Sets

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

The months in a year are January, February, March, ..., December.

Given any object, we can definitely say whether it is a month in a year or not.

∴ The collection is well-defined.

Hence, the collection of all the months in a year is a set.

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

Given any river, we can definitely say whether it flows in India or not.

∴ The collection is well-defined.

Hence, the collection of all the rivers flowing in India is a set.

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

The planets in the solar system are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune.

Given any object, we can definitely say whether it is a planet in the solar system or not.

∴ The collection is well-defined.

Hence, the collection of all the planets in the solar system is a set.

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

Here, the term 'interesting' is vague, since a drama may be interesting to one person but not to another.

So, we are not definite as to which dramas are there in the collection.

∴ The collection is not well-defined.

Hence, the collection of all the interesting dramas written by Shakespeare is not a set.

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

Here, the term 'short' is vague, as no definite height is given below which a boy is to be called short.

So, we cannot judge definitely as to which boys of the class are there in the collection.

∴ The collection is not well-defined.

Hence, the collection of all the short boys of your class is not a set.

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

The letters of the English alphabet which precede K are A, B, C, D, E, F, G, H, I and J.

Given any letter, we can definitely say whether it precedes K or not.

∴ The collection is well-defined.

Hence, the collection of all the letters of English Alphabet which precede K is a set.

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

Given any dog, we can definitely say whether it is a pet dog in Meerut or not.

∴ The collection is well-defined.

Hence, the collection of all the pet dogs in Meerut is a set.

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

Here, the term 'dishonest' is vague, since a dealer may be considered dishonest by one person but not by another.

So, we are not definite as to which dealers are there in the collection.

∴ The collection is not well-defined.

Hence, the collection of all the dishonest dealers in Delhi is not a set.

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

Given any student of the school, we can definitely say whether his or her age exceeds 12 years or not.

∴ The collection is well-defined.

Hence, the collection of all the students of your school with age exceeding 12 years is a set.

A collection is a set only if it is well-defined, i.e., we can tell beyond doubt whether a given object belongs to the collection or not.

The height of Meena is a definite height and the height of every girl of her class can be compared with it.

So, given any girl of Meena's class, we can definitely say whether she is taller than Meena or not.

∴ The collection is well-defined.

Hence, the collection of all the girls of Meena's class, who are taller than Meena, is a set.

If x is an element of a set A, we say that x belongs to A and we write x ∈ A.

Hence, p is an element of set A is written as p ∈ A.

If x is not an element of a set B, we say that x does not belong to B and we write x ∉ B.

Hence, q does not belong to set B is written as q ∉ B.

If x is a member of a set C, we write x ∈ C.

As both a and b are members of set C, we write a ∈ C and b ∈ C.

Hence, a and b are members of set C is written as a, b ∈ C.

Two finite sets B and C are said to be equivalent, if n(B) = n(C), and it is written as B ↔ C.

Hence, B and C are equivalent sets is written as B ↔ C.

The number of distinct elements contained in a finite set E is called the cardinal number of E and is denoted by n(E).

Hence, cardinal number of set E is 15 is written as n(E) = 15.

A set consisting of no element at all is called an empty set and it is denoted by φ.

A set which has at least one element is called a non-empty set.

Hence, A is an empty set and B is a non-empty set is written as A = φ and B ≠ φ.

The set of all whole numbers is denoted by W and the set of all natural numbers is denoted by N.

As 0 belongs to W, we write 0 ∈ W and as 0 does not belong to N, we write 0 ∉ N.

Hence, 0 is a whole number, but 0 is not a natural number is written as 0 ∈ W but 0 ∉ N.

B = {x | x ∈ W, x ≤ 6}

In roster form, we make a list of the elements of the set and put it within braces { }.

The whole numbers which are less than or equal to 6 are 0, 1, 2, 3, 4, 5 and 6.

Hence, B = {0, 1, 2, 3, 4, 5, 6}.

C = {x | x is a factor of 32}

In roster form, we make a list of the elements of the set and put it within braces { }.

32 = 1 × 32 = 2 × 16 = 4 × 8

So, the factors of 32 are 1, 2, 4, 8, 16 and 32.

Hence, C = {1, 2, 4, 8, 16, 32}.

E = {x | x = (2n + 1), n ∈ W, n ≤ 4}

In roster form, we make a list of the elements of the set and put it within braces { }.

The whole numbers which are less than or equal to 4 are 0, 1, 2, 3 and 4.

Putting n = 0, 1, 2, 3 and 4 in (2n + 1), we get

2 × 0 + 1 = 1, 2 × 1 + 1 = 3, 2 × 2 + 1 = 5, 2 × 3 + 1 = 7 and 2 × 4 + 1 = 9.

Hence, E = {1, 3, 5, 7, 9}.

F = {x | x = n², n ∈ N, 2 ≤ n ≤ 5}

In roster form, we make a list of the elements of the set and put it within braces { }.

The natural numbers from 2 to 5 are 2, 3, 4 and 5.

Putting n = 2, 3, 4 and 5 in n², we get

2² = 4, 3² = 9, 4² = 16 and 5² = 25.

Hence, F = {4, 9, 16, 25}.

G = $\bigg\lbrace x\ |\ x = \frac{n}{(n + 3)},\ n \in N \text{ and } n \le 5\bigg\rbrace$

In roster form, we make a list of the elements of the set and put it within braces { }.

The natural numbers which are less than or equal to 5 are 1, 2, 3, 4 and 5.

Putting n = 1, 2, 3, 4 and 5 in $\frac{n}{(n + 3)}$, we get

$\frac{1}{1 + 3} = \frac{1}{4}, \frac{2}{2 + 3} = \frac{2}{5}, \frac{3}{3 + 3} = \frac{3}{6}, \frac{4}{4 + 3} = \frac{4}{7} \text{ and } \frac{5}{5 + 3} = \frac{5}{8}$.

Hence, G = $\mathbf{\bigg\lbrace\frac{1}{4},\ \frac{2}{5},\ \frac{3}{6},\ \frac{4}{7},\ \frac{5}{8}\bigg\rbrace}$.

H = {x | x is a two-digit number, the sum of whose digits is 8}

In roster form, we make a list of the elements of the set and put it within braces { }.

A two-digit number is of the form (10a + b), where a is a whole number from 1 to 9 and b is a whole number from 0 to 9.

Here, a + b = 8. So, the possible pairs of a and b are

a = 1, b = 7; a = 2, b = 6; a = 3, b = 5; a = 4, b = 4; a = 5, b = 3; a = 6, b = 2; a = 7, b = 1 and a = 8, b = 0.

So, the required two-digit numbers are 17, 26, 35, 44, 53, 62, 71 and 80.

Hence, H = {17, 26, 35, 44, 53, 62, 71, 80}.

I = {x | x ∈ N, x is divisible both by 4 and 6 and x ≤ 60}

In roster form, we make a list of the elements of the set and put it within braces { }.

A number divisible both by 4 and 6 is divisible by their L.C.M.

L.C.M. of 4 and 6 = 12

So, the multiples of 12 which are less than or equal to 60 are 12, 24, 36, 48 and 60.

Hence, I = {12, 24, 36, 48, 60}.

J = $\lbrace x\ |\ x = \frac{1}{n},\ n \in N \text{ and } n \le 5\rbrace$

In roster form, we make a list of the elements of the set and put it within braces { }.

The natural numbers which are less than or equal to 5 are 1, 2, 3, 4 and 5.

Putting n = 1, 2, 3, 4 and 5 in $\frac{1}{n}$, we get

$\frac{1}{1} = 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \text{ and } \frac{1}{5}$.

Hence, J = $\mathbf{\bigg\lbrace1,\ \frac{1}{2},\ \frac{1}{3},\ \frac{1}{4},\ \frac{1}{5}\bigg\rbrace}$.

L = {x | x is a letter of the word 'careless'}

In roster form, we make a list of the elements of the set and put it within braces { }.

The letters of the word 'careless' are c, a, r, e, l, e, s, s.

As every repeated element is taken only once in a set, writing each letter once and only once, we get c, a, r, e, l and s.

Hence, L = {c, a, r, e, l, s}.

A = {5, 6, 7, 8, 9, 10, 11, 12}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

Each element of A is a natural number lying between 4 and 13.

Hence, A = {x | x ∈ N, 4 < x < 13}.

B = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

48 = 1 × 48 = 2 × 24 = 3 × 16 = 4 × 12 = 6 × 8

So, the elements of B are exactly the factors of 48.

Hence, B = {x | x ∈ N, x is a factor of 48}.

C = {11, 13, 17, 19, 23, 29, 31, 37}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

Each element of C is a prime number and these are all the prime numbers lying between 10 and 40.

Hence, C = {x | x is a prime number, 10 < x < 40}.

D = {21, 23, 25, 27, 29, 31, 33, 35, 37}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

Each element of D is an odd number and these are all the odd numbers lying between 20 and 38.

Hence, D = {x | x is an odd number, 20 < x < 38}.

