the English alphabet}
C = {z : z is an odd natural number}.
Example: Write the set of all vowels in the English alphabet which precede q.
Solution: The vowels which precede q are a, e, i, o. Thus A = {a, e, i, o} is the set of all vowels
in the English alphabet which precede q.
Example: Write the set of all positive integers whose cube is odd.
Solution: The cube of an even integer is also an even integer. So, the members of the required
set can not be even. Also, cube of an odd integer is odd. So, the members of the required set are
all positive odd integers. Hence, in the set builder form we write this set as {x : x is an odd
positive integer} or equivalently as
{2k + 1 : k ≥ 0, k is an integer}
Example: Write the set of all real numbers which can not be written as the quotient of two
integers in the set builder form.
Solution: We observe that the required numbers can not be rational numbers because a rational
number is a number in the form , where p, q are integers and q ≠ 0. Thus, these must be real
and irrational. Hence, in set builder form we write this set as
{x : x is real and irrational}

7. Example: Write the set in the set builder form.
Solution: Each member in the given set has the denominator one more than the numerator. Also,
the numerators begin from 1 and do not exceed 6. Hence, in the set builder form the given set is
Example: Match each of the sets on the left described in the roster form with the same set on the
right described in the set builder form:
i) { L, I, T, E) a) {x : x is a positive integer and is a divisor of 18}
ii) {0) b) {x : x is an integer and x2 – 9 = 0}
iii) {1, 2, 3, 6, 9, 18} c) {x : x is an integer and x + 1 = 1}
iv) {3, – 3} d) {x : x is a letter of the word LITTLE}
Solution: Since in (d), there are six letters in the word LITTLE and two letters T and L are
repeated, so (i) matches (d). Similarly (ii) matches (c) as x + 1 = 1 implies x = 0. Also, 1, 2, 3, 6,
9, 18 are all divisors of 18. So,
(iii) matches (a). Finally, x2 – 9 = 0 implies. x = 3, –3. So, (iv) matches (b).
Example: Write the set {x : x is a positive integer and x2 < 40} in the roster form.
Solution: The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the roster form is {1, 2, 3,
4, 5, 6}.
1.3 The Empty Set
Consider the set
A = {x : x is a student of Class XI presently studying in a school}
We can go to the school and count the number of students presently studying in Class XI in the
school. Thus, the set A contains a finite number of elements.
Consider the set {x : x is an integer, x2 + 1 = 0}. We know that there is no integer whose square
is –1. So, the above set has no elements.
We now define set B as follows:

8. B = {x : x is a student presently studying in both Classes X and XI}.
We observe that a student cannot study simultaneously in both Classes X and XI. Hence, the set
B contains no element at all.
Definition: A set which does not contain any element is called the empty set or the null set or the
void set.
According to this definition B is an empty set while A is not. The empty set is denoted by the
symbol ‘ . We give below a few examples of empty sets.
i) Let P = {x: 1 < x < 2, x is a natural number }.
Then P is an empty set, because there is no natural number between
1 and 2.
ii) Let Q = {x : x2 – 2 = 0 and x is rational}.
Then, Q is the empty set, because the equation x2 - 2 = 0 is not satisfied by any rational
number x.
iii) Let R = {x : x is an even prime number greater than 2}
Then R is the empty set, because 2 is the only even prime number.
iv) Let S = {x : x2 = 4, and x is an odd integer}. Then, S is the empty set, because equation
x2 = 4 is not satisfied by any value of x which is an odd integer.
1.4 Finite and Infinite Sets
Let A = {1, 2, 3, 4, 5), B = {a, b, c, d, e, f} and C = {men in the world}.
We observe that A contains 5 elements and B contains 6 elements. How many elements does C
contain ? As it is, we do not know the exact number of elements in C, but it is some natural
number which may be quite a big number. By number of elements of a set A, we mean the
number of distinct elements of the set and we denote it by n(A). If n(A) is a natural number, then
A is a finite set, otherwise the set A is said to be an infinite set. For example, consider the set, N,
of natural numbers. We see that n(N), i.e., the number of elements of N is not finite since there is
no natural number which equals n(N). We, thus, say that the set of natural number is an infinite
set.
Definition: A set which is empty or consists of a definite number of elements is called finite.
Otherwise, the set is called infinite.
We shall denote several set of numbers by the following symbols:

9. N : the set of natural numbers
Z : the set of integers
Q : the set of rational numbers
R : the set of real numbers
Z+ : the set of positive Integers
Q+ : the set of positive rational numbers
R+ : the set of positive real numbers
We consider some examples:
1. Let M be the set of days of the week. Then M is finite.
2. Q, the set of all rational numbers is infinite.
3. Let S be the set of solution (s) of the equation x2 – 16 = 0. Then S is finite.
4. Let G be the set of all points on a line. Then G is infinite.
When we represent a set in the roster form, we write all the elements of the set within braces { }.
It is not always possible to write all the elements of an infinite set within braces { } because the
number of elements of such a set is not finite. However, we represent some of the infinite sets in
the roster form by writing a few elements which clearly indicate the structure of the set followed
(or preceded) by three dots.
For instance, {1, 2, 3, 4, … } is the set of natural numbers, {1, 3, 5, 7, 9, .. . } is the set of odd
natural numbers and {…, – 3, –2, –1, 0, 1, 2, 3, … } is the set of integers. But the set of real
numbers cannot be described in this form, because the elements of this set do not follow any
particular pattern.
1.5 Equal and Equivalent Sets
Given two sets A and B. If every element of A is also an element of B and if every element of B is
also an element of A, the sets A and B are said to be equal. Clearly, the two sets have exactly the
same elements.
Definition: Two sets A and B are said to be equal if they have exactly the same elements and we
write A = B. Otherwise, the sets are said to be unequal and we write A ≠ B
We consider the following examples:
1. Let A = {1, 2, 3, 4, } and B = {3, 1, 4, 2).
2. Then A = B.

10. 3. Let A be the set of prime numbers less than 6 and P the set of prime factors of 30.
Obviously, the set A and P are equal, since 2, 3 and 5 are the only prime factors of 30 and
are less than 6.
Let us consider two sets L = {1, 2, 3, 4} and M = {1, 2, 3, 8}. Each of them has four elements but
they are not equal.
Definition: Two finite sets A and B are said to be equivalent if they have the same number of
elements. We write A ≈ B.
For example, let A = {a, b, c, d, e} and B = {1, 3, 5, 7, 9}. Then A and B are equivalent sets.
Obviously, all equal sets are equivalent, but all equivalent sets are not equal.
Example: Find the pairs of equal sets, if any, giving reasons:
A = {0}, B = {x : x > 15 and x < 5}, C = {x : x – 5 = 0}, D = {x:x2 = 25}
E = {x : x is a positive integral root of the equation x2 – 2x – 15 = 0}
Solution: Since 0 ∈ A and 0 does not belong to any of the sets B, C, D and E. Therefore, A B,
A C, A D, A E. B = but none of the other sets are empty. Hence B C, B D and B E. C
= {5} but , hence C D. Since E = {5}), C = E. D = {–5, 5} and E = {5}. Therefore D
E. Thus, the only pair of equal sets is C and E.
1.6 Subsets
Consider the sets S and T, where S denotes the set of all students in your school and T denotes
the set of all students in your class. We note that every element of T is also an element of S. We
say that T is a subset of S.
Definition: If every element of a set A is also an element of a set B, then A is called a subset of B
or A is contained in B. We write it as A B.
If at least one element of A does not belong to B, then A is not a subset of B. We write it as A
B.
We may note that for A to be a subset of B, all that is needed is that every element of A is in B. It
is possible that every element of B may or may not be in A. If it so happens that every element of
B is also in A, then we shall also have B A. In this case, A and B are the same sets so that we
have A B and B A which implies A = B.

11. It follows from the definition that every set A is a subset of itself, i.e., A A. Since the empty set
has no elements, we agree to say that is a subset of every set. We now consider some
examples
1. The set Q of rational numbers is a subset of the set R of real numbers and we write Q R.
2. If A is the set of all divisors of 56 and B the set of all prime divisors of 56, then B is a
subset of A, and we write B A.
3. Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6}, then A B and B
A and hence A = B.
4. Let A = {a, e, i, o, u}, B = {a, b, c, d}. Then A is not a subset of B. Also B is not a subset
of A. We write A B and B A.
5. Let us write down all the subsets of the set {1, 2}. We know is a subset of every set. So
is a subset of {1, 2}. We see that {1}, {2} and {l, 2} are also subsets of {1,2}. Thus the
set {1,2} has, in all, four subsets, viz. , {1}, {2} and {1,2}.
Definition: Let A and B be two sets. If A B and A ≠ B, then A is called a proper subset of B and
B is called a superset of A. For example, A= {1, 2, 3} is a proper subset of B = {1, 2, 3, 4}.
Definition: If a set A has only one element, we call it a singleton. Thus {a } is a singleton.
1.7 Power Set
In example (v) of Section 1.6, we found all the subsets of the set {1, 2}, viz., , {1}, {2} and {1,
2}. The set of all these four subsets is called the power set of {1, 2}.
Definition: The collection of all subsets of a set A is called the power set of A. It is denoted by
P(A). In P(A), every element is a set.
Example (v) of section 1.6, if A = {1, 2}, then P(A)={ , {1}, {2}, {1,2}. Also, note that, n[P(A)]
= 4 = 22.
In general, if A is a set with n(A) = m, then it can be shown that
n[P(A)] = 2m > m = n(A).
1.8 Universal Set

12. If in any particular context of sets, we find a set U which contains all the sets under consideration
as subsets of U, then set U is called the universal set. We note that the universal set is not unique.
For example, for the set Z of all integers, the universal set can be the set Q of rational numbers
or, for that matter, the set R of real numbers.
For another example, in the context of human population studies, the universal set consists of all
the people in the world.
Example: Consider the following sets : , A = {1, 3), B = {1, 5, 9},
C = {1, 3, 5, 7, 9}, Insert the correct symbol or between each pair of sets
(i) — B, (ii) A — B (iii) A — C (iv) B — C.
Solution:
1. B as is a subset of every set.
2. A B as 3 A and 3 B..
3. A C as 1, 3 A also belongs to C.
4. B C as each element of B also belongs to C.
Example: Let A = {1, 2, 3, 4}, B = {1, 2, 3} and C = {2, 4}. Find all sets X such that
(i) X B and X C (ii) X A and X B.
Solution:
i) X B means that X is a subset of B, and the subsets of B are , {1}, {2}, {3}, {1,2}, {1,3},
{2,3} and {1,2,3} . X C means that X is a subset of C, and the subsets of C are , {2}, {4}
and {2, 4}. Thus, we note that X B and X C means that X is a subset of both B and C.
Hence, X = , {2}.
ii) X A, X B means that X is a sub set of A but X is not a subset of B. So, X is one of
these {4}, {1,2,4}, {2,3,4}, {l,3,4}, {1,4}, {2,4}, {3,4}, {1,2,3,4}.
Note:
A set can easily have some elements which are themselves sets. For example, {1, {2,3}, 4} is a set
having {2,3} as one element which is a set and also elements 1,4 which are not sets.

