Sets Class 9 Notes Advanced Maths Chapter 1

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Class 9 Maths Advanced Chapter 1 Sets Notes

Class 9 Advanced Maths Chapter 1 Notes

Sets and Their Representations

A well-defined collection of objects is called set. Sets are usually denoted by capital letters A, B, X, Y etc. The elements of set are represented by small letters a, b, x, y, z etc.

Suppose, a is an element of set A then we write a ∈ A (a belongs to A) and if b is not an element of set A then we write b ∉ A (b does not belong to A).

Here, the word ‘belongs to’ is denoted by ∈.
e.g. If A = {1, 2, 3, 4} then 2 ∈ A and 5 ∉ A.
Some examples of sets are

• N → The set of all natural numbers.
• W → The set of all whole number.
• Z → The set of all integers.
• Q → The set of all rational numbers.
• R → The set of real numbers (rational and irrational numbers).
• Z⁺ → The set of positive integers.
• Q⁺ → The set of positive rational numbers.
• R⁺ → The set of positive real numbers (positive rational and irrational numbers).

Example \ 01
Which of the following are sets? Justify your answer.
(i) The collection of all the months of a year beginning with the letter J.
(ii) The collection of ten most talented writers of India.
(iii) A collection of novels written by the writer Munshi Premchand.
(iv) A collection of most dangerous animals of the world.
Solution:
(i) We are sure that the members of this collection are January, June and July.
So, this collection is well-defined.
Hence, it is a set.

(ii) A writer may be most talented for one person and may not be for other. Therefore, we cannot definitely decide, which writer will be there in the collection.
So, this collection is not well-defined.
Hence, it is not a set.

(iii) Here, we can definitely decide whether a given novel is written by Munshi Premchand or not.
So, this collection is well-defined.
Hence, it is a set.

(iv) The term most dangerous is vague term. An animal may be most dangerous for one person and may not be for the other. So, it is not well-defined.
Hence, it is not a set.

Example\ 02
If A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} then insert appropriate symbol e or g in each of the following blank spaces.
(i) 4 _____ A
(ii) -4 _____ A
(iii) 12 _____ A
(iv) 9 _____ A
(v) 0 _____ A
(vi) -2 _____ A
Solution:
Given, A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(i) Since, 4 is an element of A, therefore 4 ∈ A.
(ii) Since, -4 is not an element of A, therefore -4 ∉ A.
(iii) Since, 12 is not an element of A, therefore 12 ∉ A.
(iv) Since, 9 is an element of A, therefore 9 ∈ A.
(v) Since, 0 is an element of A, therefore 0 ∈ A.
(vi) Since, -2 is not an element of A, therefore -2 ∉ A.

Representations of Sets

1. Roster Form or Tabular Form or Listing Method
In this form, all the elements of a set are listed, the elements are being separated by commas and are enclosed within curly braces { }.
This method is also known as tabular form, e.g. The set of all natural numbers less than 10 is represented in roster form as {1, 2, 3, 4, 5, 6, 7, 8, 9}.

2. Set-Builder Form or Rule Method
In this form, all the elements of set a possess a single common property p(x), which is not possessed by any other element outside the set.
In such a case, the set is described by{x : p(x) holds}.
e.g. Set of all natural numbers less than 10
i.e. A = {x : x ∉ N and x < 10}.

Example \ 03
Write the following sets in roster form.
(i) Set of all vowels in English alphabet, which precedes s.
(ii) Set of all natural numbers x such that 4x + 9 < 50.
(iii) Set of all letters in MATHEMATICS.
(iv) {x : x = \(\frac{n}{1+n^2}\) and 1 ≤ n ≤ 3, where n ∈ N}.
(v) {a_n : n ∈ N, a_n+2 = a_n+1 + a_n and a₁ = a₂ = 1}.
Solution:
(i) The vowels which precedes s are a, e, i and 0.
So, the required set is {a, e, i, 0}.

(ii) We have, 4x + 9 < 50
⇒ 4x + 9 – 9 < 50 – 9 [subtract 9 from both sides]
⇒ 4x < 41
⇒ x < \(\frac{41}{4}\)
∴ x < 10.25
Since, x is a natural number, so x can take values 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
So, the required set is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

(iii) The letters in the word MATHEMATICS (without repeating any letter) are M, A, T, H, E, I, C, S. So, the required set is {M, A, T, H, E, I, C, S}.

