Question

Let A be Q\{1} Define * on A by x * y = x + y – xy. Is * binary on A? If so, examine the commutative and associative properties satisfied by * on A

Answer 1

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

L.H.S: 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)

R.H.S: (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)