I = {9, 16, 25, 36, 49, 64, 81, 100}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

9 = 3², 16 = 4², 25 = 5², 36 = 6², 49 = 7², 64 = 8², 81 = 9² and 100 = 10²

So, each element of I is the square of a natural number n, where n takes the values from 3 to 10.

Hence, I = {x | x = n², n ∈ N, 3 ≤ n ≤ 10}.

J = {-2, 2}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

(-2)² = 4 and (2)² = 4

So, each element of J is an integer whose square is 4.

Hence, J = {x | x ∈ Z, x² = 4}.

K = {0}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

The only element of K is 0, which is the only whole number less than 1.

Hence, K = {x | x ∈ W, x < 1}.

L = { }

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

L has no element at all, so we describe it by a property which no object satisfies.

As there is no natural number which is less than 1,

Hence, L = {x | x ∈ N, x < 1}.

M = {a, b, c, d, e, f, g, h, i}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

The elements of M are the first nine letters of the English alphabet.

Hence, M = {x | x is one of the first nine letters of the English alphabet}.

P = $\bigg\lbrace \frac{2}{7},\ \frac{3}{8},\ \frac{4}{9},\ \frac{5}{10},\ \frac{6}{11},\ \frac{7}{12},\ \frac{8}{13},\ \frac{9}{14} \bigg\rbrace$

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

The numerators are 2, 3, 4, ..., 9 and each denominator is 5 more than its numerator.

Taking the numerator as (n + 1), the denominator becomes (n + 6), where n = 1, 2, 3, ..., 8.

Hence, P = $\mathbf{\bigg\lbrace x\ |\ x = \frac{n + 1}{n + 6},\ n \in N \text{ and } n \le 8 \bigg\rbrace}$.

S = {Atlantic, Arctic}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

The oceans of the world are Pacific, Atlantic, Indian, Southern and Arctic, out of which only Atlantic and Arctic begin with the letter A.

Hence, S = {x | x is an ocean whose name begins with the letter A}.

T = {Mars, Mercury}

In set-builder form, we list the property satisfied by all the elements of the set and put it within braces.

The planets of the solar system are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune, out of which only Mercury and Mars begin with the letter M.

Hence, T = {x | x is a planet of the solar system whose name begins with the letter M}.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

Although the leaves on a tree are very large in number, the process of counting them will surely come to an end.

Hence, the set of leaves on a tree is a finite set.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

The set of all counting numbers is N = {1, 2, 3, 4, 5, ....} and the process of counting its elements never comes to an end.

Hence, the set of all counting numbers is an infinite set.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

{x | x ∈ N, x > 1000} = {1001, 1002, 1003, ....} and the process of counting its elements never comes to an end.

Hence, {x | x ∈ N, x > 1000} is an infinite set.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

{x | x ∈ W, x < 5000} = {0, 1, 2, 3, ...., 4999}, which has 5000 elements. So, the process of counting its elements comes to an end.

Hence, {x | x ∈ W, x < 5000} is a finite set.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

{x | x ∈ Z, x < 4} = {...., -3, -2, -1, 0, 1, 2, 3} and the process of counting its elements never comes to an end.

Hence, {x | x ∈ Z, x < 4} is an infinite set.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

An unlimited number of triangles can be drawn in a plane, so the process of counting them never comes to an end.

Hence, the set of all triangles in a plane is an infinite set.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

There is an unlimited number of points on the circumference of a circle, so the process of counting them never comes to an end.

Hence, the set of all points on the circumference of a circle is an infinite set.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

As every repeated element is taken only once in a set, {1, 2, 3, 1, 2, 3, 1, 2, 3, ....} = {1, 2, 3}, which has only 3 elements.

Hence, {1, 2, 3, 1, 2, 3, 1, 2, 3, ....} is a finite set.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

Between any two rational numbers there lies an unlimited number of rational numbers. So, the process of counting the rational numbers lying between 2 and 3 never comes to an end.

Hence, {x | x ∈ Q, 2 < x < 3} is an infinite set.

A set consisting of no element at all is called an empty set and it is denoted by φ.

A = {x | x ∈ N, x + 5 = 5}

⇒ x + 5 = 5

⇒ x = 5 - 5

⇒ x = 0

As 0 is not a natural number, there is no natural number x satisfying x + 5 = 5.

Hence, A = {x | x ∈ N, x + 5 = 5} is an empty set.

A set consisting of no element at all is called an empty set and it is denoted by φ.

B = {x | x ∈ N, 2x + 3 = 6}

2x + 3 = 6 ⇒ 2x = 6 - 3 = 3 ⇒ x = $\frac{3}{2}$

As $\frac{3}{2}$ is not a natural number, there is no natural number x satisfying 2x + 3 = 6.

Hence, B = {x | x ∈ N, 2x + 3 = 6} is an empty set.

A set consisting of no element at all is called an empty set and it is denoted by φ.

C = {x | x ∈ W, x + 2 < 2}

x + 2 < 2 ⇒ x < 2 - 2 ⇒ x < 0

As there is no whole number which is less than 0, there is no whole number x satisfying x + 2 < 2.

Hence, C = {x | x ∈ W, x + 2 < 2} is an empty set.

A set consisting of no element at all is called an empty set and it is denoted by φ.

D = {x | x ∈ N, 1 < x ≤ 2}

The natural number which is greater than 1 and less than or equal to 2 is 2.

∴ D = {2}

As D has one element, D is not an empty set.

Hence, D = {x | x ∈ N, 1 < x ≤ 2} is not an empty set.

A set consisting of no element at all is called an empty set and it is denoted by φ.

E = {x | x ∈ N, x² + 4 = 0}

x² + 4 = 0 ⇒ x² = -4

As the square of a natural number is always positive, there is no natural number x satisfying x² + 4 = 0.

Hence, E = {x | x ∈ N, x² + 4 = 0} is an empty set.

A set consisting of no element at all is called an empty set and it is denoted by φ.

F = {x | x is a prime number, 90 < x < 96}

The numbers lying between 90 and 96 are 91, 92, 93, 94 and 95.

91 = 7 × 13, 92 = 2 × 46, 93 = 3 × 31, 94 = 2 × 47 and 95 = 5 × 19

So, none of these numbers is a prime number.

Hence, F = {x | x is a prime number, 90 < x < 96} is an empty set.

A set consisting of no element at all is called an empty set and it is denoted by φ.

G = {x | x is an even prime}

2 is a prime number and it is even.

∴ G = {2}

As G has one element, G is not an empty set.

Hence, G = {x | x is an even prime} is not an empty set.

A set consisting of no element at all is called an empty set and it is denoted by φ.

H = {0}

H is a set containing one element, namely 0. So, H is not a set consisting of no element at all.

Hence, H = {0} is not an empty set.

Two finite sets A and B are said to be equivalent, if n(A) = n(B), i.e., they have the same number of elements.

A = {2, 3, 5, 7}

As A has 4 elements, n(A) = 4.

B = {x : x is a whole number, x ≤ 4}

B = {0, 1, 2, 3, 4}

As B has 5 elements, n(B) = 5.

As n(A) ≠ n(B),

Hence, A and B are not equivalent sets.

Two finite sets C and D are said to be equivalent, if n(C) = n(D), i.e., they have the same number of elements.

C = {x : x + 2 = 2}

x + 2 = 2 ⇒ x = 2 - 2 = 0

∴ C = {0}

As C has 1 element, n(C) = 1.

D = φ

As D has no element, n(D) = 0.

As n(C) ≠ n(D),

Hence, C and D are not equivalent sets.

Two finite sets E and F are said to be equivalent, if n(E) = n(F), i.e., they have the same number of elements.

E = {x : x is a natural number, x < 4}

E = {1, 2, 3}

As E has 3 elements, n(E) = 3.

F = {x : x is a whole number, x < 3}

F = {0, 1, 2}

As F has 3 elements, n(F) = 3.

As n(E) = n(F),

Hence, E and F are equivalent sets.

Two finite sets G and H are said to be equivalent, if n(G) = n(H), i.e., they have the same number of elements.

G = {x : x is an integer, -3 < x < 3}

G = {-2, -1, 0, 1, 2}

As G has 5 elements, n(G) = 5.

H = {x : x is a factor of 16}

16 = 1 × 16 = 2 × 8 = 4 × 4

H = {1, 2, 4, 8, 16}

As H has 5 elements, n(H) = 5.

As n(G) = n(H),

Hence, G and H are equivalent sets.

A set is a well-defined collection of distinct objects.

{a, b, c, 1, 2, 3} is a collection of six distinct objects and given any object, we can definitely say whether it belongs to this collection or not.

So, {a, b, c, 1, 2, 3} is a set.

Hence, the given statement is false.

The order in which the elements of a set are listed is immaterial.

Each of 5, 7 and 9 belongs to {5, 7, 9} as well as to {9, 5, 7}.

So, every element of {5, 7, 9} is in {9, 5, 7} and every element of {9, 5, 7} is in {5, 7, 9}.

Hence, the given statement is true.

A set containing exactly one element is called a singleton set.

⇒ x + 8 = 8

⇒ x = 8 - 8

⇒ x = 0

As 0 is a whole number,

{x | x ∈ W, x + 8 = 8} = {0}, which contains exactly one element.