(iv) Given, {x : x = \(\frac{n}{1+n^2}\) and 1 ≤ n ≤ 3, where n ∈ N}.
Here, x = \(\frac{n}{1+n^2}\), 1 ≤ n ≤ 3, n ∈ N}
x = \(\frac{n}{1+n^2}\), n = 1, 2, 3
x = \(\frac{1}{1+1^2}\), \(\frac{2}{1+2^2}\), \(\frac{3}{1+3^2}\),
So, the required set is {\(\frac{1}{2}\), \(\frac{2}{5}\), \(\frac{3}{10}\)}

(v) We have, a₁ = 1, a₂ = 1
and a_n+2 = a_n+1 + a_n, ∀n ∈ N
On putting n = 1, 2, 3, … in a_n+2 = a_n+1 + a_n, we get
a₃ = a₂ + a₁ = 1 + 1 = 2
a₄ = a₃ + a₂ = 2 + 1 = 3
a₅ = a₄ + a₃ = 3 + 2 = 5
a₆ = a₅ + a₄ = 5 + 3 = 8 and so on.
So, the required set is {2, 3, 5, 8, …}.

Example \ 04
Write the following in set-builder form.
(i) A = {14, 21, 28, 35, 42, … 98}
(ii) B = {1, \(\frac{1}{4}\), \(\frac{1}{9}\), \(\frac{1}{16}\), \(\frac{1}{25}\), …}
Solution:
(i) Let x represent the elements of given set.
Given numbers are natural numbers greater than 13, less than 99 and multiples of 7.
Thus, A = {x : x is a natural number greater than 13, less than 99 and a multiple of 7}, which is the required set-builder form of given set.
This can also be written as
A = {x : x is a natural number, a multiple of 7 and 13 < x < 99}
or A = {x : x = 7n, n ∈ N and 2 ≤ n ≤ 14}.

(ii) We observe that the elements of set B are the reciprocals of the squares of all natural numbers. So, the set B in the set-builder form is
B = {x : x = \(\frac{1}{n^2}\), n ∈ N}.

Types of Sets

Empty Set A set which does not contain any element, is called an empty set or null set or void set. It is denoted by Φ or { }.
e.g. A = {x : x is a natural number less than 1}.

Singleton Set A set containing only one element, is called a singleton set.
e.g. The sets {0}, {5} and {-7} are singleton sets.

Finite Set A set which is empty or consists of a definite number of elements, is called a finite set.
e.g. The set {1, 2, 3, 4} is a finite set because it contains a definite number of elements i.e. only 4 elements.

Infinite Set A set which is not finite is called infinite set. e.g. The set of real numbers.

Equal Sets Two sets A and B are said to be equal, if they have exactly the same elements and we write it as A ≠ B.
Otherwise, two sets are said to be unequal and we write it as A ≠ B.
e.g. Let A = {a, b, c, d} and B = {c, d, b, a}.
Then, A = B because each element of set A is in set B and vice-versa.

• Do not write any repeated element in roster form.
• All the infinite sets cannot be written in roster form.
• In roster form, the order of the elements does not matter.

Example \ 05
Which of the following sets are empty sets?
(i) Set of all even natural numbers divisible by 5.
(ii) Set of all even prime numbers.
(iii) {x : x² – 2 = 0 and x is rational}
(iv) {x : x is a point common to any two parallel lines)
Solution:
(i) There are infinite even natural numbers, which are divisible by 5.
e.g. 10, 20, 30, 40, 50 etc.
Therefore, it is not an empty set.
(ii) We know that 2 is only even prime.
Therefore, it is a singleton set, not an empty set.
(iii) We have, {x : x² – 2 = 0 and x is rational}
Here, x² – 2 = 0 ⇒ x² = 2
⇒ x = ±√2 ∉ Q
Therefore, it is an empty set.

(iv) We have,
{x : x is a point common to any two parallel lines}
We know that no point is common to any two parallel lines.
Therefore, it is an empty set.

Example \ 06
Classify the given sets are finite or infinite.
(i) The set (x ∈ Z : -1 < x < 3}
(ii) A set of rational numbers.
(iii) The set of positive integers greater than 101.
(iv) Set of prime numbers less than 99.
(v) The set of animals living on Earth.
Solution:
(i) There are three numbers 0, 1, 2 between -1 and 3.
∴ {0, 1, 2} is a finite set.

(ii) There are infinite rational numbers.
∴ It is an infinite set.

(iii) There are infinite positive integers greater than 101.
∴ It is an infinite set.

(iv) There are 25 prime numbers, which are less than 99.
∴ It is a finite set.

(v) There are finite numbers of animals living on Earth.
∴ It is a finite set.

