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Karnataka State Open University

(KSOU) was established on 1st June 1996 with the assent of H.E. Governor of Karnataka

as a full fledged University in the academic year 1996 vide Government notification

No/EDI/UOV/dated 12th February 1996 (Karnataka State Open University Act – 1992).

The act was promulgated with the object to incorporate an Open University at the State level for

the introduction and promotion of Open University and Distance Education systems in the

education pattern of the State and the country for the Co-ordination and determination of

standard of such systems. Keeping in view the educational needs of our country, in general, and

state in particular the policies and programmes have been geared to cater to the needy.

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accommodation to students coming to Mysore for attending the Contact Programmes or Term-

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

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:

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

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}

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:

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:

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.

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.

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

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.

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

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}.

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.

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.

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

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.

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).

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

= 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.

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

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)}

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:

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

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

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,

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.

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

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

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

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.

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”.

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

~ 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

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

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

~ 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

~ 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.

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.

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

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)]

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 q ~p ~q (~ p) ∨ (~ q) ~ [( ~ p)] ∧

[(~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

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.

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)

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)

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 q ~q p ∨ (~q) ~ [p ∨ (~

q)]

p q ~p (~p) ~ q

T T F T F T T F F

T F

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]

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.

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:

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

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.

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

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.

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

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.

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.

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

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

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

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 × 7 4 = 4 × 7 5

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

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.

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.

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)

= (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–1

a = 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

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

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 = x2

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

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 = a

2b

2 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.

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

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.

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

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

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.

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

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

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

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:

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

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 + sin

2 θ = 1

b) . If x ≠ 0, we can divide by x2.

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

called the adjacent side. The six trigonometrical ratio of are defined as

These definitions hold good whenever θ is one of the acute angles of a right angled triangle.

The following identities should be memorized.

1) 2)

3) 4)

5) 6)

7)

Worked Examples:

1. Show that

2. Prove that

write 1 in numerator as

…………………..(1)

Now

…………………………(2)

From (1) and (2) the result follows

3. Show that cosec

LHS = cosec A – cot A

4. Show that

= 2 – 2 sin A – 2 sin A cos A + 2 cos A

= 2(1 – sin A – sin A cos A + cos A)

RHS = 2[1 – sin A + cos A – sin A cos A]

LHS = RHS

5. If show that

…….. (1)

…….. (2)

Square and add (1) and (2)

6. If then show that (x – a) (b – x) = (a – b)2 sin

2 t cos

2 t

x – a = a cos2 t + b sin

2 t – a

= b sin2 t – a + a cos

2 t

= b sin2 t – a (1 – cos

2 t)

= b sin2 t – a sin

2 t

= sin2 t. (b – a) (1)

b – x = b – (a cos2 t + b sin

2 t) = b – a cos

2 t – b sin

2 t

= b(1 – sin2 t) – a cos

2 t

= b cos2 t – a cos

2 t

= cos2 t (b – a) (2)

Multiply (1) and (2),

(x – a) (b – x) = sin2 t (b – a) . cos

2 t(b – a)

= (b – a)2 sin

2 t cos

2 t

= (a – b)2 sin

2 t cos

2 t

7. Express all the trigonometrical ratios of angle A in terms of sin A where A is acute.

I Method

From the figure

(1)

(2)

(3)

(4)

(5)

(6)

Thus we have expressed all trigonometric al ratios in terms of A.

II Method

We know that (1)

Since we have tan (2)

(3)

(4)

(5)

sin A = sin A (6)

8. Express all the trigonometrical ratios of angle A in terms of sec A

I Method

Take sec

Let PQR be the right angled triangle, in which

Now PR2 = PQ

2 – QR

2 = x

2 – 1

sec A = sec A (1)

(2)

(3)

(4)

(5)

(6)

II Method

We know that 1 + tan2 A = sec

2 A tan

2 A = sec

2 A – 1

(1)

(2)

(3)

(4)

(5)

sec A = sec A (6)

SAQ 2: If find the value of

SAQ 3: If cos θ = (m2 – n

2)/ (m

2 + n

2) find the values of cot θ and cosec θ in terms of m and n.

4.5 Trigonometrical ratios of certain standard angles

To remember the trigonometrical ratios of the standard angles the following table is useful.

Worked examples

1. Find the value of

2. Find the value of

3. Find ‘x’ from the equation,

Given equation is,

x = 1

4. Find x from (2x – 3) (cosec2 π/3 – sin

2 π/4) = x tan

2 (π/4) – sec

2 (π/6) – 2

(2x – 3) cosec2 60° – sin

2 45°) = x tan

2 45° – sec

2 30° – 2

(2x – 3) (4/3 – ½) = x – 4/3 – 2

(2x – 3) (5/6) = x – 10/3

5(2x – 3) = 6x – 20; 10x – 15 = 6x – 20; i.e., 4x = – 5.

x = – (5/4).

5. Show that

SAQ 4: Prove that sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60° =

4.6 Allied Angles

To connect the trigonometrical ratios of with those of θθθθ

Thus

Similarly

sin (90° – θ) = cos θ cosec (90° – θ) = sec θ

cos (90° – θ) = sin θ sec (90° – θ) = cosec θ

tan (90° – θ) = cot θ cot (90° – θ) = tan θ

To connect the trigonometrical ratios of 90°°°° +θθθθ with those θθθθ

sin (90° + θ) = cos θ cosec (90° + θ) = sec θ

cos (90° + θ) = –sin θ sec (90° + θ) = –cosec θ

tan (90° + θ) = –cot θ cot (90° + θ) = –tan θ

To express the trigonometrical ratios of 180 °°°° –θθθθ in terms of those of θθθθ

sin (180° – θ) = sin θ cosec (180° – θ) = cosec θ

cos (180° – θ) = –cos θ sec (180° – θ) = –sec θ

tan (180° – θ) = –tan θ cot (180° – θ) = –cot θ

To express the trigonometrical ratios of 180 +θθθθ in terms of those of θθθθ

Thus

sin (180° + θ) = sin θ

cos (180° + θ) = –cos θ

tan (180° + θ) = tan θ

Taking reciprocals,

cosec (180° + θ) = –cosec θ

sec (180° + θ) = –sec θ

cot (180° + θ) = cot θ

Coterminal angles

Two angles are said to be coterminal angles, if their terminal sides are one and the same. θ and

360° + θ are coterminal angles. – θ and 360° – θ are coterminal angles.

sin (360° – θ) = sin (– θ) = – sin θ cosec (360° – θ) = cosec (– θ) = cosec θ

cos (360° – θ) = cos (– θ) = cos θ sec (360° – θ) = sec (– θ) = –sec θ

tan (360° – θ) = tan (– θ) = – tan θ cot (360° – θ) = cot (– θ) = –cot θ

sin (360° + θ) = sin θ cosec (360° + θ) = cosec

cos (360° + θ) = cos θ sec (360° + θ) = sec θ

tan (360° + θ) = tan θ cot (360° + θ) = cot θ

Worked Examples

1. Find the value of (a) sin 120° (b) sec 300° (c) tan 240° (d) cos 1770° (e) cosec 1305°

(f) cosec (-1110°)

a) sin 120° =

b) sec 300° =

c) tan 240° =

d) cos 1770° =

e) cosec 1305° =

=

f) cosec (-1110°) =

=

2. Find the value of

tan 120° =

cot 240° =

cot 210° =

tan 240° =

sin 270° =

3. Find the value of

4. Show that

sin 135° = sin (180° – 45°) = sin 45° =

cos 480° = cos (6 × 90° – 60°) = – cos 60° = -1/2

cos 120° = -1/2

5. Find x given that

The given equation is

6. Simply:

7. Show that

8. If find the value of

OP is the bounding line for θ.

So coordinates of P = (–3, 4)

SAQ 5: If find the value of

Note: As tan θ is given to be negative, the angle lies in II quadrant and IV quadrant.

4.7 Compound Angles

The sum of difference of angles like A + B, A – B, A + B – C etc, are known as compound

angles.

The trigonometrical ratios of A + B, A – B can be expressed in term of those of A and B. It should

be noted that sin (A + B) ≠ sin A + sin B, cos (A + B) ≠ cos A + cos B etc. This can be easily

verified by the example

i)

ii)

iii)

iv)

v)

vi)

Worked Examples

1. Find the values of sin 75°, cos 75° and tan 75°.

We know that sin (A + B) = sin A cos B + cos A sin B.

Put A = 45°, B = 30°

sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

i.e.,

We know that cos (A + B) = cos A cos B – sin A Sin B.

Put A = 45°, B = 30°

cos (45° + 30°) = cos 45° cos 30° – sin 45° sin 30°

Thus, we have

Put A = 45°, B = 30

2. Find the values of sin 15°, cos 15°, tan 15°

cos 15° = cos (45° – 30°) = cos 45° cos 30° + sin 45° sin 30°

3. Show that (i)

(ii)

i)

Divide both N & D by sin A sin B

ii)

Divide both N and D by sin A sin B

4. If find

i) sin (A + B) ii) cos (A + B) iii) sin (A – B) iv) cos (A – B).

Find also the quadrants in which the angles A + B and A – B lie.

(– ve sign is taken cosine is negative in II quadrant)

( B is in the III quadrant)

Thus we have sin A = 5/13 sin B = -3/5

cos A= -12/13 cos B = -4/5

i) sin (A + B) = sin A cos B + cos A sin B

=

ii) cos (A + B) = cos A cos B – sin A sin B

=

iii) sin (A – B) = sin A cos B – cos A sin B

=

iv) cos (A – B) = cos A cos B + sin A sin B

=

Now, since sin (A + B) and cos (A + B) are both positive, A + B is in the first quadrant. Since

sin (A – B) is negative and cos (A – B) is positive,

A – B is in the fourth quadrant.

5. Prove that sin (A + B) sin (A – B) = sin2A – sin

2B

LHS = sin (A + B) sin (A – B)

= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)

= sin2 A cos

2 B – cos

2 A sin

2 B

= sin2 A (1 – sin

2 B) – (1 – sin

2 A) sin

2 B

= sin2

A – sin2 A sin

2 - sin

2 B + sin

2 A sin

2 B

= sin2 A – sin

2 B = RHS

6. If show that (1 + tan A) 1 + tan B) = 2. Deduce that

tan A + tan B = 1 – tan A tan B …… (1)

LHS = (1 + tan A) (1 + tan B)

= 1 + tan A + tan B + tan A tan B

= 1 + (1 – tan A tan B) + tan A tan B using (1)

= 2

(1 + tan A) (1 + tan B) = 2 ……(2)

Put A = B in (2). (1 + tan A)2 = 2

i.e.,

But

7. If find tan B.

7 + 21 tan B = 3 – tan B tan B = -2/11

8. Show that cot 2θ + tanθ = cosec 2θ

SAQ 6: If then show that

4.8 Multiple and Sub-multiple angle

The angles 2A, 3A, 4A etc., are called multiple angles. And etc., are called

submultiple angles.

sin 2A = 2 sin A cos A

cos 2A = cos2 A – sin

2 A = 1 – 2 sin

2 A

= 2 cos A2 – 1

tan 2A =

To prove that

i)

ii)

iii) cot A – tan A = 2 cot 2A

iv) cot A + tan A = 2 cosec 2A

v)

i)

ii)

iii)

iv)

= 2 cosec 2A = RHS

v)

4.8.1 Function of half angles

To prove that i) sin 3A = 3 sin A – 4 sin3A ii) cos 3A = 4 cos

3 A – 3 cos A

iii)

i) sin 3A = sin(2A + A

= sin 2 A cos A + cos 2 A sinA

= (2 sin A cos A) cos A + (1 – 2 sin2 A) sin A

( sin 2A = 2 sin A cos A, cos 2A = 1 – 2 sin2 A)

= 2 sin A cos2 A + sin A – 2 sin3 A

= 2 sin A(1 – sin2 A) + sin A – 2 sin3 A

= 2 sin A – 2 sin3 a + sin A – 2 sin3 A

= 3 sin A – 4 sin3 A

sin 3A = 3 sin A – 4 sin3 A

ii) cos 3A = cos(2A + A)

= cos 2 A cos A – sin 2A sin A

= (2 cos2 A – 1) cos A – 2 sin A cos A sin A

( cos 2A = 2 cos2 A – 1, sin 2A = 2 sin A cos A)

= 2 cos3 A – cos A – 2 cos A sin

2 A

= 2 cos3 A – cos A – 2 cos A (1 – cos

2 A)

= 2 cos3 A – cos A – 2 cos A + 2 cos

3 A

= 4 cos3 A – 3 cos A

cos3A = 4 cos3 A – 3 cos A

iii) tan (3A) = tan(2A + A)

=

=

=

=

=

Worked Examples

1. Show that

(by dividing both N & D by cos A)

2. Show that

3. If show that tan 2A = tan B.

Give that

i.e.,

Now,

4. If show that

5. If show that i) ii)

i) sin θ =

=

[by dividing both N & D by

=

ii) cos =

=

=

=

Aliter:

In the where C = 90°,

AC = t, BC = 1

i)

ii)

SAQ 7: Show that

4.9 Summary

In this unit we study the concept of radian measure and its applications. The basics of Trigonometric

functions and trigonometric ratios of standard angles is discussed here with examples. Allied angles

and compound angles is a explained in a simple manner with illustrative examples wherever

necessary.

4.10 Terminal Questions

1. If ABC is any triangle show that

2. Find the value of θ in between 0° and 360° and satisfying the equation

3. Show that the value of tan 3 α cot α can not lie between and 3.

4. Show that sec (45° + A) sec [45° – A) = 2 sec 2A.

5. Show that

4.11 Answers

Self Assessment Questions

1. Angle through which the wheel turns in one minutes = 25 × 2π = 50π

Angle described in one sec

s = rθ = 5π/6 × 50 = 130.9 cm

Distance covered by a point in the rim in 1 sec is 130.9 cms.

2. Since cos = (4/5) and (adj/hyp), consider a right angled triangle ABC in which C = 90°, B

= , BC = 4, AB = 5; clearly AC = 3.

cos = 4/5, sin = 3/5

cosec = 5/3, cot = 4/3

Substituting these in the given expression

3. ABC is a triangle in which and

Since cos

BC = m2 – n

2 and AB = m

2 + n

2

AC2 = AB

2 – BC

2

= (m2 + n

2)

2 – (m

2 – n

2)2

= m4

+ n4 + 2 m

2n

2 – (m

4 + n

4 – 2 m

2n

2)

= 4m2n

2

Now

4. LHS = sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°

5.

x = 5, r = 13

6.

(Divide Nr and Dr. by cos2 a)

LHS =

7.

Terminal Questions

1. In any triangle ABC we have A + B + C = 180°

Thus,

2.

We know that

Consider

…(1)

…(2)

From (1) and (2) it follows that θ = 150°, θ = 330°

3.

Since tan2 α is positive, either so x cannot lie between

4.

= sec (45° + A) cosec (45° + A)

=

=

5. LHS =

=

=

=

BT0063-Unit-05 Limits and Continuity

Unit 5 Limits and Continuity

Structure

5.1 Introduction

Objectives

5.2 The Real Number System

5.3 The Concept of Limit

5.4 Concept of Continuity

5.5 Summary

5.6 Terminal Questions

5.7 Answers

5.1 Introduction

In this chapter you will be recalling the properties of number. You will be studying the limits of

a function of a discrete variable, represented as a sequence and the limit of function of a real

variable. Both these limits describe the long term behaviour of functions. You will be studying

continuity which is essential for describing a process that goes on without abrupt changes. You

will see a good number of examples for understanding the concepts clearly. As mathematics is

mastered only by doing, examples are given for practice.

You are familiar with numbers and using them in day – to – day life. Before introducing the

concept of limits let us refresh our memory regarding various types of numbers.

Objectives:

At the end of the unit you would be able to

• understand the concept of limit.

• apply the concept of continuity in problems.

• find whether a given function is continuous or not.

5.2 The Real Number System

You are using numbers like etc. The last two numbers – 1

+ i and 2 – 3i are complex numbers. The rest of them are real numbers.

The numbers 1, 2, 3, …… are called natural numbers.

N = {1, 2, 3, ……..}

The numbers …………………….. – 3, –2, –1, 0, 1, 2, 3, …… are integers.

or

The set of quotients of two integers, the denominator not equal to 0 are called rational

numbers and the set of rational numbers is denoted by Q.

Usually these numbers are represented as points on a horizontal line called the real axis. (Refer

to Fig. 5.1)

Fig. 5.1 Representation of numbers

After representing the integers and rational numbers. So there are no gaps in the real line and

so it is called “continuum”.

We can also represent the relations “greater than” or “less than” geometrically. If a < b, then a

lies to the left of b in the real line (and b lies to the right of a).

The modulus function

The modulus function simply represents the numerical value of a number. It is defined as

follows:

For example,

Note:

SAQ: (Self Assessment Questions).

Choose the right answer

1. If a > b, then

A) Positive

B) Negative

C) Zero

2. Choose the right answer

is equal to

A)

B) 0

C) 2 (a – b)

D) 2 (b – a)

An Important Logical Symbol

In Mathematics, we use symbols instead of sentences. For example, “3 is greater than 2” is

written as 3 > 2. Throughout the test we used the symbol ⇒ (read as “implies”)

If x > 2, then x > 4 is written as (x > 2) (2x > 4).

Generally ‘If P, then Q” is written as

P Q. (P is given and Q is the conclusion)

The distance function

If a and b are two real numbers. Then the distance between a and b is defined as | a – b|. Refer

Figure 5.2. Why we choose |a – b| as the distance between a and b should be clear from figure

5.2. When b > a, then the distance is b – a; when a > b, it is a – b. Both a – b and b – a are equal

to

|a – b|. So wherever a and b are on the real line, the distance is |a – b|.

Figure 5.2 Distance function

The distance function satisfies the following properties.

1.

2.

3.

5.3 The Concept of Limit

The concept of limit is an important concept in Mathematics, which is used to describe the long

term behaviour of a phenomenon. You might have heard about the half life of a radioactive

substance. It is the time required for a radioactive substance to lose half is radioactivity. This is

used in carbon dating. Carbon dating is a method of calculating the age of a very old object by

measuring the radioactive carbon it contains. Thus the long term state of an old object is

described by the concept of limit.

Function of a discrete variable and a continuous variable

The Concept of limit is associated with functions. A function from a set A to a set B is a rule

which assigns, to each element of A a unique (one and only one) element of B. Examples of

functions are the marks of students in a class of 100 in a particular subject or the B.P. or sugar

level in blood of a particular student in the class.

There are two types of functions. The first is a function from the set N of natural numbers (i.e.,

N = {1, 2, 3,….}. If the students in a class are numbered as 1, 2, ……., 100.

Then the mark in a particular subject is a function from {1 2, ….., 100} to {0, 1, 2, …., 100) if the

marks are given as percentages. This is called a function of a discrete variable. (Don’t be afraid

of the term variable. It is a simply a symbol which can be replaced by value you choose).

Definition A function of a discrete variable is a function from N or a subset of N to the set R of

all real numbers.

The second type of functions refer to functions from R to R. It is called a function of a

continuously. So it can be treated as a function of a continuous real variable t, t denoting time.

Definition: A function of a continuous real variable or simply a function of a real variable is a

function from R to R. (or [a, b]). Here [a, b] denote the set of all real numbers between the

numbers a and b).

Functions of a discrete variable

We have defined a function of a discrete variable as a function from N

{1, 2, 3, …} or subset of N to the set R of real numbers. A convenient way to representing this

function is by listing the images of 1, 2, 3, etc. If f denotes the function then the list.

F(1), F(2), F(3), ….. ………………… (*)

Represents the function f usually f(1), f(2) are written as a1, a2 etc.

The list given in (*) is called a sequence.

In a sequence the order of the elements appearing in it is important. A common example of a

sequence is a queue you see in a reservation counter. Then a1 is the person standing in front of

the counter getting his reservation done. a2 is the person behind a1 etc. the order of persons in the

queue is important. You won’t certainly be happy if the order of the persons in the queue is

changed.

The limit of a sequence

From the above discussion, two points should be clear to you.

1. A sequence is an arrangement of real numbers as the first element, second element etc.

2. A sequence represents a function of a discrete variable.

We denote a sequence by (an) and an denotes the nth term.

Assume that you have a string of length 1 cm. Denote it by a1. Cut the string into two halves and

throw away one half. Denote the remaining half by a2. Then . Repeat the process

indefinitely.

Then etc. After (n + 1) repetitions, you are left with a string of length

Intuitively you feel that the string becomes smaller and smaller and you are left

with a string whose length is nearly zero in the long run. At the same time you realize that you

will have “some bit” of positive length at any time. Also you can make the string as small as you

please provided you repeat the process sufficient number of times. In this case we say that “an

tends to 0 as n tends to infinity” “an tends to 0” means is as small as we please/ “n lends

to infinity” means we repeat the process sufficient number of times.

Now we are in a position to define the limit of a sequence (an)

Definition: Let (an) be a real sequence. Then (an) tends to a number a, if given a positive number

∈, (pronounced as epsilon), there exists a natural number n0 such that

for all ……………. (1.1)

In this case we write We also say (an) converges to a.

Note: is the numerical value of an – a. For example | 2 | = 2 and

| – 3 | = 3, n0 is a “stage”.

n > n0 means after a certain stage simply means that an comes as close to a as we

choose.

Example: Show that

Solution:

Let ∈ be a given positive number.

when .

Let n0 be the smallest natural number

(For example, if take n0 = 148). Then

If n > n0, then Hence when

This proves (1.1) with 0 in place of a.

Hence

Example: Show that where .

Solution: As in Example,

For we require

By choosing n0 to be the smallest natural number greater than , we see that (1.1) is

satisfied for n0.

Hence

You can see that several similar sequences tend to 0. Some of them are

…… (1.2)

Algebra of limits of sequences

If (an) and (bn) are two sequences, then we can get a new sequence by “adding them”. Define cn

= an + bn. Then (cn) is a sequence and we can write (cn) = (an + (bn). We can also subtract one

sequence (bn) from another sequence (an), multiply two sequences etc. We can also multiply a

sequence (an) by a constant k.

Let us answer the following questions.

1. What happens to the limit of sum of two sequences ?

2. What happens to the limit of difference, product of two sequences ?

We summarize the results as a theorem.

Theorem(∗∗∗∗): It (an) and (bn) and two sequences converging to a and b respectively, then

a) (an + bn) → a + b

b) (an – bn) → a – b

c) (kan) → ka

d) (anbn) → ab

e) provided bn 0 for all n and b 0.

Proof: We prove only a) ∈ > 0 be a positive number.

As (an) → a, we can apply (5.5) by taking in place of ∈. Thus we get a natural number n0

such that

for all ………….. (1.3).

Similarly, using the convergence of (bn), we an get n1 such that

for all …………….. (1.4).

Let m = maximum of n0 and n1. Then (5.3) and (5.4) are simultaneously true for n > m.

Thus,

for all ………………… (1.5)

From (1.5) we get

for n > m.

Thus (1.1) holds good for the sequence (an + bn) in place of (an) and a + b (in place of a)

Hence (an + bn) → a + b

Note: The choice of m may puzzle you. When for all n > 1000, then certainly

for all n > 1001, 1002 etc. So for all n > m, m being greater than

1000.

Remark: The other subdivisions can be proved similarly. As you are more interested in

applications you need not get tied down by the technical details of the proof. In mathematics,

there is a “commandment”, Though Shall Not Divide By Zero”. If bn = 0, then is not

defined. So is when b = 0. So when you apply (2), see to it that the conditions are satisfied.

(bn ≠ 0 and

b 0). With this theoretical foundation, you are in a position to find limits of sequences.

Worked Examples

W.E.: Show that a constant sequence is convergent (A sequence

(A sequence (an) is a constant sequence if an = k for all n).

Solution an = k for all n. Consider .

As 0 < ∈, for every positive number ∈, for all n > 1, that is, n0 = 1 for all ∈ > 0. So a constant

sequence converges to its constants value.

W.E.: Find the nth term of the sequence

Solution: To discover a pattern in the terms of the sequence start from the third term.

So a positive choice is when an is written in this way

etc.