13. Example: Let A, B and C be three sets. If A B and B C, is it true that
A C? If not, give an example.
Solution: No. Let A = {1}, B = C = { { 1 }, 2}. Here A B as A = {1} and
B = C implies B C. But A C as 1 A and 1 C.
Note that an element of a set can never be a subset of it.
�
1.9 Venn
Diagrams
Most of the relationships between sets can be represented by means of diagrams. Figures
representing sets in the form of enclosed region in the plane are called Venn diagrams named
after British logician John Venn (1834—1883 A.D.). The universal set U is represented by the
interior of a rectangle.Other sets are represented by the interior of circles.
Fig. 1.1
Fig. 1.1 is a Venn diagram representing sets A and B such that A ⊂ B.
Fig. 1.2

14. In Fig.1.2, U = {1, 2, 3, …, 10} is the universal set of which A = {2,4,6,8,10} and B = {4,6} are
subsets. It is seen that B A. The reader will see an extensive use of the Venn diagrams when we
discuss the operations on sets.
�
1.10 Complement of a Set
Let the universal set U be the set of all prime numbers. Let A be the subset of U which consists
of all those prime numbers that are not divisors of 42. Thus A = {g x : x U and x is not a divisor
of 42}. We see that 2 U but 2 A, because 2 is a divisor of 42. Similarly 3 U but 3 A, and 7
U but
7 A. Now 2, 3 and 7 are the only elements of U which do not belong to A. The set of these three
prime numbers, i.e., the set {2, 3, 7} is called the complement of A with respect to U, and is
denoted by A′. So we have
A′ = {2, 3, 7}. Thus, we see that A′={x : x U and x A). This leads to the following definition.
Definition: Let U be the universal set and A is a subset of U. Then the complement of A with
respect to (w.r,t.) U is the set of all elements of U which are not the elements of A. Symbolically
we write A′ to denote the complement of A with respect to U. Thus A′ = {x:x U and x A}. It
can be represented by Venn diagram as
Fig. 1.3
The shaded portion in Fig. 1.3 represents A′.
Example: Let U = {1,2,3,4,5,6,7,8,9,10} and A= {1,3,5,7,9}. Find A′.
Solution: We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A. Hence
A′ = {2, 4, 6, 8, 10}.

15. Example: Let U be the universal set of all the students of Class XI of a
co-educational school. Let A be the set of all girls in the Class Xl. Find A′.
Solution: As A is the set of all girls, hence A′ is the set of all boys in the class.
1.11 Operations on Sets
In earlier classes, you learnt how to perform the operations of addition, subtraction,
multiplication and division on numbers. You also studied certain properties of these operations,
namely, commutativity, associativity, distributivity etc. We shall now define operations on sets
and examine their properties. Henceforth, we shall refer all our sets as subsets of some universal
set.
a) Union of Sets:
Let A and B be any two sets. The union of A and B is the set which consists of all the elements of
A as well as the elements of B, the common elements being taken only once. The symbol ‘∪‘ is
used to denote the union. Thus, we can define the union of two sets as follows.
Definition: The union of two sets A and B is the set C which consists of all those elements which
are either in A or in B (including those which are in both).
Symbolically, we write = {x:x A or x B} and usually read as
‘A union B‘.
The union of two sets can be represented by a Venn diagram as shown in Fig. 1.4.
Fig. 1.4
The shaded portion in Fig. 1.4 represents A B.
Example: Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A B.

16. Solution: We have A B = {2, 4, 6, 8, 10, 12}.
Note that the common elements 6 and 8 have been taken only once while writing A B.
Example: Let A = {a, e, i, o, u} and B = {a, i, u}. Show that A B = A.
Solution: We have A B = {a, e, i, o, u} = A.
This example illustrates that the union of a set A and its subset B is the set A itself, i.e., if ,
then A B = A.
Example: Let X = {Ram, Shyam, Akbar} be the set of students of Class XI who are in the school
Hockey team. Let Y = {Shyam, David, Ashok} be the set of students from Class XI who are in
the school Football team. Find
and interpret the set.
Solution: We have = {Ram, Shyam, Akbar, David, Ashok}. This is the set of students
from Class XI who are either in the Hockey team or in the Football team.
Example: Find the union of each of the following pairs of sets:
Solution:
1. A B = {1, 2, 3, 4, 5}
2. A = {3, 4, 5,… }, B = {1, 2, 3}. So, A B = {1, 2, 3, 4, 5,… } = Z+
3. A={1, 2, 3,. . .}, B ={x:x is a negative integer}. So A B={x:x ∈ Z, x ≠ 0}.
4. A = {2, 3, 4}, B = {5, 6, 7, 8}. So, A B = {2, 3, 4, 5, 6, 7, 8}.
b) Intersection of Sets: The intersection of sets A and B is the set of all elements which are
common to both A and B. The symbol ∩ is used to denote the intersection.
Thus, we have the following definition.

17. Definition: The intersection of two sets A and B is the set of all those elements which belong to
both A and B. Symbolically, we write A ∩ B =
{x:x ∈ A and x ∈ B} and read as ‘A intersection B’.
The intersection of two sets can be represented by a Venn diagram as shown in Fig. 1.5.
�
Fig. 1.5
The shaded portion represents A B.
If A B = φ, then A and B are said to be disjoint sets. For example, let
A = {2, 4, 6, 8} and B = {1, 3, 5, 7}. Then, A and B are disjoint sets, because there is no element
which is common to A and B. The disjoint sets can be represented by Venn diagram as shown in
Fig. 1.6.
Fig. 1.6

18. Example: Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A B.
Solution: We see that 6, 8 are the only elements which are common to both the sets A and B.
Hence A B = {6, 8}.
Example: Consider the sets X and Y of Example 17. Find X Y.
Solution: We see that the element “Shyam” is the only element common to both the sets X and
Y. Hence, X Y = { Shyam }.
SAQ 1: Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}. Find and prove that
= B.
SAQ 2: Let A = A = {x : x ∈ Z+} ; B = {x : x is a multiple of 3, }:
C = {x:x is a negative integer}; D = {x:x is an odd integer}. Find (i) A B,
(ii) , (iii) , (iv) , (v) , (vi) .
c) Difference of Sets: The difference of sets A and B, in this order, is the set of elements which
belong to A but not to B. Symbolically, we write
A — B and read as ‘A difference B’. Thus A — B = {x : x ∈ A and x ∉ B} and is represented by
Venn diagram in Fig.1.7. The shaded portion represents
A — B.
Fig. 1.7
SAQ 3: Let V = {a, e, i, o, u} and B = {a, i, k, u}. Find V – B and B – V.

19. SAQ 4: Let A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8}. Find A – B and
B – A.
1.12 Applications of Sets
Let A and B be finite sets. If , then
= n(A) + n(B) (1) �
The elements in A B are either in A or in B but not in both as
. So (1) follows immediately.
In general, if A and B are finite sets, then
n(A B) = n(A) + n(B) – n(A B) (2)
Fig. 1.8
Note that the sets A – B, A B and B – A are disjoint and their union is
A B (Fig 1.8). Therefore
n(A B) = n(A – B) + n(A B) + n(B – A)
= n(A – B) + n(A B) + n(B – A) +n(A B) – n(A B)
= n(A) + n(B) – n(A B).
which verifies (2).

20. If A, B and C are finite sets, then
n(A B C) = n(A) + n(B) + n(C) –n(A B) –n(B C)
– n(A C) + n(A B C) (3)
In fact, we have
n(A B C) = n(A) + n(B C) – n(A (B C )) [by (2)]
= n(A) + n(B) + n(C) – n(B C) – n (A (B C)) [by (2)]
Since A = (A B) (A C), we get
= n(A B) + n (A C) – n[A B A C)]
= n(A B) + n (A C) – n[A B C)]
Therefore
n(A B C) = n(A) + n(B) + n(C) – n(B C) – n(A B)
– n(A C) + n(A B C).
This proves (3).
Example: If X and Y are two sets such that n(X ∪ Y) = 50, n(X) = 28 and n(Y) = 32, find n(X
Y).
Solution: By using the formula
,
we find that

21. = 28 + 32 –50 = 10..
Alternatively, suppose , then
Fig. 1.9
n(X – Y) = 28 – k, n(Y – X) = 32 – k. (by Venn diagram in Fig 1.9)
This gives 50 = = (28 – k) + k + (32 – k).
Hence, k = 10
Example: In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teach
mathematics and 4 teach physics and mathematics. How many teach physics?
Solution Let M denote the set of teachers who teach mathematics and
P denote the set of teachers who teach physics. We are given that
= 4. Therefore
= 20 – 12 + 4 = 12.
Hence, 12 teachers teach physics.
SAQ 5: In a group of 50 people, 35 speak Hindi, 25 speak both English and Hindi and all the
people speak at least one of the two languages. How many people speak only English and not
Hindi ? How many people speak English?
1.13 Cartesian Product of Sets
Let A, B be two sets. If a ∈ A, b ∈ B, then (a, b) denotes an ordered pair whose first component
is a and the second component is b. Two ordered pairs (a, b) and (c, d) are said to be equal if and
only if a = c and b = d.