Example \ 07
Which of the following pairs of sets are equal? Justify your answer.
(i) A = {x : x is a letter of the word ‘LOYAL’},
B = {x : x is a letter of the word ‘ALLOY’}
(ii) A = {x : x ∈ Z and x² ≤ 8},
B = {x : x ∈ R and x² – 4x + 3 = 0}
Solution:
(i) Given, A = {x : x is a letter of the word ‘LOYAL’}
A = { L, O, Y, A}
and B = {x : x is a letter of the word ‘ALLOY’}
B = {A, L, O, Y}
Here, we see that both sets have exactly the same elements.
∴ A = B

(ii) Given,A = {x : x ∈ Z and x² ≤ 8}
[∵ the squares of integers 0, ±1, ±2 are less than 8]
∴ A = {-2, -1, 0, 1, 2}
and B = {x : x ∈ R
and x² – 4x + 3 = 0}
[∵ x² – 4x + 3 = 0 ⇒ (x – 1)(x – 3) = 0 ⇒ x = 1, 3]
∴ B = {1, 3}
Here, we see that set A has 5 distinct elements and set B has 2 distinct elements.
So, they do not have same elements.
∴ A ≠ B

Subsets, Power Set and Universal Set

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

We write it as A ⊆ B, which is read as A is a subset of B or A is contained in B or A ⊆ B, if x ∈ A ⇒ x ∈ B.
If A is not a subset of B then we write A ⊈ B.
e.g. {2, 3, 4, 5} is a subset of a set of natural numbers.
Every set is a subset of itself. The empty set Φ is a subset of every set.

Proper Subset

If A ⊂ B and A ≠ B then A is called a proper subset of B and B is called superset of A.
e.g. Let A = {x : x is an even natural number}
and B = {x : x is a natural number}
Then, A = {2, 4, 6, 8,…} and B = {1, 2, 3, 4, 5, …}.
A ⊂ B and A ≠ B
So, A is proper subset of B and B is superset of A.

Subsets of Set of Real Numbers

Some important subsets of set of real numbers R are

• the set of natural numbers, N = {1, 2, 3…}
• the set of whole numbers, W = {0, 1, 2, …}
• the set of integers, Z = {… -2, -1, 0, 1, 2,…}
• the set of rational numbers, Q = {x : x = \(\frac{p}{q}\), p, q ∈ Z and q ≠ 0}
• the set of irrational numbers, T = {x : x ∈ R and x ∉ Q}
• The total number of subsets and proper subsets of a finite set containing n elements is 2ⁿ and 2ⁿ – 1, respectively.
• If A ⊆ B and B ⊆ C, then A ⊆ C.
• A = B ⇔ A ⊆ B and B ⊆ A, where ‘⇔’ read as if and only if (briefly written as iff).
• Some obvious results among subsets are
  N ⊂ Z ⊂ Q, Q ⊂ R, T ⊂ R, N ⊈ T,
  where T = the set of irrational numbers.

Example \ 01
If S = {1, 3, 5, 7, 9, 11, 13} then which of the following set is a subset of S?
(i) A = {3, 5, 7}
(ii) B = {2, 5, 7}
Solution:
We have, S = {1, 3, 5, 7, 9, 11, 13}
(i) Given, A = {3, 5, 7}
Here, we see that each element of set A lies in set S.
Hence, A ⊆ S.

(ii) Given, B = {2, 5, 7}
Since, element 2 of set B is not belonging to the set S.
⇒ 2 ∈ B but 2 ∉ S
∴ B ⊈ S

Example \ 02
Insert the correct symbol ⊂ or ⊄ between each of the following pair of sets.
(i) {x : x is a student of class XI of your school} __________
{x : x is a student of your school}
(ii) {x : x is a triangle in the plane} __________
(x : x is a rectangle in the plane}
(iii) {x : x is an equilateral triangle in the plane} __________
{x : x is a triangle in the plane}
(iv) {x : x is an even natural number} __________
{x : x is an integer}
(v) {1, 4, 8} _________ {1, 2, 4, 6, 8}
Solution:
(i) {x : x is a student of class XI of your school} ⊂ {x : x is a student of your school}

(ii) {x : x is a triangle in the plane}
⊄ {x : x is a rectangle in the plane} [∵ all elements of first set are not there in the second set]

(iii) {x : x is an equilateral triangle in the plane} ⊂ {x : x is a triangle in the plane}
[∵ elements in the second set are triangles, so there will also be equilateral triangle, thus all elements of first set are there in the second set]

(iv) {x : x is an even natural number] ⊂ {x : x is an integer} [∵ integers have both even and odd natural numbers, so all elements of first set will be there in the second set]

(v) {1, 4, 8} ⊂ {1, 2, 4, 6, 8}

Cardinality of a set
The number of distinct elements in a finite set A is called cardinal number of set A and it is denoted by n(A).
e.g. If A = {-1, -2, 3, 4, 8} then n(A) = 5.