Hence, the given statement is true.

A set consisting of no element at all is called an empty set and it is denoted by φ.

The whole numbers are 0, 1, 2, 3, .... and none of them is less than 0.

So, {x | x ∈ W, x < 0} has no element at all.

Hence, the given statement is true.

A set consisting of no element at all is called an empty set and it is denoted by φ.

x + 5 = 3 ⇒ x = 3 - 5 = -2

As -2 is not a natural number, there is no natural number x satisfying x + 5 = 3.

So, {x | x ∈ N, x + 5 = 3} has no element at all.

Hence, the given statement is true.

A set consisting of no element at all is called an empty set and it is denoted by φ.

The natural number which is greater than 3 and less than or equal to 4 is 4.

So, {x | x ∈ N, 3 < x ≤ 4} = {4}, which has one element.

∴ {x | x ∈ N, 3 < x ≤ 4} ≠ φ

Hence, the given statement is false.

The repetition of elements in a set is meaningless and so while finding the cardinal number of a set we do not consider the elements that are repeated.

The letters of the word 'MEERUT' are M, E, E, R, U, T.

Writing each letter once and only once, we get

A = {M, E, R, U, T}

As A has 5 elements, n(A) = 5 and not 6.

Hence, the given statement is false.

The number of distinct elements contained in a finite set A is called the cardinal number of A and is denoted by n(A).

A = {x | x ∈ N, 8 < x < 13}

A = {9, 10, 11, 12}

As A has 4 elements, n(A) = 4.

B = {x | x ∈ Z, -3 ≤ x < 1}

B = {-3, -2, -1, 0}

As B has 4 elements, n(B) = 4.

∴ n(A) = n(B)

Hence, the given statement is true.

Two sets A and B are said to be equal, if every element of A is in B and every element of B is in A.

If n(A) = n(B), then A and B are only equivalent sets and two equivalent sets need not be equal.

For example, let A = {1, 2, 3} and B = {a, b, c}.

Here n(A) = n(B) = 3, but 1 ∈ A and 1 ∉ B, so A ≠ B.

Hence, the given statement is false.

The number of distinct elements contained in a finite set B is called the cardinal number of B and is denoted by n(B).

There are 26 letters in the English alphabet, out of which a, e, i, o and u are the 5 vowels and the remaining letters are consonants.

∴ Number of consonants = 26 - 5 = 21

So, n(B) = 21.

Hence, the given statement is true.

A prime factor of a number is a factor of that number which is a prime number.

24 = 2 × 2 × 2 × 3

So, the prime factors of 24 are 2 and 3.

∴ {x | x is a prime factor of 24} = {2, 3} ≠ {2, 3, 4, 6, 8, 12, 24}

Hence, the given statement is false.

If every element of A is in B and there exists at least one element in B which is not there in A, then A is called a proper subset of B and we write A ⊂ B.

Hence, set A is a proper subset of set B is written as A ⊂ B.

If D is a subset of C, then we say that C is a superset of D and we write C ⊇ D.

Hence, set C is a superset of set D is written as C ⊇ D.

If set B contains set A, then every element of A is in B, i.e., B is a superset of A and we write B ⊇ A.

Hence, set B contains set A is written as B ⊇ A.

If A is not a subset of B, we write A ⊄ B and if B is not a subset of A, we write B ⊄ A.

Hence, neither A is a subset of B, nor B is a subset of A is written as A ⊄ B and B ⊄ A.

A = {all quadrilaterals}, B = {all rectangles}, C = {all squares} and D = {all rhombuses}

B ⊂ C means every rectangle is a square.

A rectangle whose adjacent sides are unequal, for example a rectangle of length 5 cm and breadth 3 cm, is not a square.

So, there is a rectangle which is not a square.

∴ B ⊄ C

Hence, the given statement is false, as every rectangle is not a square.

A = {all quadrilaterals}, B = {all rectangles}, C = {all squares} and D = {all rhombuses}

Every square is a rectangle whose adjacent sides are equal, but every rectangle is not a square.

∴ C ⊂ B

Every rectangle is a quadrilateral, but every quadrilateral is not a rectangle.

∴ B ⊂ A

∴ C ⊂ B ⊂ A

Hence, the given statement is true, as every square is a rectangle and every rectangle is a quadrilateral.

A = {all quadrilaterals}, B = {all rectangles}, C = {all squares} and D = {all rhombuses}

Every square is a rhombus whose angles are right angles, but every rhombus is not a square.

∴ C ⊂ D

Every rhombus is a quadrilateral, but every quadrilateral is not a rhombus.

∴ D ⊂ A

∴ C ⊂ D ⊂ A

Hence, the given statement is true, as every square is a rhombus and every rhombus is a quadrilateral.

A = {all quadrilaterals}, B = {all rectangles}, C = {all squares} and D = {all rhombuses}

D ⊂ C means every rhombus is a square.

A rhombus whose angles are not right angles is not a square.

So, there is a rhombus which is not a square.

∴ D ⊄ C

Hence, the given statement is false, as every rhombus is not a square.

A = {all quadrilaterals}, B = {all rectangles}, C = {all squares} and D = {all rhombuses}

Every rectangle is a quadrilateral, so B is a subset of A.

∴ A ⊇ B

Every square is a rectangle, so C is a subset of B.

∴ B ⊇ C

∴ A ⊇ B ⊇ C

Hence, the given statement is true, as the set of quadrilaterals contains the set of rectangles which contains the set of squares.

A = {all quadrilaterals}, B = {all rectangles}, C = {all squares} and D = {all rhombuses}

A ⊆ B means every quadrilateral is a rectangle.

A quadrilateral whose angles are not right angles, for example a trapezium, is not a rectangle.

∴ A ⊄ B

Hence, the given statement is false, as every quadrilateral is not a rectangle.

A = {all triangles}, B = {all isosceles triangles} and C = {all equilateral triangles}

B ⊂ C means every isosceles triangle is an equilateral triangle.

An isosceles triangle whose sides are 5 cm, 5 cm and 3 cm does not have all its sides equal, so it is not an equilateral triangle.

∴ B ⊄ C

Hence, the given statement is false, as every isosceles triangle is not an equilateral triangle.

A = {all triangles}, B = {all isosceles triangles} and C = {all equilateral triangles}

An equilateral triangle has all its three sides equal, so it has at least two sides equal and hence it is an isosceles triangle. But every isosceles triangle is not an equilateral triangle.

∴ C ⊂ B

Every isosceles triangle is a triangle, but every triangle is not an isosceles triangle.

∴ B ⊂ A

∴ C ⊂ B ⊂ A

Hence, the given statement is true, as every equilateral triangle is an isosceles triangle and every isosceles triangle is a triangle.

A = {1, 2}

The symbol ⊂ is used between two sets.

Here 1 is an element of A and not a set, so we cannot write 1 ⊂ A.

Hence, the given statement is not true.

A = {1, 2}

The elements of A are 1 and 2.

As {1} is a set and it is not an element of A,

∴ {1} ∉ A

Hence, the given statement is not true.

A = {1, 2}

The elements of A are 1 and 2.

As 1 is an element of A,

∴ 1 ∈ A

Hence, the given statement is true.

A = {1, 2}

The elements of A are 1 and 2.

As φ is not an element of A,

∴ φ ∉ A

Hence, the given statement is not true.

A = {1, 2}

The empty set is a subset of every set.

Also, A is a non-empty set, so φ ≠ A.

∴ φ ⊂ A

Hence, the given statement is true.

A = {1, 2}

The elements of A are 1 and 2.

As {1} is a set and it is not an element of A,

∴ {1} ∉ A

Hence, the given statement is true.

The symbol ⊂ is used between two sets. Here a is an element of {a, b, c} and not a set, so we cannot write a ⊂ {a, b, c}.

Hence, the given statement is not correct.

The elements of {a, b, c} are a, b and c. As {a} is a set and it is not an element of {a, b, c},

∴ {a} ∉ {a, b, c}

Hence, the given statement is not correct.

If every element of a set P is in a set Q and there exists at least one element in Q which is not there in P, then P ⊂ Q. The only element of {a} is a and a ∈ {a, b, c}. Also, b ∈ {a, b, c} but b ∉ {a}.

∴ {a} ⊂ {a, b, c}

Hence, the given statement is correct.

The elements of {a, b, c} are a, b and c. As φ is not an element of {a, b, c},

∴ φ ∉ {a, b, c}

Hence, the given statement is not correct.

The empty set is a subset of every set. Also, {a, b, c} is a non-empty set, so φ ≠ {a, b, c}.

∴ φ ⊂ {a, b, c}

Hence, the given statement is correct.

The only element of {φ} is φ. As φ ∉ {a, b, c}, every element of {φ} is not in {a, b, c}.

∴ {φ} ⊄ {a, b, c}

Hence, the given statement is not correct.

The elements of {{a}, b} are {a} and b. As a is not an element of {{a}, b},

∴ a ∉ {{a}, b}

Hence, the given statement is not correct.

The only element of {a} is a and the elements of {{a}, b} are {a} and b. As a ∉ {{a}, b}, every element of {a} is not in {{a}, b}.