To find the limit of the sequence (if it exists), with

Taking and cn = 1, we get an = 2bn + cn

As and

. Hence the given sequence converges to 1.

W.E.: Evaluate

Solution As we know that and we try to write the nth

term of the given

sequence in terms of

As and (1) → 1,

Similarly,

By theorem (∗∗∗∗),

Note: It is difficult to prove that a given sequence is not convergent. For proving convergence

we start with ∈ > 0 and find a stage no satisfying (∗). To prove that a sequence is not convergent

we have to prove that (∗) does not hold good for every stage for a particular ∈ > 0 and this is

certainly difficult. However we can prove that certain sequences are not convergent indirectly as

the following example shows.

W.E.: Show that the sequence 1, –1, 1, –1, 1, –1 , …… is not convergent.

Solution: We can write the sequence as (an) where

Suppose (an) → a for some real number a. the number has to satisfy one and only one of the

following conditions: a < –1, –1 < a < 1, a > 1.

(See Fig. 5.3 representing these cases).

Fig. 5.3: Illustration of W.E.

In case if n is odd. So we cannot prove (∗) in this case

(for ∈ > 2).

In case if n is even and we cannot prove (∗) for ∈ > 2.

In case 2, if a is closer to 1, then for even n. If a is close to –1, then for

odd n. If a = 1, then for even n. So (∗) cannot hold good for ∈ > 2 or ∈ = 2. So the

given sequence is not convergent.

Try to answer the following questions before you proceed to the next section.

S.A.Q. 3: Which of the following sets can be arranged as a sequence?

a) The passengers in a 3 ties coach

b) The people attending a meeting in a beach

c) The people living in Karnataka

d) The students of I M.Sc. Biotechnology in a college

The Limit of a Function of a Real Variable

You are now familiar with natural numbers and real numbers. The natural numbers appear as

“discrete” points along the real line and we are able to fix some element say 1 as the first natural

number, 2 as the second natural number etc. So the natural numbers appear as the terms of a

sequence. But it is not possible to arrange the real numbers as a sequence. If a real number a is

the nth

element and b is the (n+1)th

element, where will you place . It appears between the

nth

element and the (n+1)th

element. If you take as the (n+1)th element, where will you

place So you feel initiatively that real numbers can not be arranged as a

sequence.

When we consider numbers between a and b. We consider points lying between the points

representing the numbers a and b. The numbers lying between two numbers a and b from an

“interval”. So “interval” on the real line is the basic concept. Usually we define a function of a real

variable on an interval. We define various types of “intervals” as follows (refer to Table 5.1)

Table 5.1 Intervals

Note: Here (a represents inclusion of all numbers > a. [a represents the inclusion of all numbers

> a].

is not a number. It simply represents the inclusion of all “large” –ve numbers.

represents the inclusion of all “large” positive real numbers.

(a, b) and [a, b] are called open interval and closed interval.

So it is natural to represent R as the interval

S.A.Q.4

a) Find all natural numbers in the intervals [3, ∞), (3, ∞), (– ∞, 3) and

(– ∞, 3]

b) Find all integers lying in the intervals given in a

c) Find all numbers in [2, 2], (2, 2)

Example; Represent as an interval

Solution: represents two inequalities x – 3 < 2, – (x – 3) < 2.

When x – 3 < 2, x < 5

When – (x – 3) < 2, x – 3 > – 2, or x > 3 – 2 = 1

Hence the given set is {1, 5}

We can also arrive at this interval geometrically (see Fig 5.4)

Figure 5.4

S.A.Q.5: Represent the sets ,

as intervals.

Now we have enough background to define the limit of a function f of a real variable x. In the

case of functions of a discrete variable or sequence, we defined . This limit

represented the long term behaviour of f. In the case of a function of a real variable, we can

discuss the behaviour of f(x), when the variable x comes close to a real number a. In other ways

we will be defining We want to write the statement “x

comes close to a” rigorously. The geometric representation of real numbers can be used for this

purpose. When do you say that your house is near your college? When the distance between your

house and your college is small. In the same way, we can say that “x is close to “when | x – a| is

small. If “smallness” is defined by a distance of say 0.1, then x is close to a if

|x – a| < 0.1 of course the measure of “smallness” is relative. For a person living in Mangalore,

Manipal is not near Mangalore. For a person living in US, Mangalore and Manipal are near to

each other. So “smallness” is decided by the choice of a positive number ∈ (This was done in

defining the limit of a sequence also)

Before giving a rigorous definition of limit, let us consider two examples.

Consider F(x) = 1 + x. Let us try to see what happens when x is close to 1. We evaluate f(x),

when x = 0.9, 0.99, 0.999, 1.1, 1.01, 1.001

F(0.9) = 1.9 f(0.99) = 1.99 f(0.999) = 1.999

F(1.1) = 2.1 f(1.01) = 2.01 f(1.001) = 2.001

Note all these values are near the value 2.

Consider another function

(Why don’t we define g(1) ? If we put x = 1 in then we get which is not defined).

As in the case of f(x), we compute some function values.

g(0.999) = 1.999 g(1.1) = 2.1

g(1.01) = 2.01 g(1.001) = 2.001

Note: These values are close to 2. Hence we can say that x close to

1 both f(x) and g(x) are closed to 2 and we can take 2 as

or .

Now let us formulate a rigorous definition of

Definition: Let f be a function of a real variable. Then if given a positive number ∈

There exists a positive number δ such that

…………. (1.6)

Let us analyze the definition.

We have two choice of positive numbers (∈ and δ) and two conditions

Given any positive number ∈ there exists a positive number δ

such that the condition P Implies the condition

The choice of δ depends on the given number ∈. The function f need not be defined at x = a.

The condition P says that x is close to a.

The condition Q says that f(x) is close to I

We can also express the definition geometrically.

Given ∈ > 0, there exists δ > 0 such that

In figure 5.5, the point (a, I) is measured as 0, meaning that the functional value of f at x = a is

not known or defined.

Figure 5.5: Definition of limit of function

The images of points in under the function f is a subset of the interval (I –

∈, I + ∈) along the vertical axis.

W.E.: Evaluate

Solution: Here f(x) = 2x – 6

Choose any ∈ > 0. We have to choose a δ such that (5.5) is satisfied. We have to guess the value

of I. When x comes close to 3.2x – 6 should come close to 2(3) – 6 = 0. Take I = 0. Then

So, choose then

Hence,

Note: f(3) = 2(3) – 6 = 0. In this case the function f is defined at x = 3 and f(3) coincides with

.

W.E.: Evaluate

Solution: Let . As in the previous problem, we can guess the value of I it is

Hence for a given , the corresponding δ is chosen as and condition (1.6) holds good.

Hence

W.E.: Evaluate

Solution: Proceed as in the previous example. In this problem and

General Remark While evaluating if f(a) is defined or it is not of the form and f(x)

is defined by a single expression, it will turn out that If f(a) is not defined or f is

given by two different expressions. Then we have to guess the value of the limit and prove

condition (1.6).

In some problems, f(x) may be given as quotient of two expressions but it may reduce to an

easier function on simplification. In such cases the problem will reduce to an easier one.

W.E.: Evaluate

Solution: When x = 2, f(x) is of the form So we try to see when x – 2 is a factor of

2x2 – 3x – 2, 2x

2 – 3x – 2 = 2x

2 – 4x + x – 2 = 2x (x – 2) + 5 (x – 2) = (2x + 5) (x – 2).

When x → 2, x does not assume the value 2. So,

= 2x + 1 on canceling x – 2, since x – 2 ≠ 0.

So the given limit reduces to

(as in W.E.)

Algebra of Limits of Functions

It is not necessary that we use the ∈–δ definition for every problem. We study important

properties of limits of functions as a theorem. We can evaluate limits using this theorem (noted

as a proposition).

Proposition:

a)

b) , where k is a constant

c)

d)

e)

f) when a > 0

Theorem: Let k be a constant, f and g functions having limit at a and n a positive integer. Then

the following hold good.

a)

b)

c)

d)

e)

f) provided

g)

h) provided is positive

You need not prove these results. It is enough if you clearly understand the theorem and

proposition and apply them for evaluating limits.

Self Assessment Questions

SAQ 6: Evaluate the following limits

a) b) c)

d) e) f)

g) h)

SAQ 7: Evaluation the following limits

a) b) c)

d) e) f)

g) h)

S.A.Q.8 If evaluate the following

a) b)

c) d)

e) if m is positive f) if I, m > 0.

5.4 Concept of Continuity

In mathematics and sciences, we use the word “continuous” to describe a process that goes on

without abrupt changes. For example, the growth of a plant, the water level in a tank and the

speed of a moving car in a four-base highway are exhibiting continuous behaviour.

Before defining continuous functions, let us look at the graphs of three functions.

Figure 5.6: Two discontinuous functions and a continuous function

The first graph has a break at x = a in the second graph also there is a break at x = a. If you

ignore the point corresponding to x = a, there is no break for the break occurs at x = a the third

function has no break. So it should be intuitively clear to you that the first two functions are not

continuous while the third function is continuous.

Let us formulate a rigorous definition of continuity.

Definition: Let f be a function of a real variable defined in an open interval containing a. Then f

is continuous at a if

Note: In order to define continuity at a, we need three conditions.

1) exists

2) f(a) is defined

3)

Even if one of them fails, then the function f is not continuous at a.

Now look at Fig. 5.4. The first function say f has no limit at a, i.e., does not exist. For

the second function, exists but The third function is continuous.

Example: Define f as follows:

Is f continuous at 2?

Solution:

= 4

Note: We can cancel (x – 2) since it is non zero.

So exists.

But f(2) = 3 ≠ 4

Hence the function f is not continuous 2.

Example: Define f as follows.

Is f continuous at 2?

Solution: From above example, we have A s f(2) = 4, f is continuous at 2.

Sometimes function may be defined by two different expressions. In such cases the following

method of proving continuity will be useful. For that we need the concept of left limit and right

limit.

Definition: Let a function f be defined in an interval (b, a) where b < a). Then

(called the left limit) if given ∈ > 0. There exists δ > 0 such that

………… (1.7).

Definition; Let a function f be defined in an interval (a, b), where b > a. Then

called the right limit if given ∈ > 0, there exists δ > 0 such that

………….. (1.8)

Note: We can prove that (1.7) and (1.8) implies (1.6). Hence if the left and right limits exist and

are equal then

When a function is defined by two different expressions, we have to evaluate the left and right

limits.

if both the left and right limits exist and are equal.

Worked Examples:

W.E.: Test whether f is continuous at x = 3 where f is defined by

Solution: As f is defined by two expressions one for and another for (3, ∞), we evaluate

the left and right limits.

As the left and right limits are not equal, does not exist. Hence f is not continuous at 3.

When a function f is continuous at every point of an interval (a, b) we say that the function is

continuous on (a, b). In particular if a function is continuous at every real number. Then we say

that a function is continuous on R.

Using propositions, we can prove that the functions, k (a constant), x, x2, ……..x

n, where n > 2

are continuous on R.

The function is continuous on (0, ∞)

The function is continuous on

Using previous theorem, we can prove the following theorem.

Theorem:

a) If f and g are continuous at a, then

i) f(x) + g(x) ii) f(x) – g(x) iii) f(x) g(x) are continuous at a

b) If f and g are continuous at a and g(a) ≠ 0, then is continuous at a

c) If f is continuous at a, f(x) > 0 for x in an open interval containing a, is continuous

at a.

Worked Examples

W.E.: Test the continuity of the function f at all real points where f is defined by

Solution: If a < 3, then f(x) is defined by the expression x2 in an open interval containing a. So f

is continuous for all a < 3.

So it remains to test continuity only at 3.

As does not exist. So f is not continuous at 3.

Thus f is continuous at all real points except 3.

W.E.: Test the continuity of the function f where f is defined by

Solution: When x < 2, |x – 2| is negative. So |x – 2| = – (x – 2)

When x > 2, |x – 2| is positive. So |x – 2| = x – 2

f(2) = 0. Thus

As in the previous worked example, f is continuous for all a < 2 and all a > 2.

As does not exist. So f is continuous at all points except 2.

Not all functions are continuous. In some cases may not exist. You may ask a question:

how to establish that does not exist. One sample case is where the left and right limits

exist but are not equal. We have examples for this case. The worst case is when neither of the two

limits exist. For proving the non-existence of limits, we use the following theorem.

Theorem: A function f is continuous at a if and only if the following conditions holds good.

To prove the non existence of the limit it is enough to construct a sequence (xn) converging to a

such that f(xn) does not converge to f(a).

W.E. : Show that the function f defined by

is not continuous at 0.

Solution: Let Then (obvious) f(xn) = n and so f(xn) does not converge to 0

since f(xn) indefinitely increases and so cannot approach 0.

Self Assessment Questions

S.A. Q. 9 Verify whether the following functions f is continuous at a

a)

b)

c)

d)

e)

f)

g)

h)

S.A.Q. 10 Show that the function f defined by

is not continuous at 1.

S.A.Q. 11 Show that the following functions are continuous on R.

a)

b)

c)

d)

5.5 Summary

In this unit we studied the basics of real number system then the concept of limit was discussed

which was further extended to the concept of continuity. All definitions and properties of the

above mentioned concepts is given very clearly with sufficient number of examples wherever

necessary.

5.6 Terminal Questions

1. Find all natural numbers in the following intervals

2. Find when

a) b) c)

d) e)

3. Show does not exist when

a) an = n b) an = 2n c) an = n!

4. Evaluate when

a) b) c)

d) e) f)

g) h) i) j)

k)

5. Evaluate when f(x) is equal to

a) b) c)

d) e) f)

6. Evaluate the following limits

a) b)

c) d)

e) f)

g) h)

7. If and evaluate

a) b)

c) (when I, m, > 0)

d) when m > 0

8. Show that the following functions are continuous at a

a) for all x in R, a = 0

b) for all x in R a = 0

c) for all x in R a = 0

9. Show that the following functions are continuous at a

a)

b)

c)

10. Show that the following functions are not continuous at a

a)

b)

c)

d)

e)

5.7 Answers

Self Assessment Questions

1. A

2. A

3. a) and d) can be arranged as a sequence according to chart and attendance

register

4. a) {3, 4, 5, …..} b) {4, 5, 6, ……}

c) {……. –3, –2, –1, 0, 1, 2,} d) {…. –3, –2, –1, 0, 1, 2, 3}

5. [1, 5], (–∞, 1) ∪ (5, ∞), (–∞, 1] ∪ [5, ∞)

6. a) 1 b) 1 c) 100 d)

e) 1 f) –1 g) 2 h) 9

7. a) 3 b) c) 2 d) 4

e) 3 (Hint: x3 – 1 = (x – 1) (x

2 + x + 1)) f) g) 1

h)

8. a) 2I + 3m b) I2 – m

2 c) d)

e) f)

9. a), b), c), d) continuous e) discontinuous f), g) continuous,

d) not continuous

10. Take but f(xn) = n f(xn) does not converge. So f is not continuous at

1.

11. a) For a < 1, f(x) = x2 – 6x + 5. So f is continuous for a < 1. As f(x) = 2x

2 – 2, f is

continuous for a > 1.

Hence

Also f(d) = 1 – 6 + 5 = 0

b), c), d) Similar.

Terminal Questions

1. a) {4, 5, 6, 7, 8} b) {5, 6, 7, 8, 9} c) {5, 6, 7, 8}

d) {4, 5, 6, 7, 8, 9}

e) {4, 5, 6, …..} f) 10, 11, 12,…..} g) {9, 10, 11, …..} h) {9, 10, 11, …..

}

2. a) 0 b) 1 c) 0 d) 0 e) 0

3. a) As n increases an increases. So cannot be made less than a fixed number ∈.

So does not exists.

4. a) b) 1 c) 3 d)

e) – 1 f) 1 g) h)

i) Write an as

j)

k)

5.

a) 2a3 b) a

3 + a

4 c) 2a

3 – a

4 + a

5 d) (2 + 3a) (4 + 5a

2)

e) f)

6. a) 2(–1)2 – 1 = 1 b) (2 + 1) (3 – 2) = 3 c) d)

e) f) g) 1

h) Hence answer is

7. a) Im b) (l + 2m) (2I – m) c) d)

10. a) f(2) = 7. So f is not continuous at x=2.

b) But f(1) = 1

c) Let limit at 1 = 8; right limit at 1 = 2

d) The left and right limits are 1 and –1 respectively.

e) The left and right limits are –1 and 1.

BT0063-Unit-05 Limits and Continuity

Unit 5 Limits and Continuity

Structure

5.1 Introduction

Objectives

5.2 The Real Number System

5.3 The Concept of Limit

5.4 Concept of Continuity

5.5 Summary

5.6 Terminal Questions

5.7 Answers

5.1 Introduction

In this chapter you will be recalling the properties of number. You will be studying the limits of

a function of a discrete variable, represented as a sequence and the limit of function of a real

variable. Both these limits describe the long term behaviour of functions. You will be studying

continuity which is essential for describing a process that goes on without abrupt changes. You

will see a good number of examples for understanding the concepts clearly. As mathematics is

mastered only by doing, examples are given for practice.

You are familiar with numbers and using them in day – to – day life. Before introducing the

concept of limits let us refresh our memory regarding various types of numbers.

Objectives:

At the end of the unit you would be able to

• understand the concept of limit.

• apply the concept of continuity in problems.

• find whether a given function is continuous or not.

5.2 The Real Number System

You are using numbers like etc. The last two numbers – 1

+ i and 2 – 3i are complex numbers. The rest of them are real numbers.

The numbers 1, 2, 3, …… are called natural numbers.

N = {1, 2, 3, ……..}

The numbers …………………….. – 3, –2, –1, 0, 1, 2, 3, …… are integers.

or

The set of quotients of two integers, the denominator not equal to 0 are called rational

numbers and the set of rational numbers is denoted by Q.

Usually these numbers are represented as points on a horizontal line called the real axis. (Refer

to Fig. 5.1)

Fig. 5.1 Representation of numbers

After representing the integers and rational numbers. So there are no gaps in the real line and

so it is called “continuum”.

We can also represent the relations “greater than” or “less than” geometrically. If a < b, then a

lies to the left of b in the real line (and b lies to the right of a).

The modulus function

The modulus function simply represents the numerical value of a number. It is defined as

follows:

For example,

Note:

SAQ: (Self Assessment Questions).

Choose the right answer

1. If a > b, then

A) Positive

B) Negative

C) Zero

2. Choose the right answer

is equal to

A)

B) 0

C) 2 (a – b)

D) 2 (b – a)

An Important Logical Symbol

In Mathematics, we use symbols instead of sentences. For example, “3 is greater than 2” is

written as 3 > 2. Throughout the test we used the symbol ⇒ (read as “implies”)

If x > 2, then x > 4 is written as (x > 2) (2x > 4).

Generally ‘If P, then Q” is written as

P Q. (P is given and Q is the conclusion)

The distance function

If a and b are two real numbers. Then the distance between a and b is defined as | a – b|. Refer

Figure 5.2. Why we choose |a – b| as the distance between a and b should be clear from figure

5.2. When b > a, then the distance is b – a; when a > b, it is a – b. Both a – b and b – a are equal

to

|a – b|. So wherever a and b are on the real line, the distance is |a – b|.

Figure 5.2 Distance function

The distance function satisfies the following properties.

1.

2.

3.

5.3 The Concept of Limit

The concept of limit is an important concept in Mathematics, which is used to describe the long

term behaviour of a phenomenon. You might have heard about the half life of a radioactive

substance. It is the time required for a radioactive substance to lose half is radioactivity. This is

used in carbon dating. Carbon dating is a method of calculating the age of a very old object by

measuring the radioactive carbon it contains. Thus the long term state of an old object is

described by the concept of limit.

Function of a discrete variable and a continuous variable

The Concept of limit is associated with functions. A function from a set A to a set B is a rule

which assigns, to each element of A a unique (one and only one) element of B. Examples of

functions are the marks of students in a class of 100 in a particular subject or the B.P. or sugar

level in blood of a particular student in the class.

There are two types of functions. The first is a function from the set N of natural numbers (i.e.,

N = {1, 2, 3,….}. If the students in a class are numbered as 1, 2, ……., 100.

Then the mark in a particular subject is a function from {1 2, ….., 100} to {0, 1, 2, …., 100) if the

marks are given as percentages. This is called a function of a discrete variable. (Don’t be afraid

of the term variable. It is a simply a symbol which can be replaced by value you choose).

Definition A function of a discrete variable is a function from N or a subset of N to the set R of

all real numbers.

The second type of functions refer to functions from R to R. It is called a function of a

continuously. So it can be treated as a function of a continuous real variable t, t denoting time.

Definition: A function of a continuous real variable or simply a function of a real variable is a

function from R to R. (or [a, b]). Here [a, b] denote the set of all real numbers between the

numbers a and b).

Functions of a discrete variable

We have defined a function of a discrete variable as a function from N

{1, 2, 3, …} or subset of N to the set R of real numbers. A convenient way to representing this

function is by listing the images of 1, 2, 3, etc. If f denotes the function then the list.

F(1), F(2), F(3), ….. ………………… (*)

Represents the function f usually f(1), f(2) are written as a1, a2 etc.

The list given in (*) is called a sequence.

In a sequence the order of the elements appearing in it is important. A common example of a

sequence is a queue you see in a reservation counter. Then a1 is the person standing in front of

the counter getting his reservation done. a2 is the person behind a1 etc. the order of persons in the

queue is important. You won’t certainly be happy if the order of the persons in the queue is

changed.

The limit of a sequence

From the above discussion, two points should be clear to you.

1. A sequence is an arrangement of real numbers as the first element, second element etc.

2. A sequence represents a function of a discrete variable.

We denote a sequence by (an) and an denotes the nth term.

Assume that you have a string of length 1 cm. Denote it by a1. Cut the string into two halves and

throw away one half. Denote the remaining half by a2. Then . Repeat the process

indefinitely.

Then etc. After (n + 1) repetitions, you are left with a string of length

Intuitively you feel that the string becomes smaller and smaller and you are left

with a string whose length is nearly zero in the long run. At the same time you realize that you

will have “some bit” of positive length at any time. Also you can make the string as small as you

please provided you repeat the process sufficient number of times. In this case we say that “an

tends to 0 as n tends to infinity” “an tends to 0” means is as small as we please/ “n lends

to infinity” means we repeat the process sufficient number of times.

Now we are in a position to define the limit of a sequence (an)

Definition: Let (an) be a real sequence. Then (an) tends to a number a, if given a positive number

∈, (pronounced as epsilon), there exists a natural number n0 such that

for all ……………. (1.1)

In this case we write We also say (an) converges to a.

Note: is the numerical value of an – a. For example | 2 | = 2 and

| – 3 | = 3, n0 is a “stage”.

n > n0 means after a certain stage simply means that an comes as close to a as we

choose.

Example: Show that

Solution:

Let ∈ be a given positive number.

when .

Let n0 be the smallest natural number

(For example, if take n0 = 148). Then

If n > n0, then Hence when

This proves (1.1) with 0 in place of a.

Hence

Example: Show that where .

Solution: As in Example,

For we require

By choosing n0 to be the smallest natural number greater than , we see that (1.1) is

satisfied for n0.

Hence

You can see that several similar sequences tend to 0. Some of them are

…… (1.2)

Algebra of limits of sequences

If (an) and (bn) are two sequences, then we can get a new sequence by “adding them”. Define cn

= an + bn. Then (cn) is a sequence and we can write (cn) = (an + (bn). We can also subtract one

sequence (bn) from another sequence (an), multiply two sequences etc. We can also multiply a

sequence (an) by a constant k.

Let us answer the following questions.

1. What happens to the limit of sum of two sequences ?

2. What happens to the limit of difference, product of two sequences ?

We summarize the results as a theorem.

Theorem(∗∗∗∗): It (an) and (bn) and two sequences converging to a and b respectively, then

a) (an + bn) → a + b

b) (an – bn) → a – b

c) (kan) → ka

d) (anbn) → ab

e) provided bn 0 for all n and b 0.

Proof: We prove only a) ∈ > 0 be a positive number.

As (an) → a, we can apply (5.5) by taking in place of ∈. Thus we get a natural number n0

such that

for all ………….. (1.3).

