Operations on Sets

Given,

A = {0, 1, 2, 3, 4, 5, 6, 7, 8},

B = {3, 5, 7, 9, 11},

C = {0, 5, 10, 20}

(i) A ∪ B

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

A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8} ∪ {3, 5, 7, 9, 11}

Hence, A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11}.

(ii) A ∪ C

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

A ∪ C = {0, 1, 2, 3, 4, 5, 6, 7, 8} ∪ {0, 5, 10, 20}

Hence, A ∪ C = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 20}.

(iii) B ∪ C

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

B ∪ C = {3, 5, 7, 9, 11} ∪ {0, 5, 10, 20}

Hence, B ∪ C = {0, 3, 5, 7, 9, 10, 11, 20}.

(iv) A ∩ B

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

A ∩ B = {0, 1, 2, 3, 4, 5, 6, 7, 8} ∩ {3, 5, 7, 9, 11}

Hence, A ∩ B = {3, 5, 7}.

(v) A ∩ C

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

A ∩ C = {0, 1, 2, 3, 4, 5, 6, 7, 8} ∩ {0, 5, 10, 20}

Hence, A ∩ C = {0, 5}.

(vi) B ∩ C

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

B ∩ C = {3, 5, 7, 9, 11} ∩ {0, 5, 10, 20}

Hence, B ∩ C = {5}.

Cardinal numbers :

From part (iii), B ∪ C = {0, 3, 5, 7, 9, 10, 11, 20}.

As B ∪ C has 8 elements, n(B ∪ C) = 8.

From part (iv), A ∩ B = {3, 5, 7}.

As A ∩ B has 3 elements, n(A ∩ B) = 3.

From part (v), A ∩ C = {0, 5}.

As A ∩ C has 2 elements, n(A ∩ C) = 2.

As B ∩ C has only 1 element, n(B ∩ C) = 1.

Hence, n(B ∪ C) = 8, n(A ∩ B) = 3, n(A ∩ C) = 2 and n(B ∩ C) = 1.

A = {0, 1, 4, 7} and ξ = {x | x ∈ W, x ≤ 10}

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

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

A′ = ξ - A

A′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {0, 1, 4, 7}

Hence, A′ = {2, 3, 5, 6, 8, 9, 10}.

A = {consonants} and ξ = {alphabets of English}

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

A′ = ξ - A

A′ = {alphabets of English} - {consonants}

Hence, A′ = {vowels}.

A = {boys in class VIII of all schools in Bengaluru} and ξ = {students in class VIII of all schools in Bengaluru}

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

A′ = ξ - A

A′ = {students in class VIII of all schools in Bengaluru} - {boys in class VIII of all schools in Bengaluru}

Hence, A′ = {girls in class VIII of all schools in Bengaluru}.

A = {letters of KALKA} and ξ = {letters of KOLKATA}

Writing each element of the set once and only once, we get

A = {K, A, L} and ξ = {K, O, L, A, T}

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

A′ = ξ - A

A′ = {K, O, L, A, T} - {K, A, L}

Hence, A′ = {O, T}.

A = {odd natural numbers} and ξ = {whole numbers}

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

A′ = ξ - A

A′ = {whole numbers} - {odd natural numbers}

Hence, A′ = {even whole numbers}.

A = {x : x ∈ N and 3 < x < 7}

A = {4, 5, 6}

B = {x : x ∈ W and x ≤ 4}

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

(i) A ∪ B

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

A ∪ B = {4, 5, 6} ∪ {0, 1, 2, 3, 4}

Hence, A ∪ B = {0, 1, 2, 3, 4, 5, 6}.

(ii) A ∩ B

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

A ∩ B = {4, 5, 6} ∩ {0, 1, 2, 3, 4}

Hence, A ∩ B = {4}.

(iii) A - B

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

A - B = {4, 5, 6} - {0, 1, 2, 3, 4}

Hence, A - B = {5, 6}.

(iv) B - A

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

B - A = {0, 1, 2, 3, 4} - {4, 5, 6}

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

P = {x : x ∈ W and x < 6}

P = {0, 1, 2, 3, 4, 5}

Q = {x : x ∈ N and 4 ≤ x < 9}

Q = {4, 5, 6, 7, 8}

(i) P ∪ Q

P ∪ Q - contains all the elements in set P and Q.

P ∪ Q = {0, 1, 2, 3, 4, 5} ∪ {4, 5, 6, 7, 8}

Hence, P ∪ Q = {0, 1, 2, ..., 8}.

(ii) P ∩ Q

P ∩ Q - contains all the common elements in set P and Q.

P ∩ Q = {0, 1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8}

Hence, P ∩ Q = {4, 5}.

(iii) P - Q

P - Q - contains all the elements in set P but not in Q.