22. In the ordered pair (a, b), the order in which the elements a and b appear in the bracket is
important. Thus (a, b) and (b, a) are two distinct ordered pairs if a ≠ b. Also, an ordered pair (a,
b) is not the same as the set {a, b}.
Definition: The set of all ordered pairs (a, b) of elements is called the Cartesian
Product of sets A and B and is denoted by A x B. Thus
Let A = {a1, a2}, B = {b1, b2, b3}. To write the elements of A x B, take a1
∈ A and write all elements of B with a1, i.e., (a1, b1), (a, b2), (a1, b3). Now take
a2
ε A and write all the elements of B with a2, i.e., (a2, b1), (a2, b2), (a2, b3). Therefore, A x B will
have six elements, namely, (a1, b1), (a1, b2), (a1, b3), (a2, b1), (a2, b2), (a2, b3).
Remarks:
1. If A = or B = , then A × B =
2. If A ≠ and B ≠ , then . Thus, if and only if A and B ≠ . Also, A
B B A.
3. If the set A has m elements and the set B has n elements, then A × B has mn elements.
4. If A and B are non-empty sets and either A or B is an infinite set, so is A x B.
5. If A = B, then A B is expressed as A2.
6. We can also define, in a similar way, ordered triplets. If A, B and C are three sets, then (a
,b, c), where a ∈ A, b ∈ B and c ∈ C, is called an ordered triplet. The Cartesian Product
of sets A, B and C is defined as
A B C = {(a, b, c): a ∈ A, b ∈ B, c ∈ C}. An ordered pair and ordered triplet are also called 2-
tuple and 3-tuple, respectively. In general, if
A1, A2,.. ., An are n sets, then (a1,a2,…, an) is called an n-tuple where
ai
∈ Ai, i = 1, 2 n and the set of all such n-tuples, is called the Cartesian product of A1, A2, ……, An.
It is denoted by A1 x A2 x. . .x An. Thus
A1
× A2 × ….. × An = {(a1, a2, …. an): a1 ∈ A1, 1≤ i ≤ n}}.
Example: Find x and y if (x + 2, 4) = (5, 2x + y).
Solution: By definition of equal ordered pairs, we have

23. x+2=5 (1)
2x + y = 4 (2)
Solving (1) and (2), we get x = 3, y = –2.
Example: Let A = {1, 2, 3} and B = {4, 5}. Find A x B and B x A and show that A × B ≠ B × A.
Solution: We have
= {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
and = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}
Note that and (1, 4) ∉ B × A. Therefore, .
Example: Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find
i) ii)
iii) iv)
Solution:
i) We have . Therefore, = {(1, 4), (2, 4), (3, 4)}.
ii) We note that
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
and = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Therefore = {(1, 4), (2, 4), (3, 4)}.
iii) Clearly = {3, 4, 5, 6}. Thus
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3),
(3, 4), (3, 5), (3, 6)}

24. iv) In view of (ii), we see that
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3,
5), (3, 6)}.
In view of the assertion in Example 3 above, we note that
and
SAQ 6: Let A and B be two sets such that n(A) = 5 and n(B) = 2.
If (a1, 2), (a2, 3), (a3, 2), (a4, 3), (a5, 2) are in A × B and a1, a2, a3, a4 and
a5 are distinct. Find A and B.
1.14 Summary
This unit tells us about sets and their representations. We study the concepts of Empty sets,
Finite and Infinite sets, Equal sets. All the concepts discussed is well illustrated by standard
examples. The different operations on sets like complement of Set, Operation on Sets and
Applications of sets is discussed here.
1.15 Terminal Questions
1. Which of the following pairs of sets are equal ? Justify your answer.
i) A, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”
ii) A= and B = .
2. State which of the following sets are finite and which are infinite:

25. 3. If A and B are two non-empty sets such that , show that
A=B
1.16 Answer
Self Assessment Questions
1. We have A ∩ B = {2, 3, 5, 7} = B.
We note that if B ⊂ A , then A ∩ B = B.
2. A = {x:x is a positive integer}, B = {3n : n ∈ Z};
1. A B = {3, 6, 9, 12,…} = {3n:n ∈ Z+}.
2. A C =
3. A D = {1, 3, 5, 7,…}
4. B C = {– 3, –6, –9,. . . } = {3n : n is a negative integer}
5. B D = {. . ., –15, –9, –3, 3, 9, 15,…}
6. C D = {–1, –3, –5, –7,…}
3. We have V – B = {e, o}, since the only elements of V which do not belong to B
are e and o. Similarly B – V = {k}
4. We have A – B = {1, 3, 5}, as the only elements of A which do not belong to B
are 1, 3 and 5. Similarly, B – A = {8}.
We note that
5. Let H denote the set of people speaking Hindi and E the set of people speaking
English. We are given that = 50, n(H) = 35,
= 25. Now
= n(H) + n(E – H).
So 50 = 35 + n(E – H), i.e. , n(E – H) = 15.
Thus, the number of people who speak only English but not Hindi is 15.
Also, n(H ∪ E) = n(H) + n(E) – n(H E) implies

26. 50 = 35 + n(E) – 25,
which gives n(E) = 40.
Hence, the number of people who speak English is 40.
6. Since and n(A) = 5, A = {a1, a2, a3, a4, a5}. Also and
n(B) = 2. Therefore, B = {2, 3}.
Terminal Quesitons
1. i) A = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then A, B are equal sets as
repetition of elements in a set do not change a set. Thus A = {A, L,
O, Y} = B.
ii) A = {–2, –1, 0, 1, 2,}, B = (1, 2). Since 0 ∈ A and 0∉ B, A and B are not
equal sets.
2. i) Given set = {1, 2}. Hence, it is finite.
ii) Given set = {2}. Hence, it is finite.
iii) Given set = . Hence, it is finite.
1. The given set is the set of all prime numbers and since the set of prime numbers is
infinite, hence the given set is infinite.
2. Since there are infinite number of odd numbers, hence the given set is infinite
3. Let a ∈ A. Since B ≠ , there exists . Now, implies
. Therefore, every element in A is in B giving . Similarly, . Hence
A=B
BT0063-Unit-02-Mathematical Logic
Unit 2 Mathematical Logic
Structure
2.1 Introduction

27. Objectives
2.2 Statements
2.3 Basic Logical Connectives
2.4 Conjunction
2.5 Disjunction
2.6 Negation
2.7 Negation of Compound Statements
2.8 Truth Tables
2.9 Tautologies
2.10 Logical Equivalence
2.11 Applications
2.12 Summary
2.13 Terminal Questions
2.14 Answers
2.1 Introduction
Logic is the study of general patterns of reasoning, without reference to particular meaning or
contexts. If an object is either black or white, and if it is not black, then logic leads us to the
conclusion that it must be white. Observe that logical reasoning from the given hypotheses
cannot reveal what ‘black’ or ‘white’ mean, or why an object can not be both.
Logic can find applications in many branches of sciences and social sciences. Logic, infact is the
theoretical basis for many areas of computer science such as digital logic circuit design,
automata theory and artificial intelligence.
In this chapter, we shall learn about statements, truth values of a statement, compound
statements, basic logical connectives, truth tables, tautologies, logical equivalence, duality,

28. algebra of statements, use of Venn diagrams in logic and finally, some simple applications of
logic in switching circuits.
Objectives:
At the end of the unit you would be able to
• understand the ideas in Mathematical Logic
• identify a proposition
• apply the concept of Mathematical Logic in circuits
2.2 Statements
A statement is a sentence which is either true or false, but not both simultaneously.
Note: A sentence which is both true and false simultaneously is not a statement, rather, it is a
paradox.
Example:
(a) Each of the following sentences:
i) New Delhi is in India.
ii) Two plus two is four.
iii) Roses are red.
iv) The sun is a star.
v) Every square is a rectangle.
is true and so each of them is a statement.
(b) Each of the following sentences:
i) The earth is a star.
ii) Two plus two is five.
iii) Every rectangle is a square.
iv) 8 is less than 6.
v) Every set is a finite set.