Example \ 03
Find the cardinality of the following sets.
(i) {Φ}
(ii) {{1}, {2}, {1, 2}}
(iii) {{a}, {to}, {a, b}}
Solution:
(i) Let A = {Φ}
Since, the given set A contains only one element.
∴ n(A) = 1

(ii) Let B = {{ 1}, {2}, {1, 2}}
Since, the number of elements in set B are 3.
∴ n(B) = 3

(iii) Let C = {{a}, {b}, {a, b}}
Since, the number of elements in set C are 3.
∴ n(C) = 3

Example \ 04
Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. Find the values of m and n.
Solution:
Let the two sets be AandB such that n (A) = m and n (B) = n.
Then, the number of subsets of set A = 2^m
and number of subsets of set B = 2ⁿ.
According to the given condition,
2^m = 112 + 2ⁿ
=> 2^m – 2ⁿ = 16 x 7 = 2⁴(2³ – 1) = 2⁷ – 2⁴
On comparing both sides, we get
2^m = 2⁷ and 2ⁿ = 2⁴
∴ m = 7 and n = 4

Power Set

The collection of all subsets of a set, say A, is called the power set of A and it is denoted by P{A). In P(A), every element is a set.
e.g. Let A = {3, 4, 5}
Then, P(A) = { Φ {3}, {4}, {5}, {3,4}, {3, 5}, {4, 5}, {3,4, 5}}
(i) If set A has n elements then the total number of elements in its power set is 2ⁿ.
(ii) If A is an empty set Φ then P (A) has just one element namely Φ. Thus, P(A) = {Φ}.

Method to Write the Power Set of a Given Set
Let a set having n elements be given then for writing its power set, we use the following steps.
Step I Write all the possible subsets having single element of given set.

Step II Write all the possible subsets having two elements at a time of given set.

Step III Write all the possible subsets having three elements at a time of given set. Repeat this process for writing all possible subsets having n elements at a time, as the given set has n elements.

Step IV Write a set using the subsets obtained in steps I, II and III along with Φ as its elements.
This gives the required power set of the given set.

Example \ 05
If A = {1, 2, 3} then find the power set of A.
Solution:
All possible subsets of A having single element are {1}, {2}, {3}.
All possible subsets of A having two elements at a time are {1, 2}, {2, 3}, {3, 1}.
All possible subsets of A having three elements at a time is {1, 2, 3}.
Thus, the required power set is
{{1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1, 2, 3}, Φ}.

Example /06
Show that n {P [P (P (Φ))]} = 4.
Solution:
We have, P(Φ) = {Φ}
∴ P(P(Φ)) = {Φ, {Φ}}
⇒ P[P(P(Φ))] = {Φ, {Φ}, {{Φ}}, {Φ, {Φ}}}
Hence, number of elements in P [P(P(Φ))] is 4
i.e. n{P[P(P (Φ))]} = 4.

Example /07
For any two sets A and B, prove that P(A) = P(B) implies A = B.
Solution:
Let P(A) = P(B)
Now, as A ⊆ A, therefore A ∈ P(A)
⇒ A ∈ P(B) [∵ P(A) = P(B)]
⇒ A ⊆ B …… (i)
Again, B ⊆ B ⇒ B ∈ P(B)
⇒ B ∈ P(A) [∵ P(A) = P(B)]
⇒ B ⊆ A …….. (ii)
From Eqs. (i) and (ii), we get
A = B

Universal Set

If there are some sets under consideration then a set can be chosen arbitrarily, which is a superset of each one of the given sets. Such a set is known as the universal set and it is denoted by U.
e.g. Let A = {2, 4, 6},B = {1, 3, 5} and C = {0, 7}.
Then, U = {0, 1, 2, 3, 4, 5, 6, 7} is an universal set.

Venn Diagrams and Operations on Sets

Venn Diagrams

A Venn diagram is a pictorial representation of sets. In Venn diagrams, the universal set is represented by a rectangular region and its subsets are represented by circles or a closed geometrical figure inside the universal set.

Also, an element of a set is represented by a point within the circles of set.
e.g. (i) If A = {1, 2, 3, 5, 9}, B = {4, 6, 7} and
U = { 1, 2, 3, 4, 5, 6, 7, 8, 9} then

Here, there is no element common in set A and B.

(ii) If A = {1, 2, 3, 4, 5}, B = {1, 2, 3} and U = {1, 2, 3, 4, 5} then

∴ all the elements of B are common in set A
∴ B ⊂ A

(iii) If U = {1, 2, 3, 4, 5}, A = {1, 2}, B = {1, 2, 3} and C = (1, 2, 3, 4, 5} then

Here, A ⊂ B ⊂ C

(iv) If U = {3, 4, 6, 7, 10, 12}, A = {3, 4, 6, 7} and B = {4, 6, 10, 12} then

Here, elements 4 and 6 are common in both sets.