P - Q = {0, 1, 2, 3, 4, 5} - {4, 5, 6, 7, 8}

Hence, P - Q = {0, 1, 2, 3}.

(iv) Q - P

Q - P - contains all the elements in set Q but not in P.

Q - P = {4, 5, 6, 7, 8} - {0, 1, 2, 3, 4, 5}

Hence, Q - P = {6, 7, 8}.

Proper superset :

From part (i), P ∪ Q = {0, 1, 2, ..., 8} and from part (ii), P ∩ Q = {4, 5}.

As every element of P ∩ Q = {4, 5} is contained in P ∪ Q = {0, 1, 2, ..., 8} and P ∪ Q ≠ P ∩ Q,

Hence, yes, P ∪ Q is a proper superset of P ∩ Q.

Writing each element of the set once and only once, we get

A = {letters of word INTEGRITY}

A = {I, N, T, E, G, R, Y}

B = {letters of word RECKONING}

B = {R, E, C, K, O, N, I, G}

(i) A ∪ B

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

A ∪ B = {I, N, T, E, G, R, Y} ∪ {R, E, C, K, O, N, I, G}

Hence, A ∪ B = {I, N, T, E, G, R, Y, C, K, O}.

(ii) A ∩ B

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

A ∩ B = {I, N, T, E, G, R, Y} ∩ {R, E, C, K, O, N, I, G}

Hence, A ∩ B = {I, N, E, R, G}.

(iii) A - B

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

A - B = {I, N, T, E, G, R, Y} - {R, E, C, K, O, N, I, G}

Hence, A - B = {T, Y}.

(iv) B - A

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

B - A = {R, E, C, K, O, N, I, G} - {I, N, T, E, G, R, Y}

Hence, B - A = {C, K, O}.

Verification :

Here n(A) = 7, n(B) = 8, n(A ∪ B) = 10, n(A ∩ B) = 5, n(A - B) = 2 and n(B - A) = 3.

(a) n(A) + n(B) - n(A ∩ B) = 7 + 8 - 5 = 10 = n(A ∪ B)

Hence, n(A ∪ B) = n(A) + n(B) - n(A ∩ B).

(b) n(A ∪ B) - n(B) = 10 - 8 = 2 = n(A - B)

and n(A) - n(A ∩ B) = 7 - 5 = 2 = n(A - B)

Hence, n(A - B) = n(A ∪ B) - n(B) = n(A) - n(A ∩ B).

(c) n(A ∪ B) - n(A) = 10 - 7 = 3 = n(B - A)

and n(B) - n(A ∩ B) = 8 - 5 = 3 = n(B - A)

Hence, n(B - A) = n(A ∪ B) - n(A) = n(B) - n(A ∩ B).

(d) n(A - B) + n(B - A) + n(A ∩ B) = 2 + 3 + 5 = 10 = n(A ∪ B)

Hence, n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B).

ξ = {natural numbers between 10 and 40}

ξ = {11, 12, 13, ..., 39}

A = {multiples of 5}

A = {15, 20, 25, 30, 35}

B = {multiples of 6}

B = {12, 18, 24, 30, 36}

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

A ∪ B = {15, 20, 25, 30, 35} ∪ {12, 18, 24, 30, 36}

A ∪ B = {12, 15, 18, 20, 24, 25, 30, 35, 36}

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

A ∩ B = {15, 20, 25, 30, 35} ∩ {12, 18, 24, 30, 36}

A ∩ B = {30}

Hence, A ∪ B = {12, 15, 18, 20, 24, 25, 30, 35, 36} and A ∩ B = {30}.

(ii) Here n(A) = 5, n(B) = 5, n(A ∪ B) = 9 and n(A ∩ B) = 1.

⇒ n(A) + n(B) - n(A ∩ B) = 5 + 5 - 1 = 9 = n(A ∪ B)

Hence, n(A ∪ B) = n(A) + n(B) - n(A ∩ B).

ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {1, 2, 3, 4, 6, 7, 8}

B = {4, 6, 8}

(i) A′

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

A′ = ξ - A

A′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 6, 7, 8}

Hence, A′ = {5, 9}.

(ii) B′

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

B′ = ξ - B

B′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {4, 6, 8}

Hence, B′ = {1, 2, 3, 5, 7, 9}.

(iii) A ∪ B

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

A ∪ B = {1, 2, 3, 4, 6, 7, 8} ∪ {4, 6, 8}

Hence, A ∪ B = {1, 2, 3, 4, 6, 7, 8}.

(iv) A ∩ B

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

A ∩ B = {1, 2, 3, 4, 6, 7, 8} ∩ {4, 6, 8}

Hence, A ∩ B = {4, 6, 8}.

(v) A - B

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

A - B = {1, 2, 3, 4, 6, 7, 8} - {4, 6, 8}

Hence, A - B = {1, 2, 3, 7}.

