Question

For any two sets A and B, prove that
(i) (A ∪ B) – B = A – B

(ii) A – (A ∩ B) = A – B

(iii) A – (A – B) = A ∩ B

(iv) A ∪ (B – A) = A ∪ B

(v) (A – B) ∪ (A ∩ B) = A

Answer 1

(i) (A ∪ B) – B = A – B

Consider LHS (A ∪ B) – B

= (A – B) ∪ (B – B)

= (A – B) ∪ ϕ (here, B – B = ϕ)

= A – B (here, x ∪ ϕ = x for any set)

= RHS

Thus proved.

(ii) A – (A ∩ B) = A – B

Consider LHS A – (A ∩ B)

= (A – A) ∩ (A – B)

= ϕ ∩ (A – B) (here, A - A = ϕ)

= A – B

= RHS

Thus proved.

(iii) A – (A – B) = A ∩ B

Consider LHS A – (A – B)

Suppose, x ∈ A – (A – B) = x ∈ A and x ∉ (A – B)

x ∈ A and x ∉ (A ∩ B)

= x ∈ A ∩ (A ∩ B)

= x ∈ (A ∩ B)

= (A ∩ B)

= RHS

Thus proved.

(iv) A ∪ (B – A) = A ∪ B

Consider LHS A ∪ (B – A)

Suppose, x ∈ A ∪ (B – A) ⇒ x ∈ A or x ∈ (B – A)

⇒ x ∈ A or x ∈ B and x ∉ A

⇒ x ∈ B

⇒ x ∈ (A ∪ B) (here, B ⊂ (A ∪ B))

This is true for all x ∈ A ∪ (B – A)

∴ A ∪ (B – A) ⊂ (A ∪ B)…… (1)

Conversely,

Suppose x ∈ (A ∪ B) ⇒ x ∈ A or x ∈ B

⇒ x ∈ A or x ∈ (B – A) (here, B ⊂ (A ∪ B))

⇒ x ∈ A ∪ (B – A)

∴ (A ∪ B) ⊂ A ∪ (B – A)…… (2)

From the equation 1 and 2 we get,

A ∪ (B – A) = A ∪ B

Thus proved.

(v) (A – B) ∪ (A ∩ B) = A

Consider LHS (A – B) ∪ (A ∩ B)

Suppose, x ∈ A

Now either x ∈ (A – B) or x ∈ (A ∩ B)

⇒ x ∈ (A – B) ∪ (A ∩ B)

∴ A ⊂ (A – B) ∪ (A ∩ B)…. (1)

Conversely,

Suppose x ∈ (A – B) ∪ (A ∩ B)

⇒ x ∈ (A – B) or x ∈ (A ∩ B)

⇒ x ∈ A and x ∉ B or x ∈ B

⇒ x ∈ A

(A – B) ∪ (A ∩ B) ⊂ A………. (2)

∴ From the equation (1) and (2), We get

(A – B) ∪ (A ∩ B) = A

Thus proved.