29. is false and so each of them is a statement.
Example:
a) Each of the sentences:
i) Open the door.
ii) Switch on the fan.
iii) Do your homework.
can not be assigned true or false and so none of them is a statement. Infact, each of them is
a command.
b) Each of the sentences:
i) Did you meet Rahman?
ii) Where are you going?
iii) Have you ever seen Taj Mahal?
can not be assigned true or false and so none of them is a statement. Infact, each of them is
a question.
c) Each of the sentences:
i) May you live long!
ii) May God bless you!
can not be assigned true or false and so none of them is a statement. Infact, each of them is
optative.
d) Each of the sentences:
i) Hurrah! We have won the match.
ii) Alas! I have failed.
can not be assigned true or false and so none of them is a statement. In fact, each of them is

30. exclamatory.
e) Each of the sentences:
i) Good morning to all.
ii) Wish you best of luck.
can not be assigned true or false and so none of them is a statement. In fact, each them is a
wish.
f) Each of the sentences:
i) Please do me a favour. T
ii) Give me a glass of water.
can not be assigned true or false and so none of them is a statement. In fact, each them is a
request.
g) Each of the following sentences:
i) x is a natural number
ii) He is a college student.
is an open sentence because the truth or falsity of (xv) depends on x and that of xvi) depends
on the reference for the pronoun he. We may observe that for some values of x like x = 1,
2,….. etc, (xv) may be true and for some other values like etc, (xv) is false.
Similarly, (xvi) may be true or false. However, at a particular point of time or situation they
are either true or false. Since, we are interested only in the fact that it is true or false,
sentences (xv) and (xvi) can be considered as statements.
Note: The statements (xv) and (xvi) in Example 2 are also called open statements.
It is useful to have some notation to represent statements. Let us represent the statements by
lower case letter like p, q, r, s, ….. Thus, a statement ‘New Delhi is city may be represented or
denoted by p and we write

31. p : New Delhi is a city.
similarly, we may denote a statement ‘2 + 3 = 6’ by q and write
q : 2 + 3 = 6.
Truth value of a statement: The truth or falsity of a statement is called its truth value. Every
statement must be either true or false. No statement can be both true and false at the same time.
If a statement is true, we say that its truth value is TRUE or T and if it is false we say that its
truth value is FALSE or F.
Example: The statements in Example 1(a) have the truth value T while the statements in
Example 1(b) have the truth value F.
Compound statements: A statement is said to be simple, if it cannot be broken down into two or
more sentences. The statements that we considered in Example 1(a) and (b) are all simple
statements.
New statements that can be formed by combining two or more simple statements are called
compound statements. Thus, a compound statement is the one which is made up of two or more
simple statements.
Example:
a) The statement “Roses are red and Violets are blue” is a compound statement which is a
combination of two simple statements “Roses are red” and “Violets are blue”.
b) The statement “Gita is sick or Rehana is well” is a compound statement made up of two
simple statements “Gita is sick” and “Rehana is well”.
c) The statement “It is raining today and 2 + 2 = 4” is a compound statement composed of two
simple statements “It is raining today” and
“2 + 2 = 4”.
Simple statements, which on combining, form compound statements, are called sub-statements
or component statements of the compound statements. The compound statements S consisting of

32. sub-statements
p, q, r,… is denoted by S (p, q, r,…).
A fundamental property of a compound statements is that its truth value is completely
determined by the truth value of each of its sub-statements together with the way in which
they are connected to form the compound statement.
2.3 Basic Logical Connectives
There are many ways of combining simple statements to form compound statements. The words
which combine simple statements to form compound statements are called connectives. In the
English language, we combine two or more statements to form a new statement by using the
connectives ‘and’, ‘or’, etc. with a variety of meanings. Because the use of these connectives in
English language is not always precise and unambiguous, it is necessary to define a set of
connectives with definite meanings in the language of logic, called object language. We now
define connectives for object language which corresponds to the connectives discussed above.
Three basic connectives (logical) are conjunction which corresponds to the English word ‘and’ ;
disjunction which corresponds to the word ‘or’ ; and negation which corresponds to the word
‘not’.
Throughout we use the symbol ‘ ’ to denote conjunction ; ‘∨’ to denote disjunction and the
symbol ‘~‘ to denote negation.
Note:. Negation is called a connective although it does not combine two or more statements. In
fact, it only modifies a statement.
2.4 Conjunction
If two simple statements p and q are connected by the word ‘and’, then the resulting compound
statement “p and q” is called a conjunction of p and q and is written in symbolic form as “p ∧
q“.
Example: Form the conjunction of the following simple statements:
p : Dinesh is a boy.

33. q : Nagma is a girl.
Solution: The conjunction of the statement p and q is given by
p ∧ q : Dinesh is a boy and Nagma is a girl.
Example: Translate the following statement into symbolic form
“Jack and Jill went up the hill.”
Solution: The given statement can be rewritten as
“Jack went up the hill and Jill went up the hill”
Let p : Jack went up the hill and q : Jill went up the hill.
Then the given statement in symbolic form is p ∧ q.
Example: Write the truth value of each of the following four statements:
i) Delhi is in India and 2 + 3 = 6.
ii) Delhi is in India and 2 + 3 = 5.
iii) Delhi is in Nepal and 2 + 3 = 5.
iv) Delhi is in Nepal and 2 + 3 = 6.
Solution: In view of (D1) and (D2) above, we observe that statement (i) has the truth value F as
the truth value of the statement “2 + 3 = 6” is F. Also, statement (ii) has the truth value T as
both the statement “Delhi is in India” and “2 + 3 = 5” has the truth value T. Similarly, the truth
value of both the statements (iii) and (iv) is F.
2.5 Disjunction
If two simple statements p and q are connected by the word ‘or’, then the resulting compound
statement “p or q” is called disjunction of
p and q and is written in symbolic form as “p ∧ q”.

34. Example: Form the disjunction of the following simple statements:
p : The sun shines.
q : It rains.
Solution: The disjunction of the statements p and q is given by
p ∨ q : The sun shines or it rains.
Example: Write the truth value of each of the following statements:
i) India is in Asia or 2 + 2 = 4.
ii) India is in Asia or 2 + 2 =5.
iii) India is in Europe or 2 + 2 = 4.
iv) India is in Europe or 2 + 2 = 5.
Solution: In view of (D3) and (D4) above, we observe that only the last statement has truth value
F as both the sub-statements “India is in Europe” and “2 + 2 = 5” have the truth value F. The
remaining statements (i) to (iii) have the truth value T as at least one of the sub-statements of
these statements has the truth value T.
2.6 Negation
An assertion that a statement fails or denial of a statement is called the negation of the statement.
The negation of a statement is generally formed by introducing the word “not” at some proper
place in the statement or by prefixing the statement with “It is not the case that” or “It is false
that”.
The negation of a statement p in symbolic form is written as “~ p”.
Example: Write the negation of the statement
p : New Delhi is a city.
Solution: The negation of p is given by

35. ~ p : New Delhi is not a city
or ~ p : It is not the case that New Delhi is a city.
or ~ p : It is false that New Delhi is a city
Example: Write the negation of the following statements:
p : I went to my class yesterday.
q:2+3=6
r : All natural numbers are integers.
Solution: Negation of the statement p is given by
~ p : I did not go to my class yesterday.
or
It is not the case that I went to my class yesterday.
or
It is false that I went to my class yesterday.
or
I was absent from my class yesterday.
The negation of the statement q is given by
~q : 2 + 3 ≠ 6
or
It is not the case that 2 + 3 = 6

36. or
It is false that 2 + 3 = 6
The negation of the statement r is given by
~ r : Not all natural numbers are integers.
or
There exists a natural number which is not an integer.
or
it is not the case that all natural numbers are integers.
or
It is false that all natural numbers are integers.
Regarding the truth value of the negation ~ p of a statement p. we have
(D5) : ~ p has truth value T whenever p has truth value F.
(D6) : ~ p has truth value F whenever p has truth value T.
Example: Write the truth value of the negation of each of the following statements::
i) p : Every square is a rectangle.
ii) q : The earth is a star.
iii) r :2 + 3 < 4
Solution: In view of (D5) and (D6), we observe that the truth value of ~p is F as the truth value
of p is T. Similarly, the truth value of both ~q and ~r is T as the truth value of both statements q
and r is F
2.7 Negation of compound statements

37. I) Negation of conjunction: Recall that a conjunction p ∧ q consists of two sub-statements p
and q both of which exist simultaneously. Therefore, the negation of the conjunction would
mean the negation of at least one of the two sub-statements. Thus, we have
(D7): The negation of a conjunction p ∧ q is the disjunction of the negation of p and the
negation of q. Equivalently, we write
~ ( p ∧ q) = ~ p v ~ q
Example: Write the negation of each of the following conjunctions:
a) Paris is in France and London is in England.
b) 2 + 3 = 5 and 8 < 10.
Solution:
(a) Write p : Paris is in France and q : London is in England.
Then, the conjunction in (a) is given by p ∧ q.
Now ~ p : Paris is not in France, and
~ q : London is not in England.
Therefore, using (D7), negation of p ∧ q is given by
~ p ∧ q = Paris is not in France or London is not in England.
(b) Write p : 2+3 = 5 and q :8 < 10.
Then the conjunction in (b) is given by p ∧ q.
Now ~ p : 2 + 3 ≠ 5 and
Then, using (D7), negation of p ∧ q is given by

38. ~ p ∧ q = 2 + 3 ≠ 5 or
(II) Negation of disjunction: Recall that a disjunction p ∨ q is consisting of two sub-statements
p and q which are such that either p or q or both exist. Therefore, the negation of the
disjunction would mean the negation of both p and q simultaneously. Thus, in symbolic
form, we have
(D8): The negation of a disjunction p ∨ q is the conjunction of the negation of p and the
negation of q. Equivalently, we write
~ (p∨ q) = ~ p ∧ ~ q
Example: Writ the negation of each of the following disjunction:
a) Ram is in class X or Rahim is in Class XII
b) 7 is greater than 4 or 6 is less than 7
Solution:
a) Let p : Ram is in class X and q : Rahim is in Class XII.
Then, the disjunction in (a) is given by p ∨ q.
Now ~ p : Ram is not in Class X.
~ q : Rahim is not in Class XII.
Then, using (D8), negation of p ∨ q is given by
~ p ∨ q : Ram is not in Class X and Rahim is not in Class XII.
b) Write p : 7 is greater than 4, and q : 6 is less than 7.
Then, using (D8), negation of p ∨ q is given by

39. ~ p ∨ q : 7 is not greater than 4 and 6 is not less than 7.
(III) Negation of a negation: As already remarked the negation is not a connective but a
modifier. It only modifies a given statement and applies only to a single simple statement.
Therefore, in view of (D5) and (D6), for a statement p, we have
(D9) : Negation of negation of a statement is the statement itself Equivalently, we write
~ (~p) = p
Example: Verify (D9) for the statement
p : Roses are red.
Solution: The negation of p is given by
~ p : Roses are not red.
Therefore, the negation of negation of p is
~ (~ p) : It is not the case that Roses are not red.
or
It is false that Roses are not red.
or
Roses are red.
Many statements, particularly in mathematics, are of the type “If p then q”. Such statements
are called conditional statements and are denoted by p → q read as ‘p implies q’.
Another common statement is of the form “p if and only if q”. Such statements are called bi-
conditional statements and are denoted by
p ↔ q.