(vi) B - A

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

B - A = {4, 6, 8} - {1, 2, 3, 4, 6, 7, 8}

Hence, B - A = φ.

(vii) (A ∩ B)′

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

(A ∩ B)′ = ξ - (A ∩ B)

(A ∩ B)′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {4, 6, 8}

Hence, (A ∩ B)′ = {1, 2, 3, 5, 7, 9}.

(viii) A′ ∪ B′

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

A′ ∪ B′ = {5, 9} ∪ {1, 2, 3, 5, 7, 9}

Hence, A′ ∪ B′ = {1, 2, 3, 5, 7, 9}.

Verification :

(a) From part (vii), (A ∩ B)′ = {1, 2, 3, 5, 7, 9} ...............(1)

From part (viii), A′ ∪ B′ = {1, 2, 3, 5, 7, 9} ...............(2)

From (1) and (2), we can see

LHS = RHS

Hence, (A ∩ B)′ = A′ ∪ B′.

(b) Here n(A) = 7, n(A′) = 2 and n(ξ) = 9.

n(A) + n(A′) = 7 + 2 = 9 = n(ξ)

Hence, n(A) + n(A′) = n(ξ).

(c) Here n(A ∩ B) = 3, n((A ∩ B)′) = 6 and n(ξ) = 9.

n(A ∩ B) + n((A ∩ B)′) = 3 + 6 = 9 = n(ξ)

Hence, n(A ∩ B) + n((A ∩ B)′) = n(ξ).

(d) Here n(A - B) = 4, n(B - A) = 0, n(A ∩ B) = 3 and n(A ∪ B) = 7.

n(A - B) + n(B - A) + n(A ∩ B) = 4 + 0 + 3 = 7 = n(A ∪ B)

Hence, n(A - B) + n(B - A) + n(A ∩ B) = n(A ∪ B).

ξ = {x : x ∈ W, x ≤ 10}

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {x : x ≥ 5}

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

B = {x : 3 ≤ x < 8}

B = {3, 4, 5, 6, 7}

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

A′ = ξ - A

A′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 7, 8, 9, 10}

A′ = {0, 1, 2, 3, 4}

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

B′ = ξ - B

B′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7}

B′ = {0, 1, 2, 8, 9, 10}

(A ∪ B)′ = A′ ∩ B′

Taking LHS : (A ∪ B)′

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

A ∪ B = {5, 6, 7, 8, 9, 10} ∪ {3, 4, 5, 6, 7}

A ∪ B = {3, 4, 5, 6, 7, 8, 9, 10}

(A ∪ B)′ = ξ - (A ∪ B)

(A ∪ B)′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7, 8, 9, 10}

(A ∪ B)′ = {0, 1, 2} ...............(1)

Taking RHS : A′ ∩ B′

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

A′ ∩ B′ = {0, 1, 2, 3, 4} ∩ {0, 1, 2, 8, 9, 10}

A′ ∩ B′ = {0, 1, 2} ...............(2)

From (1) and (2), we can see

LHS = RHS

Hence, (A ∪ B)′ = A′ ∩ B′.

ξ = {x : x ∈ W, x ≤ 10}

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {x : x ≥ 5}

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

B = {x : 3 ≤ x < 8}

B = {3, 4, 5, 6, 7}

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

A′ = ξ - A

A′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 7, 8, 9, 10}

A′ = {0, 1, 2, 3, 4}

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

B′ = ξ - B

B′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7}

B′ = {0, 1, 2, 8, 9, 10}

(A ∩ B)′ = A′ ∪ B′

Taking LHS : (A ∩ B)′

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

A ∩ B = {5, 6, 7, 8, 9, 10} ∩ {3, 4, 5, 6, 7}

A ∩ B = {5, 6, 7}

(A ∩ B)′ = ξ - (A ∩ B)

(A ∩ B)′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 7}

(A ∩ B)′ = {0, 1, 2, 3, 4, 8, 9, 10} ...............(1)

Taking RHS : A′ ∪ B′

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

A′ ∪ B′ = {0, 1, 2, 3, 4} ∪ {0, 1, 2, 8, 9, 10}

A′ ∪ B′ = {0, 1, 2, 3, 4, 8, 9, 10} ...............(2)

From (1) and (2), we can see

LHS = RHS

Hence, (A ∩ B)′ = A′ ∪ B′.

ξ = {x : x ∈ W, x ≤ 10}

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {x : x ≥ 5}

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

B = {x : 3 ≤ x < 8}

B = {3, 4, 5, 6, 7}

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

B′ = ξ - B

B′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7}

B′ = {0, 1, 2, 8, 9, 10}

A - B = A ∩ B′

Taking LHS : A - B

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

A - B = {5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7}

A - B = {8, 9, 10} ...............(1)

Taking RHS : A ∩ B′

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

A ∩ B′ = {5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 8, 9, 10}

A ∩ B′ = {8, 9, 10} ...............(2)

From (1) and (2), we can see

LHS = RHS

Hence, A - B = A ∩ B′.