Similarly, using the convergence of (bn), we an get n1 such that

for all …………….. (1.4).

Let m = maximum of n0 and n1. Then (5.3) and (5.4) are simultaneously true for n > m.

Thus,

for all ………………… (1.5)

From (1.5) we get

for n > m.

Thus (1.1) holds good for the sequence (an + bn) in place of (an) and a + b (in place of a)

Hence (an + bn) → a + b

Note: The choice of m may puzzle you. When for all n > 1000, then certainly

for all n > 1001, 1002 etc. So for all n > m, m being greater than

1000.

Remark: The other subdivisions can be proved similarly. As you are more interested in

applications you need not get tied down by the technical details of the proof. In mathematics,

there is a “commandment”, Though Shall Not Divide By Zero”. If bn = 0, then is not

defined. So is when b = 0. So when you apply (2), see to it that the conditions are satisfied.

(bn ≠ 0 and

b 0). With this theoretical foundation, you are in a position to find limits of sequences.

Worked Examples

W.E.: Show that a constant sequence is convergent (A sequence

(A sequence (an) is a constant sequence if an = k for all n).

Solution an = k for all n. Consider .

As 0 < ∈, for every positive number ∈, for all n > 1, that is, n0 = 1 for all ∈ > 0. So a constant

sequence converges to its constants value.

W.E.: Find the nth term of the sequence

Solution: To discover a pattern in the terms of the sequence start from the third term.

So a positive choice is when an is written in this way

etc.

To find the limit of the sequence (if it exists), with

Taking and cn = 1, we get an = 2bn + cn

As and

. Hence the given sequence converges to 1.

W.E.: Evaluate

Solution As we know that and we try to write the nth

term of the given

sequence in terms of

As and (1) → 1,

Similarly,

By theorem (∗∗∗∗),

Note: It is difficult to prove that a given sequence is not convergent. For proving convergence

we start with ∈ > 0 and find a stage no satisfying (∗). To prove that a sequence is not convergent

we have to prove that (∗) does not hold good for every stage for a particular ∈ > 0 and this is

certainly difficult. However we can prove that certain sequences are not convergent indirectly as

the following example shows.

W.E.: Show that the sequence 1, –1, 1, –1, 1, –1 , …… is not convergent.

Solution: We can write the sequence as (an) where

Suppose (an) → a for some real number a. the number has to satisfy one and only one of the

following conditions: a < –1, –1 < a < 1, a > 1.

(See Fig. 5.3 representing these cases).

Fig. 5.3: Illustration of W.E.

In case if n is odd. So we cannot prove (∗) in this case

(for ∈ > 2).

In case if n is even and we cannot prove (∗) for ∈ > 2.

In case 2, if a is closer to 1, then for even n. If a is close to –1, then for

odd n. If a = 1, then for even n. So (∗) cannot hold good for ∈ > 2 or ∈ = 2. So the

given sequence is not convergent.

Try to answer the following questions before you proceed to the next section.

S.A.Q. 3: Which of the following sets can be arranged as a sequence?

a) The passengers in a 3 ties coach

b) The people attending a meeting in a beach

c) The people living in Karnataka

d) The students of I M.Sc. Biotechnology in a college

The Limit of a Function of a Real Variable

You are now familiar with natural numbers and real numbers. The natural numbers appear as

“discrete” points along the real line and we are able to fix some element say 1 as the first natural

number, 2 as the second natural number etc. So the natural numbers appear as the terms of a

sequence. But it is not possible to arrange the real numbers as a sequence. If a real number a is

the nth

element and b is the (n+1)th

element, where will you place . It appears between the

nth

element and the (n+1)th

element. If you take as the (n+1)th element, where will you

place So you feel initiatively that real numbers can not be arranged as a

sequence.

When we consider numbers between a and b. We consider points lying between the points

representing the numbers a and b. The numbers lying between two numbers a and b from an

“interval”. So “interval” on the real line is the basic concept. Usually we define a function of a real

variable on an interval. We define various types of “intervals” as follows (refer to Table 5.1)

Table 5.1 Intervals

Note: Here (a represents inclusion of all numbers > a. [a represents the inclusion of all numbers

> a].

is not a number. It simply represents the inclusion of all “large” –ve numbers.

represents the inclusion of all “large” positive real numbers.

(a, b) and [a, b] are called open interval and closed interval.

So it is natural to represent R as the interval

S.A.Q.4

a) Find all natural numbers in the intervals [3, ∞), (3, ∞), (– ∞, 3) and

(– ∞, 3]

b) Find all integers lying in the intervals given in a

c) Find all numbers in [2, 2], (2, 2)

Example; Represent as an interval

Solution: represents two inequalities x – 3 < 2, – (x – 3) < 2.

When x – 3 < 2, x < 5

When – (x – 3) < 2, x – 3 > – 2, or x > 3 – 2 = 1

Hence the given set is {1, 5}

We can also arrive at this interval geometrically (see Fig 5.4)

Figure 5.4

S.A.Q.5: Represent the sets ,

as intervals.

Now we have enough background to define the limit of a function f of a real variable x. In the

case of functions of a discrete variable or sequence, we defined . This limit

represented the long term behaviour of f. In the case of a function of a real variable, we can

discuss the behaviour of f(x), when the variable x comes close to a real number a. In other ways

we will be defining We want to write the statement “x

comes close to a” rigorously. The geometric representation of real numbers can be used for this

purpose. When do you say that your house is near your college? When the distance between your

house and your college is small. In the same way, we can say that “x is close to “when | x – a| is

small. If “smallness” is defined by a distance of say 0.1, then x is close to a if

|x – a| < 0.1 of course the measure of “smallness” is relative. For a person living in Mangalore,

Manipal is not near Mangalore. For a person living in US, Mangalore and Manipal are near to

each other. So “smallness” is decided by the choice of a positive number ∈ (This was done in

defining the limit of a sequence also)

Before giving a rigorous definition of limit, let us consider two examples.

Consider F(x) = 1 + x. Let us try to see what happens when x is close to 1. We evaluate f(x),

when x = 0.9, 0.99, 0.999, 1.1, 1.01, 1.001

F(0.9) = 1.9 f(0.99) = 1.99 f(0.999) = 1.999

F(1.1) = 2.1 f(1.01) = 2.01 f(1.001) = 2.001

Note all these values are near the value 2.

Consider another function

(Why don’t we define g(1) ? If we put x = 1 in then we get which is not defined).

As in the case of f(x), we compute some function values.

g(0.999) = 1.999 g(1.1) = 2.1

g(1.01) = 2.01 g(1.001) = 2.001

Note: These values are close to 2. Hence we can say that x close to

1 both f(x) and g(x) are closed to 2 and we can take 2 as

or .

Now let us formulate a rigorous definition of

Definition: Let f be a function of a real variable. Then if given a positive number ∈

There exists a positive number δ such that

…………. (1.6)

Let us analyze the definition.

We have two choice of positive numbers (∈ and δ) and two conditions

Given any positive number ∈ there exists a positive number δ

such that the condition P Implies the condition

The choice of δ depends on the given number ∈. The function f need not be defined at x = a.

The condition P says that x is close to a.

The condition Q says that f(x) is close to I

We can also express the definition geometrically.

Given ∈ > 0, there exists δ > 0 such that

In figure 5.5, the point (a, I) is measured as 0, meaning that the functional value of f at x = a is

not known or defined.

Figure 5.5: Definition of limit of function

The images of points in under the function f is a subset of the interval (I –

∈, I + ∈) along the vertical axis.

W.E.: Evaluate

Solution: Here f(x) = 2x – 6

Choose any ∈ > 0. We have to choose a δ such that (5.5) is satisfied. We have to guess the value

of I. When x comes close to 3.2x – 6 should come close to 2(3) – 6 = 0. Take I = 0. Then

So, choose then

Hence,

Note: f(3) = 2(3) – 6 = 0. In this case the function f is defined at x = 3 and f(3) coincides with

.

W.E.: Evaluate

Solution: Let . As in the previous problem, we can guess the value of I it is

Hence for a given , the corresponding δ is chosen as and condition (1.6) holds good.

Hence

W.E.: Evaluate

Solution: Proceed as in the previous example. In this problem and

General Remark While evaluating if f(a) is defined or it is not of the form and f(x)

is defined by a single expression, it will turn out that If f(a) is not defined or f is

given by two different expressions. Then we have to guess the value of the limit and prove

condition (1.6).

In some problems, f(x) may be given as quotient of two expressions but it may reduce to an

easier function on simplification. In such cases the problem will reduce to an easier one.

W.E.: Evaluate

Solution: When x = 2, f(x) is of the form So we try to see when x – 2 is a factor of

2x2 – 3x – 2, 2x

2 – 3x – 2 = 2x

2 – 4x + x – 2 = 2x (x – 2) + 5 (x – 2) = (2x + 5) (x – 2).

When x → 2, x does not assume the value 2. So,

= 2x + 1 on canceling x – 2, since x – 2 ≠ 0.

So the given limit reduces to

(as in W.E.)

Algebra of Limits of Functions

It is not necessary that we use the ∈–δ definition for every problem. We study important

properties of limits of functions as a theorem. We can evaluate limits using this theorem (noted

as a proposition).

Proposition:

a)

b) , where k is a constant

c)

d)

e)

f) when a > 0

Theorem: Let k be a constant, f and g functions having limit at a and n a positive integer. Then

the following hold good.

a)

b)

c)

d)

e)

f) provided

g)

h) provided is positive

You need not prove these results. It is enough if you clearly understand the theorem and

proposition and apply them for evaluating limits.

Self Assessment Questions

SAQ 6: Evaluate the following limits

a) b) c)

d) e) f)

g) h)

SAQ 7: Evaluation the following limits

a) b) c)

d) e) f)

g) h)

S.A.Q.8 If evaluate the following

a) b)

c) d)

e) if m is positive f) if I, m > 0.

5.4 Concept of Continuity

In mathematics and sciences, we use the word “continuous” to describe a process that goes on

without abrupt changes. For example, the growth of a plant, the water level in a tank and the

speed of a moving car in a four-base highway are exhibiting continuous behaviour.

Before defining continuous functions, let us look at the graphs of three functions.

Figure 5.6: Two discontinuous functions and a continuous function

The first graph has a break at x = a in the second graph also there is a break at x = a. If you

ignore the point corresponding to x = a, there is no break for the break occurs at x = a the third

function has no break. So it should be intuitively clear to you that the first two functions are not

continuous while the third function is continuous.

Let us formulate a rigorous definition of continuity.

Definition: Let f be a function of a real variable defined in an open interval containing a. Then f

is continuous at a if

Note: In order to define continuity at a, we need three conditions.

1) exists

2) f(a) is defined

3)

Even if one of them fails, then the function f is not continuous at a.

Now look at Fig. 5.4. The first function say f has no limit at a, i.e., does not exist. For

the second function, exists but The third function is continuous.

Example: Define f as follows:

Is f continuous at 2?

Solution:

= 4

Note: We can cancel (x – 2) since it is non zero.

So exists.

But f(2) = 3 ≠ 4

Hence the function f is not continuous 2.

Example: Define f as follows.

Is f continuous at 2?

Solution: From above example, we have A s f(2) = 4, f is continuous at 2.

Sometimes function may be defined by two different expressions. In such cases the following

method of proving continuity will be useful. For that we need the concept of left limit and right

limit.

Definition: Let a function f be defined in an interval (b, a) where b < a). Then

(called the left limit) if given ∈ > 0. There exists δ > 0 such that

………… (1.7).

Definition; Let a function f be defined in an interval (a, b), where b > a. Then

called the right limit if given ∈ > 0, there exists δ > 0 such that

………….. (1.8)

Note: We can prove that (1.7) and (1.8) implies (1.6). Hence if the left and right limits exist and

are equal then

When a function is defined by two different expressions, we have to evaluate the left and right

limits.

if both the left and right limits exist and are equal.

Worked Examples:

W.E.: Test whether f is continuous at x = 3 where f is defined by

Solution: As f is defined by two expressions one for and another for (3, ∞), we evaluate

the left and right limits.

As the left and right limits are not equal, does not exist. Hence f is not continuous at 3.

When a function f is continuous at every point of an interval (a, b) we say that the function is

continuous on (a, b). In particular if a function is continuous at every real number. Then we say

that a function is continuous on R.

Using propositions, we can prove that the functions, k (a constant), x, x2, ……..x

n, where n > 2

are continuous on R.

The function is continuous on (0, ∞)

The function is continuous on

Using previous theorem, we can prove the following theorem.

Theorem:

a) If f and g are continuous at a, then

i) f(x) + g(x) ii) f(x) – g(x) iii) f(x) g(x) are continuous at a

b) If f and g are continuous at a and g(a) ≠ 0, then is continuous at a

c) If f is continuous at a, f(x) > 0 for x in an open interval containing a, is continuous

at a.

Worked Examples

W.E.: Test the continuity of the function f at all real points where f is defined by

Solution: If a < 3, then f(x) is defined by the expression x2 in an open interval containing a. So f

is continuous for all a < 3.

So it remains to test continuity only at 3.

As does not exist. So f is not continuous at 3.

Thus f is continuous at all real points except 3.

W.E.: Test the continuity of the function f where f is defined by

Solution: When x < 2, |x – 2| is negative. So |x – 2| = – (x – 2)

When x > 2, |x – 2| is positive. So |x – 2| = x – 2

f(2) = 0. Thus

As in the previous worked example, f is continuous for all a < 2 and all a > 2.

As does not exist. So f is continuous at all points except 2.

Not all functions are continuous. In some cases may not exist. You may ask a question:

how to establish that does not exist. One sample case is where the left and right limits

exist but are not equal. We have examples for this case. The worst case is when neither of the two

limits exist. For proving the non-existence of limits, we use the following theorem.

Theorem: A function f is continuous at a if and only if the following conditions holds good.

To prove the non existence of the limit it is enough to construct a sequence (xn) converging to a

such that f(xn) does not converge to f(a).

W.E. : Show that the function f defined by

is not continuous at 0.

Solution: Let Then (obvious) f(xn) = n and so f(xn) does not converge to 0

since f(xn) indefinitely increases and so cannot approach 0.

Self Assessment Questions

S.A. Q. 9 Verify whether the following functions f is continuous at a

a)

b)

c)

d)

e)

f)

g)

h)

S.A.Q. 10 Show that the function f defined by

is not continuous at 1.

S.A.Q. 11 Show that the following functions are continuous on R.

a)

b)

c)

d)

5.5 Summary

In this unit we studied the basics of real number system then the concept of limit was discussed

which was further extended to the concept of continuity. All definitions and properties of the

above mentioned concepts is given very clearly with sufficient number of examples wherever

necessary.

5.6 Terminal Questions

1. Find all natural numbers in the following intervals

2. Find when

a) b) c)

d) e)

3. Show does not exist when

a) an = n b) an = 2n c) an = n!

4. Evaluate when

a) b) c)

d) e) f)

g) h) i) j)

k)

5. Evaluate when f(x) is equal to

a) b) c)

d) e) f)

6. Evaluate the following limits

a) b)

c) d)

e) f)

g) h)

7. If and evaluate

a) b)

c) (when I, m, > 0)

d) when m > 0

8. Show that the following functions are continuous at a

a) for all x in R, a = 0

b) for all x in R a = 0

c) for all x in R a = 0

9. Show that the following functions are continuous at a

a)

b)

c)

10. Show that the following functions are not continuous at a

a)

b)

c)

d)

e)

5.7 Answers

Self Assessment Questions

1. A

2. A

3. a) and d) can be arranged as a sequence according to chart and attendance

register

4. a) {3, 4, 5, …..} b) {4, 5, 6, ……}

c) {……. –3, –2, –1, 0, 1, 2,} d) {…. –3, –2, –1, 0, 1, 2, 3}

5. [1, 5], (–∞, 1) ∪ (5, ∞), (–∞, 1] ∪ [5, ∞)

6. a) 1 b) 1 c) 100 d)

e) 1 f) –1 g) 2 h) 9

7. a) 3 b) c) 2 d) 4

e) 3 (Hint: x3 – 1 = (x – 1) (x

2 + x + 1)) f) g) 1

h)

8. a) 2I + 3m b) I2 – m

2 c) d)

e) f)

9. a), b), c), d) continuous e) discontinuous f), g) continuous,

d) not continuous

10. Take but f(xn) = n f(xn) does not converge. So f is not continuous at

1.

11. a) For a < 1, f(x) = x2 – 6x + 5. So f is continuous for a < 1. As f(x) = 2x

2 – 2, f is

continuous for a > 1.

Hence

Also f(d) = 1 – 6 + 5 = 0

b), c), d) Similar.

Terminal Questions

1. a) {4, 5, 6, 7, 8} b) {5, 6, 7, 8, 9} c) {5, 6, 7, 8}

d) {4, 5, 6, 7, 8, 9}

e) {4, 5, 6, …..} f) 10, 11, 12,…..} g) {9, 10, 11, …..} h) {9, 10, 11, …..

}

2. a) 0 b) 1 c) 0 d) 0 e) 0

3. a) As n increases an increases. So cannot be made less than a fixed number ∈.

So does not exists.

4. a) b) 1 c) 3 d)

e) – 1 f) 1 g) h)

i) Write an as

j)

k)

5.

a) 2a3 b) a

3 + a

4 c) 2a

3 – a

4 + a

5 d) (2 + 3a) (4 + 5a

2)

e) f)

6. a) 2(–1)2 – 1 = 1 b) (2 + 1) (3 – 2) = 3 c) d)

e) f) g) 1

h) Hence answer is

7. a) Im b) (l + 2m) (2I – m) c) d)

10. a) f(2) = 7. So f is not continuous at x=2.

b) But f(1) = 1

c) Let limit at 1 = 8; right limit at 1 = 2

d) The left and right limits are 1 and –1 respectively.

e) The left and right limits are –1 and 1.

MH0047-Unit-07-Promotional Tools

Unit-07-Promotional Tools

Structure

7.1 Introduction

7.2 Promotion Mix

7.3 Summary

7.4 Terminal Questions

7.1 Introduction

A hospital’s promotion strategy is two pronged. One is, what you learned in the last unit, through

the hospital promotional activities like healthcare seminars, corporate empanelments, medical

camps and others. Second, is promoting a hospital through specifically designed and developed

promotional tools or devices. In this unit you will be learning about this second aspect of

hospital’s promotion.

Learning Objective

In this unit you will learn about the different kinds of hospital promotional tools, their design and

development and the mode of their proper use to get the desired results.

7.2 Promotion Mix

The different promotional tools employed by various hospitals are as follows:

1. Media Coverage / News Releases:

The kind of stiff competition that exists among the healthcare providers today necessitates the

presence of strong relations between any hospital and the media. Churning out genuine press or

news releases by the hospital in leading local / national dailies is indeed a very fine promotional

tool and does a world of good to the hospital’s marketing. Hospitals are an enterprise where

something or the other keeps happening that needs to be communicated to the outside world, not

only for their promotion or publicity, but also for people’s awareness. If a hospital is conducting

or has already conducted a camp, if it has acquired a corporate empanelment or accreditation or

tie-ups with leading health insurance companies, or a unique / rare / complicated procedure or

surgery has got successfully transpired; all that needs to get published as news releases in dailies.

Here the choice of dailies by the hospital is an important aspect that needs to be considered. If

the hospital is primarily catering to local population, the news needs to get published in the local

vernaculars. If the hospital is catering to a wide population base, cutting across different regions

and states, then national newspapers need to be covered. Press releases are not only helpful in

increasing the visibility of the hospital to the target clientele, but at the same time, they also

widen the horizons of customer’s knowledge. The idea behind such press releases is to market

the hospital in the form of news and not as an advertisement, because the former evokes a better

response among the readers and is more cost effective. Giving news will hardly cost a hospital

anything, but giving an advertisement will turn out to be an expensive affair. Hence, for small

hospitals with budgetary constraints, it is suggested that only news should rule the roost and not

advertisements.

The press releases in the print media should also be replicated in electronic media, in the sense

that every single major / complicated procedure that gets transpired in a hospital should find

itself as news in local / national television channels. Here also, just like the choice of

newspapers, the choice of TV channels depends on the target market segment. The reasons are

also the same as that of press releases; to market the hospital in the form of news and not as an

advertisement, because the former evokes a better response among the viewers and is more cost

effective. Giving news will hardly cost a hospital anything but giving an advertisement in on TV

channels will turn out to be an expensive affair.

Last but not the least, it must be understood that in order to gain good mileage from media, it is

important that the hospital maintains strong rapport with them for long term business relations.

Light pampers like calling over media personnel for a cup of tea, sending them greetings on their

birthdays / anniversaries / festivals, giving them and their dependents a concession in case they

get some procedure done in the hospital, a monthly call to enquire about their well being and

indulging in a general chit chat are some of the tricks that should be resorted to.

2. Preventive Healthcare articles:

Giving preventive healthcare articles in newspapers is yet another very effective and ethical

marketing tool, because when people read an article from a particular hospital, a positive

favorable impression of the hospital gets created in the minds of the readers. They feel

empowered because of their increased sense of knowledge and this mindset goes a long way in

creating the much needed goodwill of the hospital. Let us suppose that there is an orthopaedic

hospital that wants to market itself through the tool of healthcare articles. Following are the

important guidelines that the hospital should keep in mind while promoting itself:

i) The articles should pertain to general and specific nuances of orthopaedics which could be of

value to the readers i.e. the present / future customers of the hospital. General interests can

encompass issues like what are the different kinds of orthopaedic pains, their causes, home tips /

precautions to ward them off; myths and realities of orthopaedics with regard to patient

knowledge; the kind of first aid to be given following a fracture or an accident; tips for

strengthening back, shoulders and other orthopaedic sites; harmful effects of self medication or

getting treated by quacks.

ii) Specific articles will include educating readers on different orthopaedic procedures, their

mode of application, pre and post operative precautions, estimated cost of the procedures, which

kind of patients can be benefited from them and other related aspects. For example: What is Hip

Replacement Surgery all about? What kind of people are at risk? What kind of patients can get

treated through it? How is it done? What is the nature of pre and post operative care involved?

How much does it cost ? (While answering questions like these, exact price should not be quoted

but just a range should be mentioned, say 1-2 lac)

iii) The articles have to be in very simple, lucid language because they are meant for general

public and not for some medical or para medical professionals. Because the public’s

comprehension of medical science and its related glossary is generally not good, it is very

important that every medical term or procedure should be simplified within brackets ().

iv) While promoting a hospital through healthcare articles, it has to be kept in mind, that at the

end of the day, what it is that the hospital wants to promote ? Does it want to promote its own

name or the doctor who has penned down the article? For a small single specialty hospital, where

the director happens to be the owner of the hospital or the head of the speciality, it may be

worthwhile to promote the doctor’s name. In such cases the articles should carry a photograph of

the doctor concerned (wearing doctor’s coat with a stethoscope dangling over the neck), his / her

e-mail ID, the hospital’s name, it’s contact numbers and it’s website.

However, if it is a big single or a multi speciality hospital with many doctors, then it is the name

of the hospital that has to be promoted through articles and not the individual doctors, because

individual doctors come and go, they may leave the hospital and join its competitors and if a

hospital starts promoting these individual doctors, the prospective customers will get aligned

with those doctors, which would be damaging to the health of the promoting hospital; something

that the hospital would not like to see happening. The essence in promoting the hospital through

healthcare articles is to align the prospective customers with the hospital and not its doctors. So

in cases like these, the articles should carry the hospital’s name, it’s contact numbers, it’s website

and the name of the department with which the article is associated and not the name of doctor(s)

who have penned the articles.

v) Another important aspect that needs to be taken into consideration is the choice of newspapers

in which the articles need to get published. That will depend on the hospital’s target market

segment. If the hospital is primarily catering to local population, the articles should get surfaced

in the local vernaculars. If the hospital is catering to a wide population base cutting across

different regions and states, then national newspapers need to be covered. It is also equally

important to understand as to where in the selected newspapers, should the articles get published.

Generally all the newspapers carry out weekly health supplements. It is there where the articles

should surface.

3. HEALTH CARDS:

Promoting a hospital through its exclusive health cards has recently caught the fancy of many

hospitals. Health cards are basically discount entitlements provided by the hospital to its

customers. Their advantages are twin fold for the hospital. First, the hospital gets promoted

through them and second, they are helpful in bringing about an increase in the sales / revenue of

the hospital.