Example \ 01
Draw the Venn diagrams to illustrate the following relationship among sets E, M and U, where E is the set of students studying English in a school, M is the set of students studying Mathematics in the same school and U is the set of all students in that school.
(i) All the students who study Mathematics also study English but some students, who study English do not study Mathematics.
(ii) Not all students study Mathematics but every student studying English studies Mathematics.
Solution:
Given, E = Set of students studying English
M = Set of students studying Mathematics
and U = Set of all students.
(i) Since, all of the students who study Mathematics also study English but some students who study English do not study Mathematics. \
∴ M ⊂ E ⊂ U
Through Venn diagram, we represent it as

(ii) Since, every student studying English studies Mathematics.
∴ E ⊂ M ⊂ U
Through Venn diagram, we represent it as

Operations on Sets

Union of Sets

Let A and B be any two sets. The union of A and B is the set of all those elements, which belong to either A or B or both. The symbol ∪ is used to denote the union.
Symbolically, it is denoted by A ∪ B and read as A union B.
∴ A ∪ B = {x : x ∈ A or x ∈ B}
e.g. Let A = {2, 3} and B = {3, 4, 5}
Then, A ∪ B = {2, 3, 4, 5}
The union of sets A and B is represented by the following Venn diagram.

The shaded portion represents A ∪ B. It is evident from the definition that A ⊂ A ∪ B and B ⊂ A ∪ B.
For any three sets A, Band C, we have

• A ∪ Φ = A
• U ∪ A = U
• A ∪ A = A
• A ∪ B = B ∪ A
• (A ∪ B) ∪ C = A ∪ (B ∪ C)

Example \ 02
Find the union of each of the following pairs of sets.
(i) A = [a, e, i, 0, u), B = [a, c, d]
(ii) A = {1, 3, 5}, B = {2, 4, 6}
(iii) A = {x : x is a natural number and 1 < x ≤ 5}
and B = {x : x is a natural number and 5 < x ≤ 10}
Solution:
(i) Given, A = {a, e, i, 0, u} and B = {a, c, d}
A ∪ B = {a, e, i, 0, u} ∪ {a, c, d]
= {a, c, d, e, i, 0, u}

(ii) Given, A = {1, 3, 5} and B = {2, 4, 6}
A ∪ B = {1, 3, 5} ∪ {2, 4, 6}
= {1, 2, 3, 4, 5, 6}

(iii) Given, A = {x : x is a natural number and 1 < x ≤ 5}
= {2, 3, 4, 5}
and B = {x : x is a natural number and 5 < x ≤ 10}
= {6, 7, 8, 9, 10}
∴ A ∪ B = {2, 3, 4, 5} ∪ {6, 7, 8, 9, 10}
= {2, 3, 4, 5, 6, 7, 8, 9, 10}
= {x : x is a natural number and 1 < x ≤ 10}

Intersection of Sets

Let A and B be any two sets. The intersection of A and B is the set of all those elements, which belong to both A and B. It is denoted by A ∩ B and read as A intersection B. The symbol n is used to denote the intersection.
∴ A ∩ B = {x : x ∈ A and x ∈ B}
e.g. Let A = {2, 3, 4, 5} and B = {1, 3, 6, 4}
Then, A ∩ B = {3, 4}

The intersection of sets A and B is represented by the following Venn diagram.

The shaded portion represents A ∩ B.
It is evident from definition that A ∩ B ⊂ A and A ∩ B ⊂ B.
For any three sets A, Band C, we have

• A ∩ Φ = Φ
• U ∩ A = A
• A ∩ A = A
• A ∩ B = B ∩ A
• (A ∩ B) ∩ C = A ∩ (B ∩ C)

Example \ 03
Find the intersection of each of the following pairs of sets.
(i) A = {1, 3, 5, 7, 9}, B = {2, 3, 6, 8, 9}
(ii) A = {e, f, g}, B = Φ
(iii) A = {x : x = 3n, n ∈ Z], B = {x : x = 4n, n ∈ Z)
Solution:
(i) Given, A = {1, 3, 5, 7, 9} and B = {2, 3, 6, 8, 9}
∴ A ∩ B = {3, 9}
[∵ 3 and 9 are only elements, which are common]

(ii) Given, A = {e, f, g} and B = Φ
∴ A ∩ B = Φ [∵ there is no common element]

(iii) Let x ∈ A ∩ B
⇒x ∈ A and x ∈ B
⇒ x is a multiple of 3 and x is a multiple of 4.
⇒ x is a multiple of 3 and 4 both x is a multiple of 12.
⇒ x = 12n, n ∈ Z
∴ A ∩ B = {x : x = 12n, n ∈ Z}

Example \ 04
If A = {4, 5, 7, 8, 10}, B = {4, 5, 9} and C = {1, 4, 6, 9} then verify that
(i) (A ∩ B) ∩ C = A ∩ (B ∩ C)
(ii) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(iii) A ∩ (B ∪ C) = {A ∩ B) ∪ (A ∩ C)
Solution:
Given, A = {4, 5, 7, 8, 10}, B = {4, 5, 9} and C = {1, 4, 6, 9}
(i) Now, A ∩ B = {4, 5, 7, 8, 10} ∪ {4, 5, 9} = {4, 5}
∴ LHS = (A ∩ B) ∩ C
= {4, 5} ∩ {1, 4, 6, 9} = {4} ……… (i)
Now, B ∩ C = {4, 5, 9} ∪ {1, 4, 6, 9} = {4, 9}
∴ RHS = A ∩ (B ∩ C)
= {4, 5, 7, 8, 10} ∩ {4, 9} = {4} ……… (ii)
From Eqs. (i) and (ii), we get
LHS = RHS = {4}
Hence, (A ∩ B) ∩ C = A ∩ (B ∩ C).