ξ = {x : x ∈ W, x ≤ 10}

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {x : x ≥ 5}

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

B = {x : 3 ≤ x < 8}

B = {3, 4, 5, 6, 7}

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

A′ = ξ - A

A′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 7, 8, 9, 10}

A′ = {0, 1, 2, 3, 4}

B - A = B ∩ A′

Taking LHS : B - A

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

B - A = {3, 4, 5, 6, 7} - {5, 6, 7, 8, 9, 10}

B - A = {3, 4} ...............(1)

Taking RHS : B ∩ A′

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

B ∩ A′ = {3, 4, 5, 6, 7} ∩ {0, 1, 2, 3, 4}

B ∩ A′ = {3, 4} ...............(2)

From (1) and (2), we can see

LHS = RHS

Hence, B - A = B ∩ A′.

Given:

n(A) = 20

n(B) = 16

n(A ∪ B) = 30

∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

⇒ 30 = 20 + 16 - n(A ∩ B)

⇒ n(A ∩ B) = 36 - 30

⇒ n(A ∩ B) = 6

Hence, n(A ∩ B) = 6.

Given:

n(ξ) = 20

n(A′) = 7

∵ n(A) + n(A′) = n(ξ)

Putting the values, we get

⇒ n(A) + 7 = 20

⇒ n(A) = 20 - 7

⇒ n(A) = 13

Hence, n(A) = 13.

Given:

n(ξ) = 40

n(A) = 20

n(B′) = 16

n(A ∪ B) = 32

∵ n(B) + n(B′) = n(ξ)

Putting the values, we get

⇒ n(B) + 16 = 40

⇒ n(B) = 40 - 16

⇒ n(B) = 24

∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

⇒ 32 = 20 + 24 - n(A ∩ B)

⇒ n(A ∩ B) = 44 - 32

⇒ n(A ∩ B) = 12

Hence, n(B) = 24 and n(A ∩ B) = 12.

Given:

n(ξ) = 32

n(A) = 20

n(B) = 16

n((A ∪ B)′) = 4

(i) ∵ n(A ∪ B) + n((A ∪ B)′) = n(ξ)

Putting the values, we get

⇒ n(A ∪ B) + 4 = 32

⇒ n(A ∪ B) = 32 - 4

⇒ n(A ∪ B) = 28

Hence, n(A ∪ B) = 28.

(ii) ∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

⇒ 28 = 20 + 16 - n(A ∩ B)

⇒ n(A ∩ B) = 36 - 28

⇒ n(A ∩ B) = 8

Hence, n(A ∩ B) = 8.

(iii) ∵ n(A - B) = n(A) - n(A ∩ B)

Putting the values, we get

⇒ n(A - B) = 20 - 8

⇒ n(A - B) = 12

Hence, n(A - B) = 12.

Given:

n(ξ) = 40

n(A′) = 15

n(B) = 12

n((A ∩ B)′) = 32

(i) ∵ n(A) + n(A′) = n(ξ)

Putting the values, we get

⇒ n(A) + 15 = 40

⇒ n(A) = 40 - 15

⇒ n(A) = 25

Hence, n(A) = 25.

(ii) ∵ n(B) + n(B′) = n(ξ)

Putting the values, we get

⇒ 12 + n(B′) = 40

⇒ n(B′) = 40 - 12

⇒ n(B′) = 28

Hence, n(B′) = 28.

(iii) ∵ n(A ∩ B) + n((A ∩ B)′) = n(ξ)

Putting the values, we get

⇒ n(A ∩ B) + 32 = 40

⇒ n(A ∩ B) = 40 - 32

⇒ n(A ∩ B) = 8

Hence, n(A ∩ B) = 8.

(iv) ∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

⇒ n(A ∪ B) = 25 + 12 - 8

⇒ n(A ∪ B) = 37 - 8

⇒ n(A ∪ B) = 29

Hence, n(A ∪ B) = 29.

(v) ∵ n(A - B) = n(A) - n(A ∩ B)

Putting the values, we get

⇒ n(A - B) = 25 - 8

⇒ n(A - B) = 17

Hence, n(A - B) = 17.

(vi) ∵ n(B - A) = n(B) - n(A ∩ B)

Putting the values, we get

⇒ n(B - A) = 12 - 8

⇒ n(B - A) = 4

Hence, n(B - A) = 4.

Given:

n(A - B) = 12

n(B - A) = 16

n(A ∩ B) = 5

(i) ∵ n(A - B) = n(A) - n(A ∩ B)

Putting the values, we get

⇒ 12 = n(A) - 5

⇒ n(A) = 12 + 5

⇒ n(A) = 17

Hence, n(A) = 17.