40. Regarding the truth values of p → q and p ↔ q , we have
a) the conditional p → q is false only if p is true and q is false. Accordingly, if p is
false then p → q is true regardless of the truth value of q.
b) the bi-conditional p ↔ q is true whenever p and q have the same truth values otherwise it
is false.
One may verify that p → q = (~ p) ∨ q
2.8 Truth Tables
A truth table consists of rows and columns. The initial columns are filled with the possible truth
values of the sub-statements and the last column is filled with the truth values of the compound
statement S (the truth values of S depends on the truth values of the sub-statements entered in the
initial columns)
Example: Construct the truth table for ~p.
Solution: Note that one simple statement ~p is consisting of only one simple statement p.
Therefore, there must be 2’ (= 2) rows in the truth table. It is necessary to consider all possible
truth values of p.
In view of (D5) above, recall that p has the truth value T if and only if ~p has the truth value F.
Therefore, the truth table for ~p is given by
Table 21 Truth table for ~ p
p ~p
T F
F T
Example: Construct the truth table for p ∧ (~p)
Solution: Note that the compound statement p ∧ (~p) is consisting of only one simple statement
p. Therefore, there must be 2’ (= 2) rows in the truth table. It is necessary to consider all possible
truth values of p.

41. Table 2.2
p ~p p ∧ (~p)
T
F
Step 1: Enter all possible truth values of p. namely, T and F in the first column of the truth table
(Table 2.2).
Table 2.3
p ~p p ∧ (~ p)
T F
F T
Step 2: Using (D5) and (D6), enter the truth values of ~ p in the second column of the truth table
(Table 2.3).
Table 2.4
p ~p p ∧ (~ p)
T F F
F T F
Step 3: Finally, using (D2) enter the truth values of p ∧ (~ p) in the last column of the truth table
(Table 2.4)
Example: Construct the truth table for p ∧ q.
Solution: The compound statement p∧q is consisting of two simple statements p and q.
Therefore, there must be 22(= 4) rows in the truth table of p ∧q. Now enter all possible truth
values of statements p and q namely TT, TF, FT and FF in first two columns of Table 2.5.
Table 2.5

42. P q p∧q
T T
T F
F T
F F
Then, in view of (D1) and (D2) above, enter the truth values of the compound statement p ∧ q in
the truth table (Table 18.6) to complete the truth table.
Table 2.6: Truth table for p ∧ q
P q p∧q
T T T
T F F
F T F
F F F
Example: Construct the truth table for p ∨ q.
Table 2.7: Truth table for p ∨ q.
P q p∨q
T T T
T F T
F T T
F F F
Solution: In view of (D3) and (D4) above, recall that the compound statement p ∨ q has the truth
value F if and only if both p and q have the truth value F; otherwise p ∨ q has truth value T.
Thus, the truth table for p ∨ q is as given in Table 2.7.
a) ~ [p ∧ (~q)]
b) (p ∧q) ∧ (~ p)
c) ~[(~p) ∨ (~q)]

43. Solution:
a) Truth table for ~ [p ∧ (~ q)] is given by
Table 2.8: Truth table for ~ [p ∧ (~ q)]
p q ~p p ∧ (~q) ~[p ∧ (~q)]
T T F F T
T F T T F
F T F F T
F F T F T
b) Truth table for (p ∧ q) ∧ (~p) is given by
Table 2.9: Truth table for (p ∧ q) ∧ (~ p)
p q p∧q ~p (p∧q) ∧ (~ p)
T T T F F
T F F F F
F T F T F
F F F T F
c) Truth table for ~ [(~p) v (~q)] is given by
Table 2.10 : Truth table for ~ [(~p) ∨ (~q)]
~ [( ~ p)] ∧
p q ~p ~q (~ p) ∨ (~ q)
[(~q)]
T T F F F T
T F F T T F
F T T F T F
F F T T T F
2.9 Tautologies
A statement is said to be a tautology if it is true for all logical possibilities. In other words, a
statement is called tautology if its truth value is T and only T in the last column of its truth table.
Analogously, a statement is said to be a contradiction if it is false for all logical possibilities. In

44. other words, a statement is called contradiction if its truth value is F and only F in the last
column of its truth table. A straight forward method to determine whether a given statement is
tautology (or contradiction) is to construct its truth table.
Example: The statement p ∨ (~p) is a tautology since it contains T in the last column of its truth
table (Table 2.11)
Table 2.11: Truth table for p ∨ (~p)
p ~p p ∨ (~p)
T F T
F T T
Example: The statement p ∧ (~p) is a contradiction since it contains F in the last column of its
truth table (Table 2.12)
Table 2.12: Truth table for p ∧ (~ p)
p ~p p ∧ (~p)
T F F
F T F
Remark: The negation of a tautology is a contradiction since it is always false, and the negation
of a contradiction is a tautology since it is always true.
SAQ 1: Show that
a) ~ [p∨ (~p)] is a contradiction.
b) ~ [p ∧ (~p)] is a tautology.
Example: Show that
a) (p ∨ q) ∨ (~ p) is a tautology.
b) (p ∧ q) ∧ (~ p) is a contradiction.

45. Solution:
a) The truth table for (p ∨ q) ∨ (~ p) is given by
Table 2.15: Truth table for (p ∨ q) ∨ (~ p)
P q p∨q ~p (p ∨ q) ∨ (~ p)
T T T F T
T F T F T
F T T T T
F F F T T
Since the truth table for (p ∨ q) ∨ (~ p) contains only T in the last column, it follows that (p ∨
q) ∨ (~ p) is a tautology.
b) Recall Table 2.9 which is the truth table for (p ∧ q) ∧ (~ p) and observe that it contains only
F in the last column. Therefore, (p ∧ q) ∧ (~ p) is a contradiction.
2.10 Logical Equivalence
Two statements S1 (p, q, r, …) and S2 (p, q, r, …) are said to be logically equivalent, or simply
equivalent if they have the same truth values for all logical possibilities is denoted by
S1 (p, q, r,…) ≡ S2 (p, q, r,…).
In other words, S1 and S2 are logically equivalent if they have identical truth tables (by identical
truth tables we mean the entries in the last column of the truth tables are same).
Example: Show that ~ p ∧ q is logically equivalent to (~p) ∨ (~ q).
Solution: The truth tables for both the statements are
Table 2.16: Truth table for ~ (p ∧ q) Table 2.17: Truth table for (~ p) ∨ (~q)
p q p∧q ~(p ∧ q) p q ~p ~q (~ p) ∨ (~q)

46. T T T F T T F F F
T F F T T F F T T
F T F T F T T F T
F F F T F F T T T
Now, observe that the entries (truth values) in the last column of both the tables are same. Hence,
the statement ~(p ∧ q) is equivalent to the statement (~ p) ∨ (~q).
Remark: Consider the statements:
p : Mohan is a boy.
q : Sangita is a girl.
Now, we have
~(p ∧ q) ≡ (~ p) ∨ (~q).
Therefore, the statement
“It is not the case that Mohan is a boy and Sangita is a girl”
has the same meaning as the statement
“Mohan is not a boy or Sangita is not a girl”.
Example: Let
p : The South-West monsoon is very good this year and
q : Rivers are rising.
Give verbal translation of ~ [(~p) ∨ (~q)].
Solution: we have
~(p ∧ q) ≡ (~ p) ∨ (~q)

47. Therefore, the statement ~ [(~p) ∨ (~q)] is the same as the negation of the statement ~(p ∧ q)
which is the same as the conjunction p ∧ q. Thus, the verbal translation for ~ [(~p) ∨ (~q)] is
“The South-West monsoon is very good this year and rivers are rising”
Example: Prove the following:
a) ~ [p ∨ (~ q)] ≡ (~p) ∧q
b) ~ [(~ p) ∧ q] ≡ p ∨ (~q)
c) ~ (~p) ≡ p
Solution:
a) The truth tables for ~ [p ∨ (~q)] and (~p) ∧ q are given by
Table 2.18: Truth table for~ [p ∨ (~q)] Table 2.19: Truth table for (~p) ∧ q
~ [p ∨ (~
p q ~q p ∨ (~q) p q ~p (~p) ~ q
q)]
T T F T F T T F F
F
T T T F T F F F
F
T F F T F T T T
F F
T T F F F T F
The last column of the two tables are the same.
b) It follows in view of the truth Table 2.20
Table 2.20: Truth table for p ∨ (~q) and ~ [(~ p) ∧ q]

48. c) The assertion follows in view of Table 2.21
Table 2.21: Truth table for ~(~p)
2.11 Applications
The logic that we have discussed so far is called two-value logic because we have considered
only those statements which are having truth values True or False. A similar situation exists in
various electrical and mechanical devices. Claude Shannon, in late 1930’s, was first to notice an
analogy between the operations of switching devices and the operations of logical connectives.
He used this analogy with great success to solve problems of circuit design.
Observe that an electric switch which is used for turning ‘on’ and ‘off’ an electric light is a two-
state device. We shall now explain various electric networks with the help of logical
connectives. For this, first we discuss how an electric switch works. Observe that, in Fig. 2.1, we
have shown two positions of a simple switch.
Fig. 2.1
In (a) when switch is closed (i.e. on), current can flow from one terminal to the other. In (b),
when the switch is open (i.e. off), current can not flow.