(ii) Here, B ∩ C = {4, 9}
∴ LHS = A ∪ (B ∩ C)
= {4, 5, 7, 8, 10} ∪ {4, 9}
= {4, 5, 7, 8, 9, 10} ………. (iii)
Now, A ∪ B = {4, 5, 7, 8, 10} ∪ {4, 5, 9}
= {4, 5, 7, 8, 9, 10}
and A ∪ C = {4, 5, 7, 8, 10} ∪ {1, 4, 6, 9}
= {1, 4, 5, 6, 7, 8, 9, 10}
∴ RHS = (A ∪ B) ∩ (A ∪ C)
= {4, 5, 7, 8, 9, 10} ∩ {1, 4, 5, 6, 7, 8, 9, 10}
= {4, 5, 7, 8, 9, 10} ……. (iv)
From Eqs. (iii) and (iv), we get
LHS = RHS = {4, 5, 7, 8, 9, 10}
Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(iii) Now, B ∪ C = {4, 5, 9} ∪ {1, 4, 6, 9}
= {1, 4, 5, 6, 9}
LHS = A ∩ (B ∪ C)
= {4, 5, 7, 8, 10} ∩ {1, 4, 5, 6, 9} = {4, 5} …….. (v)
Now, A ∩ B = {4, 5, 7, 8, 10} ∩ {4, 5, 9} = {4, 5}
and A ∩ C = {4, 5, 7, 8, 10} ∩ {1, 4, 6, 9} = {4}
RHS = (A ∩ B) u (A ∩ C)
= {4, 5} ∪ {4}
= {4, 5} ……… (vi)
From Eqs. (v) and (vi), we get
LHS = RHS = {4, 5}
Hence, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

Disjoint Sets

Two sets A and B are said to be disjoint sets, if they have no common element
i.e. A ∩ B = Φ.
e.g. Let A = {2, 4, 6} and B = {1, 3, 5}.
Then, A ∩ B = Φ.
Hence, A and B are disjoint sets said to be intersectiong or overlapping sets. The disjoint sets A and B can be represented by the Venn diagram.

Example \ 05
Which of the following pairs of sets are disjoint?
(i) A = {1, 2, 3, 4, 5, 6} and B = {x : x is a natural number and 4 ≤ x ≤ 6}
(ii) A = {x : x is the boy of your school)
and B = (x : x is the girl of your school)
Solution:
(i) Given, A = {1, 2, 3, 4, 5, 6} and B = {4, 5, 6}
∴ A ∩ B = {1, 2, 3, 4, 5, 6} ∩ {4, 5, 6}
= {4, 5, 6} ≠ Φ
Hence, this pair of set is not disjoint.

(ii) Here, A = {b₁, b₂, …….. b_n}
and B = {g₁ g₂, … g_m},
where b₁, b₂, …..,b_n are the boys and g₁, g₂, ….. , g_m are the girls of school.
Clearly, A ∩ B = Φ
Hence, this pair of set is disjoint set.

Difference of Sets

Let A and B be any two sets. The difference of sets A and B in this order is the set of all those elements of A which do not belong to B.

It is denoted by A – B and read as A minus B.
The symbol ‘-‘ is used to denote the difference of sets.
∴ A – B = {x : x ∈ A and x ∉ B}
Similarly, B – A = {x : x ∈ B and x ∉ A}
e.g. Let A = {1, 2, 3, 4, 5} and B = {3, 5, 7, 9}.
Then, A – B = {1, 2, 4} and B – A = {7, 9}.
The shaded portion represents the difference of two sets A and B.

Example \ 06
(i) If X = [a, b, c, d, e, f} and Y = { f, b, d, g, h, k} then find X- Y and Y – X.
(ii) If A = {1, 2, 3, 4, 5} and B = {2, 4, 6} then find A – B and B – A.
Also, represent each of these by Venn diagram.
Solution:
(i) Given, X = {a, b, c, d, e, f} and Y = {f, b, d, g, h, k]
∴ X – Y = {a, b, c, d, e, f} – {f, b, d, g, h, k} = {a, c, e}

[only those elements of X which do not belong to Y ]
andY-X= {f, b, d, g, h, k} – {a, b, c, d, e, f} = {g, h, k}

[only those elements of Y, which do not belong to X ]

