Question

7. If &= 1. 2. 3. ... 9, A = {1, 2, 3, 4, 6, 7, 8} and B = {4, 6, 8), then find. (i) A'
(ii) B'
(iii) AUB. (iv) A n B (v) A-B
(vi) B-A
(vii) (A n B)'. (viii) A' U B'
Also verify that:
(a) (ANB)' = A' U B'
(b) n(A) + n(A') = n(&)
(c) n(A B) + n((ANB)') = n(&) (d) n(A-B) + n(B - A) + n(A nB)n(AU B)​

Answer 1

Solution :

Given,

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

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

B = {4, 6, 8}

(i) A'

= S - A

= { x | x ∈ S but x ∉ A }

= { 9 }

(ii) B'

= S - B

= { x | x ∈ S but x ∉ B }

= { 1, 2, 3, 5, 7, 9 }

(iii) A ∪ B

= { x | x ∈ A or x ∈ B }

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

(iv) A ∩ B

= { x | x ∈ A and x ∈ B }

= { 4, 6, 8 }

(v) A - B

= A ∩ B'

= { x | x ∈ A and x ∈ B' }

= { 1, 2, 3, 5, 7 }

(vi) B - A

= B ∩ A'

= { x | x ∈ B and x ∈ A' }

= { } or Φ, as there is no common element

(vii) (A ∩ B)'

= S - (A ∩ B)

= { x | x ∈ S but x ∉ (A ∩ B) }

= { 1, 2, 3, 5, 7, 9}

(viii) A' ∪ B'

= { x | x ∈ A' or x ∈ B' }

= { 1, 2, 3, 5, 7, 9 }

Verification :

(a) We have found

(A ∩ B)' = { 1, 2, 3, 5, 7, 9 }

A' ∪ B' = { 1, 2, 3, 5, 7, 9 }

∴ (A ∪ B)' = A' ∪ B'

(b) Here, n(A) = 8, n(A') = 1 and n(S) = 9

∴ n(A) + n(A') = 8 + 1 = 9 = n(S)

(c) Now, n(A ∩ B) + n(A ∩ B)'

= 3 + 6 = 9 = n(S)

(d)

∴ n(A - B) + n(B - A) + n{(A ∩ B) ∩ (A ∪ B)}

= 5 + 0 + 3 = 8