The operating mechanism behind the health card generally involves one primary card holder and

his / her three designated beneficiaries who will be able to avail facilities offered by the hospital

at discounted rates. These health cards generally have a validity period of one year and, if the

customer desires, they can be renewed. Usually these cards are nominally priced and once the

customers purchase these cards, they along with their dependents become eligible to avail

discounted services like two to three free OPD consultations in a year, discounts on the health

checkups, diagnostics, investigations and total billing (for indoor patients).

It is not necessary that apart from the primary card holder, there has to be three beneficiaries.

The number of beneficiaries of the health card and the extent of its discounted services vary from

hospital to hospital. But care and caution need to be exercised while pricing and promoting these

health cards. They need to be priced and promoted in such a fashion that there should not be any

resistance in their purchase by the customers. They can and are being priced at a nominal value

of Rs.365. 365 is a magical figure. There are 365 days in a year and by being priced at Rs. 365,

the hospital is sending out a signal that good health and healthcare facilities are available at Rs.1

per day for the entire family.

Further it has to be understood that health cards are internal promotional tools and are promoted

from within the hospital premises. They can be marketed to the prospects through verbal

communication by the hospital’s staff, particularly the front office; a hospital can have posters or

banners promoting the cards being displayed at hospital’s reception, canteen, wards and patients

rooms. They can be promoted on the hospital’s website also. It may not be a bad idea to come

out with a press release as and when they are about to be launched, enumerating their

characteristic details. It may be mentioned that not only these health cards are internal

promotional tools; they are also kind of reverse promotional tools: First the hospital promotes

them and then the hospital gets promoted by them.

Following is sample of a typical health card:

IDE 1:

SIDE 2:

6. EMERGENCY CARDS:

Just like Health Cards there are Emergency Cards through which a hospital or its services can be

promoted. These cards are helpful in promoting a hospital by highlighting its emergency or

trauma and ambulance services and they particularly find favour with emergency and trauma

centers.

Any hospital that has or specializes in managing trauma and other emergencies, promotion of the

hospital through specially designed EMERGENCY CARDS will help in boosting its business

and goodwill among the target clientele. The essence is, as and when there is a trauma or any

other emergency, the promoting hospital should come across as the most favored healthcare

provider among the public. If the hospital manages to have an increase in its trauma and

emergency cases, financially it will turn out to be a fine proposition, because there are lesser

chances of patients dropping out from the hospital following an emergency / trauma incident.

A typical EMERGENCY CARD will have the size of a normal visiting card and will be printed

on both sides. One side will carry a helpline number along with broaching of the facility of

AMBULANCE. On its reverse will be the detailed address of the promoting hospital, along with

a brief on the services being provided by the hospital.

In order to make success out of promoting a hospital through EMERGENCY CARDS, it is

important to know and understand to how they will be promoted. The EMERGENCY CARDS

should always be available at the reception and every single person coming to the hospital

whether an in patient or an out patient, attendants / visitors of the patients should be handed over

these cards. The front office should make it a point that the cards should only be handed over to

the recipients. At no point of time should the cards be stapled with the registration form and

given to the patients / attendants, because then they will not create the desired effect and the

desired effect is that whenever there is any emergency or trauma, the promoting hospital should

be in the minds of the people. This can happen only if the cards stay in their purses or pockets; so

it always makes sense to hand over these cards to the customers, rather than clubbing them with

their registration or discharge cards.

Following is a sample of an EMERGENCY CARD for an orthopedic setup:

5. Website:

In today’s internet savvy world a well designed website acts as a powerful promotional tool for

leveraging branding of the hospitals. However, a website has to be carefully designed, keeping in

mind the nature of services it provides, the kind of market segment it caters to and the

psychological makeup of it customers. .

We all know that the information present in nearly all the websites is distributed over different

pages. The case of a hospital’s website is no different but following information should be kept

in mind while drafting a hospital’s website:

HOME PAGE: This page will have the links to all other pages. Apart from having the links,

there may be a photograph of the hospital in the background as a water mark. In fact this water

mark may figure in all the pages. Also the hospital’s punch line should appear on the top of all

pages in a moving / flashing mode and the hospital’s Emergency Helpline Number should appear

on the bottom of all pages in a moving / flashing mode. The various links / pages that should

feature in the HOME PAGE are:

ABOUT US: This link can contain an abstract on the promoters of the hospital, as well as

information pertaining to entrepreneurial history of the hospital such as the operating philosophy

behind setting up the hospital (e.g.: that it was set up to provide high quality, affordable and

comprehensive clinical care to remove the sufferings of ailing mankind), the number of years in

service and total number of beds. A word or two about the location of the hospital in terms of

how far is it from railway station, bus stand or airport can also be mentioned. In case there are

any firsts associated with hospital like being the first private orthopaedic center in a particular

city or state; they should also be mentioned in this page. If the promoter happens to be a doctor

then the doctor’s qualification / degree(s), years of experience, number of procedures performed,

awards / felicitations received, lectures given, conferences attended and things like that can be

broached.

THE TEAM: This one can contain an account of the different doctors working for the hospital;

their names, designations, qualifications / degree(s), years of experience, number of procedures

performed, awards / felicitations received, lectures given, conferences attended etc. This

information should be provided along with the doctors’ photographs. Then there can be a group

photograph of different doctors serving in a hospital along with the rest of the employees. Apart

from the clinical fraternity, a photographic account of other managerial and pararamedical

employees can be provided with their names, designations, qualifications / degree(s) and years of

experience

VISUAL TOUR: The ‘VISUAL TOUR’ page will act as a snap book of the hospital. It will

contain pictures of the hospital’s building, reception, patient rooms, I.C.U., Operation Theatre,

Ambulance, Physiotherapy Room, Laboratory, Canteen etc.

CLINICAL SERVICES: This page will contain a detailed list of the various clinical procedures

/ facilities being provided by the hospital. From the patient’s perspective, this link is by far the

most important of all links of the hospital’s website and also the most bungled one.

· One look at this page will reveal the naivety of the hospital promoters. A lot of emphasis is laid

on the hospital’s state of the art machines and their clinical details and the clinical procedures

being carried out and their methodology. Why it is naïve is because the patients and the target

customers are not interested in which machine is installed at your hospital or what is the

procedure that your hospital is an expert in. They just want to know how you can you make a

difference in their quality of life.

· For example, let us take a gall bladder stone patient who wants to get laparoscopic surgery done

for the stones. What do you think is more important for her- the make and configuration of the

equipment being promoted at the website or the total experience of what she will go through

during the surgery? Patients need access to information like: 1. Is the operation absolutely

necessary? Can they avoid the surgery? Will there be any consequences of not getting the

surgery done? 2. How big is the surgery? Is it too complicated? 3. Will it be painful? How much

pain do they have to endure and for how many days? 4. How many days they have to stay in the

hospital? Can they afford to stay away from work for these many days? 5. Is the doctor

concerned skilled in this surgery? Does he operate routinely? Has anyone in circle of friends or

relatives had an operation at this hospital or by this doctor? 6. How much will it cost? Where will

the money come from? 7. Is the hospital environment friendly and nice? Will they have a

comfortable experience here? These are the points that need to be taken into consideration while

developing the page on clinical services. The point is, a patient is more concerned about the

overall experience he will get in the hospital including safety and pain, rather than the machines

and the way the procedure is carried out. A recent survey has revealed that a patient is driven to a

hospital mostly by the treatment he gets as a person from the doctors and staff of the hospital. He

wants to be understood fully and to be informed fully.

So in a hospital’s promotional website, a hospital should tell the people about what they will get

in terms of quality of life, if and when they get treated at its place. It should tell them what they

will go through in terms of body pain and sensations during and after the procedure. It should tell

them how much it will cost and why that cost is worth incurring. Just harping about ones’

machines, procedures, techniques and their technicalities in website and other promotional tools

does not really work. Nobody cares for that.

TESTIMONIALS: As the name suggests this page will contain testimonials / proofs from the

patients being treated by the hospital. The testimonials can be in the form of written

documentation or video clips and their content should include name and address of the patients,

as well their verbal excerpts, giving an account of their respective conditions before coming to

hospital and improvement that they have had following treatment at the hospital

=> IN-PATIENT FACILITIES: This page will contain an account of various kinds of patient

rooms available to patients. The prices of different rooms along with their additions / deletions

viz: attached washroom, T.V., A.C., cooler etc should be mentioned. Also, the facility of canteen

and local / STD phone calls should be broached.

CORPORATE SERVICES: This link should have a list of all the corporates / organizations

with whom the hospital has got empanelled for providing various services to their respective

employees, as well as their dependents. The general modus operandi through which beneficiaries

(employees / their dependents) will avail services of the hospital should be mentioned. An

accepted convention in corporate empanelment is , that prior to availing the services of any

hospital, the beneficiary is required to show his / her ID card which acts as a proof of him being

an employee of that particular organization. If there are some special conditions with regard to

particular organization(s), the same should be highlighted.

HEALTH INSURANCE: This should contain the list of various Health Insurance companies

with which the hospital has tied up for providing cashless service to the respective policy holders

of the insurance companies. Also the list of various Third Part Administrators (TPA’S) should

also figure out in the link.

HEALTH CARD: This page can contain an account of the hospital’s Health Card along with its

terms / benefits of use. Also a sample of the card should be shown on the website itself.

E-HEALTH: This page can be divided into two links: 1. “Q & A with the hospital’s doctor(s)”

and “Doctor Cam 2 Cam”. Q & A with the doctor(s) will have a provision that any web surfer

can seek information / solicit advice / question the doctor(s) anything regarding his or her

clinical concerns, through a pre designed query form which will be available at the link, and the

doctor(s) concerned will provide a prompt reply to it. It should be seen that the reply to all the

queries should be answered within the first 68 hours of their receipt. Further, the website surfers

should also be made aware of the fact that their respective queries will be answered within the

first 68 hours of receipt.

“Doctor Cam 2 Cam” can be yet another innovative marketing strategy whereby the website

visitors will get to interact directly with the doctor(s) of the hospital on a webcam and they can

sort out all their queries in consultation with the doctor(s). The day(s) and timings of “Doctor

Cam 2 Cam” should be arranged when the doctor(s) are relatively free and the same should be

broached on the website for the seekers information. Also, the day E-HEALTH goes on floor, its

corresponding press release should figure in dailies.

HOSPITAL IN NEWS: This page will contain a soft copy of all the press clippings that

covered happenings / achievements / accomplishments of the hospital. The clippings should be

arranged in a chronological order beginning with the latest and ending at the oldest.

NEWSLETTER: This move is designed to empower the customers with useful information

pertaining to the specialties that the hospital is engaging in, as well as to create a favorable

impression of the hospital in their minds. A system has to be designed in the website that

whosoever is desirous of having the news letter should feed his / her respective E-mail ID and

the letter should get dispatched in no time. The content of the newsletter can be the same as that

of preventive healthcare articles which already may have appeared in the dailies. For a start, it is

suggested that the newsletter can be made a monthly affair at least for the first year. Preparing

newsletters will not involve any additional hard workbecause only 12 of them have to be framed

(1 per month) and that too will be a straight lift from 12 preventive healthcare articles that would

have had already appeared in various vernaculars.

=> CONTACT US: This link should have all the contact numbers / addresses of the hospital. It

should cover detailed address of the hospital, it’s phone numbers, emergency helpline number,

E-mail ID’s of the doctor(s) as well as that of the hospital. This page can also contain a location

map to the hospital.

The above mentioned guidelines for website development of the hospital have been prepared,

keeping in mind customers’ / patients’ requirements and concerns. The website developers

should not deviate much from the framed plot. They should bring in their share of expertise to

leverage what has been discussed above, so that the hospital gets the best and reaps the best.

6. Boards and Hoardings:

Boards, hoardings, banners, pamphlets are the perhaps the oldest promotional tools being utilized

by the hospital to promote its name in the market. No doubt they serve the purpose, but if they

are amateurishly developed and designed and placed at wrong locations, the entire exercise is a

waste of time and money.

On an average, a city person is exposed to 300 advertising messages in a day. He whips past

hoardings on the road, he reads through ads in the newspaper, he overlooks promotional

messages on TV and he indifferently listens to the radio in his car which plays the local FM

channel. Perhaps the biggest challenge for any business today is to be seen and/or heard in an

overcrowded marketplace. Everyone is fighting for that one moment of attention from the target

customer. Most of the times hospitals are found wondering as to why their boards and

advertisements fail at getting noticed. The commonest plain reason is that they mar their chances

of success by being boastful and indulging in self praise.

Do you indulge in self praise? Well, once in a while, most of us do. We want to tell others about

us, else they may make a wrong impression. The same logic goes while we want to tell them

about our hospital and its services.

A hospital will have to learn to resist the temptation of telling more and more about its services.

Remember, in advertising, less is more and more is less. A hoarding in any case will be read by

commuters. They will not have the time to stand and read about all your latest equipments and

the list of services you offer. Let’s face it, “NO ONE HAS THE TIME”. People hate confusion

and complexities. In any case they are being bombarded by hundreds of messages everyday

through various mediums. A hospital may think that if it does not write all the things, then

people will not know about it. Here is the lesson to be learnt, “TO MAKE SOMETHING

IMPRESSIVE, YOU DON’T HAVE TO MAKE IT EVERLASTING”.

Keep it short and simple. Pictures speak a thousand words. So add a picture. Forget the list of

services and equipment. Concentrate on what the customers will get rather than what you have to

offer. Yes, it takes a lot of effort and time to be an expert copywriter [that is why the advertising

agencies make so much money]. A hospital cannot turn into an ace creative director in one

month. But sure it can at least get some return on its money by making its message short, simple

and relevant. You can even tickle the funny bone by using some humor. Just be careful that your

message should not be lost among the humor of the ad.

Some messages on safe driving [if you are a trauma centre], women education [if you are a Ob-

Gyn centre], quitting smoking and exercising [if you are a cardiac centre] etc., should be put on

the highways in form of hoardings. Not only these messages are for public good, they also create

a good image about your hospital. Once again, the location of the message is important.

It’s now time to be introduced to a very interesting marketing tool. It is called converting your

disadvantage into an advantage. Most of the hospitals lose sleep over what they think is a

weakness of their organization. Well, it is possible to convert the weakness into a strength and

boards and hoardings can go a long way in bringing about that much needed turnaround.

A doctor client of a hospital consultancy was worried that his hospital is viewed as the most

expensive hospital in the target market. He said, "We are sure to miss out on many cases because

we are expensive.’ Indeed, when the consultancy compared the prices, the services were

expensive. The surgery charges, ICU, investigations did cost more to the patients. He had more

number of doctors and expensive high quality machines, so his costs were also high. The hospital

infrastructure was maintained diligently, and that added to the costs as well. He had more

number of nurses and cleaners per patient. All in all he had ensured a good quality service, but at

a cost.

The dilemma was that he could either compromise on the service quality and reduce the prices;

or he could reduce the prices and earn fewer profits but maintain the quality. Well, he did neither

of those. The consultancy told him to not dodge the fact that his place is expensive. In fact, it

encouraged the hospital to make a statement and propagate that they are expensive.

An advertisement campaign was created for them with a headline- ‘We are the most expensive

hospital in the city.’ The tag line was, ‘But who says quality is cheap?’ In the middle of both

these lines was a picture of a happy smiling family in a plush, private room of the hospital. There

were other messages too. Through boards, hoardings and other mediums, the hospital told the

people that they deserve what they pay for. A premium brand associated with top quality service

was developed. This message was spread through boards, hoardings and other media all over the

area. Within a year, the disadvantage was converted into an advantage. The hospital today is

viewed as a premium place with high quality service. When a person gets his family member

treated there, he is making a statement that he will buy the best for his loved one. Needless to say

that the sales have increased and so have the profits in one year’s time.

The lesson from this story is- If you are candid about your ’so called disadvantage’ you can

actually convert it into an advantage, provided you promote it right.

Apart from the above broached promotional tools, a hospital can resort to some other

promotional measures like ensuring that there are certain take aways for its patients when they

go home. They may be key chains, pens, greeting cards, diaries, etc. A hospital should also

engage itself in some social activities like training volunteers in first aid in case of accidents,

giving away school books and school uniforms to school going children of underprivileged

families. Needless to say, that when such social work makes headlines in the papers, people

remember the hospital and the hospital does not need to spend money on advertising too.

7.3 Summary

It may be summarized that the above mentioned promotional tools point to one thing. They

emphasize the role of being creative to be seen in the market place and being empathetic to the

needs of the customers. A hospital has to be visible without sounding boisterous. In this era of

competition, the hidden good doers will not survive for long. Only the hospitals with properly

laid out promotional strategy will.

7.4 Terminal Questions

1. Write a note on any 5 components of promotion mix.

2. Develop a promotion mix for a hospital located in the urban place and compare it with a

promotion mix of a hospital located in the rural place.

BT0063-Unit-07-Integrations

Unit 7 Integrations

Structure

7.1 Introduction

Objectives

7.2 Integration of Standard Functions

7.3 Rules of Integration

7.4 More Formulas in Integration

7.5 Definite Integrals

7.6 Summary

7.7 Terminal Questions

7.8 Answers

7.1 Introduction

Most of the mathematical operations we come across occur in inverse pairs. For example,

addition and subtraction, multiplication and division, squaring and taking square roots are such

pairs. In this chapter we study integration as the inverse operation of differentiation. Integrals

also have independent interpretation. It generalizes the process of summation. It can be used for

evaluating the area under the graph of a function.

Objectives:

At the end of the unit you would be able to

· integrate standard functions

· apply the concept of definite integrals in the process of summation

7.2 Integration of standard function

Definition: If f(x) is a function of a real variable, then g(x) is the integral of f(x) if

It is denoted by

Remark: Integration is the operation of determining a function whose derivative is the given

function.

Note: Clearly

…… (7.1)

…… (7.2)

(7.1) and (7.2) can be stated as follows:

Differential coefficient of integral of f(x) = integral of differential coefficient of f(x) = f(x)

Remark: If f(x) = x2 and g(x) = x

2 + 2, then That is, the derivative of a

function remains the same if a constant is added to it.

So while writing the integral of f(x), we need to add c, an arbitrary constant (by arbitrary

constant we mean any real value).

In table 7.1 we list the integrals of some of standard function.

Table 7.1: Table of integrals

We get formulas in Table 7.1 by simply reading Table 2.1 backwards.

We frequently apply constant function rule (Rule 1) and sum and difference rule (Rule 2) for

integration which are similar to differentiation. However we do not have “product rule and

quotient rule” in integration. So problems in integration are more difficult their problems in

differentiation.

Rule 1: (Constant function rule)

where c is a constant

Rule 2: (Sum and difference rule)

Example: Evaluate when f(x) equals

a) b)

c) d)

e)

Solution:

a)

b)

(by rules 1 and 2)

(Note: We use k for arbitrary constant since c appears in the function)

c)

(by rule 2)

(by rule 1)

d)

(by rule 2)

(by rule 1)

e)

(by rule 2)

(by rule 1)

S.A.Q.1: Integrate

7.3 Rules of Integration

We have the following rules for integrating complicated function.

Rule 1: Constant function rule

Rule 2: Sum and difference rule

Rule 3: By substitution

Rule 4: Integration by parts

Sum and difference rules:

We have already given rules 1 and 2 and used it for integrating functions in 7.1. In trigonometry

some products can be expressed as um or difference of two simpler trigonometric functions. The

following formulas are useful in integration.

2 Sin A Cos B = Sin (A + B) + Sin (A – B)

2 Cos A Sin B = sin (A + B) – Sin (A – B)

2 Cos A Cos B = Cos (A + B) – Cos (A – B)

2 sin A Sin B = Cos (A – B) – Cos (A + B)

Worked Example: Evaluate

Solution:

Worked Example: Integrate Sin 10x Sin 2x w.r.t. x

Solution:

Worked Example: Evaluate

Solution:

Integration by substitution:

Now we state rule 3.

Rule 3 (Substitution rule) If x can be written as a function g(t) of t, then

Proof: As

Hence (Don’t think that this step is obtained by cross multiplication. dx and dt are

called differentials and is the relation connecting the differentials).

So .

Note If f(x) cannot be integrated using known formulas, we try to write f(x)dx as g(t)dt. Part of

f(x) is written as f(g(t)) and the remaining part together with dx is written as

Example: Evaluate

Solution: Let t = ax + b. Then So dt = adx

Note: If then

…………………. (7.3)

So using table (7.1) we get a list of formulas for etc.

For example:

Example: Integrate the following w.r.t. x

a) b)

c) d)

Solution:

a)

b)

c)

d)

Working Example 4: Integrate the following w.r.t. x

a) b) c) Sin3x d) e) Sin

2 3x

Solution: Denote the required integrals by 1

a) (say)

Put x + a = t2 so x = t

2 – a, dx = 2t dt

b) (say)

Put a + bx = t2

c)

Let t = cosx; dt = –sinx dx

d) (say)

Put a + bx = t. So

e)

Your skill in integration lies in spotting a suitable variable t – g(x) so that can be

“located” in f(x) dx. We illustrate the method of substitution by more examples, so that you can

master this technique thoroughly.

Hereafter I stand for the integral to be evaluated.

Example: Integrate the following functions w.r.t. x

a) b) c)

Solution:

a) If t = xn then dt = nx

n – 1 dx and this appears in . So let

t = xn, then dt = nx

n-1 dx.

b) (Say)

Put log x = t so

c) (Say)

Put tan–1

x = t So

Working Example: Integrate the following w.r.t. x

a) b) c)

Solution:

a) (Say)

Put x3 = t so 3x

2 dx = dt;

b) (Say)

c) (Say)

Put

Working Example: Evaluate

a) b) c)

d) e) f)

Solution:

a)

Let y = cos ; dy = – sin d

b)

Let y = sec + tan ; dy = (sec tan + sec2 ) d

c) (Say)

Put y = x + sinx; dy = (1 + cos x) dx

d) (Say)

Put

Also

e) I = log sin + c (check it your self)

f) I = – log (cosec + cot ) + c (Check)

Note: Remember the integrals of tan , sec , cot and cosec .

Working Example: Integrate the following w.r.t. x

a) b)

Solution:

a)

Let

b)

Let numerator = I (denominator) + (denominator)

Put

We have 41 + 5m = 2 and 51 – 4m = 3

20t + 25m = 10

20t – 16 m = 12

S.A.Q.2: Integrate the following functions w.r.t. x

a) b) c)

d) e)

S.A.Q.3: Evaluate the following integrals

a) b)

c) d)

(Hint: b) c)

d)

S.A.Q. 4: Integrate the following functions w.r.t. x

a) b) c) d)

S.A.Q.5: Integrate the following functions w.r.t. x

a) b) c) d)

(Hint: b) c) d)

S.A.Q. 6: Integrate the following functions w.r.t. x

a) b) c) d)

S.A.Q. 7: Integrate the following functions w.r.t. x

a) b) c) d)

Integration by parts

By now you might have noticed that there are not many general rules for integration as in the

case of differentiation. Even simple functions like logx cannot be integrated with the rules we

have come across so far.

In this section we give a method of integration called “Integration by parts”. Using this method

we can integrate many functions.

Rule 4: If u and v are functions of x,

….. (7.4)

This follows by integrating the product rule

Integrating both sides we get

(7.4) Follows from the above identity while applying integration by parts, we “locate” dv in f(x)

dx. That is we integrate that part of f(x) and term it as v. The remaining part is taken as u.

Working Example: Evaluate

Solution: Let u = log x dv = dx

Then

V = x.

Working Example: Evaluate

Solution: Let and dv = dx

v = x

Working Example: Evaluate

a) b)

Solution

a)

Let

Put 1 – x2 = t

2 in the second term; x

2 = 1 – t

2; – 2x dx = –2tdt; xdxv= –tdt

Hence

b)

Let u = x dv = sin2 x dx

Then du = dx

7.4 More formulas in integration

Using the method of substitution or integration by parts, we can derive the following formulas

(see table 7.2)

Table 7.2 Additional formulas for integration

Now we are in a position to integrate functions having quadratic factors or square root of

quadratic factors in numerator or denominator of functions.