(ii) ∵ n(B - A) = n(B) - n(A ∩ B)

Putting the values, we get

⇒ 16 = n(B) - 5

⇒ n(B) = 16 + 5

⇒ n(B) = 21

Hence, n(B) = 21.

(iii) ∵ n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B)

Putting the values, we get

⇒ n(A ∪ B) = 12 + 16 + 5

⇒ n(A ∪ B) = 33

Hence, n(A ∪ B) = 33.

(i) If A, B are two sets, then A ∪ B = {x | x ∈ A or x ∈ B}.

(ii) If A, B are two sets, then A ∩ B = {x | x ∈ A and x ∈ B}.

(iii) If A, B are two sets, then A - B = {x | x ∈ A and x ∉ B}.

(iv) A and B are disjoint sets if and only if A ∩ B = φ.

(v) A and B are overlapping sets if and only if A ∩ B ≠ φ.

(vi) The set {x : x ∈ W, x < 3} in the roster form = {0, 1, 2}.

(vii) If A is any set, then A ∪ φ = A and A ∪ ξ = ξ.

(viii) If ξ = {all digits in our number system} and A = {1, 2, 3, 4, 5}, then A′ = {0, 6, 7, 8, 9}.

(ix) If A is any set and A′ is its complement, then A ∪ A′ = ξ and A ∩ A′ = φ.

(i) True

The complement of a set A is the set consisting of all those elements of ξ which do not belong to A, i.e., A′ = {x; x ∈ ξ and x ∉ A}.

(ii) False

A ∩ B = {0, 1, 2, 3, 4, 5} ∩ {0, 3, 5, 7} = {0, 3, 5}

As 7 ∈ B but 7 ∉ A ∩ B,

∴ A ∩ B ≠ B

(iii) False

A ∪ B = {0, 1, 2, 3, 4, 5} ∪ {0, 3, 5, 7} = {0, 1, 2, 3, 4, 5, 7}

As 7 ∈ A ∪ B but 7 ∉ A,

∴ A ∪ B ≠ A

(iv) True

ξ = {all non-zero digits in our number system} = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {multiples of 2} = {2, 4, 6, 8}

B = {multiples of 3} = {3, 6, 9}

∴ A ∩ B = {2, 4, 6, 8} ∩ {3, 6, 9} = {6}

A = {x | x is a colour of rainbow}

A = {violet, indigo, blue, green, yellow, orange, red}

B = {white, red, green}

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

A ∩ B = {violet, indigo, blue, green, yellow, orange, red} ∩ {white, red, green}

A ∩ B = {green, red}

Hence, option 4 is the correct option.

P ∪ Q - contains all the elements in set P and Q.

P ∪ Q = {-1, 0, 1, 2, 5} ∪ {3, 5, 7}

P ∪ Q = {-1, 0, 1, 2, 3, 5, 7}

Hence, option 3 is the correct option.

A - B is the set consisting of all those elements which belong to A but do not belong to B.

∴ A - B = {x | x ∈ A and x ∉ B}

Hence, option 3 is the correct option.

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

∴ A ∩ φ = φ

Hence, option 2 is the correct option.

As every element of A belongs to the universal set ξ, the elements common to A and ξ are all the elements of A.

∴ A ∩ ξ = A

Hence, option 1 is the correct option.

ξ = W = {0, 1, 2, 3, ...}

A = {x | x ∈ W and x ≤ 10}

A = {0, 1, 2, ..., 10}

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

A′ = ξ - A

A′ = {0, 1, 2, 3, ...} - {0, 1, 2, ..., 10}

A′ = {11, 12, 13, ...}

∴ A′ = {x | x ∈ W and x ≥ 11}

Hence, option 4 is the correct option.

ξ = {x | x ∈ N, x ≤ 12}

ξ = {1, 2, 3, ..., 12}

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

A = {3, 6, 9, 12}

B = {x | x is a multiple of 4}

B = {4, 8, 12}

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

A ∩ B = {3, 6, 9, 12} ∩ {4, 8, 12}

A ∩ B = {12}

Hence, option 3 is the correct option.

ξ = {all digits in our number system}

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {x | x is prime}

A = {2, 3, 5, 7}

B = {x | x is even}

B = {0, 2, 4, 6, 8}

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

B - A = {0, 2, 4, 6, 8} - {2, 3, 5, 7}

B - A = {0, 4, 6, 8}

Hence, option 2 is the correct option.

Given:

n(A) = 22

n(B) = 18

n(A ∪ B) = 35

∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

⇒ 35 = 22 + 18 - n(A ∩ B)

⇒ n(A ∩ B) = 40 - 35

⇒ n(A ∩ B) = 5

Hence, option 2 is the correct option.

