Question

(AUB)' = A' nB'
solve this with example:​

Answer 1

Answer:

Let P = (A U B)' and Q = A' ∩ B'

Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)'

⇒ x ∉ (A U B)

⇒ x ∉ A and x ∉ B

⇒ x ∈ A' and x ∈ B'

⇒ x ∈ A' ∩ B'

⇒ x ∈ Q

Therefore, P ⊂ Q …………….. (i)

Again, let y be an arbitrary element of Q then y ∈ Q ⇒ y ∈ A' ∩ B'

⇒ y ∈ A' and y ∈ B'

⇒ y ∉ A and y ∉ B

⇒ y ∉ (A U B)

⇒ y ∈ (A U B)'

⇒ y ∈ P

Therefore, Q ⊂ P …………….. (ii)

Now combine (i) and (ii) we get; P = Q i.e. (A U B)' = A' ∩ B'

Answer 2

Answer:

(AUB)' = A' n B'

Step-by-step explanation:

Example:

Let:

U = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

A = [1, 2, 3, 4]

B = [4, 5, 6, 7]

Solve:

(AUB) = [1, 2, 3, 4, 5, 6, 7]

(AUB)' = [8, 9, 10]

A' = [5, 6, 7, 8, 9, 10]

B' = [1, 2, 3, 8, 9, 10]

A' n B' = [5, 6, 7, 8, 9, 10] n [1, 2, 3, 8, 9, 10]

          = [8, 9, 10]

Therefore (AUB)' = A' n B'