Question

Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities: 

(i) A × (B ∩ C) = (A × B) ∩ (A × C) 

(ii) A × (B – C) = (A × B) – (A × C)

(iii) (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B)

Answer 1

Given: A = {a, b, c, d,}, B = {c, d, e} and C = {d, e, f, g} 

(i) Need to prove: A × (B ∩ C) = (A × B) ∩ (A × C) 

Left hand side, 

(B ∩ C) = {d, e} 

⇒ A × (B ∩ C) = {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)} 

Right hand side,

(A × B) = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)} 

(A × C) = {(a, d), (a, e), (a, f), (a, g), (b, d), (b, e), (b, f), (b, g), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g)} 

Now, 

(A × B) ∩ (A × C) = {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)} 

Here, right hand side and left hand side are equal. 

That means, A × (B ∩ C) = (A × B) ∩ (A × C) [Proved] 

(ii) Need to prove: A × (B – C) = (A × B) – (A × C) 

Left hand side, 

(B – C) = {c} 

⇒ A × (B – C) = {(a, c), (b, c), (c, c), (d, c)} 

Right hand side, 

(A × B) = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)} 

(A × C) = {(a, d), (a, e), (a, f), (a, g), (b, d), (b, e), (b, f), (b, g), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g)} 

Therefore, (A × B) – (A × C) = {(a, c), (b, c), (c, c), (d, c)} 

Here, right hand side and left hand side are equal. 

That means, A × (B – C) = (A × B) – (A × C) [Proved]

(iii) Need to prove: 

(A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B) 

Left hand side, 

(A × B) = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)} 

(B × A) = {(c, a), (c, b), (c, c), (c, d), (d, a), (d, b), (d, c), (d, d), (e, a), (e, b), (e, c), (e, d)} 

Now, (A × B) ∩ (B × A) = {(c, c), (c, d), (d, c), (d, d)} 

Right hand side, 

(A ∩ B) = {c, d} 

So, (A ∩ B) × (A ∩ B) = {(c, c), (c, d), (d, c), (d, d)} 

Here, right hand side and left hand side are equal. 

That means, (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B) [Proved]