If A ∪ B = A ∩ B, then every element of A belongs to A ∪ B = A ∩ B and hence belongs to B. Similarly, every element of B belongs to A. So A = B.

∴ Statement I is true.

By definition, A ∪ B = {x | x ∈ A or x ∈ B} and A ∩ B = {x | x ∈ A and x ∈ B}.

∴ Statement II is true.

Hence, option 3 is the correct option.

A = {x | x is a digit in the number system}

A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

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

B = {3, 6, 9}

A - B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} - {3, 6, 9}

A - B = {0, 1, 2, 4, 5, 7, 8} ≠ {1, 2, 4, 5, 7, 8}

∴ Statement I is false.

Statement II: By definition, A - B = {x | x ∈ A and x ∉ B}, not {x | x ∈ A or x ∉ B}.

∴ Statement II is false.

Hence, option 4 is the correct option.

As the empty set φ has no elements, A ∪ φ = A and A ∩ φ = φ.

∴ Statement I is true.

As A ∩ φ = φ, sets A and φ have no element in common, so A and φ are disjoint sets.

∴ Statement II is true.

Hence, option 3 is the correct option.

ξ = {x : x ∈ N, x < 25}

ξ = {1, 2, 3, ..., 24}

A = {x : x is a composite number}

A = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24}

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

A′ = ξ - A

A′ = {1, 2, 3, ..., 24} - {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24}

A′ = {1, 2, 3, 5, 7, 11, 13, 17, 19, 23}

As A′ contains 1 and all the prime numbers less than 25,

Hence, A′ = {x : x ∈ N, x = 1 or x is a prime number less than 25} and A′ = {1, 2, 3, 5, 7, 11, 13, 17, 19, 23}.

ξ = {x | x ∈ I, -5 ≤ x ≤ 5}

ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

A = {-5, -3, 0, 3, 5}

B = {-4, -2, 0, 2, 4}

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

A ∪ B = {-5, -3, 0, 3, 5} ∪ {-4, -2, 0, 2, 4}

A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5}

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

A ∩ B = {-5, -3, 0, 3, 5} ∩ {-4, -2, 0, 2, 4}

A ∩ B = {0}

Hence, A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5} and A ∩ B = {0}.

ξ = {x | x ∈ I, -5 ≤ x ≤ 5}

ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

A = {-5, -3, 0, 3, 5}

B = {-4, -2, 0, 2, 4}

From part (i), A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5} and A ∩ B = {0}.

Here n(A) = 5, n(B) = 5, n(A ∪ B) = 9 and n(A ∩ B) = 1.

n(A) + n(B) - n(A ∩ B) = 5 + 5 - 1 = 9 = n(A ∪ B)

Hence, n(A ∪ B) = n(A) + n(B) - n(A ∩ B).

ξ = {x | x ∈ I, -5 ≤ x ≤ 5}

ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

A = {-5, -3, 0, 3, 5}

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

A′ = ξ - A

A′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} - {-5, -3, 0, 3, 5}

A′ = {-4, -2, -1, 1, 2, 4}

Here n(A) = 5, n(A′) = 6 and n(ξ) = 11.

n(A) + n(A′) = 5 + 6 = 11 = n(ξ)

Hence, A′ = {-4, -2, -1, 1, 2, 4} and n(A) + n(A′) = n(ξ).

ξ = {x | x ∈ I, -5 ≤ x ≤ 5}

ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

B = {-4, -2, 0, 2, 4}

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

B′ = ξ - B

B′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} - {-4, -2, 0, 2, 4}

B′ = {-5, -3, -1, 1, 3, 5}

Here n(B) = 5, n(B′) = 6 and n(ξ) = 11.

n(ξ) - n(B) = 11 - 5 = 6 = n(B′)

Hence, B′ = {-5, -3, -1, 1, 3, 5} and n(B′) = n(ξ) - n(B).

ξ = {x | x ∈ I, -5 ≤ x ≤ 5}

ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

A = {-5, -3, 0, 3, 5}

B = {-4, -2, 0, 2, 4}

From part (iii), A′ = {-4, -2, -1, 1, 2, 4} and from part (iv), B′ = {-5, -3, -1, 1, 3, 5}.

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

A ∪ A′ = {-5, -3, 0, 3, 5} ∪ {-4, -2, -1, 1, 2, 4}

A ∪ A′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} = ξ

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

A ∩ A′ = {-5, -3, 0, 3, 5} ∩ {-4, -2, -1, 1, 2, 4}

A ∩ A′ = φ

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

B ∪ B′ = {-4, -2, 0, 2, 4} ∪ {-5, -3, -1, 1, 3, 5}

B ∪ B′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} = ξ

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

B ∩ B′ = {-4, -2, 0, 2, 4} ∩ {-5, -3, -1, 1, 3, 5}

B ∩ B′ = φ

Hence, A ∪ A′ = ξ, A ∩ A′ = φ, B ∪ B′ = ξ and B ∩ B′ = φ.

