(i) Let a,b ∈ A (i.e.) a ≠ ±1 , b ≠ 1 

Now a * b = a + b – ab 

If a + b – ab = 1 ⇒ a + b – ab – 1 = 0 

(i.e.) a(1 – b) – 1(1 – b) = 0 

(a – 1)(1 – b) = 0 ⇒ a = 1, b = 1 

But a ≠ 1 , b ≠ 1 

So (a – 1) (1 – 6) ≠ 1 

(i.e.) a * b ∈ A. So * is a binary on A. 

To verify the commutative property:

Let a, b ∈ A (i.e.) a ≠ 1 , b ≠ 1 

Now a * b = a + b – ab and b * a = b + a – ba 

So a * b = b * a ⇒ * is commutative on A.

To verify the associative property: 

Let a, b, c ∈ A (i.e.) a, b, c ≠ 1 

To prove the associative property we have to prove that 

a * (b * c) = (a * b) * c

LHS: b * c = b + c – bc = D(say) 

So a * (b * c) = a * D = a + D – aD 

= a + (b + c – bc) – a(b + c – bc) 

= a + b + c – bc – ab – ac + abc 

= a + b + c – ab – bc – ac + abc … (1) 

RHS: (a * b) = a + b – ab = K(say) 

So (a * b) * c = K * c = K + c – Kc 

= (a + b – ab) + c – (a + b – ab) c 

= a + b – ab + c – ac – bc + abc 

= a + b + c – ab – bc – ac + abc ... (2) 

(ii) To verify the identity property: 

Let a ∈ A (a ≠ 1)

If possible let e ∈ A such that 

a * e = e * a = a 

To find e: 

a * e = a

(i.e.) a + e – ae = a

e(1 - a) = 0 

e = 0/(1 - a) = 0 (because a ≠ 1) 

So, e = (≠ 1) ∈ A 

(i.e.) Identity property is verified. 

To verify the inverse property: 

Let a ∈ A (i.e. a ≠ 1) 

If possible let a’ ∈ A such that 

To find a’: 

a * a’ = e 

(i.e.) a + a’ – aa’ = 0 

⇒ a'(1 – a) = – a

a' = -a/(1 - a) = a/(a - 1) A (because a ≠ 1)

So, a' ∈ A

⇒ For every a ∈ A there is an inverse a’ ∈ A such that

a* a’ = a’ * a = e 

⇒ Inverse property is verified.