Integral of functions of the form

Method

Step 1: Let numerator = A (Denominator) + (Denominator)

Step 2: Find the values of A and B

Step 3: Split the function and integrate

We have already seen this in W.E. 7(b)

Completion of squares

All the subsequent methods use a technique called “Completing the square”. It is simply writing

a quadratic expression in the form, a2 + x

2,

a2 – x

2, x

2 – a

2.

Example: Consider x2 + 3x + 2. We can write

Example: Consider 2 – 7x – x2

Integration of functions of the form

Method:

Step 1: Write

Step 2: Find the values of A and B

Step 3: Split the function and integrate

Working Example: Evaluate

Solution:

Here

Let x + 4 = A (–2x + 6) + B = –2 Ax + 6A + B

1 = –2A; 4 = 6A + B

Integration of functions of the form

Method: Write in the form and integrate.

Working Example: Evaluate

a) b)

Solution:

a)

b)

Integration of functions of the form

Method:

Step 1: Write

Step 2: Find the values of A and B

Step 3: Split the function and integrate

Working Example: Find

Solution:

Let

Integrals of functions of the form

Method: Write in form and integrate

Working Example: Evaluate

Solution:

Integration of functions of the form

Method:

Step 1: Write

Step 2: Find the values of A and B

Step 3: Split the function and integrate

Working Example: Find

Solution: Let

3x – 2 = 2Ax + A + B

3 = 2A; –2 = A + B

Integration of functions of the form

Method: Write in the form and integrate.

Working Example: Find

Solution:

Integration using partial fractions

Consider the function Expanding the denominator as we can

integrate the function using the method given in 7.3.3. We can also integrate the same function

using partial fractions.

Now it is easy to integrate . We illustrate this method using an example.

Working Example: Evaluate

Solution: Let us write the given function as

Then

Multiplying both sides by (x – 1) (x – 2), we get

7x – 6 = A(x – 2) + B(x – 1) ……………… (*)

Put x = 1 in (*). We get 7(1) – 6 = A(1 – 2)

That is, 1 = – A of

Put x = 2 in (*). We get 7(2) – 6 = B(2 – 1)

That is, 8 = B

= log (x – 1) + 8 log (x – 2) + c

7.5 Definite integrals

is called the indefinite integral in integral calculus. You will be curious to know why it

is called indefinite integral. Rieman defined a definite integral first. It is the form

If

Note that the definite integral is a real number whereas is a function.

Working Example 18: Evaluate

Solution:

(Check)

Hence

Note: The arbitrary constant c gets cancelled while finding g(0) – g(a).

S.A.Q 8: Integrate the following functions w.r.t. x

a) b)

S.A.Q 9: Integrate the following functions w.r.t. x

a) b)

S.A.Q. 11: Integrate the following functions w.r.t. x

a) b)

S.A.Q. 12: Integrate the following functions w.r.t. x

a) b)

S.A.Q. 13: Integrate the following functions w.r.t. x

a) b)

S.A.Q. 14: Evaluate the following definite integrals

a) b)

7.6 Summary

In this unit we studied the standard forms of integration. The different rules of integration is

given and explained with the help of standard examples. All the formula concerned with

integration is followed by good examples. The concept of definite integral and its application is

explained clearly with good examples.

7.7 Terminal Questions

1. Integration of the following w.r.t. x

a) b)

c) d)

e)

2. Evaluate the following integrals

a) b)

c) d)

3. Integrate the following functions w.r.t. x

a) b) c) d)

(Hint: a) sin2x + cos

2 x=1 b) sin

2x=1 – cos

2x c) sin2x=2 sin x cos x

d) Expand and integrate.)

4. Integrate the following functions w.r.t. x

a) b)

c) d)

5. Integrate the following functions w.r.t. x

a) b)

c) d)

6. Integrate the following functions w.r.t. x

a) b) x sin2x

c) d)

(Hint: a) b) u = x c) u = tan-1

x d) u = log x)

7. Integrate the following functions w.r.t. x

a) x tan2 x b) e

x(sin x + cos x) c) x

2 sin x

(Hint: a) u = x b) use integration by parts to and proceed c) take u =

x2, evaluate

8. Integrate the following functions w.r.t.x

a) b)

9. Integrate the following functions w.r.t. x

a) b)

10. Integrate the following functions w.r.t. x

a) b)

11. Integrate the following functions w.r.t. x

a) b)

12. Integrate the following function w.r.t. x

a) b)

13. Evaluate the following integrals

a) b)

(Hint: x2 – 1 = (x + 1) (x – 1). Write the given function as )

14. Evaluate the following integrals

a) b)

7.8 Answers

Self Assessment Question

1.

2. a) b)

c) d)

e)

3. a) b)

c) d)

4. a) b)

c) d)

5. a) b)

c) d)

6. a) b)

c)

d)

7. a) b)

c) d)

8. a)

b)

9. a) b)

10. a) b)

11. a) b)

12. a) b)

13. a) b)

14. a) The indefinite integral is 2 log (x + 3) – log (x + 2) + c

Answer: 2 log 4 – 3 log 3 + log 2

b) Answer:

Terminal Questions:

1. a) b)

c) d)

e)

2. a) b)

c) d)

3. a) b)

c) d)

4. a) sin (x3) b) – cos (log x)

c) log (log x) d) tan (xex)

5. a) b)

c) d)

6. a) b)

c) d)

7. a) b)

c)

8. a)

b)

9. a) b)

10. a)

b)

11. a) b)

12. a)

b)

13. a)

b)

14. a) The indefinite integral is (Put t = x2 and integrate).

Answer:

b) is the indefinite integral (Put t = x4 and integrate).

Answer:

BT0063-Unit-08-Differential Equations

Unit 8 Differential Equations

Structure

8.1 Introduction

Objectives

8.2 First Order Differential Equations

8.3 Practical Approach to Differential Equations

8.4 First Order and First Degree Differential Equations

8.5 Homogeneous Equations

8.6 Linear Equations

8.7 Bernoulli’s Equation

8.8 Exact Differential Equations

8.9 Summary

8.10 Terminal Questions

8.11 Answers

8.1 Introduction

Differential equation is a branch of Mathematics which finds its application in a variety of fields.

First of all a given problem is converted to differential equations, which is then solved and the

solution to the problem is found out.

Objectives:

At the end of the unit you would be able to

• to find the solution of Differential Equations

• apply differential equations in practical situations

8.2 First order Differential Equations

Definitions

A differential equation is an equation which involves differential coefficients or differentials.

Thus,

i) ii)

iii) iii)

v)

are all examples of differential equations.

An ordinary differential equation is that in which all the differential coefficients have

reference to a single independent variable. Thus the equation (i) to (iv) are all ordinary

differential equations.

A partial differential equation is that in which there are two or more independent variables and

partial coefficients with respect to any of them. Thus equation (v) is an example for the partial

differential equation.

The order of a differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it, after

the equation has been expressed in a form free from radicals and fractions as far as the

derivatives are concerned.

Thus from the examples above,

i) is of the order and first degree;

ii) is the second order and first degree;

iii) can be written as and is clearly a first order but second degree

equation.

8.3 Practical Approach to Differential Equations

Differential equations arise from many problems in oscillations of mechanical, electrical

systems, bending of beams, conduction of heat, velocity of chemical reactions etc., and as much

play a very important role in al modern scientific and engineering studies.

The approach of an engineering student to the study of differential equations has got to be

practical unlike that a student of mathematics, who is only interested in solving the differential

equations without knowing as to how the differential equations are formed and how their

solutions are physically interpreted.

Thus for the applied mathematics, the study of the differential equation consists of 3 phases:

i) Formation of the differential equation from the given problem

ii) Solution of this differential equation, evaluating the arbitrary constants from the given

conditions, and

iii) Physical interpretation of the solution

Formation of a differential equation

An ordinary differential equation is formed in an attempt to eliminate certain arbitrary constants

from a relation in the variables and constants. In applied mathematics, every geometrical or

physical problem when translated into mathematical symbols gives rise to a differential equation.

Examples

From the differential equation of simple harmonic motion given by t,

Solution: To eliminate the constant A and ∝, differentiating it twice, we have,

and

thus, is the desired differential equation.

Example: Form the differential equation of all circles of radius a.

Solution: The general equation of a circle with centre at (h, k) is given by

(x – h)2 + (y – k)

2 = a

2 …….. (1)

Where h and k, the co-ordinates of the centre, and a are the constants. Differentiate twice, we

have,

and

Then,

And x – h = -(y – k)

Substituting these in (1), and simplifying, we get,

It states that the radius of curvature of a circle at any point is constant

Solution of a differential equation

A solution (or integral) of a differential equation is a relation between the variables which

satisfies the given differential equation.

For example, ———— (1)

Is a solution of ———— (2)

The general or complete solution of a differential equation is that in which the number of

arbitrary constants is equal to the order of the differential equation. Thus, (1) is a general solution

of (2) as the number of arbitrary constants (A, α) is the same as the order of (2).

A particular solution is that which can be obtained from the general solution by giving

particular values to the arbitrary constants

For example,

is the particular solution of the equation (2) as it can be derived from the general solution by

putting

8.4 First Order and First Degree Differential Equations

One can represent the general and particular solution of a differential equation geometrically. But

it is not possible to solve a family of parabola

y = x2 + c, in this approach. We shall, however, discuss some special methods of solution which

are applied to the following types of equations:

(i) Equations where variables are separable

(ii) Homogeneous equations

(iii) Linear equations

(iv) Exact equations

Variables separable

If in any equation if it is possible to collect all functions of x and dx on one side and all the

functions of y and dy on the other side, then the variables are said to be separable. Thus the

general form of such equation is f(y), dy = ϕ(x) dx.

Integrating both the sides, we get,

as its solution.

Example:

Solve

Solution: Given equation is

Integrating on both the sides,

8.5 Homogeneous Equations

Homogeneous Equations are of the form

Where f(x, y) and ϕ(x, y) are homogeneous functions of the same degree in x and y.

To solve a homogeneous equation,

i) Put y = vx, then

ii) Separate the variables v and x and then integrate.

Example:

Solve

Solution: The given equation is which is homogeneous in x and y.

Put y = vx then

Then the given problem becomes,

or

Separating the variables, we get,

Integrating both the sides

or

or

or

or

or

Hence the required solution is

Equations reducible to homogeneous form

The equations of the form ———- (1)

can be reduced to the homogeneous form as follows:

Case 1: when

Putting x = X + h, y = Y + k (h, k being constants)

So that, dx = dX = dY, becomes,

———– (2)

Choose h, k so that (2) may becomes homogeneous.

Put ah + bk + c = 0,

and

So that,

or

———- (3)

Thus when then (2), becomes

——— (4)

Which is a homogeneous in X, Y and can be solved by putting Y = vX.

Case II: When

i.e. the above method fails as h and k become infinite or in determinant

Now,

and (1) becomes

Put ax + by = t, so that

or

therefore we have

so that the variables are separable. In this solution, putting t = ax + by, we get the required

solution of (1).

Examples

1. Solve

Solution: The given equation is [this is where

--------- (1)

Putting x = X + h, y = Y + k (h, k being constants)

So that dx = dX, dy = dY, (1) becomes

--------- (2)

Put x + h – 2 = 0 and k – h – 4 = 0,

So that, h = –1, k = 3.

Therefore (2) becomes,

Which is homogeneous in X and Y.

Put Y = vX, then

Therefore (3) becomes,

Or

Integrating both the sides

Or

Or

Or log (X2 + 2XY – Y

2) = –2c

Or X2 + 2XY – y

2 = e – 2c =

Putting X = x – h = x + 1, and Y = y – k – 3, equation (4) becomes,

Which is the required solution.

8.6 Linear Equations

A differential equation is said to be linear if the different variable and its differential coefficients

occur only in the first degree and are not multiplied together.

Thus the standard form of a linear equation of the first order, commonly known as Leibnitz’s

linear equation, is

---------- (1)

Where P, Q are any functions of x.

To solve the equation, multiply both the sides by so that we get,

Integrating both the sides

As the required solution

Example

1. Solve

Solution: The given equation is,

Dividing both the sides by (x + y), given equation becomes,

---------- (1)

Which is Leibnitz’s equation.

Here

Therefore,

And

Thus the solution of (1) is,

8.7 Bernoulli’s Equation

The equation -------- (1)

Where P, Q are functions of x, is reducible to the Leibnitz’s linear equation and is usually called

the Bernoulli’s equation.

To solve (1), divide both the sides of yn, so that

-------- (2)

Put so that

(2) Becomes,

Which is Leibnitz’s linear in z and can be solved easily.

Examples

1. Solve

Solution: The given equation is, -------- (1)

Dividing throughout by xy6,

-------- (2)

Put y–5

= z, so that

Therefore (2) becomes,

Or --------- (3)

Which is Leibnitz’s linear in z, and the intermediate form I.F. is,

I.F. =

Therefore the solution of (3) is,

Or [since z = y–5

]

Dividing both sides by we get,

1 = (2.5 + cx2) x

3 y

5

8.8 Exact Differential Equations

1. Definition: A differential equation of the form

Is said to be exact if its left hand member is the exact differential of some function u(x, y).

i.e.,

Its solution therefore, is u(x, y) = c.

2. Theorem: The necessary and sufficient condition for the differential equation Mdx + Ndy =

0 to be exact is

Condition is necessary:

The equation M(x, y) dx + N(x, y) dy = 0 will be exact, if Mdx + Ndy

≡ du —– (1)

Where u is some function of x and y.

But ——- (2)

Equating coefficients of dx and dy in (1) and (2), we get

and

and

But, (Assumption)

Which is the necessary condition of exactness.

Condition is sufficient:

i.e., if then Mdx + Ndy = 0 is exact.

Let where u is supposed constant while performing integration. Then

——- (3)

since (given)

Or and

Integrating both the sides, w.r.t x (taking y as constant

, where f(y) is a function of y alone. ——– (4)

[by (3) and (4).]

——– (5)

Which shows that Mdx + Ndy = 0 is exact.

Method of solution: By equation (5), equation Mdx + Ndy = 0 becomes

Integrating

But =terms of N not containing x.

The solution of Mdx + Ndy = 0 is

(terms of N not containing x) dy = c

Provided

Examples

1. Solve

Solution: Given equation can be written as,

(y cos x + sin y + y) dx + (sin x + x cos y + x) dy = 0

Here,

M = y cos x + sin y + y

N = sinx + x cos y + x

Thus the equation is exact and its solution is

i.e.,

Or y sin x + (sin y + y)x = c

Self Assessment Questions

1. Solve

2. Solve

3. Solve

8.9 Summary

In this unit we study the first order differential equations. The practical approach to differential

equations is clearly explained with suitable examples. Solving first order first degree differential

equation by the variable separable method is discussed here. Equations reducible to

homogeneous form is discussed here with example. The linear equations, Bernoulli’s equation

and exact differential equation is solved here in a simple manner with proper examples.

8.10 Terminal Questions

1. Derive the necessary and sufficient condition for the differential equation Mdx + Ndy to be

exact

2. Briefly describe Bernoulli’s equation

8.11 Answers

Self Assessment Questions

1. Dividing throughout by cos2y,

——– (1)

Put tan y = z, so that

Therefore equation (1) becomes,

Which is Leibnitz’s linear equation in z.

Therefore the solution is,

Replacing z by tan y, we get

Which is the required form

2. This equation contains y2 and tan

–1 y and is, therefore not a linear in y, but since only x

occurs, it can be written as,

Which is leibnitz’s equation in x.

Therefore, we have intermediately form I.F. as,

Thus the solution is,

Or

Therefore

Or

3. The given equation is, ——– (1)

Putting 2x + 3y = t, so that

(1) becomes,

Integrating both the sides,

Or

Or

Putting t = 2x + 3y, we have

14(2x + 3y) – 9 log (14x + 21y + 22) = 49x + 49 c

Or

Which is the required solution

Terminal Questions

1. Refer to Section 16.8

2. Refer to Section 16.7

BT0063-Unit-09-Complex Numbers

Unit 9 Complex Numbers

Structure

9.1 Introduction

Objectives

9.2 Complex Numbers

9.3 Conjugate of a Complex Number

9.4 Modulus of a Complex Number

9.5 Geometrical Representation of Complex Number

9.6 Exponential Form of a Complex Number

9.7 De Moivere’s* Theorem

9.8 nth

Roots of a Complex Number

9.9 Summary

9.10 Terminal Questions

9.11 Answers

9.1 Introduction

We recall that, if x and y are real numbers and then x + iy is called a complex number.

The complex numbers were first introduced by Cardan (1501 – 1576). Two hundred years later

Euler (1707 – 1783) and John Bernoulli recognized the complex numbers introduced by Cardan

and studied their properties in detail. In 1983, Sir William Rowan Hamilton (1805 – 1865) an

Irish mathematician introduced the complex number as an ordered pair of real numbers. In this

chapter we begin the study of complex numbers as ordered pairs.

Objectives:

At the end of the unit you would be able to

• understand the concept of complex numbers.

• apply De Moivere’s Theorem in finding the roots of complex numbers.

9.2 Complex Numbers

Let C denote the set of all ordered pairs of real numbers.

That is,

On this set C define addition “+” and multiplication “.” by,

(x1, y1) + (x2 + y2) = (x1 + x2, y1 + y2) … (1)

(x1, y1) . (x2, y2) = (x1x2 – y1y2, x1y2 + x2y1) … (2)

Then the elements of C which satisfy the above rules of addition and multiplication are called

complex numbers. If z = (x, y) is a complex number then x is called the real part and y is called t

he imaginary part of the complex number z and they are denoted by x = Re z and y = Im z. If (x1,

y1) and (x2, y2) are two complex numbers then (x1, y1) = (x2, y2) if and only if

x1 = x2 and y1 = y2.

(a) Properties of addition

1. Closure law: If z1 = (x1, y1), z2 = (x2, y2) then from (1) z1 + z2 = (x1, y2) + (x2, y2)

= (x1 + x2, y1 + y2), which is also an ordered pair of real numbers. Hence .

Therefore for every .

2. Commutative law: z1 + z2 = z2 + z1 for every

Consider z1 + z2 = (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2)

= (x2 + x1, y2 + y1) = (x2, y2) + (x1, y1) = z2 + z1.

3. Associative law: z1 + (z2 + z3) = (z1 + z2) + z3 for every

Proof of this is similar to (2)

4. Existence of identity element: There exists an element such that,

(x, y) + (0, 0) = (x + 0, y + 0) = (x, y).

for every . Here (0, 0) is called the additive identity element

of C.

5. Existence of inverse: For every there exists such that

(x, y) + (–x, –y) = (x – x, y – y) = (0, 0).

Hence (–x, –y) is the additive inverse of (x, y).

Thus we have shown that the set C is an abelian group w.r.t. the addition of complex

numbers defined by (1).

(b) Properties of multiplication

1. Closure law: If z1 = (x1, y1), z2 = then from (2)

z1z2 = (x1, y1) (x2, y2) = (x1x2 – y1y2, x1y2, x1y2 + x2y1), which is also an ordered pair of real

numbers. Hence z1z2

is also a complex number.

Thus for every .

2. Commutative law: z1z2 = z2z1

for every .

Now z1z2 = (x1, y1) (x2, y2) = (x1x2 – y1y2, x1y2 + x2y1) ….. (i)

and z2z1 = (x2, y2) (x1, y1) = (x2x1 – y2y1, x2y1 + x1y2)

= (x1x2 – y1y2, x1y2 + x2y1) ….. (ii)

From (i) and (ii) z1z2 = z2z1.

3. Associative law: z1(z2z3) = (z1z2) z3, for every Proof is similar to (2)

4. Existence of identity element: There exists such that

(x, y) (1, 0) = (x . 1 – y . 0, x . 0 + 1 . y) = (x, y) for every .

Here (1, 0) is called the multiplicative identity element.

5. Existence of inverse: Let z = (x, y) (0, 0), be a complex number. Let (u, v) be the

inverse of (x, y).

Then (u, v) . (x, y) = (1, 0), the identity element.

i.e. (ux – vy, uy + vx) = 1, 0).

Hence ux – vy = 1, and uy + vx = 0.

Solving for u and v, we get,

Hence is the multiplicative inverse of (x, y).

Thus we have shown that the set of non-zero complex numbers forms an abelian group w.r.t. the

multiplication defined by (2).

Also we can prove that the multiplication is distributive over addition.

(c) Distributive law: For all

1. z1 (z2 + z3) = z1z2 + z1z3 (left distributive law)

2. (z2 + z3) z1 = z2z1 + z3z1 (right distributive law)

The complex numbers whose imaginary parts are equal to zero possess the following properties.

(x1, 0) + (x2, 0) = (x1 + x2, 0).

and (x1, 0) . (x2, 0) = (x1 x2, 0).

Which are essentially the rules for addition and multiplication of real numbers. We identify the

complex number (x, 0) with the real number x. Denote the complex number (0, 1) by i.

Now i2 = I . I = (0, 1) (0, 1) = (0 . 0 – 1 . 1, 0 . 1 + 1 . 0)

= (–1, 0) = –1.

Hence i2 = –1.

With this convention we shall show that the ordered pair (x, y) is equal to x + iy.

For, (x, y) = (x, 0) + (0, y)

= (x, 0) + (0, 1) (y, 0)

= x + iy

Since (x, 0) = x, (y, 0) = y and (0, 1) = i.

Because of the extreme manipulative convenience we shall continue to use the notation x + iy for

the complex number (x, y).

9.3 Conjugate of a Complex Number

Let z = x + iy be a complex number. Then the complex number x – iy is called the complex

conjugate or simply, the conjugate of z and is denoted by

Thus, if z = x + iy then

For example, if z = 3+4i then

Clearly

and

Also,

= x2 – i

2y

2

=x2 + y

2, which is a real number.

Thus the product of two conjugate complex numbers is a real number.

Theorem: For all

1.

i.e., the conjugate of a sum is equal to the sum of the conjugates.

2.

i.e., the conjugate of a product is equal to the product of the conjugates.

3.

i.e., the conjugate of a quotient is equal to the quotient of then conjugates.

9.4 Modulus of a Complex Number

If z = x + iy is a complex number then is called the modulus or absolute value of z and

is denoted by | z |.

Thus

Clearly | z | is a non-negative real number i.e.,

then

We can easily verify the following:

1. 2. 3.

Theorem: For all

1.

i.e., modulus of a product is equal to the product of their moduli.

2.

i.e., modulus of a quotient is equal to the quotient of the moduli.

3.

4.

9.5 Geometrical Representation of Complex Number

A complex number x + iy can be represented by a point P(x, y) in the Cartesian plane with x as

the abscissa and y as the ordinate. Thus every point on the x-axis corresponds to a real number

and every point on the y-axis corresponds to a pure imaginary number (iy) and vice versa. Hence

x-axis is called the real axis and y-axis, the imaginary axis. And the plane whose points are

represented by complex numbers is called the complex plane or Argand plane named after the

French mathematician J.R. Argand (1768 – 1822). Although the geometric representation of

complex numbers is usually attributed to J.R. Argand but it was Casper Wessel of Norway (1745

– 1818) who first gave the geometric representation of complex numbers.

Now draw PM perpendicular to the x-axis. Let and OP = r. Clearly OM = x and MP = y.

Now

Hence x + iy = r (cos θ + i sin θ)

Thus every complex number z = x + iy can be represented in the form This form

of a complex number is called the polar form or the trigonometric form.

Squaring and adding the equations given in (1), we get

which is the modulus of the complex number z = x + iy.

Thus represents the distance of the point z from the origin.

The angle θ is called the argument or the amplitude of z and is denoted by

θ = arg z or θ = amp z.

Since sin (2nπ + θ) = sin θ, cos (2nπ + θ) = cos θ, when n is any integer, θ is not unique. The

value of θ satisfying –π < θ

<

π is called the principal value of the argument.

Note:

1. If z1 = x1 + iy1

and z2 = x2 + iy2

Then z1 – z2 = (x1 – x2) + i(y1 – y2).

which is the distance between the points z1 and z2.

2. cos θ + I sin θ is briefly denoted by cis θ

Theorem: 1.

2.