49. Let us now consider the example of an electric lamp controlled by switch. Such a circuit is given
in Fig. 2.2.
Fig. 2.2
Observe that when the switch s is open, no current flows in the circuit and therefore, the lamp
is ‘off’. But when switch s is closed, the lamp is ‘on’. Thus the lamp is on if and only if the switch
s is closed.
If we denote the statements as
p : The switch s is closed
l : The lamp l is ‘on’
then, by using logic, the above circuit can be expressed as p ≡ l.
Next, consider an extension of the above circuit in which we have taken two switches s1 and s2 in
series as shown in Fig. 2.3.
Fig. 2.3
here, observe that the lamp is ‘on’ if and only if both the switches s1 and s2 are closed.
If we denote the statements as:

50. p : the switch s1 is closed.
q : the switch s2 is closed.
l : the lamp l is ’on’.
then the above circuit can be expressed as p ∧ q ≡1.
Now, we consider a circuit in which two switches s1 and s2 are connected in parallel (Fig. 2.4).
Fig. 2.4
SAQ 2: Express the following circuit in Fig. 2.5 in symbolic form of logic.
Fig. 2.5
2.12 Summary
In this unit we study the truth values of a statements. The different basic logical connectives are
discussed in detail with some standard examples. Compound statements and the negation are

51. clearly explained . The concept of Tautology, Contradiction and Logical Equivalence is discussed
in detail with example wherever necessary. The applications of mathematical logic to switching
circuits is dealt with standard examples.
2.13 Terminal Questions
1. Define Tautology and Contradiction
2. Draw the truth tables of Conjunction, disjunction and Biconditional
2.14 Answers
Self Assessment Questions
1. a) The truth table of ~ [p∨ (~p)] is given by
Table 2.13: Truth table for ~ [p∨ (~p)]
P ~p p ∨ (~p) ~ [p ∨ (~p)]
T F T F
F T T F
Since it contains only F in the last column of its truth table, it follows that
~ [p ∨ (~ p)] is a contradiction.
b) The truth table of ~ [p ∧ (~ p)] is given by
Table 2.14: Truth table for ~ [p ∧ (~ p)]
P ~p p ∧ (~ p) ~ [p ∧ (~ p)]
T F F T
F T F T
Since it contains only T in the last column of its truth table, it follows that ~ [p ∧ (~ p)] is
a tautology.
2. Observe that the lamp is ‘on’ if and only if either s1 and s2 both are closed or s1 and s2 both
are open or only s1 is closed.

52. If we denote the statements as
p : The switch s1 is closed
q : The switch s2 is closed
l : The lamp l is ‘on’
then
~p: The switch s1 is open.
or
The switch s1 is closed.
~ q: The switch s2 is open.
or
The switch s2 is closed.
Therefore, the circuit in Fig. 2.5 in symbolic form of logic may be expressed as
p ∨ [(~ p) ∧ (~ q)] ∨ (p ∧ q) ≡1
BT0063-Unit-03-Modern Algebra
Unit 3 Modern Algebra
Structure
3.1 Introduction
Objectives
3.2 Binary Operation

53. 3.3 Addition Modulo n
3.4 Multiplication Modulo n
3.5 Semigroup
3.6 Properties of Groups
3.7 Subgroup
3.8 Summary
3.9 Terminal Questions
3.10 Answers
3.1 Introduction
The theory of groups which is a branch of Abstract Algebra is of paramount importance in the
development of mathematics.
The idea of group was first given by the French Mathematician Evariste Galois in 1832 who died
at the age of 21 years in a duel. The group theory was later developed by an English
Mathematician Arthur Cayley. He defined the notion of an abstract group with a general
structure which could be applied to numerous particular cases. The theory of groups has
applications in Quantum Mechanics and other branches of mathematics.
Objectives:
At the end of the unit you would be able to
• apply the concepts of Algebraic Structure in practical problems
• understand Binary Operations and its applications in group theory
3.2 Binary Operation
Let G be a non-empty set. Then G × G = {(x, y): x, y ∈ G}. A function of
G × G in to G is said to be a binary operation on the set G. The image of an ordered pair (x, y) under f is
denoted by x f y.

54. The symbols +, x, 0, *, …. Are very often used as the binary operations on a set.
Thus * is a binary operation on the set G if for every a, b∈G implies a * b∈ G.
Hence a binary operation * combines any two elements of G to give an element of the same set
G.
Examples:
1. If Z is the set of integers then usual addition (+) is the binary operation on Z. For if M
and n are two integers then m + n is again an integer i.e. for every m, n ∈ Z, m + n ∈ Z.
In particular – 5, 3 ∈ Z, implies – 5 + 3 = –2 ∈ Z, etc.
Similarly the usual multiplication is the binary operation on the set Q of rationals, for
the product of two rational numbers is again a rational number.
2. Let E be the set of even integers. i.e., E = {0, ±2, ±4, ±6, ….} and O be the set of odd
integers i.e. O = {±1, ±3, ±5, ….}. Clearly the usual addition is the binary operation on E
whereas it is not a binary operation on O. Because the sum of two even integers is even but
the sum of two odd integers is not an odd integer.
Also the usual subtraction is not a binary operation on the set N of natural numbers.
Algebraic Structure
A non-empty set with one or more binary operations is called an algebraic structure. If * is a binary
operation on G then (G, *) is an algebraic structure.
For example the set of integers Z is an algebraic structure with usual addition as the binary operation.
Similarly (Q, .), (E, +) are algebraic structures.
Group
A non-empty set G is said to be a group with respect to the binary operation * if the following axioms
are satisfied.
1. Closure law. For every a, b ∈ G, a * b ∈ G.
2. Associative law. For every a, b, c ∈ G

55. a * (b * c) = (a * b) * c
3. Existence of identity element. There exists an element e ∈ G such that
a * e = e * a = a for every a ∈ G.
Here e is called the identity element
4. Existence of inverse. For every a ∈ G there exists an element b ∈ G such that
a * b = b * a = e. Here b is called the inverse of a and is denoted by
b = a–1. A group G with respect to the binary operation * is denoted by (G, *). If in a group (G, *), a *
b = b * a for every a, b, ∈ G then G is said to be commutative or Abelian group named after
Norwegian mathematician Niels Henrik Abel (1802 – 1820).
Finite and Infinite Groups
A group G is said to be finite if the number of elements in the set G is finite, otherwise it is said to be an
infinite group. The number of elements in a finite group is said to be the order of the group G and is
denoted by O(G).
Example: Prove that the set Z of integers is an abelian group with respect to the usual addition as the
binary operation.
1. Closure law. We know that the sum of two integers is also an integer. Hence for every m, n ∈ Z, m
+ n ∈ Z.
2. Associative law. It is well known that the addition of integers is associative. Therefore (m + n) + p =
m + (n + p) for every m, n, p ∈ Z.
3. Existence of identity element. There exists 0 ∈ Z such that
m + 0 = 0 + m = m for every m ∈ Z. Hence 0 is called the additive identity.

56. 4. Existence of inverse. For every m ∈ Z there exists – m ∈ Z such that
m + (–m) = (–m) + m = 0.
Here – m is called the additive inverse of m or simply the negative of m. Therefore (Z, +) is a
group.
5. Commutative law. We know that the addition of integers is commutative i.e., m + n = n + m for
every m, n ∈ Z. Hence (Z, +) is an abelian group. Since there are an infinite elements in Z, (Z, +) is an
infinite group.
Similarly we can prove that the set Q of rationals, the set R of reals and the set C of complex
numbers are abelian groups with respect to usual addition.
Example: Prove that the set Q0 of all non-zero rationals forms an abelian group with respect to usual
multiplication as the binary operation.
Now Q0 = Q – {0}
Solution:
1. Closure law. Let a, b ∈ Q0 i.e. a and b are two non-zero rationals. Then their product a b is also a
non-zero rational. Hence a b ∈ Q0.
Since a, b are two arbitrary elements of Q0, we have for every
a, b, ∈ Q0, ab ∈ Q0.
2. Associative law. We know that the multiplication of rationals is associative. i.e.,, a(b c) = (a b) c for
every a, b, c ∈ Q0.
3. Existence of identity element. There exists 1 ∈ Q0 such that
a.1 = 1 . a = a for every a ∈ Q0. Here 1 is called the multiplicative identity element.
4. Existence of inverse. Let a ∈ Q0. Then a is a non-zero rational. Therefore exists and is also a
rational ≠ 0.