∴ {a} ⊄ {{a}, b}

Hence, the given statement is not correct.

The elements of {{a, b}, c} are {a, b} and c. As {a, b} is an element of {{a, b}, c},

∴ {a, b} ∈ {{a, b}, c}

Hence, the given statement is correct.

φ is a set containing no element at all. {0} is a set containing one element, namely 0. As 0 ∈ {0} but 0 ∉ φ,

∴ φ ≠ {0}

Hence, the given statement is not true.

φ is a set containing no element at all. {φ} is a set containing one element, namely φ. As φ ∈ {φ} but φ has no element at all,

∴ φ ≠ {φ}

Hence, the given statement is not true.

The only element of {0} is 0. As φ is not an element of {0},

∴ φ ∉ {0}

Hence, the given statement is not true.

The only element of {φ} is φ. As φ is an element of {φ},

∴ φ ∈ {φ}

Hence, the given statement is true.

The elements of {φ, {0}} are φ and {0}. As φ is an element of {φ, {0}},

∴ φ ∈ {φ, {0}}

Hence, the given statement is true.

The only element of {φ} is φ and the only element of {0} is 0. As φ ∉ {0}, every element of {φ} is not in {0}.

∴ {φ} ⊄ {0}

Hence, the given statement is not true.

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

Here 1 ∈ A but 1 ∉ B, so A ⊄ B.

Also, 3 ∈ B but 3 ∉ A, so B ⊄ A.

So, there are two sets A and B such that neither A ⊆ B nor B ⊆ A.

Hence, the given statement is not true.

A finite set has a countable number of elements.

Every element of a subset of a finite set is also an element of that finite set.

So, a subset of a finite set cannot have more elements than the finite set itself and hence the process of counting its elements surely comes to an end.

Hence, the given statement is true.

Let N = {1, 2, 3, 4, 5, ....} be the set of all natural numbers, which is an infinite set.

Let A = {1, 2, 3}.

Every element of A is in N, so A is a subset of N.

But A has only 3 elements, so A is a finite set.

So, there is a subset of an infinite set which is finite.

Hence, the given statement is not true.

If A is a proper subset of B, then every element of A is in B and there exists at least one element in B which is not there in A.

The empty set φ has no element at all, so there can be no set which has fewer elements than φ.

∴ φ has no proper subset.

Hence, the given statement is not true.

The set of all possible subsets of a set A is called the power set of A, denoted by P(A).

If A contains n elements, then A has 2ⁿ subsets and all of these subsets are the members of P(A).

So, P(A) consists of the 2ⁿ subsets of A.

Hence, the given statement is true.

(i) A is the set of letters in the word 'seed'.

The letters of the word 'seed' are s, e, e, d.

As every repeated element is taken only once in a set, writing each letter once and only once, we get

Hence, A = {s, e, d}.

(ii) A = {s, e, d}

As A has 3 elements,

Hence, n(A) = 3.

(iii) If a set contains n elements, then it has 2ⁿ subsets.

Here n(A) = 3.

∴ Number of subsets of A = 2³ = 8

Hence, the number of subsets of A is 8.

(iv) If a set contains n elements, then it has (2ⁿ - 1) proper subsets.

Here n(A) = 3.

∴ Number of proper subsets of A = 2³ - 1 = 8 - 1 = 7

Hence, the number of proper subsets of A is 7.

A = {4, 9}

If a set contains n elements, then it has 2ⁿ subsets.

As A contains 2 elements, there are 2² = 4 subsets of A.

The empty set is a subset of every set and every set is a subset of itself.

So, all possible subsets of A are A₁ = φ, A₂ = {4}, A₃ = {9} and A₄ = {4, 9}.

Hence, all possible subsets of A are φ, {4}, {9} and {4, 9}.

B = {2, 3, 8}

If a set contains n elements, then it has 2ⁿ subsets.

As B contains 3 elements, there are 2³ = 8 subsets of B.

The empty set is a subset of every set and every set is a subset of itself.

So, all possible subsets of B are B₁ = φ, B₂ = {2}, B₃ = {3}, B₄ = {8}, B₅ = {2, 3}, B₆ = {3, 8}, B₇ = {2, 8} and B₈ = {2, 3, 8}.

Hence, all possible subsets of B are φ, {2}, {3}, {8}, {2, 3}, {3, 8}, {2, 8} and {2, 3, 8}.

C = {0, 1, 2}

If a set contains n elements, then it has 2ⁿ subsets.

As C contains 3 elements, there are 2³ = 8 subsets of C.

The empty set is a subset of every set and every set is a subset of itself.

So, all possible subsets of C are C₁ = φ, C₂ = {0}, C₃ = {1}, C₄ = {2}, C₅ = {0, 1}, C₆ = {1, 2}, C₇ = {0, 2} and C₈ = {0, 1, 2}.

Hence, all possible subsets of C are φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2} and {0, 1, 2}.

A = {0, 5}

The set of all possible subsets of a set A is called the power set of A, denoted by P(A).

As A contains 2 elements, there are 2² = 4 subsets of A.

All possible subsets of A are φ, {0}, {5} and {0, 5}.

∴ P(A) = Set of all possible subsets of A

Hence, P(A) = {φ, {0}, {5}, {0, 5}}.

B = {7, 9}

The set of all possible subsets of a set B is called the power set of B, denoted by P(B).

As B contains 2 elements, there are 2² = 4 subsets of B.

All possible subsets of B are φ, {7}, {9} and {7, 9}.

∴ P(B) = Set of all possible subsets of B

Hence, P(B) = {φ, {7}, {9}, {7, 9}}.

C = {2, 4, 6}

The set of all possible subsets of a set C is called the power set of C, denoted by P(C).

As C contains 3 elements, there are 2³ = 8 subsets of C.

All possible subsets of C are φ, {2}, {4}, {6}, {2, 4}, {4, 6}, {2, 6} and {2, 4, 6}.

∴ P(C) = Set of all possible subsets of C

Hence, P(C) = {φ, {2}, {4}, {6}, {2, 4}, {4, 6}, {2, 6}, {2, 4, 6}}.

A = {1, {2}}

Let A = {1, a}, where a = {2}.

The set of all possible subsets of a set A is called the power set of A, denoted by P(A).

As A contains 2 elements, there are 2² = 4 subsets of A.

All possible subsets of A are A₁ = φ, A₂ = {1}, A₃ = {a} and A₄ = {1, a}.

∴ P(A) = {φ, {1}, {a}, {1, a}}

Replacing a by {2}, we get

Hence, P(A) = {φ, {1}, {{2}}, {1, {2}}}.

ξ = {x : x ∈ N, x < 50}

ξ = {1, 2, 3, ...., 49}

A = {x : x² ∈ ξ}

Here x² ≤ 49, i.e., x ≤ 7.

As 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36 and 7² = 49 all belong to ξ, while 8² = 64 does not belong to ξ,

∴ A = {1, 2, 3, 4, 5, 6, 7}

B = {x : x = n², n ∈ N}

Putting n = 1, 2, 3, 4, 5, 6 and 7 in n², the squares which belong to ξ are 1, 4, 9, 16, 25, 36 and 49.

∴ B = {1, 4, 9, 16, 25, 36, 49}

C = {x : x is a factor of 36}

36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6

∴ C = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Here n(A) = 7, n(B) = 7 and n(C) = 9.

(i) A ⊆ B

As 2 ∈ A but 2 ∉ B, every element of A is not in B.

∴ A ⊄ B

Hence, the given statement is false.

(ii) A = B

Two sets A and B are said to be equal, if every element of A is in B and every element of B is in A.

As 2 ∈ A but 2 ∉ B,

∴ A ≠ B

Hence, the given statement is false.

(iii) A ↔ B

Two finite sets A and B are said to be equivalent, if n(A) = n(B).

As n(A) = 7 and n(B) = 7, so n(A) = n(B).

∴ A ↔ B

Hence, the given statement is true.

(iv) B ↔ C

Two finite sets B and C are said to be equivalent, if n(B) = n(C).

As n(B) = 7 and n(C) = 9, so n(B) ≠ n(C).

∴ B and C are not equivalent sets.

Hence, the given statement is false.

(v) n(A) < n(C)

As n(A) = 7 and n(C) = 9 and 7 < 9,

∴ n(A) < n(C)

Hence, the given statement is true.

A = {a, b, c, d}

B = {b, c, e}

A ∪ B - contains all the elements in set A and B.

A ∪ B = {a, b, c, d} ∪ {b, c, e}

Hence, A ∪ B = {a, b, c, d, e}.

B = {b, c, e}

C = {a, b, e}

B ∪ C - contains all the elements in set B and C.

B ∪ C = {b, c, e} ∪ {a, b, e}

Hence, B ∪ C = {a, b, c, e}.

A = {a, b, c, d}

C = {a, b, e}

A ∪ C - contains all the elements in set A and C.

A ∪ C = {a, b, c, d} ∪ {a, b, e}

Hence, A ∪ C = {a, b, c, d, e}.

A = {a, b, c, d}

B = {b, c, e}

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {a, b, c, d} ∩ {b, c, e}

Hence, A ∩ B = {b, c}.