Theorem: ����

1. amp z1z2 = amp z1 + z1 + amp z2

2.

Remark:

1. To find the amplitude of a complex number we use the following rule:

For example,

1. if then

2. If then

3. then

4. if then

2. The value of the amplitude θ must satisfy the equations and where

Some times we combine these equations dividing one by another. In that case we get

or Because of the difference in principal values of the

value of the argument is not necessarily the principal value of For example,

and so but

9.6 Exponential Form of a Complex Number

If x is real, it can be proved, in the advanced mathematics that the functions ex, sin x, cos x etc.

can be expressed in the form of an infinite series.

i.e. … (1)

… (2)

… (3)

Assuming that (1) holds good for a complex number also, replacing x by ix in (1) we get,

= cos x + i sin x

Thus

This is called the Euler’s formula.

We know that a complex number z = x + iy can be expressed in the polar form as

where

The complex number can also be written in the form

This is called the exponential form of a complex number.

Example: Express the following complex numbers in the polar form and hence find their

modulus and amplitude.

1) 2) 1 – i 3)

Solution:

1) Let .

On equating the real and imaginary parts, we get

and r sin θ = 1.

Squaring and adding,

Hence

Hence

Therefore modulus = 2 and amp

2. Let 1 – I = r (cos θ + i sin θ)

r cos θ = 1, r sin θ = −1

.

Therefore

i.e. Modulus =

3. Let

Hence r cos

Hence

Modulus = 2, amp

Example: If a = cos θ + i sin θ, prove that

Solution:

Multiplying and dividing b y i,

Self Assessment Questions

1. Find the smallest positive integer n such that

Example: If prove that

Solution:

Now . Taking the conjugate on both sides

We get,

Multiplying,

9.7 De Moivere’s* Theorem

If n is any integer then

And if n is a rational fraction say then has q values and one of its values is

Proof: Case (i) Let n be a positive integer.

In this case we shall prove (1) by mathematical induction.

Hence (1) is true for n = 1. Assume that (1) is true for n = m,

i.e., (Induction hypothesis)

Multiplying both sides of (2) by we get

(cos θ + i sin θ)m+1 = (cos mθ + i sin mθ) cos θ + i sin θ)

Hence the theorem is true for n = m + 1.

Hence by mathematical induction the theorem is true for all positive

integers n.

Case (ii). Let n be a negative integer.

n = – m, where m is a positive integer.

Consider

Case (iii). Let n be a rational fraction i.e., where p and q are integers and q > 0.

Let

= cos p + i sin p

zq = (cos + i sin )

p, from Cases (i) and (ii),

which is an algebraic equation of degree q. Hence from fundamental theorem of algebra it has q

roots. Therefore taking qth

root on both sides, we get

Hence has q values and one of them is

This completes the proof of the theorem.

Self Assessment Questions

2. Simplify

Example: Prove that where n is any integer

Solution:

Expressing in the polar form, we get

Taking conjugate on both sides, we get

L.H.S. =

Using De Moivre’s theorem

since cos 2nπ = 1

= R.H.S.

Example: Prove that

Where n is any integer.

Solution:

From De Moivre’s theorem.

9.8 nth

Roots of a Complex Number

If zn = a, where a is a non-zero complex number and n, is a positive integer then z is called the

nth

root of a. Since the given equation is of degree n, there are n roots of the equation. Hence

solving zn = a, we obtain n, n

th roots of a.

Example: Find the cube roots of and represent them on the Argand plane. Also find their

continued product.

Let

Squaring and adding r2 cos

2

θ + r2 sin

2

θ = 3+1

r2 = 4, r = 2

Hence;

(Principal value)

Substituting n = 0, 1, 2 (or any three consecutive values of n), we obtain the cube roots of

They are �

i.e.,

To represent these roots on the Argand plane consider a circle whose centre is at the origin and

whose radius is Since modulus of each of these roots is these roots lie on the circle.

In above figure the points A, B, C represent the cube roots of

Since are the vertices of an equilateral triangle

Continued product =

Note:

1. The cube roots of unity are where Also,

2. The fourth roots of unity are

3. In general the nth roots of unity are where

9.9 Summary

In this we discuss about the concept of complex numbers in detail. The idea of representing a

complex number in Polar Form is explained in a simple manner. The method of finding the roots

of a Complex Numbers using De Moivere’s Theorem is well illustrated.

9.10 Terminal Questions

1. State and prove De Movere’s Theorem

2. Find the Cube roots of the Complex Numbers 1+i and express it in the Argand Diagram

9.11 Answers

Self Assessment Questions

1. Now

Therefore

Now by inspection n = 4 is the smallest positive integer such that

2. Given expression

= cis 21

= cos 21 + i sin 21

BT0063- Unit-10 -Matrices and Determinants

Unit 10 Matrices and Determinants

Structure

10.1 Introduction

Objectives

10.2 Definition of a Matrix

10.3 Operations on Matrices

10.4 Square Matrix and Its Inverse

10.5 Determinants

10.6 Properties of Determinants

10.7 The Inverse of a Matrix

10.8 Solution of Equations Using Matrices and Determinants

10.9 Solving equations using determinants

10.10 Summary

10.11 Terminal Questions

10.12 Answers

10.1 Introduction

The theory of matrices, introduced by French mathematician Cayley in 1957, is presently a

powerful tool in the study of branches of Mathematics, Physical sciences, biological sciences and

business applications. The concept was initially developed for solving equations.

Objectives:

At the end of the unit you would be able to

• solve determinant using their properties

• find the solution of equations using matrices and determinant

10.2 Definition of a Matrix

Definition: A matrix A is a rectangular array of numbers arranged as m horizontal lists, called

rows, each list having n elements; the vertical lists are called columns.

It is written as

Note: The element in the ith

row and jth

column is aij. So A is also written as

or simply A is called an m × n matrix. We also write the (I, j)th

entry as (A)ij.

When m = n, A is called a square matrix (also called an n – square matrix A)

Example ,

are matrices respectively.

Definition: Two matrices are equal if (i) the number of rows of A and B are

the same (ii) the number of columns of A and B are the same (iii) for all i, j.

Example: Find the values of a, b, c, d if

Solution: Equating the corresponding entries of the matrix, we get

2a + 2b = 6 2a – 2b = 2

2c + d = 14 c – d = 10

By adding the first two equations we get 4a = 8. so a = 2.

2b = 6 – 2a = 6 – 4 = 2. So b = 1

Similarly,

3c = 24. So c = 8, d = 14 – 2c = 14 – 16 = –2

S.A.Q.1 How many entries are there in an m × n matrix ?

10.3 Operations on Matrices

In this section we define sum of two matrices, difference of two matrices, the scalar multiple of a

matrix A by a scalar (real number) k.

An m × n matrix is said to be of size (m, n)

Definition: If are two m × n matrices then A + B is defined as an m × n

matrix as follows:

Note: For getting the sum of A and B, we add to an entry in A, the entry in B in the same place.

We can add only two matrices of the same size.

Definition: If are two m × n matrices then A – B is defined as an m × n

matrix as follows:

Example: If and find A + B, A – B, 2A + 3B

Solution:

We list some special matrices in the next example.

Example:

1. is called the zero matrix.

We can write it as O. We can have zero matrix of any order.

1. is called then n – square unit matrix.

(Note: The number of rows is equal to the number of columns in the unit matrix. It is

denoted by In or simply I when n is understood.)

1. is called a diagonal matrix. A square matrix is diagonal matrix if only the

entries on the diagonal are nonzero and other entries are s.

2. A diagonal matrix having the same number along the diagonal is called a scalar matrix. A

scalar matrix is simply kIn for some scalar k and some positive integer n.

Definition: If is an matrix then the transpose of A denoted by is defined as

Note: The transpose of an matrix is an matrix.

Just as addition of real numbers is commutative, the addition of matrices is also commutative.

As far as addition is concerned matrices behave like numbers. The following theorem lists

properties of addition of matrices.

Theorem: If A, B, C and matrices and k and I are scalars.

1. A + (B + C) = (A + B) + C

2. A + 0 = 0 + A where 0 is the m × n zero matrix

3. A + (–A) = (–A) + A = 0 (Here –A denotes (–1) A)

4. A + B = B + A

5. k(A + B) = kA + kB

6. (k + I) A = kA + IA

7. (kI) A = k (lA) = l(kA)

8. lA = A

9. (A + B)T = A

T + B

T

10.

11.

The theorem can be proved by using the definition of sum of two matrices and scalar multiple of

a matrix.

If A and B are two matrices AB is defined only when the number of columns of A = number of

rows of B. We defined AB in Definition 4.7.

Definition: If A is an m × n matrix and B is an n × p matrix then the product of A and B,

denoted by AB, is an m × p matrix and is defined by

for

Note (AB)ik

can be understood as follows.

is the ith row of A, Is the k

th column of B and both these have n elements.

For calculating (AB)ik, multiply the respective elements of ith row of A and k

th column of B and

add them. The resulting number is (AB)ik.

Example: Find AB when

and

Solution A is a 2 × 3 matrix and B is a 3 × 2 matrix. So AB is a 2 × 2 matrix.

S.A.Q. 2 Find BA for matrices A and B given in above example

We have seen that A + B = B + A when A and B are matrices of the same size. But AB ≠ BA in

general. It can happen that one of the products is defined whereas the other product is not

defined. Let us illustrate this with an example.

Example: Find two matrices A and B such that

a) AB is defined but BA is not

b) BA is defined but AB is not

c) Both are defined but AB ≠ BA

d) Both are defined and AB = BA

Solution:

1. Assume and

Then AB is defined as the number of columns of A = 3 = number of rows of B.

Number of columns of B = 3 ≠ number of rows of A. Hence BA is not defined.

2. If and then BA is defined, as number of columns of B = 3 =

number of rows of A.

Number of columns of A = 1 ≠ number of rows of rows of B. Hence AB is not defined.

3. Assume and

Hence

4. Consider and

Then

Thus AB = BA

Theorem 4.2 lists the properties of multiplication.

Theorem: If A is an m × n matrix and B is an n × p matrix and k is any scalar, then

1. (AB)T = B

TA

T

2. 3. k(AB) = (kA)B = A(kB)

4. OA = O, BO = O where the four zero matrices are k × m, k × n, p × t and n × t matrices

respectively (for some k and t).

The proof of theorem 10.2 follows from the definition of product of two matrices.

We can define the product of 3 matrices A, B, C when

……………… (10.1)

The following theorem describes the properties of product of three matrices. Theorem 10.3 Let

A, B, C be 3 matrices. Then the following hold good whenever the sums and products of

matrices appearing below are defined.

a) (AB) C = A(BC) (Associative law)

b) A(B + C) = AB + AC (Left distributive law)

c) (B + C) A = BA + CA (Right distributive law)

Theorem 10.3 follows from the definition. The proof is technical in nature and it is enough if you

remember these properties.

S.A.Q. 3 If and find A + B, 2A – 3B, 3B – 2A, (A – B)T and (B – A)

T.

S.A.Q. 4 If and

Verify that (A + B)T = A

T + B

T

S.A.Q. 5 Find a matrix A such that

S.A.Q. 6 If and find X and Y.

S.A.Q. 7 A matrix A is said to be symmetric if A = AT. Show that A + A

T is symmetric for a 3 ×

3 matrix A

S.A.Q. 8 If and

Show that Does BA exist ?

S.A.Q.9 If

show that (A – I) (A + 2I) =

10.4 Square Matrix and its Inverse

We know that a square matrix is an n × n matrix for some integer n. The set of n × n square

matrices satisfy some additional properties.

In theorem 4.2 we saw that AIn = InA = A for any n × n square matrix A. We can multiply 2n × n

matrices and the product is an n × matrix. In general we can define AA, AAA etc.

We define powers of a square matrix as follows. We define A0 = In,

….. (10.2)

The set of all n × n matrices satisfy the properties of indices (powers) of numbers.

Note Am

An = A

m+n = A

n . A

m ……… (10.3)

If a is a non zero real number then we know that

A similar property holds good for some square matrices. In the case of numbers is called the

reciprocal of a. But in the case of matrices it is called the inverse of a square matrix.

Definition: A square matrix A is invertible (or non singular) if there exists a square matrix B

such that AB = BA = In ………….. (10.4) B is called the inverse of A and is denoted by A – 1.

Note: If A has an inverse then A is called as invertible matrix.

Example: Let If then

Hence is the inverse of A.

Theorem: If B is the inverse of a square matrix A then A is the inverse of the matrix B.

Proof: As B is the inverse of A, by definition (4.8), AB = BA = In ………. (10.5)

So BA = AB = In ………………………….. (10.6)

From (10.6) we see that A satisfies the condition for the inverse of B. Hence A is the inverse of

B.

Note In the case of numbers is the reciprocal of a and a is the reciprocal of . Theorem 10.4

guarantees this property for square matrices.

We are going to see a method for finding the inverse of a matrix. However you will have a

formula for the inverse of a 2 × 2 matrix (Example 10.7)

Example: If then ………….. (10.7)

For the present, you can verify that

S.A.Q. 10 If show that

S.A.Q. 11 Verify that is the inverse of

10.5 Determinants

The determinant of an n – square matrix A is a unique number associated with A and is denoted

by det (A) or | A|. | A | is called a determinant of order n.

If then | A | is denoted by

As the definition of | A | is complex for a general n – square matrix A, we define determinant of

orders 1, 2, 3 and then extend it for a general n – square matrix A.

Evaluation of determinants

Definition: The determinants of orders 1, 2, 3 are defined as follows

a)

b)

(We can understand the determinant in the following way).

We (1) multiply the elements in the diagonal from left to right (ii) multiply the elements in the

diagonal from right to left (iii) subtract product got in (ii) from the product got in (i)

c)

Note: We can calculate the value of a determinant of order 3 as follows:

1. Consider the first element a11 in the first row. Attach the sign + (plus)

2. Delete the row and column in which a11 appears; that is first row and first column. We

get

3. Multiply + a11 and the value of

4. Consider the second element a12 in first row. Attach the sign – (minus)

5. Delete the row and second column in which a12 appears; that is the first row and second

column. We get

6. Multiply – a12 and the value of

7. Consider the third element a13 in the first row. Attach the sign + (plus)

8. Delete the row and column in which a13 appears; that is the first row and third column.

We get

9. Multiply + a13 and the value of

Add the values got in steps 3, 6 and 9. This is the value of the given determinant.

Note: We usually denote a determinant by the symbol ∆ (read as Delta).

Example: Evaluate the determinant

Solution:

= 0 – 2[1(4) – 2(7)] + 3[1(0) – 2(4)]

= –2(–10) – 24

= 20 – 24

= –4

Thus

Evaluation of a determinant in term of any row or column

Recall the definition 10.9(c). we obtained

The 2 – order determinants are called the minors of a11, a12, a13.

We can denote the minors by M11, M12, and M13. If we attach the signs, these are called cofactors

of a11, a12, a13. We denote them by A11, A12, A13. Then (10.7) can be written as

We can also define minors of a21, a22, a23, a31, a32, a33 in a similar manner. For example.

(M23 is got by deleting the second row and third column of | A |)

The cofactors can be defined in a similar manner using the rule of signs given by (10.8)

………………………….. (10.8)

Any cofactor is got by multiplying the minor and its sign given in (4.8). For example cofactor for

a32 is – M32.

So

Thus we can expand ∆ in term of any row or column in a similar way. (10.9) gives the expansion

in term of various rows and columns.

…………………………………. (10.9)

You may wonder why so many expansions given in (10.9) are necessary. If a row or column has

many zeros then evaluating by the elements of that row or column makes the evaluation simpler.

Example:: Evaluate

Solution: As the second row has two zeros, we expand by the elements of the second row.

(Note: For (2, 1) position, the sign is –. See (4.8)).

= – 4[2(5) – 4(4)]

= – 4(10 – 16)

= – 4(–6)

= 240

The evaluation of a determinant of order n is similar.

For example, if

Then | A | = a11A11 + a12A12 + a13A13 + a14A14

The signs of cofactors can be defined by

………………………… (10.10)

Example: Evaluate

Solution As the third row has two zeros we expand by the elements of third row. The signs of

cofactors of A determined by using (10.10)

| A | = a31A31 + a32A32 + a33A33 + a34A34

= 4A31 + 0A32 + 0A33 + 1A34

= 4[4(–12) –7 (–8) + 2 (–4)] –1[1 (–4) –4 (–2) + 7(0)]

= 4(–56 + 56) –1 (4)

= 4(0) – 4

= – 4

Thus | A | = –4

S.A.Q. 12: Evaluate the following determinants

a) b) c)

S.A.Q. 13: Evaluate the following determinants

a) b) c)

10.6 Properties of Determinants

In this section we list some properties of determinants. These properties enable us to evaluate a

determinant in an easier way.

Property 1: If then

Let A11, A12, a13 denote the cofactors of a11, a12, a13 in ∆. These are also t he cofactors of ma11,

ma12, ma13, in m∆.

So = ma11 A11 + ma12A12 + ma13A13

= m (a11A11 + a12A12 + a13A13)

Note: This property holds good when any row or column of ∆ is multiplied by m. This property

essentially means that any common factor of a row or column can be taken outside the

determinant.

Remark: If A is a matrix then mA is got by multiplying each entry of A by m. In the case of

determinant m∆ is got by multiplying the entries of a single row or column by m.

Example:

Property 2: If |A| = det (a), then det (AT) = |A|

Proof: If then

So the second order minors of A and AT have the same value. As the sign of a cofactor is the

same in both A and AT, the value det (A

T) through expanding along the first column is equal to

det (A) through expanding along the first row.

Hence dt (AT) = |A|

Property 3: If two rows or columns of a determinant ∆ are interchanged then the value of the ∆

is unchanged but the sign is changed.

Proof: Let If the second and third rows are interchanged we get

= a11 (a32a23 – a22a33) – a12 (a31a23 – a21a33) + a13 (a31a22 – a32a21)

Property 4: If a determinant ∆ has two identical rows then the value of the determinant is 0.

Proof: If we interchange two identical rows then the value of the new determinant is . (By

property 3). But the new determinant is the same as

So

Example: Evaluate

Solution: The first and fourth rows of ∆ are identical. By property 4, ∆ = 0.

Property: The value of a determinant remains the same when multiple of some rows are added

to a particular row. The same is true for columns.

Note: For example,

Since we add k times the second row and l times the third row to the first row.

Example: Evaluate

Solution The first entry in the first row (R1) is 4. To make it 0, we subtract 4 times the third row

(R3). Thus R1 of ∆ is replaced by R1 – 4R3 (This is indicated on the right of the determinant).

Similarly if we subtract 2R3 from R2 we get 0, as the first element in R2. In the resulting

determinant the first column has two zeros and a one. This makes the evaluation (along the first

column) easier.

(by expanding along the first column)

= 0 + 0 +1 (4 – 0)

= 4

Example: Evaluate

Solution

= a (bc – 0)

= abc.

Example: Show that

Solution

= (a – b) (a – c) (a + c) – (a – c) (a – b) (a + b)

= (a – b) (a – c) [a + c – a – b]

= (a – b) (a – c) (c – b)

= (a – b) [– (c – a)] [–(b – c)]

= (a – b) (b – c) (c – a)

S.A.Q. 14 Evaluate

S.A.Q. 15 Prove that

S.A.Q. 16 Evaluate

S.A.Q. 17 Evaluate the following determinants

a) b)

10.7 The Inverse of a Matrix

In this section we give a method of finding the inverse of a matrix.

Definition: If then the adjoint matrix of A (denoted by Adj A) is given by Adj

Example 10.17 Find the adjoint of matrix

Solution

We use Adj A for evaluating the inverse of a matrix.

A square matrix is invertible if and only if When the inverse of a matrix A is

given by

………………………. (10.11)

If |A| = 0, the matrix a is called singular; otherwise it is non singular. So a matrix is invertible if

and only if it is nonsingular.

Example: Find the inverse of

Solution A11 = d, A12 = –c, A21 = –b, A22 = a.

So Adj

|A| = ad – bc. Hence

………. (10.12)

Example: Find the inverse of

Solution

Adj

|A| = a11A11 + a12A12 + a13A13

= 1 (0) = 1(3) + 1(3)

= 6

S.A.Q.18 Find the inverse of

S.A.Q. 19 Test whether A-1

exists when

S.A.Q. 20 Find the inverse of

10.8 Solution of Equations using Matrices and Determinants

Matrices are useful in representation of data. For example if we want to classify the students of a

class in terms of gender and their grades then we can use a matrix for representing the

information. Suppose we have three grades A, B, C. Then the matrix represents.

The classified data:

a11 denotes the number of male students who got grade A.

a12 denotes the number of male students who got grade B

a13 denotes the number of male students who got grade C.

a21 denotes the number of female students who got grade A.

a22 denotes the number of female students who got grade B.

a23 denotes the number of female students who got grade C.

Solving linear equations using matrices

We can also use matrices for solving n equations in n variables. The ideas is to represent n

equations as a single matrix equation and then solve the matrix equation. The next example

illustrates this.

Example: Solve x + y = 3

2x + 3y = 8

Solution The given system of equations is equivalent to the single matrix equation AX = B

where

Multiplying both side of AX = B by A–1

. we get X = A–1

B. By (10.12)

Hence x = 1, y = 2

Example: Solve the equations

x + y + z = 6

x + 2y + 3z = 14

–x + y – z = –2

Solution: The given systems of equations is equivalent to A X = B where

|A|=a11A11 + a12A12 + a13A13

= 1(–5) + 1(–2) + 1(3)

= –5 – 2 + 3

= –7 + 3

= –4.

Hence x = 1, y = 2, z = 3

10.9 Solving equations using determinants

We can also solve a system of n linear equations in n variables using determinants. The method

is provided by Cramer’s rule. Cramer’s rule for three equations in three variables.

Consider the system of three linear equation in three variables x, y, z.

a11x + a12y + a13z = b1

a21x + a22y + a23z = b2

a31x + a32y + a33z = b3

Let ∆ be the co-efficient determinant i.e., the determinant of the coefficients of the variables x, y,

z such that

(If ∆ = 0 the system has no unique solution).

By Cramer’s rule we have

or

Note: As in the previous section the system of equations Ax = B. Then |A| = ∆. Now ∆1 is

obtained by replacing the first column of ∆ by B. Similarly ∆2 and ∆3 are obtained by replacing

the second and third columns of ∆ by B respectively.

Example: Solve

2x + 3y + 4z = 20

x + y + 2z = 9

3x + 2y + z = 10

Solution: In this example,

∆ = 2(1 – 4) – 3 (1 – 6) + 4(2 – 3) = 5

∆1 = 20(1 – 4) – 3(9 – 20) + 4 (18 – 10) = 5

∆2 = 2(9 – 20) – 20(1 – 6) + 4 (10 – 27) = 10

∆3 = 2(10 – 18) – 3 (10 – 27) + 20(2 – 3) = 15

Hence

S.A.Q. 21. Solve the following system of equations using matrices

a) 2x + 3y – z = 9 b) 2x – y + 3z = – 9

x + y + z = 9 x + y + z = 6

3x – y – z = –1 x – y + z = 2

S.A.Q. 22 Solve the following system of equations using Cramer’s rule

a) 5x – 6y + 4z = 15 b) x + y + z = 9

7x + 4y – 3z = 19 2x + 5y + 7z = 52

2x + y + 6z = 46 2x + y – z = 0

10.10 Summary

In this unit we discuss about the concept of matrices and determinants. The different types

matrices is defined, the concept of inverse of matrix is well defined with good examples,

Determinants, the different properties of determinants and solving equations using matrices and

determinants is explained with the help of standard examples.

10.11 Terminal questions

1. Find the values of x, y, z and t satisfy the matrix relationship

2. Find the values for x, y, z that satisfy the matrix relationship

3. Find a matrix A satisfying

4. If and

Show that AB = AC. (In the case of real numbers, ab = ac will imply that b = c. But this is

not so for matrices as this example shows)

1. If and evaluate AB – BA.

2. If , show that

3. If , show that A2 – 2A – 5I = 0

4. If find AAT and A

TA

5. If Evaluate A2 and A

3.

6. If find A.

7. If find A

8. , find A.

