Question

Discuss the commutativity and associativity of binary operation '*' defined on A = Q − {1} by the rule a * b= a − b + ab for all, a, b ∊ A. Also find the identity element of * in A and hence find the invertible elements of A.

Answer 1

Given, * is a binary operation on Q − {1} defined by a*b = a- b + ab

Commutativity:

For any a, b∈A, we have 

a * b = a − b + ab and b *a = b − a + ba

Since, a − b + ab ≠ b − a + ab

∴ a * b ≠ b * a

So, * is not commutative on A

Associativity:

Let a, b, c ∈ A
(a * b) *c = (a − b + ab) * c

⇒(a * b) * c = (a − b + ab) − c + (a − b + ab)c

⇒(a * b) * c = a − b + ab − c + ac − bc + abc

a * (b * c) = a * (b − c + bc)

⇒a * (b * c) = a − (b − c + bc) + a(b − c + bc)

⇒a * (b * c) = a − b + c − bc + ab − ac + abc

⇒ (a * b) * c ≠ a *(b * c)

So, * is not associative on A

Identity Element

Let e be the identity element in A, then

a * e = a = e * a          ∀a ∈ Q − {1}

⇒ a − e + ae = a

⇒(a − 1)e = 0

⇒e = 0 (As a ≠ 1)

So, 0 is the identity element in A.

Inverse of an Element

Let a be an arbitrary element of A and b be the inverse of a. Then,

a * b = e = b * a

⇒ a * b = e

⇒a − b + ab = 0   [∵ e = 0]

⇒a = b(1 − a)

`=>b= a/(1-a)`

Since, b ∈ Q - 1

So, every element of A is invertible.