57. Also for every a ∈ Q0.
is the multiplicative inverse of a.
Therefore (Q0, .) is a group.
Further, it is well-known that the multiplication of rationals is commutative i.e., ab = ba for
every a, b ∈ Q0.
Hence (Q0, .) is an abelian group.
Similarly we can show that the set R0 of non-zero reals and the set C0 of non-zero complex
numbers are abelian groups w.r.t. usual multiplication.
1. The set N of natural numbers is not a group w.r.t. usual addition, for there does not exist the
identity element 0 in N and the additive inverse of a natural number is not a natural number i.e., for
example
2 ∈ N but – 2 ∉ N. Also N is not a group under multiplication because
5 ∈ N but
2. The set of integers is not a group under multiplication for 2 ∈ Z but
3. The set of rationals, reals and complex numbers (including 0) do not form groups under
multiplication for multiplicative inverse of 0 does not exist.
SAQ 1: Prove that the fourth roots of unity form an abelian group with respect to multiplication.
3.3 Addition Modulo n
Let n be a positive integer a and b be any two integers. Then “addition modulo n of two integers a and
b”, written a + n b, is defined as the least non-negative remainder when a + b is divided by n. If r is the
remainder when a + b is divided by n, then

58. A + n b = r where 0 ≤ r < n.
In other words, if a + b ≡ r (mod n), 0 ≤ r < n. Then a + n b = r.
For example,
7 + 5 10 = 2 since 7 + 10 = 17 ≡ 2 (mod 5)
15 + 7 11 = 5 since 15 + 11 = 26 ≡ 5 (mod 7)
17 + 8 21 = 38 since 17 + 21 = 38 ≡ 6 (mod
12 + 5 8 = 0 since 12 + 8 = 20 ≡ 0 (mod 5)
1 + 7 1 = 2 since 1 + 1 = 2 ≡ 2 (mod 7)
Properties:
1. Commutative since a + b and b + a leave the same remainder when divided by n, a + n b = b + n a.
For example 5 + 7 6 = 4 = 6 + 7 5
2. Associative since a + (b + c) and (a + b) + c leave the same remainder when divided by n, a + n (b + n
c) = (a + n b) + n c.
For example 4 +6 (3 + 6 5) = (4 + 6 3) + 6 5
Example: Prove that the set Z4 = {0, 1, 2, 3} is an abelian group w.r.t. addition modulo 4.
Solution: Form the composition table w.r.t. addition modulo 4 as below:
+4 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 2 0 1 2

59. Since 1 + 3 = 4 ≡ 0 (mod 4), 3 + 3 = 6 ≡ 2 (mod 4) 2 + 3 = 5 ≡ 1 (mod 4) etc.
1. Closure law. From the above composition table for all a, b ∈ G, a +4 b also belongs to Z4.
2. Associative law. Since a + (b + c) and (a + b) + c leave the same remainder when divided by 4, we
have
(a + 4 (b +4 c) = (a +4 b) +4 c.
3. Existence of identity element. From the above table, we observe that
0 ∈ Z4 satisfies a + 4 0 = 0 +4 a = a for every a ∈ Z4.
0 is the identity element.
4. Existence of inverse. From the above table, the inverses of 0, 1, 2, 3 are respectively 0, 3, 2, 1
because 0 +4 0 = 0, 1 + 4 3 = 0, 2 +4 2 = 0, and
3 + 41 = 0.
Hence (z4, +4) is a group
Further, since a + b and b + a leave the same remainder when divided by 4, a + 4 b = b +4 a.
(Z4, +4) is an abelian group.
Similarly, we can show that the set of remainders of 5 viz.
Z5 = {0, 1, 2, 3, 4} from an abelian group under addition (mod 5).
In general the set of remainders of a positive integer m.
Zm = {0, 1, 2, …. (m –1) form an abelian group under addition
(mod m).
3.4 Multiplication modulo n
Let n be a positive integer an a, b any two integers. Then multiplication modulo n of two integers a and
b, written a ×n b, is defined as the least non-negative remainder when ab is divided by n. If r is the

60. remainder when ab is divided by n. If r is the remainder when ab is divided by n then a ×n b = r, where 0
≤ r < n. In other words, if ab ≡ r (mod n), 0 ≤ r < n then a xn b = r.
For example,
7×5 3 = 1 since 7 . 3 = 21 ≡ 1 (mod 5)
9 ×7 5 = 3 since 9 . 5 = 45 ≡ 3 (mod 7)
12 ×8 7 = 4 since 12 . 7 = 84 ≡ 4 (mod
2 ×7 3 = 6 since 2 . 3 = 6 ≡ 6 (mod 7)
14 ×46 = 0 since 14 . 6 = 84 ≡ 0 (mod 4)
Properties
1. Commutative: Since ab and ba leave the same remainder when divided by n,
a ×n b = b ×n a
For example 5×74=4×75
2. Associative: Since a(bc) and (ab)c leave the same remainder when divided by n
a ×n (b ×n c) = (a ×n b) ×n c
For example 3 ×7 (4 ×7 5) = (3 ×7 4) ×7 5
Example: Prove that the set is an abelian group under multiplication modulo 5.
Solution: Form the composition table w.r.t. multiplication modulo 5 as below:
x5 1 2 3 4
1 1 2 3 4
2 2 4 1 3

61. 3 3 1 4 2
4 4 3 2 1
Since 2 . 3 = 6 ≡ 1 (mod 5)
2 . 4 = 8 ≡ 3 (mod 5)
4 . 4 = 16 ≡ 1 (mod 5) etc.
1. Closure law. Since all the elements entered in the above table are the elements of closure law
holds good i.e. for all a, b ∈ G, a ×5 b also belongs to
2. Associative law. Since a (bc) and (ab) c leave the same remainder when divided by 5 we have for
every a, b, c ∈
a × 5 (b × 5 c) = (a × 5 b) × 5 6.
3. Existence of identity element. From the above table, we observe that
1∈ satisfies a ×5 1 = 1 × 5 a = a for every a ∈ .
1 is the identity element.
4. Existence of inverse. Also the inverses of 1, 2, 3, 4 are respectively
1, 3, 2, 4 because 1 × 5 1 = 1, 2 × 5 3 = 1, 3 × 5 2 = 1, and 4 × 5 4 = 1.
Therefore ( x5) is an abelian group.
Similarly, we can show that the non-zero remainders of 7 viz.
= {1, 2, 3, 4, 5, 6} form an abelian group under multiplication
(mod 7). In general, the non-zero remainders of a positive integer
p viz. = {1, 2, 3, …… (p – 1)} form a group under multiplication
(mod p) if and only if p is a prime number.

62. Note: The set = {1, 2, 3, 4, 5} does not form a group under multiplication (mod 6) for 2, 3 ∈G, but 2
× 6 3 = 0 ∉ G. This is because 6 is not a prime number.
3.5 Semigroup
A non-empty set G is said to be a semigroup w.r.t. the binary operation if the following axioms are
satisfied.
1. Closure: For every a, b, ∈ G, a * b ∈ G
2. Associative: For every a, b, c ∈ G, a * (b * c) = (a * b) * c.
Examples:
1. The set N of all natural numbers under addition is a semigroup because for every a, b, c ∈ N
(i) a + b ∈ N, and (ii) a + (b + c) = (a + b) + c. The set, N is semigroup under multiplication also.
2. The set Z of integers is a semigroup under multiplication because for every a, b ∈ Z, a + b ∈
Z and for every a, b, c ∈ Z, a(bc) = (ab) c. Note that every group is a semigroup but a semigroup
need not be a group. For example, the set N of all natural numbers is a semigroup under
multiplication (also under addition) but it is not a group. Similarly Z, the set of integers is an
example of a semigroup but not a group under multiplication.
3.6 Properties of Groups
For the sake of convenience we shall replace the binary operation * by dot . in the definition of the
group. Thus the operation dot . may be the operation of addition or multiplication or some other
operation. In what follows by ab we mean a . b or a * b. With this convention, we rewrite the definition
of the group.
Definition: A non-empty set G is said to be a group w.r.t. the binary operation. if the following axioms
are satisfied.
1. Closure property: For every a, b ∈ G, ab ∈ G
2. Associative property: For every a, b, c ∈ G, a (bc) = (ab) c.

63. 3. Existence of identity element: There exists an element e ∈ G such that ae = ea = a for every a ∈ G.
Here e is called the identity element.
4. Existence of inverse: For every a ∈ G there exists an element b ∈ G such that ab = ba = e. Here b is
called the inverse of a i.e., b = a–1 Further,
5. If ab = ba for every a, b ∈ G then G is said to be an abelian group or a commutative group.
Theorem: The identity element in a group is unique.
Proof: Let e and be the two identity elements of a group G. Then by definition, for every a ∈ G.
ae = ea = a
and
Substitute in (1) and a = e in (2). Then we obtain
and
Hence
The identity element in a group is unique.
Theorem: In a group G the inverse of an element is unique
Proof: Let b and c be the two inverses of an element a in G.
Then by definition ab = ba = e
ac = ca = e
Now consider, b = be
= b(ac)

64. = (ba) c
= ec
=c
Therefore inverse of every element in a group is unique
Theorem: If a is any element of a group G, then (a–1)–1 = a.
Proof: Since a–1 is the inverse of a, we have aa–1 = a–1a = e
This implies that a is an inverse of a–1, but inverse of every element is unique
Thus the inverse of the inverse of every element is the same element.
Theorem: If a and b are any two elements of a group G then
Proof:
Consider, (ab) (b–1 a–1) =
=
=
= aa–1
=e
Similarly we can prove that
Hence

65. Therefore is the inverse of ab,
i.e.,
Corollary: If a, b, c belong to a group G then (abc)–1 = c–1 b–1 a–1 etc.
Note: If (ab)–1 = a–1 b–1 for all a, b ∈ G, the G is abelian.
For, (ab)–1 = a–1 b–1 implies
i.e.
= ba for all a, b ∈ G
Hence G is abelian.
Theorem: (Cancellation laws).
If a, b, c are any three elements of a group G, then
ab = ac implies b = c (left cancellation law)
ba = ca implies b = c (right cancellation law)
Proof: Since a is an element of a group G, there exists a–1 ∈ G there exists a–1 ∈ G such that aa–1 = a–1 a =
e, the identity element
Now ab = ac
⇒
⇒
⇒ eb = ec
⇒ b=c

66. Similarly ba = ca
⇒
⇒
⇒ be = ce
⇒ b=c
Theorem: If a and b are any two elements of a group G, then the equations ax = b and ya = b have
unique solutions in G.
Proof:
i) Since
Now and b ∈ G implies (closure axiom) and
Hence x = a–1 b satisfies the equation ax = b and hence is a solution. If x1, x2 are the two
solutions of the equation, ax = b then ax1 = b and ax2 = b.
ax1 = ax2
x1 = x 2
Hence the solution is unique.
ii) Also b ∈ G, a–1 ∈ G implies ba–1 ∈ G and
y = ba–1 satisfies the equation ya = b and hence is a solution. If y1, y2 are two solutions of the
equation ya = b then y1a = b and y2a = b
y1a = y2a
y1 = y2

67. Therefore the solution is unique
SAQ 2: Prove that in a group G if a2 = a then a = e, the identity element.
Note: Any element a which satisfies a2 = a is called the idempotent element in a group. Thus e is the
only idempotent element in G.
Example: If in group G, (ab)2 = a2b2 for every a, b ∈ G prove that G is abelian.
Solution:
Now
⇒ (ab) (ab) = (a . a) (b . b)
⇒ a[b(ab)] = a[a(bb)] (Associative)
⇒ b (ab) = a (bb) (Left cancellation law)
⇒ (ba) b = (ab) b (Associative)
⇒ ba = ab (Right cancellation law)
Hence G is an abelian group.
Example: Show that if every element of a group G is its own inverse then G is abelian.
Solution: Let a, b ∈ G then a–1 = a and b–1 = b
Clearly ab ∈ G (ab)–1 = ab by hypothesis
i.e. b–1 a–1 = ab
i.e. ba = ab since b–1 = b, a–1 = a
G is abelian.