B = {b, c, e}

C = {a, b, e}

B ∩ C - contains all the common elements in set B and C.

B ∩ C = {b, c, e} ∩ {a, b, e}

Hence, B ∩ C = {b, e}.

A = {a, b, c, d}

C = {a, b, e}

A ∩ C - contains all the common elements in set A and C.

A ∩ C = {a, b, c, d} ∩ {a, b, e}

Hence, A ∩ C = {a, b}.

A = {2, 3, 4, 6}

B = {5, 7, 8}

A ∪ B - contains all the elements in set A and B.

A ∪ B = {2, 3, 4, 6} ∪ {5, 7, 8}

Hence, A ∪ B = {2, 3, 4, 5, 6, 7, 8}.

B = {5, 7, 8}

C = {2, 7, 8, 9}

B ∪ C - contains all the elements in set B and C.

B ∪ C = {5, 7, 8} ∪ {2, 7, 8, 9}

Hence, B ∪ C = {2, 5, 7, 8, 9}.

A = {2, 3, 4, 6}

C = {2, 7, 8, 9}

A ∪ C - contains all the elements in set A and C.

A ∪ C = {2, 3, 4, 6} ∪ {2, 7, 8, 9}

Hence, A ∪ C = {2, 3, 4, 6, 7, 8, 9}.

A = {2, 3, 4, 6}

B = {5, 7, 8}

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {2, 3, 4, 6} ∩ {5, 7, 8}

As there is no element common to A and B,

Hence, A ∩ B = φ.

A = {2, 3, 4, 6}

C = {2, 7, 8, 9}

A ∩ C - contains all the common elements in set A and C.

A ∩ C = {2, 3, 4, 6} ∩ {2, 7, 8, 9}

Hence, A ∩ C = {2}.

B = {5, 7, 8}

C = {2, 7, 8, 9}

B ∩ C - contains all the common elements in set B and C.

B ∩ C = {5, 7, 8} ∩ {2, 7, 8, 9}

Hence, B ∩ C = {7, 8}.

A = {1, 4, 7, 8}

B = {4, 6, 8, 9}

A - B - contains all the elements in set A but not in B.

A - B = {1, 4, 7, 8} - {4, 6, 8, 9}

Hence, A - B = {1, 7}.

A = {1, 4, 7, 8}

B = {4, 6, 8, 9}

B - A - contains all the elements in set B but not in A.

B - A = {4, 6, 8, 9} - {1, 4, 7, 8}

Hence, B - A = {6, 9}.

ξ = {13, 14, 15, 16, 17, 18, 19, 20, 21}

A = {13, 17, 19}

A′ - contains all the elements in universal set which are not in A.

A′ = ξ - A

A′ = {13, 14, 15, 16, 17, 18, 19, 20, 21} - {13, 17, 19}

Hence, A′ = {14, 15, 16, 18, 20, 21}.

ξ = {13, 14, 15, 16, 17, 18, 19, 20, 21}

B = {14, 16, 18, 20}

B′ - contains all the elements in universal set which are not in B.

B′ = ξ - B

B′ = {13, 14, 15, 16, 17, 18, 19, 20, 21} - {14, 16, 18, 20}

Hence, B′ = {13, 15, 17, 19, 21}.

ξ = {x | x ∈ Z, -4 ≤ x ≤ 4}

ξ = {-4, -3, -2, -1, 0, 1, 2, 3, 4}

A = {x | x ∈ W, x < 4}

A = {0, 1, 2, 3}

A′ - contains all the elements in universal set which are not in A.

A′ = ξ - A

A′ = {-4, -3, -2, -1, 0, 1, 2, 3, 4} - {0, 1, 2, 3}

Hence, A′ = {-4, -3, -2, -1, 4}.

ξ = {x | x ∈ Z, -4 ≤ x ≤ 4}

ξ = {-4, -3, -2, -1, 0, 1, 2, 3, 4}

B = {x | x ∈ N, 2 < x ≤ 4}

B = {3, 4}

B′ - contains all the elements in universal set which are not in B.

B′ = ξ - B

B′ = {-4, -3, -2, -1, 0, 1, 2, 3, 4} - {3, 4}

Hence, B′ = {-4, -3, -2, -1, 0, 1, 2}.

ξ = {x | x ∈ N, x is a factor of 144}

144 = 1 × 144 = 2 × 72 = 3 × 48 = 4 × 36 = 6 × 24 = 8 × 18 = 9 × 16 = 12 × 12

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}

A = {x | x ∈ N, x is a factor of 24}

24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6

A = {1, 2, 3, 4, 6, 8, 12, 24}

A′ - contains all the elements in universal set which are not in A.

A′ = ξ - A

A′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 8, 12, 24}

Hence, A′ = {9, 16, 18, 36, 48, 72, 144}.

ξ = {x | x ∈ N, x is a factor of 144}

144 = 1 × 144 = 2 × 72 = 3 × 48 = 4 × 36 = 6 × 24 = 8 × 18 = 9 × 16 = 12 × 12

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}

B = {x | x ∈ N, x is a factor of 36}

36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

B′ - contains all the elements in universal set which are not in B.

B′ = ξ - B

B′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 9, 12, 18, 36}

Hence, B′ = {8, 16, 24, 48, 72, 144}.

ξ = {x | x ∈ N, x is a factor of 144}

144 = 1 × 144 = 2 × 72 = 3 × 48 = 4 × 36 = 6 × 24 = 8 × 18 = 9 × 16 = 12 × 12

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}

C = {x | x ∈ N, x is a factor of 48}

48 = 1 × 48 = 2 × 24 = 3 × 16 = 4 × 12 = 6 × 8

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

C′ - contains all the elements in universal set which are not in C.

C′ = ξ - C

C′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

Hence, C′ = {9, 18, 36, 72, 144}.

A = {x | x ∈ N, x is a factor of 24}

A = {1, 2, 3, 4, 6, 8, 12, 24}

B = {x | x ∈ N, x is a factor of 36}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

A ∪ B - contains all the elements in set A and B.

A ∪ B = {1, 2, 3, 4, 6, 8, 12, 24} ∪ {1, 2, 3, 4, 6, 9, 12, 18, 36}

Hence, A ∪ B = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}.

B = {x | x ∈ N, x is a factor of 36}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {x | x ∈ N, x is a factor of 48}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

B ∪ C - contains all the elements in set B and C.

B ∪ C = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∪ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

Hence, B ∪ C = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}.

A = {x | x ∈ N, x is a factor of 24}

A = {1, 2, 3, 4, 6, 8, 12, 24}

C = {x | x ∈ N, x is a factor of 48}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

A ∪ C - contains all the elements in set A and C.

A ∪ C = {1, 2, 3, 4, 6, 8, 12, 24} ∪ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

Hence, A ∪ C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}.

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}

A = {x | x ∈ N, x is a factor of 24}

A = {1, 2, 3, 4, 6, 8, 12, 24}

B = {x | x ∈ N, x is a factor of 36}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

B′ - contains all the elements in universal set which are not in B.

B′ = ξ - B

B′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 9, 12, 18, 36}

B′ = {8, 16, 24, 48, 72, 144}

A ∩ B′ - contains all the common elements in set A and B′.

A ∩ B′ = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {8, 16, 24, 48, 72, 144}

Hence, A ∩ B′ = {8, 24}.

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}

B = {x | x ∈ N, x is a factor of 36}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {x | x ∈ N, x is a factor of 48}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

C′ - contains all the elements in universal set which are not in C.

C′ = ξ - C

C′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

C′ = {9, 18, 36, 72, 144}

B ∩ C′ - contains all the common elements in set B and C′.

B ∩ C′ = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∩ {9, 18, 36, 72, 144}

Hence, B ∩ C′ = {9, 18, 36}.

A = {x | x ∈ N, x is a factor of 24}

A = {1, 2, 3, 4, 6, 8, 12, 24}

C = {x | x ∈ N, x is a factor of 48}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

C - A - contains all the elements in set C but not in A.

C - A = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} - {1, 2, 3, 4, 6, 8, 12, 24}

Hence, C - A = {16, 48}.

A = {x | x ∈ N, x is a factor of 24}

A = {1, 2, 3, 4, 6, 8, 12, 24}

B = {x | x ∈ N, x is a factor of 36}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {x | x ∈ N, x is a factor of 48}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

B ∩ C - contains all the common elements in set B and C.

B ∩ C = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∩ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

B ∩ C = {1, 2, 3, 4, 6, 12}

A - (B ∩ C) - contains all the elements in set A but not in B ∩ C.

A - (B ∩ C) = {1, 2, 3, 4, 6, 8, 12, 24} - {1, 2, 3, 4, 6, 12}

Hence, A - (B ∩ C) = {8, 24}.

A = {1, 2, 3, 4, 6, 8, 12, 24}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

B ∪ C - contains all the elements in set B and C.

B ∪ C = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∪ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

B ∪ C = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}

A ∩ (B ∪ C) - contains all the common elements in set A and B ∪ C.

A ∩ (B ∪ C) = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}

A ∩ (B ∪ C) = {1, 2, 3, 4, 6, 8, 12, 24} = A

Hence, the given statement is true.

