Question

If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8} X = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10} then verify the following:
i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ B) ∩ (B ∪ C)
ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
iii) (A ∪ B)' = A' ∩ B'
iv) (A ∩ B)' = A' ∪ B'
v) A = (A ∩ B) ∪ (A ∩ B')
vi) B = (A ∪ B) ∪ (A' ∩ B)
vii) n (A ∪ B) = n(A) + n(B) - n(A ∩ B)

Answer 1

Answer:

Step-by-step explanation:

Hi,

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

X = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

i)A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Consider L.H.S = A ∪ (B ∩ C)

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

= {1, 2, 3, 4} ∪ {4, 5, 6}

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

Consider R.H.S = (A ∪ B) ∩ (A ∪ C)

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

(A ∪ C) = {1, 2, 3, 4}  ∪ {4, 5, 6, 7, 8} = {1, 2, 3,4, 5, 6, 7, 8}

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

Thus, L.H.S = R.H.S

ii)A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Consider L.H.S = A ∩ (B ∪ C)

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

= {1, 2, 3, 4} ∩ {3, 4, 5, 6, 7, 8}

= {3, 4}

Consider R.H.S = (A ∩ B) ∪ (A ∩ C)

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

(A ∩ C) = {1, 2, 3, 4}  ∩ {4, 5, 6, 7, 8} = {4}

(A ∩ B) ∪ (A ∩ C) = { 3, 4}

Thus, L.H.S = R.H.S

(iii)(A ∪ B)' = A' ∩ B'

Consider L.H.S = (A ∪ B)'

= X - (A ∪ B)

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

= (1, 2, 3, 4, 5, 6, 7, 8, 9, 10}  - {1, 2, 3, 4, 5, 6}

= {7, 8, 9, 10}

Consider R.H.S = A' ∩ B'

A' = X - A

= (1, 2, 3, 4, 5, 6, 7, 8, 9, 10}  - {1, 2, 3, 4)

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

B' = X - B

= (1, 2, 3, 4, 5, 6, 7, 8, 9, 10}  -{3, 4, 5, 6}

= {1, 2, 7, 8, 9, 10}

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

= { 7, 8, 9, 10}

Thus, L.H.S = R.H.S.

iv)(A ∩ B)' = A' ∪ B'

Consider L.H.S = (A ∩ B)'

= X - (A ∩ B)

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

= (1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 4}

= (1, 2, 5, 6, 7, 8, 9, 10}

Consider R.H.S = A' ∪ B'

A' = X - A

= (1, 2, 3, 4, 5, 6, 7, 8, 9, 10}  - {1, 2, 3, 4)

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

B' = X - B

= (1, 2, 3, 4, 5, 6, 7, 8, 9, 10}  -{3, 4, 5, 6}

= {1, 2, 7, 8, 9, 10}

A' ∪ B' = {5, 6, 7, 8, 9, 10} ∪ {1, 2, 7, 8, 9, 10}

= { 1, 2, 5, 6, 7, 8, 9, 10}

Thus, L.H.S = R.H.S.

v) A = (A ∩ B) ∪ (A ∩ B')

Consider R.H.S = (A ∩ B) ∪ (A ∩ B')

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

(A ∩ B') = {1, 2, 3, 4} ∩ {1, 2, 7, 8, 9, 10}

= { 1, 2}

(A ∩ B) ∪ (A ∩ B') =  {3, 4} ∪ {1, 2}

= {1, 2, 3, 4}

= A

Thus, L.H.S = R.H.S

vi) B = (A ∪ B) ∪ (A' ∩ B)

Consider R.H.S =  (A ∩ B) ∪ (A' ∩ B),

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

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

(A ∩ B) ∪ (A' ∩ B) = {3, 4} ∪ {5, 6}

= {3, 4, 5, 6}

= B

vii) n (A ∪ B) = n(A) + n(B) - n(A ∩ B)

Consider L.H.S = n (A ∪ B)

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

n(A ∪ B) = 6

Consider R.H.S =  n(A) + n(B) - n(A ∩ B)

n(A) = 4

n(B) = 4

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

n(A ∩ B) = 2

So, n(A) + n(B) - n(A ∩ B)

= 4 + 4 - 2

= 6

Thus, L.H.S = R.H.S

Hope, it helps !