ξ = {x | x ∈ I, -5 ≤ x ≤ 5}

ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

A = {-5, -3, 0, 3, 5}

B = {-4, -2, 0, 2, 4}

From part (i), A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5}.

From part (iii), A′ = {-4, -2, -1, 1, 2, 4} and from part (iv), B′ = {-5, -3, -1, 1, 3, 5}.

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

(A ∪ B)′ = ξ - (A ∪ B)

(A ∪ B)′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} - {-5, -4, -3, -2, 0, 2, 3, 4, 5}

(A ∪ B)′ = {-1, 1} ...............(1)

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

A′ ∩ B′ = {-4, -2, -1, 1, 2, 4} ∩ {-5, -3, -1, 1, 3, 5}

A′ ∩ B′ = {-1, 1} ...............(2)

From (1) and (2), we can see that

(A ∪ B)′ = {-1, 1} and A′ ∩ B′ = {-1, 1}. Hence, they are equal.

ξ = {x | x ∈ I, -5 ≤ x ≤ 5}

ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

A = {-5, -3, 0, 3, 5}

B = {-4, -2, 0, 2, 4}

From part (i), A ∩ B = {0}.

From part (iii), A′ = {-4, -2, -1, 1, 2, 4} and from part (iv), B′ = {-5, -3, -1, 1, 3, 5}.

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

(A ∩ B)′ = ξ - (A ∩ B)

(A ∩ B)′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} - {0}

(A ∩ B)′ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5} ...............(1)

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

A′ ∪ B′ = {-4, -2, -1, 1, 2, 4} ∪ {-5, -3, -1, 1, 3, 5}

A′ ∪ B′ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5} ...............(2)

From (1) and (2), we can see that

(A ∩ B)′ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5} and A′ ∪ B′ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5}. Hence, they are equal.

ξ = {x | x ∈ N, x ≤ 12}

ξ = {1, 2, 3, ..., 12}

A = {prime numbers}

A = {2, 3, 5, 7, 11}

B = {odd numbers}

B = {1, 3, 5, 7, 9, 11}

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

A ∪ B = {2, 3, 5, 7, 11} ∪ {1, 3, 5, 7, 9, 11}

A ∪ B = {1, 2, 3, 5, 7, 9, 11}

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

A ∩ B = {2, 3, 5, 7, 11} ∩ {1, 3, 5, 7, 9, 11}

A ∩ B = {3, 5, 7, 11}

Hence, A ∪ B = {1, 2, 3, 5, 7, 9, 11} and A ∩ B = {3, 5, 7, 11}.

(ii) Here n(A) = 5, n(B) = 6, n(A ∪ B) = 7 and n(A ∩ B) = 4.

n(A ∪ B) + n(A ∩ B) = 7 + 4 = 11

n(A) + n(B) = 5 + 6 = 11

Hence, n(A ∪ B) + n(A ∩ B) = n(A) + n(B).

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

A′ = ξ - A

A′ = {1, 2, 3, ..., 12} - {2, 3, 5, 7, 11}

A′ = {1, 4, 6, 8, 9, 10, 12}

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

B′ = ξ - B

B′ = {1, 2, 3, ..., 12} - {1, 3, 5, 7, 9, 11}

B′ = {2, 4, 6, 8, 10, 12}

Hence, A′ = {1, 4, 6, 8, 9, 10, 12} and B′ = {2, 4, 6, 8, 10, 12}.

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

(A ∪ B)′ = ξ - (A ∪ B)

(A ∪ B)′ = {1, 2, 3, ..., 12} - {1, 2, 3, 5, 7, 9, 11}

(A ∪ B)′ = {4, 6, 8, 10, 12}

Here n(A ∪ B) = 7, n((A ∪ B)′) = 5 and n(ξ) = 12.

n(A ∪ B) + n((A ∪ B)′) = 7 + 5 = 12 = n(ξ)

Hence, (A ∪ B)′ = {4, 6, 8, 10, 12} and n(A ∪ B) + n((A ∪ B)′) = n(ξ).

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

(A ∩ B)′ = ξ - (A ∩ B)

(A ∩ B)′ = {1, 2, 3, ..., 12} - {3, 5, 7, 11}

(A ∩ B)′ = {1, 2, 4, 6, 8, 9, 10, 12} ...............(1)

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

A′ ∪ B′ = {1, 4, 6, 8, 9, 10, 12} ∪ {2, 4, 6, 8, 10, 12}

A′ ∪ B′ = {1, 2, 4, 6, 8, 9, 10, 12} ...............(2)

From (1) and (2), we can see that

(A ∩ B)′ = {1, 2, 4, 6, 8, 9, 10, 12} and A′ ∪ B′ = {1, 2, 4, 6, 8, 9, 10, 12}. Hence, they are equal.