(ii) Given, A = {1, 2, 3, 4, 5} and B = [2, 4, 6}
∴ A – B = {1, 2, 3, 4, 5} – {2, 4, 6} = {1, 3, 5}

[only those elements of A which do not belong to B]
and B – A = {2, 4, 6} – {1, 2, 3, 4, 5} = {6}

[only those elements of B, which do not belong to A]

Complement of a Set

Let U be the universal set and A be any subset of U then complement of A with respect to U is the set of all those elements of U, which are not in A. It is denoted by \(\bar{A}\), A’ or A^c and read as A complement.
Thus, \(\bar{A}\) = {x : x ∈ U and x ∉ A} or A’ = U – A
e.g.Let U ={1, 2, 3, 4, 5, 6} and A = {2, 4}.
e.g.Let U ={1, 2, 3, 4, 5, 6} and A = {2, 4}.
Then, A’ = U – A = {1, 2, 3, 4, 5, 6} – {2, 4} = {1, 3, 5, 6}
The complement of set A is represented by the following Venn diagram.

Here, the shaded portion represents the complement of set A.

Example \ 07
Let U = [ 1, 2, 3, 4, 5, 6, 7, 8, 9},
A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find
(i) A’
(ii) B’
(iii) (A ∩ C)’
(iv)(A ∪ B)’
(w)(A’)’
(vi )(B – C)’
Solution:
Given, U = [1, 2, 3, 4, 5, 6, 7, 8, 9},A = {1, 2, 3, 4},
B = [2, 4, 6, 8} and C = [3, 4, 5, 6}

(i) We have, A’ = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}

(ii) We have, B’ = U – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}

(iii) Since, A ∩ C = {3, 4}
∴ (A ∩ C)’ = U – (A ∩ C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {3, 4}
= {1, 2, 5, 6, 7, 8, 9}

(iv) Since, A ∪ B = {1, 2, 3, 4, 6, 8}
∴ (A ∪ B)’ = U – (A ∪ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8} = {5, 7, 9}

(v) We have, (A’)’ = (U – A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9} [using part (i)]
= {1, 2, 3, 4}

Alternate Method
We know that (A’)’ = A
∴ (A’)’ = {1, 2, 3, 4}

(vi) Since, B – C = {2, 8}
(B – C)’ = U – (B – C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8}
= {1, 3, 4, 5, 6, 7, 9}

Applications of Set Theory

In this topic, we will discuss some word problems related to our daily life, which are based on union and intersection of two sets. Before solving these types of problem, we should know the following formulae

If A and B are two finite sets then two cases arise.

Case I If A and B are disjoint sets i.e. there is no common element in A and B i.e. A ∩ B = Φ.
Then, n(A ∪ B) = n(A) + n(B)

Case II If A and B are not disjoint sets i.e. there are common elements in A and B.
Then, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Sets A – B, A ∩ B and B – A are disjoint and their union is A∪ B.

Important Results

(i) n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B)
(ii) n(A) = n (A – B)+ n(A ∩ B)
(iii) n(B) = n(B – A) + n(A ∩ B)
(iv) n(A’ ∪ B’ ) = n [(A ∩ B)’ ] = n (U) – n(A ∩ B)
(v) n(A’ ∩ B’) = n [(A ∪ B)’] = n(U) – n(A ∪ B)
(vi) n(A – B) = n(A) – n(A ∩ B)
(vii) If A, B and C are finite sets then
(a) n(A ∪ B ∪ C)
= n(A) + n (B) + n(C) – (A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
(b) n (A only) = n(A) – n(A ∩ B) – n(A ∩ C) + n(A ∩ B ∩ C)
(c) n(\(\bar{A}\) ∩ \(\bar{B}\) ∩ \(\bar{C}\)) = n(U) – n(A ∪ B ∪ C)

Example / 01
If X and y are two sets such that X has 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements then how many elements does Y have?
Solution:
Given,n(X) = 40, n(X ∪ Y) = 60 and n(X ∩ Y) = 10
Clearly, n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 60 = 40 + n(Y) – 10
⇒ 60 = 30 + n(Y)
⇒ n(Y) = 60 – 30
∴ (Y) = 30
Hence, Y have 30 elements.

