Question

If $A$ and $B$ are any two non-empty sets, then prove that $(A\cap B{)}^{\prime }=$

• A. $A^{\prime }\cap B^{\prime }$
• B. $A^{\prime }\cup B^{\prime }$
• C. $(A\cup B{)}^{\prime }$
• D. none

Answer 1

Answer: (B)

$A^{\prime }\cup B^{\prime }$

Let M = (A ∩ B)' and N = A' U B'

Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'

⇒ x ∉ (A ∩ B)

⇒ x ∉ A or x ∉ B

⇒ x ∈ A' or x ∈ B'

⇒ x ∈ A' U B'

⇒ x ∈ N

Therefore, M ⊂ N …………….. (i)

Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'

⇒ y ∈ A' or y ∈ B'

⇒ y ∉ A or y ∉ B

⇒ y ∉ (A ∩ B)

⇒ y ∈ (A ∩ B)'

⇒ y ∈ M

Therefore, N ⊂ M …………….. (ii)

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