ξ = {x : x ∈ N, x ≤ 12}

ξ = {1, 2, 3, ..., 12}

A = {x : x ≥ 7}

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

B = {x : 4 < x < 10}

B = {5, 6, 7, 8, 9}

(i) A′

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

A′ = ξ - A

A′ = {1, 2, 3, ..., 12} - {7, 8, 9, 10, 11, 12}

Hence, A′ = {1, 2, 3, 4, 5, 6}.

(ii) B′

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

B′ = ξ - B

B′ = {1, 2, 3, ..., 12} - {5, 6, 7, 8, 9}

Hence, B′ = {1, 2, 3, 4, 10, 11, 12}.

(iii) A ∪ B

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

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

Hence, A ∪ B = {5, 6, 7, ..., 12}.

(iv) A ∩ B

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

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

Hence, A ∩ B = {7, 8, 9}.

(v) A - B

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

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

Hence, A - B = {10, 11, 12}.

(vi) B - A

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

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

Hence, B - A = {5, 6}.

(vii) (A ∪ B)′

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

(A ∪ B)′ = ξ - (A ∪ B)

(A ∪ B)′ = {1, 2, 3, ..., 12} - {5, 6, 7, ..., 12}

Hence, (A ∪ B)′ = {1, 2, 3, 4}.

(viii) A′ ∩ B′

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

A′ ∩ B′ = {1, 2, 3, 4, 5, 6} ∩ {1, 2, 3, 4, 10, 11, 12}

Hence, A′ ∩ B′ = {1, 2, 3, 4}.

Verification :

(i) From part (vii), (A ∪ B)′ = {1, 2, 3, 4} ...............(1)

From part (viii), A′ ∩ B′ = {1, 2, 3, 4} ...............(2)

From (1) and (2), we can see

LHS = RHS

Hence, (A ∪ B)′ = A′ ∩ B′.

(ii) From part (v), A - B = {10, 11, 12} ...............(3)

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

A ∩ B′ = {7, 8, 9, 10, 11, 12} ∩ {1, 2, 3, 4, 10, 11, 12}

A ∩ B′ = {10, 11, 12} ...............(4)

From (3) and (4), we can see

LHS = RHS

Hence, A - B = A ∩ B′.

(iii) Here n(A ∪ B) = 8, n((A ∪ B)′) = 4 and n(ξ) = 12.

n(A ∪ B) + n((A ∪ B)′) = 8 + 4 = 12 = n(ξ)

Hence, n(A ∪ B) + n((A ∪ B)′) = n(ξ).

(iv) Here n(A - B) = 3, n(B - A) = 2, n(A ∩ B) = 3 and n(A ∪ B) = 8.

n(A - B) + n(B - A) + n(A ∩ B) = 3 + 2 + 3 = 8 = n(A ∪ B)

Hence, n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B).

Given:

n(ξ) = 50

n(A) = 15

n(B) = 13

n(A ∩ B) = 10

∵ n(A) + n(A′) = n(ξ)

Putting the values, we get

⇒ 15 + n(A′) = 50

⇒ n(A′) = 50 - 15

⇒ n(A′) = 35

∵ n(B) + n(B′) = n(ξ)

Putting the values, we get

⇒ 13 + n(B′) = 50

⇒ n(B′) = 50 - 13

⇒ n(B′) = 37

∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

⇒ n(A ∪ B) = 15 + 13 - 10

⇒ n(A ∪ B) = 28 - 10

⇒ n(A ∪ B) = 18

Hence, n(A′) = 35, n(B′) = 37 and n(A ∪ B) = 18.

Given:

n(ξ) = 60

n(A) = 35

n(B′) = 36

n((A ∩ B)′) = 51

(i) ∵ n(B) + n(B′) = n(ξ)

Putting the values, we get

⇒ n(B) + 36 = 60

⇒ n(B) = 60 - 36

⇒ n(B) = 24

Hence, n(B) = 24.

(ii) ∵ n(A ∩ B) + n((A ∩ B)′) = n(ξ)

Putting the values, we get

⇒ n(A ∩ B) + 51 = 60

⇒ n(A ∩ B) = 60 - 51

⇒ n(A ∩ B) = 9

Hence, n(A ∩ B) = 9.

(iii) ∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

⇒ n(A ∪ B) = 35 + 24 - 9

⇒ n(A ∪ B) = 59 - 9

⇒ n(A ∪ B) = 50

Hence, n(A ∪ B) = 50.

(iv) ∵ n(A - B) = n(A) - n(A ∩ B)

Putting the values, we get

⇒ n(A - B) = 35 - 9

⇒ n(A - B) = 26

Hence, n(A - B) = 26.