A = {1, 2, 3, 4, 6, 8, 12, 24}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

Each of 1, 2, 3, 4, 6, 8, 12 and 24 is an element of C, so every element of A is in C.

Also, 16 ∈ C but 16 ∉ A, so A ≠ C.

∴ A ⊂ C

Hence, the given statement is true.

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

As 9 ∈ B but 9 ∉ C, every element of B is not in C.

∴ B ⊄ C

Hence, the given statement is false.

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}

A = {1, 2, 3, 4, 6, 8, 12, 24}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

C′ - contains all the elements in universal set which are not in C.

C′ = ξ - C

C′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

C′ = {9, 18, 36, 72, 144}

A ∩ C′ - contains all the common elements in set A and C′.

A ∩ C′ = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {9, 18, 36, 72, 144}

As there is no element common to A and C′,

∴ A ∩ C′ = φ

Hence, the given statement is true.

B = {a, c, e, g}

C = {b, e, f, g}

B ∪ C - contains all the elements in set B and C.

B ∪ C = {a, c, e, g} ∪ {b, e, f, g} = {a, b, c, e, f, g}

C ∪ B - contains all the elements in set C and B.

C ∪ B = {b, e, f, g} ∪ {a, c, e, g} = {a, b, c, e, f, g}

∴ B ∪ C = C ∪ B = {a, b, c, e, f, g}

Hence, B ∪ C = C ∪ B.

B = {a, c, e, g}

C = {b, e, f, g}

B ∩ C - contains all the common elements in set B and C.

B ∩ C = {a, c, e, g} ∩ {b, e, f, g} = {e, g}

C ∩ B - contains all the common elements in set C and B.

C ∩ B = {b, e, f, g} ∩ {a, c, e, g} = {e, g}

∴ B ∩ C = C ∩ B = {e, g}

Hence, B ∩ C = C ∩ B.

A = {a, b, c, d, e}

B = {a, c, e, g}

C = {b, e, f, g}

B ∪ C = {a, c, e, g} ∪ {b, e, f, g} = {a, b, c, e, f, g}

∴ A ∪ (B ∪ C) = {a, b, c, d, e} ∪ {a, b, c, e, f, g} = {a, b, c, d, e, f, g}

Again, A ∪ B = {a, b, c, d, e} ∪ {a, c, e, g} = {a, b, c, d, e, g}

∴ (A ∪ B) ∪ C = {a, b, c, d, e, g} ∪ {b, e, f, g} = {a, b, c, d, e, f, g}

∴ A ∪ (B ∪ C) = (A ∪ B) ∪ C = {a, b, c, d, e, f, g}

Hence, A ∪ (B ∪ C) = (A ∪ B) ∪ C.

A = {a, b, c, d, e}

B = {a, c, e, g}

C = {b, e, f, g}

B ∩ C = {a, c, e, g} ∩ {b, e, f, g} = {e, g}

∴ A ∩ (B ∩ C) = {a, b, c, d, e} ∩ {e, g} = {e}

Again, A ∩ B = {a, b, c, d, e} ∩ {a, c, e, g} = {a, c, e}

∴ (A ∩ B) ∩ C = {a, c, e} ∩ {b, e, f, g} = {e}

∴ A ∩ (B ∩ C) = (A ∩ B) ∩ C = {e}

Hence, A ∩ (B ∩ C) = (A ∩ B) ∩ C.

A = {a, b, c, d, e}

B = {a, c, e, g}

C = {b, e, f, g}

B ∩ C = {a, c, e, g} ∩ {b, e, f, g} = {e, g}

∴ A ∪ (B ∩ C) = {a, b, c, d, e} ∪ {e, g} = {a, b, c, d, e, g}

Again, A ∪ B = {a, b, c, d, e} ∪ {a, c, e, g} = {a, b, c, d, e, g}

and A ∪ C = {a, b, c, d, e} ∪ {b, e, f, g} = {a, b, c, d, e, f, g}

∴ (A ∪ B) ∩ (A ∪ C) = {a, b, c, d, e, g} ∩ {a, b, c, d, e, f, g} = {a, b, c, d, e, g}

∴ A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) = {a, b, c, d, e, g}

Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

A = {a, b, c, d, e}

B = {a, c, e, g}

C = {b, e, f, g}

B ∪ C = {a, c, e, g} ∪ {b, e, f, g} = {a, b, c, e, f, g}

∴ A ∩ (B ∪ C) = {a, b, c, d, e} ∩ {a, b, c, e, f, g} = {a, b, c, e}

Again, A ∩ B = {a, b, c, d, e} ∩ {a, c, e, g} = {a, c, e}

and A ∩ C = {a, b, c, d, e} ∩ {b, e, f, g} = {b, e}

∴ (A ∩ B) ∪ (A ∩ C) = {a, c, e} ∪ {b, e} = {a, b, c, e}

∴ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) = {a, b, c, e}

Hence, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

ξ = {b, c, d, e, f, g}

A = {b, c, d, e}

B = {d, e, f, g}

A ∪ B = {b, c, d, e} ∪ {d, e, f, g} = {b, c, d, e, f, g}

(A ∪ B)′ - contains all the elements in universal set which are not in A ∪ B.

(A ∪ B)′ = ξ - (A ∪ B) = {b, c, d, e, f, g} - {b, c, d, e, f, g} = φ

Again, A′ = ξ - A = {b, c, d, e, f, g} - {b, c, d, e} = {f, g}

and B′ = ξ - B = {b, c, d, e, f, g} - {d, e, f, g} = {b, c}

∴ A′ ∩ B′ = {f, g} ∩ {b, c} = φ

∴ (A ∪ B)′ = (A′ ∩ B′) = φ

Hence, (A ∪ B)′ = (A′ ∩ B′).

ξ = {b, c, d, e, f, g}

A = {b, c, d, e}

B = {d, e, f, g}

A ∩ B = {b, c, d, e} ∩ {d, e, f, g} = {d, e}

(A ∩ B)′ - contains all the elements in universal set which are not in A ∩ B.

(A ∩ B)′ = ξ - (A ∩ B) = {b, c, d, e, f, g} - {d, e} = {b, c, f, g}

Again, A′ = ξ - A = {b, c, d, e, f, g} - {b, c, d, e} = {f, g}

and B′ = ξ - B = {b, c, d, e, f, g} - {d, e, f, g} = {b, c}

∴ A′ ∪ B′ = {f, g} ∪ {b, c} = {b, c, f, g}

∴ (A ∩ B)′ = (A′ ∪ B′) = {b, c, f, g}

Hence, (A ∩ B)′ = (A′ ∪ B′).

(i) A ∪ A = A.

As A ∪ A contains all the elements of A and of A, it contains exactly the elements of A.

(ii) A ∩ A = A.

As every element of A is common to A and A, A ∩ A contains exactly the elements of A.

(iii) A ∪ φ = A.

As the empty set φ has no element, A ∪ φ contains exactly the elements of A.

(iv) A ∩ φ = φ.

As the empty set φ has no element, there can be no element common to A and φ.

(v) (A ∪ B)′ = A′ ∩ B′.

This is De-Morgan's law for the complement of the union of two sets.

(vi) (A ∩ B)′ = A′ ∪ B′.

This is De-Morgan's law for the complement of the intersection of two sets.

ξ = {x | x ∈ N, 4 ≤ x < 18}

ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

A = {x | x is a multiple of 2}

A = {4, 6, 8, 10, 12, 14, 16}

B = {x | x is a multiple of 3}

B = {6, 9, 12, 15}

A ∪ B = {4, 6, 8, 10, 12, 14, 16} ∪ {6, 9, 12, 15} = {4, 6, 8, 9, 10, 12, 14, 15, 16}

(A ∪ B)′ - contains all the elements in universal set which are not in A ∪ B.

(A ∪ B)′ = ξ - (A ∪ B)

(A ∪ B)′ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17} - {4, 6, 8, 9, 10, 12, 14, 15, 16}

(A ∪ B)′ = {5, 7, 11, 13, 17}

Again, A′ = ξ - A = {5, 7, 9, 11, 13, 15, 17}

and B′ = ξ - B = {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}

∴ A′ ∩ B′ = {5, 7, 9, 11, 13, 15, 17} ∩ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17} = {5, 7, 11, 13, 17}

∴ (A ∪ B)′ = (A′ ∩ B′) = {5, 7, 11, 13, 17}

Hence, (A ∪ B)′ = (A′ ∩ B′).

ξ = {x | x ∈ N, 4 ≤ x < 18}

ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

A = {x | x is a multiple of 2}

A = {4, 6, 8, 10, 12, 14, 16}

B = {x | x is a multiple of 3}

B = {6, 9, 12, 15}

A ∩ B = {4, 6, 8, 10, 12, 14, 16} ∩ {6, 9, 12, 15} = {6, 12}

(A ∩ B)′ - contains all the elements in universal set which are not in A ∩ B.