68. 3.7 Subgroup
A non-empty subset H of a group G is said to be a subgroup of G if under the operation of G, H itself
forms a group.
If e be the identity element of a group G, Then H = { e } and H = G are always subgroups of G. These are
called the trivial or improper subgroups. If H is a subgroup of G and H ≠ {e} and H ≠ G then H is called a
proper subgroup.
Examples:
1. We know that the set Z of integers forms a group under addition. Consider a subset E = {2x : x ∈ Z} =
{0, ±2, ±4, …. } of Z. Then E also forms a group under addition.
Therefore E is a subgroup of Z.
Similarly F = {3x : x ∈ Z} = {0, ±3, ±6, ±9, ….. } is a subgroup of z.
2. Clearly the multiplicative group H = {1, –1} is a subgroup of the multiplicative group G = {1 –1, i, –i}.
3. Let G = {1, 2, 3, 4, 5, 6} be a subset of G. Now it is clear from the following composition table that H
also forms a group under x7.
X7 1 2 4
1 1 2 4
2 2 4 1
4 4 1 2
Therefore H is a subgroup of G.
Theorem: A non-empty subset H of a group G is a subgroup of G if and only if
i) for every a, b ∈ H implies ab ∈ H
ii) for every a ∈ H implies a–1 ∈ H
Note: Union of two subgroups need not be subgroups for, let H = {0, ±2, ±3, ±4, ….} and K = {0, ±e, ±6,….} be

69. two subgroups of the group of integers Z, so that
H ∪ K = {0, ±2, ±3, ±4, ±6, …. }.
Now 2, 3 ∈ H ∪ K but 2 + 3 = 5 ∉ H ∪ K because 5 is neither a multiple of 2 nor a multiple of 3.
3.8 Summary
In this unit we studied clearly that the rectangular array of numbers is denoted by matrix, also we know
that determinant is a square matrix which is associated with a real number. Then we studied that a set
which satisfies certain rules is called as a group. Here we studied sub group, semi group etc. with well
illustrated examples.
3.9 Terminal Questions
1. Prove that a non-empty subset H of a group G is a subgroup of G if and only if for every a, b ∈ H
implies ab–1 ∈ H.
2. Prove that the intersection of two subgroups of a group is again a subgroup.
3.10 Answers
Self Assessment Questions
1. Roots of the equation x4 = 1 are called the fourth roots of unity and they are 1, –1, i, – i. Let G = {1, –
1 i, – i }.
From the composition table w.r.t. usual multiplication as follows:
. 1 –1 i –i
1 1 –1 i –i
–1 –1 1 –i i
i i –i –1 1
–i –i I 1 –1
1. Closure Law. Since all the elements written in the above composition table are the elements
of G, we have for all a, b, ∈ G, ab ∈ G.

70. 2. Associative Law. We know that the multiplication of complex numbers is associative and G
is a subset of the set of complex numbers
Hence a(bc) = (ab) c for all a, b, c ∈ G.
3. existence of identity element. From the composition table it is clear that there exists 1 ∈ G
satisfying a . 1 = 1 . a = a for every a ∈ G.
Therefore 1 is the identity element.
4. Existence of inverse. From the composition table we observe that the inverses of 1, – 1, i – i
are 1, -1, -i, i.
Thus for every a ∈ G there exists a–1 ∈ G such that a a–1 = a–1 a = 1, the identity
element. Hence (G, .) is a group.
Further multiplication of complex numbers is commutative.
Therefore ab = ba for every a, b ∈ G.
Also we observe that the elements are symmetric about the principal diagonal in the
above composition table. Hence commutative law holds good.
Therefore (G, .) is an abelian group.
Note that G is a finite group of order 4.
2. Now since a = ae ⇒ a = e
BT0063-Unit-04-Trigonometry
Unit 4 Trigonometry
Structure
4.1 Introduction
Objectives
4.2 Radian or Circular Measure
4.3 Trigonometric Functions
4.4 Trigonometrical ratios of angle when is acute
4.5 Trigonometrical ratios of certain standard angles
4.6 Allied Angles

71. 4.7 Compound Angles
4.8 Multiple and Sub-multiple angle
4.9 Summary
4.10 Terminal Questions
4.11 Answers
4.1 Introduction
This unit of Trigonometry gives us an idea of circular measure. The different Trigonometric
functions are studied here. Some of the standard angles and their Trigonometric ratios are
discussed in detail. The basic knowledge allied angles and compound angles are explained in a
simple manner.
Objectives:
At the end of the unit you would be able to
• understand the concepts of Trigonometrical functions
• use allied and compound angles in calculations
4.2 Radian or Circular Measure
A radian is the angle subtended at the centre of a circle by an arc equal to the radius of the circle.
O is the centre of a circle. A and B are points on the circle such that arc AB = radius OA. Then
is called one radian or one circular measure. We write

72. Radian is a constant angle and
Consider a circle whose centre is O and radius r. A and B are points on the circle such that arc AB
= OA = r. Join OA, OB and draw OC ⊥ to OA. , right angle and arc AB = r.
We know that arc (circumference of the circle) = . In a circle the
arcs are proportionated to the angles subtended by them at the centre.
1c = 2/π ×1 right angle, which is constant
Radian is a constant angle
Further we have, π × 1C = 2 × 1 right angle
πC = 2 × 90° = 180°
Note:
i) πC = 180° mans π radians are equal to 180°
Hereafter, this is written as π = 180°.
For example and so on.
In each of these cases the unit ‘radians’ on the left side is understood.

73. ii)
(nearly)
Here π is the real number which is the ratio of circumference of a circle to its diameter.
Its approximate value is 22/7.
1 radian = (approximately)
Clearly 1 radian is < 60°
Examples:
1) Express 2.53 radians in degrees
π radians = 180°
2) Express 144° into radians
For 180° = π radians
It is better to remember the following:
1) radians
2) x radians =
Length of an arc of a circle

74. Consider a circle who centre is O and radius r. A and B are points on the circle such that arc AB
= r. P is a point on the circle such that arc PA = s and
Hence the length of an arc of a circle is equal to the product of the radius of the circle and
the angle in radians subtended at the centre by the arc.
Note:
s = the arc length of the circle; r = the radius of the circle
θ = angle in radians subtended by s at the centre
Area of a sector of a circle
The portion of the circle bounded by two radii, say, OA, OB and the arc
AB is called the sector . Consider a circle whose centre is O and radius r. Let AOB be the
sector of angle

75. Worked Examples
1. Express 792° in radians and 7π/12 in degrees
2. The angles of a triangle are in the ratio 2:3:5 find them (i) in radians
(ii) in degrees.
A:B:C=2:3:5 A = 2K, B = 3K, C = 5K
i) A + B + C = π ⇒ 10 K = π K = π/10
The angles are in radians

76. ii) A + B + C = 180° = 10K = 180° K = 18°
A = 36° B = 54° C = 90°. The angles are 36°, 54°, 90°
3. An arc of a circle subtends 15° at the centre. If the radius is 4 cms, find the length of the arc
and area of the sector formed.
θ = 15°, r = 4 cms to find s
s = rθ = 4(π/12) = π/3 = 22/21 cm
Area of the sector
4. A spaceship moves in a circular orbit of radius 7200 km round the earth. How far does it
travel while sweeping an angle of 100°?
S = rθ = (7200) (5π/9) = (800) 5π = (4000π) km.
The spaceship travels through a distance of (4000 π) km.
SAQ 1: A circular wheel is rotating at the rate of 25 revolutions per minute. If the radius of the
wheel is 50 cms, find the distance covered by a point on the rim in one second (Take π = 3.1416)
4.3 Trigonometric Functions
Consider a circle whose centre is the origin and radius is r. Let the circle cut X-axis at A and
and Y-axis at B and P(x, y) is any point on the circle. Join OP and draw PM ⊥ to X-axis. OP
= r, OM = x, MP = y, .
The six trigonometrical functions (ratios) of angle θ are defined as given below:

77. Sine of angle cosecant of angle θ = cosec
Cosine of angle secant of angle tangent of angle
cotangent of angle
Since we have . So we have
Note
i) Reciprocal relations

78. sin θ and cosec θ are reciprocal to each other.
Similarly we have,
ii)
iii) The above definitions of trigonometric functions hold good whatever may be the position of
the point P(x, y) on the circle. We shall discuss this in detail later.
iv) Identities
a)
b)
c)
a) From the figure
on dividing by r2,
But
Thus for all value of θ, cos2 θ + sin2 θ = 1
b) . If x ≠ 0, we can divide by x2.

79. But
c) If y ≠ 0 we can divide y2.
But
Thus
4.4 Trigonometrical ratios of angle when is acute
The revolving line, starting from OX rotates through an acute angle and comes to the position
OA. Draw AB ⊥ to X-axis. In the triangle OAB, The side opposite to i.e.,
AB is called opposite side. The side opposite to 90°i.e., OA is called the hypotenuse and OB is