9. Evaluate the following determinants

a) b) c)

1. Evaluate the following determinants

a) b) c)

1. Show that

2. Evaluate the following determinants

a) b)

1. Show that

2. Prove that

1. Solve the following system of equations using (i) matrices (ii) determinants

a) x + 2y – z = 3 b) 2x + 3y – z = 9

3x – y + 2z = 2 x + y + z = 9

2x – 2y + 3z = 2 3x – y – z = –1

c) a + b + z = 6 d) 2a + 3b + c = 8

a + 2b + 3c = 14 4a + b + c = 6

–a + b – z = –2 a + b + c = 3

10.12 Answers

Self Assessment Questions

1. mn entries

2.

3.

5.

6.

8. BA does not exist since the number of columns of B = 2 ≠ 3 = the number of

rows of A.

12. a) – 9 b) 2 c) 419

13. a) – 9 b) 2 c) 419

(Expand using the first column, first column and first row respectively.)

14. By row operations R2 – R1 and R3 – R1, .

So ∆ = (a – b) (b – c) (c – a) on simplification.

16. The row operations are R2 – R1 and R3 – R2. Answer is –2.

17. a) Answer 0; the row operations are R1, R, – R1, R3 – R2, R4. b) –3. The

row operations R1, R2, R3 – R2, R4, – R3 reduce ∆ to

Apply R1, R2 – R1, R3.

18.

19. As |A| = 0, A–1

does not exist.

20.

21. a) x = 2, y = 3, z = 4 b) x = 1, y = 2, z = 3

22. a) x = 3, y = 4, z = 6 b) x = 1, y = 3, z = 5

Terminal Questions

1. x = –2, y = –5, z = –8, t = –7

2. x = 4, y = 1, z = 3

3.

5.

8.

9. ,

10. (Use (10.7))

11. (Use (10.7))

12. Let

Then BAC = D. So A = B–1

DC–1

, Using (10.7),

13. a) –8 b) abc + 2fgh – af2 – bg

2 – ch

2 c) a

3 + b

3 + c

3 – 3abc

14. a) – 15 (Row operations: R1, R2, R3 –R1, R4 – R1)

b) 102 (Row operations: R1 + 2R4, R2 + R4, R3, R4

c) 87 (column operations: C1 + 4C3, C2 + 2C3, C3, C4 + 2C3).

15. Apply row operations R1, R2 – R1, R3 – R1. Expand using last column.

16. a) and b) answer 1 (Apply row operations R1, R2 – R1, R3 – R1, R4 – R1)

17. Apply R1, R2 – R1, R3 – R1, we get This is equal to (b – a) (a –

c) – (a – c) (c – a) – (b – a) (a – c) – (–1) (–1) (b – a) (a – c) = 0

18. Multiplying R1 by a, R2 by b and R3 by c and then dividing by abc we get

(By taking out abc from columns 1 and 2)

(by taking out ab + bc + ca from C3)

19. a) x = –1, y = 4, z = 4 b) x = 2, y = 3, z = 4

c) a = 1, b = 2, c = 3 c) a = 1, b = 2, c = 0

BT0063- Unit-11-Infinite Series

Unit 11 Infinite Series

Structure

11.1 Introduction

Objectives

11.2 Convergence and Divergence

11.3 Series of Positive Terms

11.4 Binomial Series

11.5 Exponential Series

11.6 Logarithmic Series

Self Assessment Questions

11.7 Summary

11.8 Terminal Questions

11.9 Answers

11.1 Introduction

Infinite series: If un is a real sequence, then an expression of the form

u1 + u2 + u3 + …. + …………. …………. + un + ………….

Which can be written also as is called an INFINITE SERIES.

Example, 1)

2) 1 + 2 + 3 + 4 + ………

Objectives:

At the end of the unit you would be able to

• understand the properties of infinite series

• test the convergence or the divergence of an infinite series

Partial Sum

The expression u1 + u2 + u3 + … + …… ….. + un + ….. involves addition of infinitely many

term. To give meaning to this expression we define its sequence of partial sums ‘Sn‘ by

Sn = u1 + u2 + u3 + … + …………… + un

We know that an infinite series is given by

are called partial sums.

S1

is called the 1st

partial sum, S2 is called the 2nd

partial sum, ……. Sn

is the nth partial sum.

The sequence (Sn) is called the sequence of partial sums, then we say converges, diverges,

Oscillates according as its sequence of partial sums Sn, converges, diverges or oscillates.

Examples

1. The expression 1+(–1) + 1 + (–1) + ….. + (–1)n+1

+ …… (i)

Or as it is usually written as 1 – 1 + 1 – 1 + 1 – 1 + ….. is a series. The meaning of

expression (i) is that from the terms 1, –1, +1, –1, ….. (–1)n+1

, ….. we form the partial sums,

S1 = 1, S2 = 1 – 1 = 0, S3 = 1 – 1 + 1 = 1, …….

S1 = 1 – 1 + ……. + (–1)n+1

= ……………… (ii)

2. The expression is a series. This series can also be written

as,

the partial sums,

General properties of Series

Following are the fundamental rules or properties of a series:

1. The coverage or divergence of an infinite series remains unaffected by the addition or

removal of a finite number of the terms; for the sum of these terms being the finite

quantity addition or removal doesnot change the nature of its sum.

2. If a series in which all the terms are positive is convergent, the series remain convergent

even when some or all of its terms are negative; for the sum is clearly the greatest when

all the terms are positive.

3. The convergence or divergence of an infinite series remains unaffected by multiplying

each term by a finite number.

11.2 Convergence and divergence

Convergence of the infinite series

A series is called convergent if the sequence of its partial sums has a finite limit, this limit is

termed as the sum of the convergent series.

An infinite series is said to be convergent if

Sn = l where l is a unique real number.

Examples

1) Show that is convergent series.

unique real number.

The given series is a convergent series.

Divergence of the infinite series

If a sequence of its partial sums has no finite limit, then the series is called divergent. A

divergent series has no sum.

An infinite series is said to be divergent.

If

Example

Show that is a divergent

Since

and hence the given series is divergent.

Necessary condition for convergence of a series

The series,

u1 + u2 + u3 + ….+ ……….. + un + ….. (1)

can converge only when the term un (the general term of the series) tends to zero.

i.e

un = 0

if the general term un

does not tend to zero, then the series diverges.

Examples:

a) The series, 0.0 + 0.44 + 0.444 + 0.4444 + ….. diverges because the general term un does

not tend to zero.

b) The series 1 – 1 + 1 – 1 + ……….. diverges because the general term un does not tend to

zero (and has no limit at all).

The remainder of a series

Let us consider and infinite series

u1 + u2 + u3 + ….+ ……. +um + um+1 + um+2 + ….. ———– (I)

If we discard first m terms of a series, we get the series,

um+1 + um+2 + ……….. ———– (II)

which converges (or diverges) if the series (I) converges (or diverges) and diverges if the series

(I) diverges. Therefore, while finding the convergence of a series we can distinguish between a

few terms.

When the series (I) converges, the sum Rm = um+1 + um+2 + …… Of series (II) is called the

remainder or (remainder term) of the first series. (R1 = u2 + u3 + … + ……… ….. is the first

remainder. R2 = u3 + u4 + …. + ……. is the second, etc.) the remainder Rm is the error

committed by substituting the partial sum Sn (or the sum S of the series (I)). The sum S of the

series and the remainder Rm are connected by

S = Sm + Rm.

As m →

∞ the reminder term of the series approaches to zero. It is of practical importance that this

approach be “sufficiently rapid”, that is, that the remainder Rm should become less than the

permissible error, for m not too great. Then we say that the series (I) converges rapidly,

otherwise that series is said to converge slowly.

For example consider the series converges very slowly. Summing the first 20

terms, we get the value of the sum of the series only to within 0.5 × 10–1

; to attain the accuracy

up to 0.5 × 10–4

; we have to take at leat 19,999 terms.

Examples

1.

Solution:

Here

all the other terms cancel]

which is a unique finite quantity.

The given series is convergent.

2. Show that the given series divergent.

Solution: Sn = –1+1 – 1+1 – 1+1 …….. to n terms

= 0 or –1 according as n is even or odd.

The given series oscillates between 2 finite values 0 and –1.

3. Test for divergence of the following series: 12 + 2

2 + 3

2 + …… +

Solution:

given series diverges to +∞.

4. Test for divergence the following series: 1 + 2 + 3 + ….. + n + …….. ∞

Solution:

given series in divergent.

5. Show that the series 1 + r + r2 + r

3 + ………. + ∞

(i) converges if | r | < 1

(ii) diverges if r > 1 and

(iii) oscillates if r < 1

Solution: Let Sn = 1 + r + r2 + r

3 + ………. + r

n – 1

Case (i), when | r | < 1, since it is a G.P. series,

given series is convergent.

Case (ii), when r = 1,

Sn = 1+ 1 + 1 + 1 + ……. + 1 = n

And,

given series is divergent.

Case (iii), when r = –1, the series becomes

Sn = 1 – 1 + 1 – 1 + …….

Which is an oscillatory series.

(ii) when r < –1, let r = –p, so that p > 1,

then

and

as

, from this n is even or odd.

Hence the series oscillates.

11.3 Series of Positive terms

If the terms of a series of are positive, then its sequence of partial sums

is monotonically increasing for for all the values of n.

Σ un of positive terms converges or diverges to according as Sn is bounded or unbounded. If Sn is

bounded then for all n, gives

Theorem1: A positive term series either converges to a positive number or diverges to ∞,

according as its sequence of partial sums is bounded or not.

Proof:

Let

is monotonically increasing.

According to we have the following 2 possibilities.

(i) is bounded, or

(ii) is unbounded above

(i) If is bounded, then is bounded above. Hence is a monotonically increasing

sequence which is bounded above.

is convergent.

unique real number.

is convergent.

(ii) If is unbounded above

The is a monotonically increasing sequence which is unbounded above

diverges to +∞

diverges to +∞

Therefore converges to diverges to ∞

Theorem 2: Necessary constant for the convergence of a series of a positive terms. If a Σ an

series is convergent then

an = 0. The converse is not true.

Theorem 3: The nature of the series is not altered by the multiplication of all the terms of the

series by the same non-zero constant C.

Theorem 4: The nature of the series is not altered by addition of a finite number of terms to the

series or by removing a finite number of terms from the beginning.

Theorem 5: If and are 2 series which converge to l and m respectively then the series

converges to l ± m.

11.4 Binomial Series

According to the Binomial theorem, we have,

If the right hand side is extended to ∞,

Become a finite series and this series is called as Binomial series.

The Binomial series is absolutely convergent if | x | < 1, and when the series is convergent, the

sum of the finite series is (1 + x)n.

Replacing x by – x,

While finding the sum of the Binomial series we can use some of the following cases.

1. When n = –1,

2. When n = –1, and x is changed to –x,

1 + x + x2 + x

3 + …….. ∞ = (1 + x)

–1

3. When n = –2,

1 – 2x + 3x2 – 4x

3 + …………..∞ = (1 + x)

–2

4. When n = –2, and x is changed to –x,

1 + 2x + 3x2 + 4x

3 + …………..∞ = (1 + x)

–2

5. When where p and q are integers and q ≠ 0, we get

6. When , x is replaced by –x, we get,

7. When we get,

8. When x is replaced by – x, we get,

Examples

1. Solve the following:

Solution: Let

S can be written as,

2. Solve:

Solution: Comparing the given series with one of the general Binomial series, we get

But the power of is not equal to the number factors. Hence we have to multiply and

divide by 3,

Let

11.5 Exponential series

The exponential function ex expressed as an infinite series in the form

This series is convergent for all values of x.

In finding the sum of the exponential series, the following are to be used.

I.

II. Putting x = 1, in form (I) we get,

III. By changing x to – x, in form (I) we get,

IV. Putting x = –1 in form (I), it gives,

V. By adding (I) and (III), results in,

VI. Subtracting (III) from (I),

VII. Putting x =1 in (V),

VIII. Putting x = 1 in (VI),

Note: In exponential series it should be carefully observed whether the summation is from 0 to ∞

or 1 to ∞ or 2 to ∞ etc.

Examples

Problem 1.

The solution is:

or

Problem 2.

Solution: The given series can be written as,

Consider

S = 6e + 2

11.6 Logarithmic Series

The series:

is called the logarithmic series and is denoted by

Theorem: If Σ un is a positive term series and (finite or infinite).

Then the series

(i) Converges if x > 1

(ii) Diverges if x < 1

(iii) May converge or diverge if x < 1

The logarithmic series convergent if –1 < x < 1. When it is convergent the sum of the

logarithmic series is given by,

1.

2. Replacing x by – x, in (1) we get,

3. Adding (1) and (2) we get,

4. Subtracting (2) from (1) we get,

Therefore, where x2 < 1

5. Putting x = 1 in (1) it gives,

Note: Usually the sum of the logarithmic series is found by resolving the nth term into partial

fractions.

Examples

1. Solve

Solution:

Let

2.

Solution:

Let

Consider,

therefore, 1 = A (2n) (2n + 1) + B(2n – 1) (2n + 1) + C(2n) (2n – 1)

Put n = 0, 1 = B(–1) ⇒ B = –1

Put m = ½, 1 = 2A or A = .

Put n = –1/2, then C = (–1) (–2) = 2 or 2C = 1.

Substituting these vales in we get,

Splitting as

Self Assessment Questions

1. Solve

2. Show that harmonic series of order converges for p > 1

and diverges for

11.7 Summary

In this unit initially we discussed about the partial sum and general properties of series. Then we

studied different rules for convergence or divergence of series. Lastly in this unit we studied

binomial series, exponential series, logarithmic series with properly illustrated examples.

11.8 Terminal Questions

1. Write the general properties of a series

2. Explain the binomial series

11.9 Answers

Self Assessment Questions

1. Let

S can be written as

Comparing with the expansion of

2. But the above test, this series will converge or diverge according as is finite. If

i.e. finite for p > 1

for p < 1

If

Therefore the series converges for p > 1 and p < 1.

Terminal Questions

1. Refer to Section 11.1.3

2. Refer to Section 11.4

BT0063- Unit-12-Probability

Unit 12 Probability

Structure

12.1 Introduction

Objectives

12.2 Concept of Probability

12.3 Sample Space and Events

12.4 Three Approaches to Probability

12.5 Kolmogorov’s Axiomatic Approach to Probability

12.6 Conditional Probability and Independence of Events

12.7 Baye’s theorem

12.8 Summary

12.9 Terminal Questions

12.10 Answers

12.1 Introductions

Even in day – to – day life uncertainly plays an important role. When we are unable to forecast

the future with certainty, we make statements like “probably it will rain in the evening”, “Ram

has a better chance of winning the elections” etc. Although we are not sure of the happening of

some event we make statements like those mentioned above.

It is interesting to note that the seed of probability theory was thrown when a French nobleman

Anokine Gombould (1607 – 1684) sought an explanation from the mathematician Blaise Pascal

(1623 – 1662) regarding the frequent occurrence of some combinations of number in the roll of

dice. Pierre de Fermat (1601 – 1655) and Blaise Pascal were working on this problem.

Another problem was posed to Blaise Pascal. If a game of change is stopped in the middle, how

should the two players divide the stake ? This was another problem leading to the concept of

probability, J. Bernoullis (1654 – 1705) first treatise on probability was published in 1718. Other

mathematicians who were instrumental in the development of probability were chebyshev (1821

– 1894). A. Markov (1856 – 1920), Liapounoff (who enunciated central limit theorem), De

Moivre, T. Bayes and P.S. Laplace.

Initially ideas of probability and statistics were used to explain natural phenomena. Now it is an

indispensable tool in many decisions regarding business also.

Objectives:

At the end of the unit you would able to

• understand the idea of Probability

• apply Baye’s theorem in problems

12.2 Concept of Probability

Consider the following statements.

1. A particular medicine is effective except for one out of 1000 patients.

2. There is likely to be moderate to heavy rains in most part of Karnataka

3. Getting a head or a tail in the toss of a coin are equally likely

4. When a single die is rolled, any number from 1 to 6 is equally likely

5. Only 3 out of 2 million parts is likely to be defective

In all the above statements the outcome is not certain. But we are able to list all possible

outcomes. For example, we are not certain about the number likely to be seen in a roll of a single

die but we know that only one of the six numbers 1, 2, 3, 4, 5, 6 will definitely be seen.

Statement 4 is about the likelihood of something to happen. From statement 1 we are not able to

say to whom the medicine is likely to be ineffective but we can say that it is ineffective only for

one patient when the medicine is administered to 1000 patients. Let us consider statement 3.

Although we cannot say whether we get a tail or head, we can say that we will be getting head

half the number of times when a coin is tossed several times.

Before proceeding to study the rigorous definition of probability let us understand that

Probability is a numerical measure of the likelihood of an event to happen.

It is a number between 0 and 1, 0 and 1 representing the impossibility and certainty.

Figure 12.1 Probability of an event

For example, non occurrence of rain is more likely in summer and occurrence of rain is more

likely in rainy season. Head or tails are equally likely in the toss of a coin.

12.3 Sample Space and Events

For defining probability we need the definition of an experiment. (to be more precise random

experiment).

Definition: An experiment is a process that generates well – defined outcomes.

When we perform an experiment we call it a trial.

For each trial there is one and only one outcome among the several well – defined outcomes.

Example: Find all the possible outcomes of the following experiments

1. Tossing a single coin

2. Tossing two coins

3. Tossing three coins

4. Roll of a single die

5. Roll of two dice

6. Play a one day cricket match

Solution: The possible outcomes are

1. H, T (H and T denotes head and tail respectively)

2. HH, HT, TH, TT

3. HHH, HHT, THH, HTH, HTT, TTH, THT, TTT

4. 1, 2, 3, 4, 5, 6

5. 11, 12, 13, 14, 15, 16

21, 22, 23, 25, 25, 26

31, 32, 33, 34, 35, 36

41, 42, 43, 44, 45, 46

51, 52, 53, 54, 55, 56

61, 62, 63, 64, 65, 66

1. win, defeat, tie

Definition: The sample space for an experiment is the set of all possible outcomes.

Example: Find the sample space for the following experiments.

1. Number of heads in a toss of two coins

2. Sum on the roll of two dice

3. Sum on the roll of three dice

4. Play a cricket game

Solution: the sample space is

a) {2, 1, 0}

b) {2, 3, 4, 5, 6, 7, 8, 10, 11, 12}

c) {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}

d) {win, defeat, tie}

Definition: An event in an experiment is a subset of the sample space.

Definition: A single outcome is called an elementary event.

Note: Any event consists of elementary events.

Example: Write down the following events as a subset of the respective sample space

a) Getting at least one head in a toss of two coins

b) Getting a sum of 6 or more in a roll of two dice

c) Getting two defective parts when five parts are inspected

d) India winning at least one in 3 matches played against Australia.

Solution:

a) {HT, TH, HH} {HT, TH, HH, TT}

b) {6, 7, 8, 9, 10, 11, 12} {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

c) {2} {0, 1, 2, 3, 4, 5}

d) {1, 2, 3} {0, 1, 2, 3}

S.A.Q. 1 Write the following events as a subset of the respective sample space.

a) Getting even number of heads in a toss of 3 coins

b) Winning in alternate matches when 5 matches are played

c) Getting a sum of 4 or 10 in a roll of two dice

d) Winning at least one match when three matches are played

12.4 Three approaches to Probability

As the concept of probability was developed for more than two centuries, the statisticians

defined it in several ways. In 1933 the Russian mathematician A. N. Kolmogorov developed

probability theory using the axiomatic approach which was used to define various mathematical

objects in the last century. The axiomatic approach unified all the three approaches of

probability. We study three approaches to probability which were developed before

Kolmogorov’s unified approach in this section. Kolmogorov’s axiomatic approach is developed

in the next section.

Classical Probability

The classical probability is also called mathematical, or a priori probability. We need a few

preliminary definitions before defining the classical probability of an event.

Definition: The outcomes of an experiment are equally likely if there is no reason to expect one

outcome in preference to other outcomes.

For example, Head or Tail are equally likely in tossing a single coin. Any number from 1 to 6 is

equally likely in the roll of a die.

Definition: Two or more events are mutually exclusive. If Head falls then Tail cannot fall and

vice versa. The appearance of 1, 2, 3, 4, 5, 6 are mutually exclusive. For when outcome of the

appearance of 1 occurs, the remaining cannot occur.

Definition: A collection of events is collectively exhaustive if they, when taken together

constitute the entire sample space.

For example, the events 1, 2, 3, 4, 5, 6 are collectively exhaustive. So also Head and Tail in the

toss of a single coin.

Definition: If the outcomes of an experiment are, equally likely, collectively exhaustive and

mutually exclusive then the probability of an event E is defined by

Note: The classical probability is called a priori probability since we are able to calculate the

probability of an event in advance (that is, without repeating the experiment). Of course this

definition is applicable only when the outcomes of an experiment are mutually exclusive,

collectively exhaustive and equally likely. In the case of the toss of a coin, if we can assume the

validity of these three conditions then we say that the coin is unbiased. Similarly we defined an

unbiased die.

Example: Find the classical probability of the following events.

a) Getting at least one Head in a toss of two coins

b) Getting a sum of 10 in a roll of two dice.

Solution: Let E denote the given event.

a) HT, TH, HH are the 3 outcomes favorable to the event E and the total number of

outcomes is 4. So

b) 46, 55 and 64 are the 3 outcomes favorable to the event and the total number of

outcomes is 36. Hence

Note A classical way of monitoring possibility is as follows: If odds in favour of E are x : y

or x to y, then

Statistical or empirical Probability

In this approach we repeat the experiment a large number of times and define the probability of

an event.

Definition: If n trials are performed and m trials are favourable to the occurrence of an event E,

then the probability of the event E is defined by Probability of

Note: We assume that such a limit exists.

Example: A bag contains 3 red balls, 4 green balls and 5 blue balls. Find the probability of

choosing 2 red balls, 1 green ball and one blue ball.

Solution: Denote the required event by E. As there are 12 balls in all, the total number of

choosing 4 balls is C (12, 4). We can choose 2 red balls from 3 red balls in C (3, 2) ways. One

green ball can be chosen in C(4, 1) ways and one blue ball can be chosen in C(5, 1) ways. By

multiplication principle, the number of outcomes favourable to E is C(3, 2) , C(4, 1) C(5, 1).

So

Example: There are 25 cards having the numbers 1, 2, …….., 25, written in them. If one card is

chosen what is the probability that the number in the card is divisible by 3 or 7.

Solution: The numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24 and those divisible by 7 are

7, 14 (and 21 which is already considered). So the number of favourable outcomes is 10.

Bayesian or subjective probability

In the last two approaches either we make certain assumptions about the outcomes or we assume

that we can have a large number of trials. When we have an experiment that can be performed

only once or only a few times, the earlier methods fail. In such cases we resort to subjective

approach.

Definition: The subjective probability of an event is the probability assigned to an event by an

individual based on the evidence available to him, if there is any.

Many of the social and managerial decisions are concerned with specific unique situations. In

such cases the decision makes has to frame subjective probability for these events.

When a new product is developed, the marketing manager makes prediction based on subjective

probability framed by him.

S.A.Q. 2 A bag contains 3 red, 6 yellow and 7 blue balls. What is the probability that the two

balls drawn are yellow and blue ?

S.A.Q. 3 A ball is drawn from a bag containing 10 black and 7 white balls. What I the

probability that it is white ?

S.A.Q. 4 One number is chosen from each of the two sets {1, 2, 3, 4, 5, 6, 7, 8, 9} and {2, 4, 6, 8,

10}. What is the probability that the sum of these two numbers is 13 ?

12.5 Kolmogorov’s Axiomatic Approach to Probability

As we saw earlier, Kolmogorov proposed axiomatic theory of probability in 1933. In axiomatic

approach to any theory, a minimal set of properties are taken as axioms (by an axiom we mean

an assumption). They are taken as a basis and all other properties are deduced from the axioms.

This is how many of modern mathematical objects are defined.

Before proceeding further, recall the definition of a sample space and events discussed in earlier

sections.

Definition: Let S be a sample spacer and β be a collection of events defined in the sample space

S.

Then the probability of an event A is defined by a function P: B → R (R is the set of all real

numbers) Satisfying the following axioms.