(A ∩ B)′ = ξ - (A ∩ B)

(A ∩ B)′ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17} - {6, 12}

(A ∩ B)′ = {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}

Again, A′ = ξ - A = {5, 7, 9, 11, 13, 15, 17}

and B′ = ξ - B = {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}

∴ A′ ∪ B′ = {5, 7, 9, 11, 13, 15, 17} ∪ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}

A′ ∪ B′ = {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}

∴ (A ∩ B)′ = (A′ ∪ B′) = {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}

Hence, (A ∩ B)′ = (A′ ∪ B′).

ξ = {x | x ∈ N, 4 ≤ x < 18}

ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

A = {x | x is a multiple of 2}

A = {4, 6, 8, 10, 12, 14, 16}

B = {x | x is a multiple of 3}

B = {6, 9, 12, 15}

A - B - contains all the elements in set A but not in B.

A - B = {4, 6, 8, 10, 12, 14, 16} - {6, 9, 12, 15} = {4, 8, 10, 14, 16}

Again, B′ = ξ - B = {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}

∴ A ∩ B′ = {4, 6, 8, 10, 12, 14, 16} ∩ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17} = {4, 8, 10, 14, 16}

∴ A - B = A ∩ B′ = {4, 8, 10, 14, 16}

Hence, A - B = A ∩ B′.

ξ = {x | x ∈ N, 4 ≤ x < 18}

ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

A = {x | x is a multiple of 2}

A = {4, 6, 8, 10, 12, 14, 16}

B = {x | x is a multiple of 3}

B = {6, 9, 12, 15}

C = {x | x ∈ N, x < 11}

C = {4, 5, 6, 7, 8, 9, 10}

B ∩ C = {6, 9, 12, 15} ∩ {4, 5, 6, 7, 8, 9, 10} = {6, 9}

∴ A ∪ (B ∩ C) = {4, 6, 8, 10, 12, 14, 16} ∪ {6, 9} = {4, 6, 8, 9, 10, 12, 14, 16}

Again, A ∪ B = {4, 6, 8, 10, 12, 14, 16} ∪ {6, 9, 12, 15} = {4, 6, 8, 9, 10, 12, 14, 15, 16}

and A ∪ C = {4, 6, 8, 10, 12, 14, 16} ∪ {4, 5, 6, 7, 8, 9, 10} = {4, 5, 6, 7, 8, 9, 10, 12, 14, 16}

∴ (A ∪ B) ∩ (A ∪ C) = {4, 6, 8, 9, 10, 12, 14, 15, 16} ∩ {4, 5, 6, 7, 8, 9, 10, 12, 14, 16}

(A ∪ B) ∩ (A ∪ C) = {4, 6, 8, 9, 10, 12, 14, 16}

∴ A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) = {4, 6, 8, 9, 10, 12, 14, 16}

Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

A = {1}

A is a set containing one element, namely 1, whereas 1 is a number and not a set.

∴ A ≠ 1, so option 1 is not correct.

The symbol ⊂ is used between two sets and 1 is not a set.

∴ 1 ⊄ A, so option 2 is not correct.

The only element of A is 1 and {1} is not an element of A.

∴ {1} ∉ A, so option 3 is not correct.

As 1 is an element of A,

∴ 1 ∈ A

Hence, option 4 is the correct option.

A = {1, 2, {3, 4}}

The elements of A are 1, 2 and {3, 4}.

As 3 is not an element of A, 3 ∉ A, so option 1 is not correct.

The only element of {1} is 1 and 1 ∈ A. Also, 2 ∈ A but 2 ∉ {1}.

∴ {1} ⊂ A

As {2} is not an element of A, {2} ∉ A, so option 3 is not correct.

The symbol ⊆ is used between two sets and 1 is not a set, so option 4 is not correct.

Hence, option 2 is the correct option.

Every set is a subset of itself, i.e., A ⊆ A, so any set A is comparable with itself.

∴ Statement (i) is correct.

A set containing exactly one element is called a singleton set and {0} contains exactly one element, namely 0.

∴ Statement (ii) is correct.

{φ} is a set containing one element, namely φ, so it is not a set consisting of no element at all.

∴ Statement (iii) is not correct.

So, the correct statements are (i) and (ii).

Hence, option 1 is the correct option.

The set of all possible subsets of a set A is called the power set of A, denoted by P(A).

If A contains n elements, then A has 2ⁿ subsets and each of these subsets is an element of P(A).

∴ Total number of elements in P(A) = 2ⁿ

Hence, option 2 is the correct option.

If a set S contains n elements, then the total number of subsets of S is 2ⁿ.

Out of these subsets, the empty set φ and the set S itself are not non-empty proper subsets of S.

∴ Number of non-empty proper subsets of S = 2ⁿ - 2

= 2 × 2^n-1 - 2 × 1

= 2(2^n-1 - 1)

Hence, option 4 is the correct option.

A set containing n elements has (2ⁿ - 1) proper subsets.

{2, 3, 5} contains 3 elements.

∴ Number of proper subsets = 2³ - 1 = 8 - 1 = 7

Hence, option 3 is the correct option.

A = {a, b}

As A contains 2 elements, there are 2² = 4 subsets of A.

All possible subsets of A are φ, {a}, {b} and {a, b}.

∴ P(A) = Set of all possible subsets of A = {φ, {a}, {b}, {a, b}}

Hence, option 4 is the correct option.

φ is a set consisting of no element at all.

0 is a number and not a set, so φ ≠ 0 and option 1 is not correct.

{0} is a set containing one element, namely 0, so φ ≠ {0} and option 2 is not correct.

{φ} is a set containing one element, namely φ, so φ ≠ {φ} and option 3 is not correct.

In roster form, an empty set is written as { }.

∴ φ = { }

Hence, option 4 is the correct option.

The elements of {a, b, c} are a, b and c.

As {a} is not an element of {a, b, c}, {a} ∉ {a, b, c}, so option 1 is not correct.

The symbol ⊆ is used between two sets and a is not a set, so option 2 is not correct.

The elements of {{a}, b} are {a} and b.

As a is not an element of {{a}, b}, a ∉ {{a}, b}, so option 3 is not correct.

So, none of the given statements is correct.

Hence, option 4 is the correct option.

{1, 2, 3} is a subset of the infinite set N = {1, 2, 3, 4, ....} and it is finite, so option 1 is not correct.

The empty set φ has no element at all, so φ has no proper subset and option 2 is not correct.

As every repeated element is taken only once in a set,

{a, b, c, 1, 2, 3, a, b, c, 1, 2, 3, ....} = {a, b, c, 1, 2, 3}, which is a finite set, so option 3 is not correct.

A subset of a finite set cannot have more elements than the finite set itself, so it is finite.

Hence, option 4 is the correct option.

A set containing exactly one element is called a singleton set.

x² = x ⇒ x² - x = 0 ⇒ x(x - 1) = 0 ⇒ x = 0 or x = 1

So, {x ∈ R : x² = x} = {0, 1}, which has two elements and option 1 is not correct.

3x = 4 ⇒ x = $\frac{4}{3}$, which is not a natural number.

So, {x ∈ N : 3x = 4} = φ and option 2 is not correct.

As the square of a real number is never negative, {x ∈ R : x² = -1} = φ and option 3 is not correct.

The only integer which is neither positive nor negative is 0.

So, {x : x is an integer which is neither positive nor negative} = {0}, which contains exactly one element.

Hence, option 4 is the correct option.

A set which consists of an uncountable number of distinct elements, is called an infinite set.

{x : x is a natural number, x < 50} = {1, 2, 3, ...., 49}, which is a finite set.

{x : x is an integer, x < 50} = {...., -3, -2, -1, 0, 1, ...., 49} and the process of counting its elements never comes to an end.

The factors of 500 are limited in number, so {x : x is an integer, x is a factor of 500} is a finite set.

{x : x is a whole number, x < 1000} = {0, 1, 2, ...., 999}, which is a finite set.

Hence, option 2 is the correct option.

A set in which the process of counting of elements surely comes to an end, is called a finite set.

{x : x is an integer, x < 1} = {...., -3, -2, -1, 0}, which is an infinite set.

Between 0 and 1 there lies an unlimited number of rational numbers, so {x : x is a rational number, 0 < x < 1} is an infinite set.

{x : x is a natural number, x > 5} = {6, 7, 8, 9, ....}, which is an infinite set.

2 is the only even prime number, so {x : x is an even prime} = {2}, which has only 1 element.

Hence, option 4 is the correct option.

A = {1, 2, 3, 4}

B = {5, 6, 7}

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {1, 2, 3, 4} ∩ {5, 6, 7}

As there is no element common to A and B,

A ∩ B = φ

Hence, option 4 is the correct option.

A = {0, 1, 3, 4}

B = {5, 6, 1, 3, 9}

C = {0, 1, 2, 3, 9, 13}

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {0, 1, 3, 4} ∩ {5, 6, 1, 3, 9} = {1, 3}

(A ∩ B) ∪ C - contains all the elements in set A ∩ B and C.

(A ∩ B) ∪ C = {1, 3} ∪ {0, 1, 2, 3, 9, 13}

(A ∩ B) ∪ C = {0, 1, 2, 3, 9, 13}

Hence, option 1 is the correct option.