Example \ 02
If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38 then find n(X ∩ Y).
Solution:
Given, n(X) = 17, n (Y) = 23
and n(X ∪ Y) = 38
Clearly, n(X ∪ Y) = n(X) + n(Y) – (X ∩ Y)
⇒ n(X ∩ Y) = n(X) + n(Y) – n(X ∪ Y)
⇒ n(X ∩ Y) = 17 + 23 – 38
= 40 – 38 = 2

Example \ 03
If n(A) = 4, n(B) = 5, n(U) = 7 and n(A ∩ B) = 2 then find the value of n (A ∪ B)’.
Solution:
Given, n(A) = 4, n(B) = 5, n(U) = 7 and n(A ∩ B) = 2
v n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
∴ n(A ∪ B) = 4 + 5 – 2 = 7
Now, n(A ∪ B)’ = n(U) – n(A ∪ B) = 7 – 7 = 0

Example \ 04
In a class of 60 students, 25 students play Cricket, 20 students play Tennis and 10 students play both the games. Then, find the number of students, who play neither games.
Solution:
Let C and T, respectively denotes the set of students, who play Cricket and Tennis and set U denotes the set of all students in a class.
Then, n(C) = 25, n(T) = 20, n(C ∩ T) = 10 and n(U) = 60.
We know that n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ n(C ∪ T) = 25 + 20 – 10
= 45 – 10 = 35
Now, the number of students, who play neither game
n(C’ ∩ T’) = n(C ∪ T)’ [by De-Morgan’s law]
= n(U) – n(C ∪ T)
= 60 – 35 = 25
Hence, 25 students play neither games.

Example \ 05
In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both newspapers. Then, find the number of persons, who read neither of the newspapers.
Solution:
Let H and E, respectively denote the set of persons, who read Hindi and English newspapers and let U set of all persons in a town.
Then, we have n (U) = 840,
n(H) = 450,
n(E) = 300
and n(H ∩ E) = 200
Clearly, n(H ∪ E) = n(H) + n(E) – n (H ∩ E)
= 450 + 300 – 200
= 750 – 200 = 550
Now, the number of persons, who read neither of the newspaper is given by
n(H’ ∩ E’) = n(H ∪ E)’ [by De-Morgan’s law]
= n(U) – n(H ∪ E)
= 840 – 550 = 290
Hence, 290 persons read neither of the newspapers.

Example \ 06
Each student in a class of 40 students study atleast one of the subjects English, Mathematics and Economics. 16 students study English, 22 Economics and 26 Mathematics. 5 study English and Economics, 14 Mathematics and Economics and 2 English, Economics and Mathematics. Find the number of students, who study English and Mathematics.
Solution:
Let A, B and C denote the set of students, who study English, Economics and Mathematics, respectively.
Then, we have total number of students, n (A ∪ B ∪ C) = 40
Number of students who study English, n (A) = 16,
Number of students who study Economics, n (B) = 22,
Number of students who study Mathematics, n (C) = 26,
Number of students who study English and Economics, n(A ∩ B) = 5
number of students who study Mathematics and Economics, n(B ∩ C) = 14
and number of students who study all subjects, n(A ∩ B ∩ C) = 2
Clearly, n(A ∪ B ∪ C) = n(A) + n(B) + n(C)
– n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
40 = 16 + 22 + 26 – 5 – 14 – n(C ∩ A) + 2
⇒ 40 = 66 – 19 – n (C ∩ A)
⇒ n(C ∩ A) = 47 – 40 = 7
Hence, the number of students, who study English and Mathematics are 7.

Example \ 07
A market research group conducted a survey of 2000 consumers and reported that 1720 consumers liked “Patanjali Ghee” and 1450 consumers liked “Amul Ghee”. What is the least number that must have liked both the product?
Solution:
Let U be the set of consumers who were participated in the survey.
Then, n(U) = 2000
Let P be the set of consumers who liked “Patanjali Ghee”.
Then, n(P) = 1720
Let A be the set of consumers who liked “Amul Ghee”.
Then, n(A) =1450
∵ n(p ∪ A) = n(P) + n(A) – n(P ∩ A)
∴ n(P ∪ A) = 1720 + 1450 – n(P ∩ A) = 3170 – n(P ∩ A)
Now, as n(P ∪ A) ≤ n(U)
∴ 3170 – n(P ∩ A) ≤ 2000
⇒ n(P ∩ A) ≥ 1170
Hence, the least number of consumers, who liked both the product is 1170.

Example \ 08
If A and Bare two sets containing 3 and 6 elements respectively, what can be the minimum number of elements in A ∪ B? Also, find the maximum number of elements in A ∪ B.
Solution:
Given, n(A) = 3 and n(B) = 6
We know that
n(A ∪ B) = n(A) + n(B) – n (A ∩ B)
Clearly, n(A ∪ B) will be maximum, when n(A ∩ B) is minimum and it will be minimum, when (A ∩ B) is maximum.
Therefore, two cases arise.
Case I
When n(A ∩ B) is minimum.
The minimum value of n(A ∩ B) = 0
∴ n(A ∪ B) = n(A) + n(B) = 3 + 6 = 9
Then, maximum value of n (A ∪ B) is 9.
Case II When n(A ∩ B) is maximum.
n(A ∩ B) will be maximum, if A ⊆ B
In this case, n(A ∩ B) = 3
∴ n(A ∪ B) = 3 + 6 – 3 = 6
Thus, the minimum number of elements in A ∪ B is 6.

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