(A1) for each satisfies

(A2) P(S) = 1

(A3) If A1, A2, ….., An, ……. Is a sequence of mutually exclusive (disjoint events in β then

Note: In most of the applications we take only a finite number of disjoint events A1, ………., An.

In this case, (A3) reduces to

(12.1)

In particular if A and B are mutually exclusive then

(12.2)

Using the axiomatic approach, we can deduce all properties of probability which hold good for

classical and statistical probabilities.

We derive a few important properties of probability using (A1) (A2) (A3) or (8.1). In most of the

proofs (8.1) or (8.2) is used.

Property 12.1 The probability of an impossible event is zero.

Proof: The impossible event is the empty set φ. We know that and the union is a

disjoint union. Then

by 8.2

Canceling P(S) on both sides. We get

Property: If is the complement of the event A, that is then

Proof: We know that (see figure 12.2)

Figure 12.2 The complementary event

So

As or

Theorem: (Addition theorem) If A and B are any two events, then

Proof: We write as a disjoint union.

Figure 12.3 as disjoint union

From Fig 12.3, we see that

and the union is disjoint. By (8.2), ……… (12.4)

Similarly

And the union is disjoint. By (8.2), ……. (12.5)

As and the union on RHS is disjoint.

by (8.4) and (8.5)

Note: (12.4) and (12.5) are useful for doing problems.

Thus (12.3) is proved

Corollary (Extended Addition theorem)

…. (12.6)

Note: Fig. 12.5 will help you to prove (12.6)

Property 12.3 If then

Proof: From Fig. 12.4, we see that

Figure 12.4, Venn Diagram for property 12.3

and the union is disjoint. Hence

by (8.2)

As

Example: If A, B, C are any three events, write the following events using and set

operations.

a) only A occurs

b) A and B occur but not C

c) All the three events occur

d) None of them occur

e) At least one of them occur

f) At least two of them occur

g) Exactly one of them occurs

h) Exactly two of them occur

Solution: We represent A, B, C and their complements in Fig. 12.5

Figure 12.5: Three events

Now using the Venn diagram given in Fig 12.5, we can represent the events as follows.

a) b)

c) d)

e) f)

g)

h)

Example: If two dice are thrown what is the probability that the sum is a) greater than 9 b)

neither 3 or 9 c) less than 4

Solution:

a) The favourable outcomes are 46, 55, 64, 56, 65, 66. So

b) Let A and B denote the events that the sum is 3 and 9 respectively. Then and

As A and B are mutually exclusive

So .

P(Sum is neither 3 or 9).

(By De Morgan’s law

c) The favourable outcomes are 11, 12, 21. So probability is

Example: A bag contains two red balls, three blue balls and five green balls. Three balls are

drawn at random. Find the probability that

a) the three balls are of different colours

b) two balls are of the same colour

c) all the three are of the same colour

Solution Let E denote the given event..

a) We can choose one red ball in C (2, 1) ways, etc. So

b) Let E1, E2, E3 denote the events that two balls among the three are red, blue and green

respectively.

Similarly

So P(E) = P(E1) + P(E2) + P(E3)

c) As there are only two red balls, the chosen three balls are of the same colour only if they

are all blue or green.

Let E1, E2 denote the events that the three balls are blue and green respectively.

So P(E) = P(E1) + P(E2)

Example: Two dice are rolled. If A is the event that the number in the first die is odd and B is

the event that the number in the second die is at least 3, find .

Solution:

The outcomes favourable to are 13, 14, 15, 16, 33, 34, 35, 36, 53, 54, 55, 56. So

(By (12.4))

(By (12.5))

S.A.Q.5 A bag contains 6 white and 10 black balls. Three balls are drawn at random. Find the

probability that

a) all the three are black

b) none of them is black

c) two of them are black

d) two of them are white

e) all the three are of the same colour

S.A.Q.6 Find the probability.

S.A.Q.7 The following table gives a distribution of wates of 1,000 workers:

Wages (in Rs.) 120 – 140 140 – 160 160 – 180 180 – 200 200 – 220 220 – 240 240 – 260

No. of workers 9 118 478 200 142 35 18

An individual is selected at random from the above group. What is the probability that his wages

are (i) under Rs. 160 (ii) above Rs. 200, (iii) between Rs. 169 to 200 ?

S.A.Q. 8 Let a sample space be S = {a1, a2, a3}. Which of the following defines probability

space on S?

i)

ii)

iii)

S.A.Q. 9 Out of numbers 1 to 100, one is selected at random. What is the probability that it is

divisible by 4 or 5?

S.A.Q. 10 The chance of an accident in a factory in a year is 1 in 5 in Bomay, 2 in 20 in Poona,

10 in 120 in Nagpur. Find the chances that a accident may happen in (i) at least one of them (ii)

all of them.

S.A.Q. 11 A person is known to hit a target in 3 out of 5 shots, whereas another person is known

to hit in 2 out of 3 shots. Find the probability that the target being hit in all when they both try.

S.A.Q. 12 A six faced dice is biased that is twice as likely to show an even number as an odd

number when it is thrown. What is the probability that the sum of the two numbers is even ?

S.A.Q. 13 The odd in favour of one student passing a test are 3 : 6. The odds against another

student passing at are 3:5. What are odds that (i) both pass, (ii) both fail ?

S.A.Q. 14 If two dice are thrown, what is the probability that the sum is (a) greater than 8, and

(b) neither 7 or 11 ?

S.A.Q. 15 Let A and B two events such that

Show that i)

ii)

S.A.Q. 16 From a group of children, 5 boys and 3 girls, three children are selected at random.

Calculate the probabilities that the selected group contain i) no girl, ii) only one girl, iii) one

particular girl, (iv) at least one girl, and v) more girls than boys.

S.A.Q. 17 According to the census Bureau, deaths in the United States occur at a rate of 2,

425,000 per year. The National Centre for Health statistics reported that the three leading causes

of death during 1997 were heart disease (725, 790), cancer (537, 390) and stroke (159, 877). Let

H, c and δ represent the events that a person dies of heart disease, cancer and stroke,

respectively.

a) Use the data to estimate P(H), P(C) and P(S).

b) Are the events H and C mutually exclusive. Find

c) What is the probability that a person dies from heart disease or cancer?

d) What is the probability that a person dies from cancer or a stroke?

e) Find the probability that someone dies from a cause other than one of these three.

12.6 Conditional Probability and Independence of Events

Let us consider a class having both boys and girls. Suppose a girl is selected. What is the

probability that the selected girl gets a first class? The required probability is the probability of B

(getting a first class) given that A(The selected student is a girl) has already happened. Such a

probability is called conditional probability.

Definition 12.12 Let A and B be two events in the same sample space and P(B) > 0. Then the

conditional probability P(A/B) is the probability for A to happen given that B has already

happened.

The following theorem gives us a method of finding the conditional probability.

Theorem 12.2 (Multiplicative law for probability and conditional probability)

For any two events A and B,

provided P(B) > 0.

= P(A) P(B/A), provided P(A) > 0.

Proof Let the total number of outcomes be N. Let nA, nB, nAB denote the number of outcomes

favourable to the events A, B and respectively. Then and

Let us calculate P(A/B) using the classical approach. As B has already happened,

the total number of outcomes is nB. The number of outcomes favourable to A given B is the

number of outcomes favourable to Hence

So proving the first identity. The second identity can be proved similarly.

Example 12.11 The following table shows the distribution of blood types in a state in India

A B AB O

Rh+ 33% 11% 4% 42%

Rh- 3% 2% 2% 3%

Find the following probabilities

a) The probability that a person has type O blood.

b) The probability that a person is Rh-

c) The probability that a married couple are both Rh+

d) The probability that a married couple have type AB blood

f) The probability that a person has type B blood given that the person is Rh+

Solution Let A, B, AB, O, Rh+ and Rh- denote the events that a person has type A blood etc.

a)

b)

= 0.03 + 0.02 + 0.02 + 0.03 = 0.10

c) P (O couple are both Rh+)

= P[(husband is Rh+) ∩ (wife is hh+)]

= P (husband is Rh+) P(wife is Rh+) assuming independent.

= [1 – P (Rh – )] [1 – P(Rh – )]

= (0.9) (0.9) = 0.81

d) P(a couple have AB)

= P(husband has AB) P(wife has AB)

= (0.04 + 0.02) (0.04 + 0.02)

= 0.0036

e)

f)

Let A be the event that it rains heavily in Sikkim and B be the event that you will score a first

class in Bio informatics. Obviously the events A and B have no dependence among themselves.

We formulate this in the following definition.

Definition: Two events A and B are independent if the occurrence or non – occurrence of one

does not affect the occurrence of the other. This happen when

and ……………….. (12.7)

Note: We know that

If P(A/B) = P(A), P(A ∩ B) = P(A) P(B)

Also, when

So P(A/B) = P(A) implies P(B/A) = P(B)

So we note the following

A and B are independent if P(A ∩ B) = P(A) P(B) ……………………… (12.8)

So (8.8) can be taken as the working definition of independence of two events.

Example: A bag has 20 blue balls and 10 green balls. Two balls are taken from the bag one after

the other. Find the probability that both are blue if.

i) The first ball is not replaced before taking out the second ball

ii) The first ball is replaced before taking out the second

Solution Let A denote the event that the first ball is blue and B be the event that the second ball

is blue. So we have to find P(A ∩ B)

As the bag has 20 blue balls and the total number of balls is 30,

i) When the first ball is not replaced and the second ball is taken out there are 19 balls. So

So P(A B) = P(B/A) P(A) (By theorem 12.2)

ii) When the first ball is replaced before the second ball is taken out, there are 20 blue balls

and 30 balls in all before the second choice

So

So

Note in the above example, A and B are not independent in (i) but independent in (ii)

Example 12.13 If A and B are independent show that

a) and independent

b) and independent

Solution As A and B are independent

We know that and the union is disjoint. Hence

A and B are independent)

= P(A) [1 – P(B)]

(by property 12.2)

Hence A and are independent

b)

= 1 – P (A B)

= 1 – [P(A) + P(B) – P(A B)] (by Addition theorem)

= 1 – [P(A) + P(B) – P(A) P(B)] (since A and B are independent)

= 1 – P(A) – P(B) + P(A) P(B)

= 1 – P(A) – P(B) [1 – P(A)]

= [1 – P(A)] [1 – P(B)]

Hence are independent.

Example: One third of the students in a class are girls and the rest are boys. The probability that

a girl gets a first class is 0.4 and that of a boy is 0.3. If a student having first class is selected,

find the probability that the student is a girl.

Solution: Let A, B and C denote the event that a student is a boy, a girl and a student having first

class. We are given the following

So Similarly

since

= by Demorgan’s law

= by Addition theorem

We are required to find P(B/C)

Example: The probability that a 60 – year old man to be alive for 5 years is 0.80 and the same

probability for a 55 – year old woman is 0.85. Find the probability that a couple of ages 60 and

50 respectively will be alive for the next 5 years.

Solution: We assume that the age expectation of the couple are independent. Let A, B denote the

probability that the husband and wife will be alive for the next 5 years.

P(both will be alive for next 5 years)

= P(A B)

= P(A) P(B)

= (0.80) (0.85)

= 0.68

We can extend the concept of independence to more than two events.

Definition: Three events A, B and C are mutually independent if the occurrence or non

occurrence of any one of the events does not affect the occurrence of other events.

Note: When A, B, C are mutually independent then A and B are independent etc. So the working

definition of 3 mutually independent events A, B, C can be given as follows.

A, B, C are mutually independent if

P(A B) = P(A) P(B), P(B C) = P(B) P(C),

P(C A) = P(C) P(A) and

P(A B C) = P(A) P(B) P(C) ….. (12.9)

Example: A difficult problem is given to the students of 1st, 2

nd and 3

rd rank by a professor. The

probability that these students solve the problem are respectively. Find the probability

that the problem is solved.

Solution Let A denote the event that A solves the problem etc. Let us find the probability that

the problem is not solved by any of them (Assume independence of A, B, C and hence

Then

Probability that the problem is solved =

S.A.Q. 18 If P(A) = 0.25, P(B/A) = 0.5 and P(A/B) = 0.25 find

S.A.Q. 19 An article consists of two parts A and B. The probabilities of defect in A and B are

0.08 and 0.04. What is the probability that the assembled part will not have any defect?

S.A.Q. 20 From a bag containing 3 red and 4 black balls two balls are drawn in succession

without replacement. Find the probability that both the balls are (i) red (ii) black (iii) of the same

colour.

12.7 Baye’s theorem

We have seen that subjective probability is used when some event may happen only once or a

few times. But after assuming subjective probability we may get some new information. This

information can be used to revise the subjective probability. Baye’s theorem is used fro revising

probability on the basis of new information.

Reverend Thomas Bayes (1702 – 1761), a Christian Priest, enunciated Baye’s theorem which has

significant applications in many areas of business administration especially marketing.

Theorem (Baye’s theorem)

If E1, E2, ……, En are mutually exclusive events with P(Ei) > 0, i = 1, 2, ….. n then for any

arbitrary event A which is a subset of such that P(A) > 0, we have

Note: P(E1). ……, P(En) are called a priori or prior probabilities. A denotes some new

information. Then we revise the probabilities P(Ei) as P(Ei/A). The revised probabilities are

called posteriori probabilities.

This process is illustrated in Fig. 12.6

Figure 12.6 Prior and posterior probabilities

Example: A company has three plants A, B and C manufacturing the same spare part in the ratio

30:45:25. The percentage of defective parts in the plants are 3%, 2% and 5% respectively. A part

is chosen at random and found to be defective. What is the probability that it is manufactured by

plants A, B or C ?

Solution Let A, B, C denote the event that it is manufactured in plants A, B and C respectively.

Let P(D) be the probability that a spare part is defective. Given

P(A) = 0.3 P(B) = 0.45, P(C) = 0.25

P(D/A) = 0.03 P(D/B) = 0.02 P(D/C) = 0.05

Probability that the chosen defective part is manufactured by plant A = P(A/D)

= 0.295

Similarly,

P(B/D) = 0.295

P(C/D) = 0.410

Example: A physician believes that the people in a particular region are prone to diseases A and

B. He estimates that P(A) = 0.6 and P(B) = 0.4. The diseases have symptoms S1, S2 and S3.

Given that the patient has the diseases, the probabilities for him to have symptoms S1, S2 and S3

are given in the following table.

Disease Symptoms

S1 S2 S3

A 0.15 0.10 0.15

B 0.80 0.15 0.03

If a patient has symptom S1, find the probability that he has disease A.

Solution We are given that

P(A) = 0.6 P(B) = 0.4

P(S1/A) = 0.15 P(S2/A) = 0.10 P(S3/A) = 0.15

P(S1/B) = 0.80 P(S2/B) = 0.15 P(S3/B) = 0.03

The required probability

= P(A/S1)

= 0.738

S.A.Q.21 The contents of vessels I, II and III are as follows:

1. white, 2 black and 1 red balls

2. white, 1 black and 1 red balls, and

4. white, 5 black and 3 red balls.

One vessel is chosen at random and two balls are drawn. They happen to be white and red. What

is the probability that they come from vessels I, II or III ?

S.A.Q. 22 A company has three machines M1, M2, M3 which produces 20%, 30% and 50% of

the products respectively. Their respective defective percentages are 7, 3 and 5. From these

products one is chosen and inspected. It is defective. What is the probability that it has been

made by machine M2 ?

12.8 Summary

In this unit we discussed about the concept of probability. The different basic term of probability

is well defined with examples. Different types of probability, Baye’s theorem and its application

is discussed with clear cut examples.

12.9 Terminal Questions

1. Five persons are selected from a group of 8 men, 6 women and 6 children. Find the

probability that 3 of the 5 persons selected are children.

2. A bag contains 5 2hite, 6 black and 3 green balls. Find the probability that a ball drawn at

random is white or green.

3. Find the probability of getting (i) only one head and (ii) at least 4 heads in six tosses of an

unbiased coin.

4. Find the chance of getting more than 15 in rolling three dice

5. A five-digit number is formed from the digits 2, 3, 5, 6, 8, no repetition being allowed.

Find the probability that the number is (i) odd (ii) even.

6. A sample space has 5 elementary events E1, E2, E3, E4, E5. If P(E3) = 0.4, P(E4) = 2P (E5)

and P(E1) = P(E2) = 0.15, determine P(E4), P(E5), P({E4, E5}), P({E1, E2, E3})

7. A problem is probability is given to three students A, B and C. The probabilities that they

solve the problem are respectively. Find the probabilities that

i) A alone solves the problem

ii) Just two of them solve the problem

iii) The problem is solved

(Mention the assumptions you make in solving the problem).

8. A large company dumps its chemical waste in a local river. The probability that either a

fish or an animal dies on drinking the water is The probability that only a fish dies is

and the probability that only an animal dies in What is the probability, (i) that both

will dies ? (ii) none of them will die ?

9. In a college, the percentage of students reading the new Indian Express, Deccan Herald

and both are 20%, 30% and 15%. Find the probability that a student of the college

i) reads at least one newspaper

ii) reads none of them

iii) reads only Deccan Herald

10. A team of 6 students is to be selected from a class consisting of 7 boys and 4 girls. Find

the probability that the team consists of a) exactly two girls b) at least 2 girls.

11. 40 candidates appeared for examination in Papers A and B. 16 students passed in paper

A, 14 passed in Paper B and 16 failed in both. If one student is selected at random what is

the probability that he

a) passed in both papers

b) failed only in A

c) failed in A or B

12. In a town, only 80% of the children born reach the age of 15 and only 85% of them reach

the age of 30. 2.5% of persons aged 30 die in one year. What is the probability that a

person will reach the age of 31.

13. If A, B, C are mutually exclusive and collectively exhaustive and 2 P(C) = 3P (A) =

6P(B) find P(A), P(B), P(C).

14. Two vessels contain 20 white, 12 red and 18 black balls; 6 white, 14 red and 30

blackballs respectively. One ball is taken out from each vessel. Find the probability that

a) both are red b) both are of the same colour

15. The probability that a candidate passes in Biostatistics is 0.6 and that the probability that

he passes in Genetics 50.5. What is the probability that he passes in only one of the

papers ? (Mention the assumption you make).

16. The following table gives the frequency distribution of 50 professors according to age

and their salaries.

Age in years Salary

10000 – 15000 15000 – 20000 20000 – 25000 25000 – 30000

20 – 30 16 6 – –

30 – 40 4 10 4 4

40 – 50 – 4 18 12

50 – 60 – – 12 10

If a professor is chosen at random find the probability that

1. he is in the age group 30 – 40 years and earns more than 20,000

2. he earns in the range 15000 – 20000 and less than 40 years old.

1. In a class of 20 boys and 40 girls, half the boys and half the girls have two – wheelers.

Find the probability that a randomly selected student is a boy or has a two wheeler.

2. Three fair (unbiased) coins are tossed. If the first coin shows a head find the probability

of getting all heads.

3. If find P B/A and P(A/B).

4. If find P(B/A) and P(B/A).

5. The records of 400 examinees are given below.

Score Educational Qualification

B.A. B.Sc B.Com Total

Below 50 90 30 60 180

Between 50 & 60 20 70 70 160

Above 60 10 30 20 60

Total 120 130 150 400

If an examinee is selected from this group find

i) The probability that he is a commerce graduate

ii) The probability that he is a science graduate given that his score is above 60

iii) The probability that his score is below 50 given that he is a B.A graduate.

1. There are 3 boxes containing respectively 1 white, 2 red, 3 black balls; 2 white, 3 red, 1

black ball; 3 white, 1 red and 2 black b alls. A box is chosen at random and from it two

balls are drawn at random. The two balls are 1 red and 1 white. What is the probability

that they come from (i) the first box (ii) second box (iii) third box?

2. An item is manufactured by three machines M1, M2 and M3. Out of the total

manufactured during a specified production period, 50% are manufactured on M1, 30%

on M2 and 20% on M3.

It is also known that 2% of the item produced by M1 and M2 are defective, white 3% of

those manufactured by M3 are defective. All the items are put into one bin. From the bin,

one item is drawn at random and is found to be defective. What is the probability that it

was made on M1, M2 or M3.

12.10 Answers

Self Assessment Questions

1. a) {TTT, HHT, HTH, THH}

b) {WLWLW, LWLWL}

c) {13, 22, 31, 46, 55, 64}

d) {WLL, LWL, LLW, WWL, WLW, LWW, WWW}

2.

3. 4. The sum 13 can be obtained from the pairs (3, 10), (5, 8), (7, 6), (9, 4). The total number

of pairs that can be chosen is 9(5) = 45.

5. Answer:

a)

b)

c)

d)

e)

6. P (no head) = Answer:

7. Total number of workers = 1000. (i) 0.127 (ii) 0.195 (iii) 0.596

8. i) As P(S) = P(a1) + P(a2) + P(a3) ≠ 1, answer is No.

ii) As

iii) P(S) = 1; Yes

9. Let A and B denote divisibility by 4, 5. Outcomes favourable to A are 4, 8, …… , 100.

So P(A) = 0.25. Similarly P(B) = 0.2 and P(A ∩ B) = 0.05. Answer: 0.4

10. i) Prob (accident occurs in none of them)

Answer:

ii)

11. Assume independence. (In case of independence, P(A ∩ B) = P(A) P(B). See later

sections).

Answer:

12. Let P (odd number) = a. Treat the die as a coin showing ‘even’ or ‘odd’. Then 2a + a = 1

or Prob (sum is even = Prob (both are even) + Prob (both are odd) =

13. For the second, odds for passing are 5 : 3. So Assume

independence i) ii) .

14. a) Favourable outcomes are: 36, 45, 54, 63, 46, 55, 64, 56, 65, 66.

Answer:

b) Answer:

15. As (i) follows.

(since P(A ∪ B) < 1 and so – P (A ∪ B) > –1). So (ii) follows.

1. (i)

(ii)

(iii)

(iv)

(v)

2. a)

b) Yes. They are also independence. Answer: P(H) P(C)

c) P(H) + P(C)

d) P(C) + P(S)

e) = P(H) + P(C) + P(S) – P(H) P(C) – P(C) P(S) – P(H)

P(S) + P(H) P(C) P(S). (We assume independence of H, C, S)

3. P(A ∩ B) = 0.125, P(B) = 0.5,

So P(A ∪ B) = 0.25 + 0.5 – 0.125 = 0.625.

4. Let x, y be the events that A and B do not have any defect respectively. P(x) = 0.92 and

P(y) = 0.96. Assuming independence, the answer is (0.92) (0.96) = 0.8832.

5. Let A and B denote the events that the ball is red in first and second attempt (i)

(ii) (iii)

6. P(E1) = P(E2) = P(E3) = 1

So

1. Answer:

Terminal Questions

1.

2.

3. One head can appear in 6 ways.

(i) (ii)

4. We should get a sum of 16, 17, 18. Answer:

5. An odd number ends in 3 or 5. Number of favourable outcomes is 2(4!).

Answer:

6. Let P(E5) = x. Then 0.15 + 0.15 + 0.4 + 2x + x = 1. Hence x = 0.1

Answers: 0.2, 0.1, 0.3, 0.7

7. a)

b)

c)

8. Given that use addition theorem,

i) ii)

9. Given that P(A) = 0.2, P(B) = 0.3 and P(A ∩ B) = 0.15

(i) P(A ∪ B) = 0.35

(ii)

(iii)

10. a)

b)

11. Given that

a) b) c)

12. P (Fifteen) = 0.8, P (thirty/ Fifteen) = 0.85 and P (thirty one/ thirty) = 0.975.

Answer: (0.8) (0.85) (0.975) = 0.663

13.

So,

14. a)

b)

15. Assume independence of B and G. Answer:

16. Total number of professors = 100

a) b) 0.16

17. Given that

Answer:

18. Let A denote getting H is the first toss and B denote getting H in second and third

tosses.

Answer:

19.

and

20.

21. (i) (ii) (iii)

22.

Hence

23. P(E1) = 0.5 P(E2) = 0.3 P(E3) = 0.2 P(A/E1) = 0.02 etc

P(E1/A) = 0.454, P(E2/A) = 0.273, P(E3/A) = 0.273

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