A = {1, 2, 3, 4, 5}

B = {2, 4, 6, 8}

C = {3, 4, 5, 6}

A ∪ B - contains all the elements in set A and B.

A ∪ B = {1, 2, 3, 4, 5} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 5, 6, 8}

(A ∪ B) ∩ C - contains all the common elements in set A ∪ B and C.

(A ∪ B) ∩ C = {1, 2, 3, 4, 5, 6, 8} ∩ {3, 4, 5, 6}

(A ∪ B) ∩ C = {3, 4, 5, 6}

Hence, option 2 is the correct option.

Here n(A) = 3 and n(B) = 6.

A ∪ B contains all the elements in set A and B, so A ∪ B cannot have fewer elements than B.

The number of elements in A ∪ B is the least when every element of A is also an element of B, i.e., when A ⊂ B.

In that case, A ∪ B = B and so n(A ∪ B) = 6.

The number of elements in A ∪ B is the greatest when A and B have no element in common.

In that case, n(A ∪ B) = n(A) + n(B) = 3 + 6 = 9.

So, the minimum and maximum number of elements in A ∪ B are 6 and 9 respectively.

Hence, option 3 is the correct option.

Here n(A) = 3 and n(B) = 6.

A ∩ B contains all the common elements in set A and B.

The number of elements in A ∩ B is the least when A and B have no element in common.

In that case, A ∩ B = φ and so n(A ∩ B) = 0.

The number of elements in A ∩ B is the greatest when every element of A is also an element of B, i.e., when A ⊂ B.

In that case, A ∩ B = A and so n(A ∩ B) = 3.

So, the minimum and maximum number of elements in A ∩ B are 0 and 3 respectively.

Hence, option 1 is the correct option.

By De-Morgan's law, (A ∪ B)′ = A′ ∩ B′ and not A′ ∪ B′.

∴ Statement 1 is not correct.

The complement of the empty set is the universal set, so φ′ = ξ.

∴ (φ′)′ = ξ′ = φ ≠ ξ

∴ Statement 2 is not correct.

By the distributive law, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

∴ Statement 3 is correct.

The complement of the universal set is the empty set, so ξ′ = φ.

∴ Statement 4 is correct.

So, statements 3 and 4 are correct.

Hence, option 4 is the correct option.

A ∩ B - contains all the common elements in set A and B.

As A ⊆ B, every element of A is also an element of B.

So, the elements common to A and B are exactly the elements of A.

∴ A ∩ B = A

Hence, option 1 is the correct option.

A ∪ B - contains all the elements in set A and B.

As A ⊆ B, every element of A is also an element of B.

So, A ∪ B contains no element other than the elements of B.

∴ A ∪ B = B

Hence, option 2 is the correct option.

Directions (Question 22 to 26)
Answer these questions on the basis of the following information:

Consider the following sets:

A = {x : x ∈ N, x is a factor of 3}

B = {x : x ∈ N, x is a factor of 5}

C = {x : x ∈ N, x is a factor of 9}

A = {x : x ∈ N, x is a factor of 3}

A = {1, 3}

B = {x : x ∈ N, x is a factor of 5}

B = {1, 5}

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {1, 3} ∩ {1, 5} = {1}

So, A ∩ B contains those natural numbers which are factors of 3 as well as of 5, i.e., factors of both 3 and 5.

As {x : x ∈ N, x is a factor of 15} = {1, 3, 5, 15} ≠ {1}, option 3 is not correct.

Hence, option 2 is the correct option.

A = {x : x ∈ N, x is a factor of 3}

A = {1, 3}

B = {x : x ∈ N, x is a factor of 5}

B = {1, 5}

A ∪ B - contains all the elements in set A and B.

A ∪ B = {1, 3} ∪ {1, 5} = {1, 3, 5}

A ∪ B contains those natural numbers, each one of which is either a factor of 3 or a factor of 5 or a factor of both.

As {x : x ∈ N, x is a factor of 15} = {1, 3, 5, 15} ≠ {1, 3, 5}, option 3 is not correct.

Hence, option 2 is the correct option.

A = {x : x ∈ N, x is a factor of 3}

A = {1, 3}

C = {x : x ∈ N, x is a factor of 9}

C = {1, 3, 9}

A ∪ C - contains all the elements in set A and C.

A ∪ C = {1, 3} ∪ {1, 3, 9} = {1, 3, 9}

∴ A ∪ C = {x : x ∈ N, x is a factor of 9}

Hence, option 2 is the correct option.

A = {x : x ∈ N, x is a factor of 3}

A = {1, 3}

C = {x : x ∈ N, x is a factor of 9}

C = {1, 3, 9}

A ∩ C - contains all the common elements in set A and C.

A ∩ C = {1, 3} ∩ {1, 3, 9} = {1, 3}

∴ A ∩ C = {x : x ∈ N, x is a factor of 3}

Hence, option 1 is the correct option.

A = {x : x ∈ N, x is a factor of 3} = {1, 3}

B = {x : x ∈ N, x is a factor of 5} = {1, 5}

C = {x : x ∈ N, x is a factor of 9} = {1, 3, 9}

I. As 3 ∈ A but 3 ∉ B, so A ⊄ B and statement I is not true.

II. Each of 1 and 3 is an element of C, so A ⊆ C and statement II is true.

III. As 5 ∈ B but 5 ∉ A, so B ⊄ A and statement III is not true.

IV. As 9 ∈ C but 9 ∉ A, so C ⊄ A and statement IV is not true.

So, only one of the given statements is true.

Hence, option 2 is the correct option.

(i) If A ⊆ B and B ⊆ A, then A = B.

As every element of A is in B and every element of B is in A, the two sets are equal.

(ii) If A = {1, 2, 3} and B = {1, 2, 3, 4}, then A - B = φ.

A - B = {1, 2, 3} - {1, 2, 3, 4} and there is no element of A which is not in B.

(iii) If ξ = {a, b, c, d, e, f} and A = {b, d, e}, then A′ = {a, c, f}.

A′ = ξ - A = {a, b, c, d, e, f} - {b, d, e} = {a, c, f}

(iv) x ∈ A ∪ B, ⇒ x ∈ A or x ∈ B (and/or)

A ∪ B is the set of all those elements, each one of which is either in A or in B or in both A and B.

(v) If A is a subset of B, then B is a superset of A.

If A ⊆ B, we write B ⊇ A and say that B is a superset of A.

(vi) The set of all possible subsets of a set is called its power set.

The power set of a set A is denoted by P(A).

(i) False

Although the persons on earth are very large in number, we know that the process of counting them will surely come to an end.

So, the set of all persons on earth is a finite set.

(ii) True

⇒ x + 4 = 4

⇒ x = 4 - 4

⇒ x = 0

So, {x : x + 4 = 4} = {0}, which contains exactly one element and hence it is a singleton set.

(iii) False

If a set contains n elements, then its power set contains 2ⁿ elements.

Here n = 3, so the power set has 2³ = 8 elements and not 6.

(iv) False

If n(A) = n(B), then A and B are equivalent sets.

Two sets A and B are equal only if every element of A is in B and every element of B is in A.

For example, if A = {1, 2} and B = {a, b}, then n(A) = n(B) but A ≠ B.

(v) True

If A = B, then A and B have exactly the same elements and so n(A) = n(B).

∴ A ↔ B

(vi) True

A non-empty set A has at least one element.

The empty set is a subset of every set and every set is a subset of itself.

So, φ and A are two different subsets of A, i.e., A has at least 2 subsets.

A = {x : x is a letter of the word GEOLOGY}

The letters of the word GEOLOGY are G, E, O, L, O, G, Y.

Writing each letter once and only once, we get

A = {G, E, O, L, Y}

As A has 5 elements, n(A) = 5.

B = {x : x is a factor of 16}

16 = 1 × 16 = 2 × 8 = 4 × 4

B = {1, 2, 4, 8, 16}

As B has 5 elements, n(B) = 5.

As n(A) = n(B), A and B are equivalent sets.

∴ Assertion (A) is true.

Two equivalent sets need only have the same number of elements, so they may well have elements in common.

For example, {1, 2} and {2, 3} are equivalent sets, but their intersection is {2}, which is not an empty set.

∴ Reason (R) is false.

Hence, option 3 is the correct option.

The complement of a set A is the set of all those elements of ξ which are not in A.

As every element of ξ is not in φ, the complement of the empty set is the universal set, i.e., φ′ = ξ.

So, the complement of an empty set does exist.

∴ Assertion (A) is false.

A′ contains all those elements of ξ which are not in A, so there can be no element common to A and A′.

∴ A ∩ A′ = φ, which is a null set.

∴ Reason (R) is true.

Hence, option 4 is the correct option.

N = {1, 2, 3, 4, 5, ....} and W = {0, 1, 2, 3, 4, 5, ....}

Every natural number is a whole number, so every element of N is in W.

∴ N ⊆ W

∴ Assertion (A) is true.

The whole numbers are 0, 1, 2, 3, 4, 5, .... and out of these only 0 is not a natural number.

∴ Reason (R) is true.

As W consists of all the natural numbers together with 0 alone, every element of N is in W and hence N ⊆ W.

So, Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.