Show that the binary operation * on A = **R** – { – 1} defined as a*b = a + b + ab for all a, b ∈ A is commutative and associative on A. Also find the identity element of * in A and prove that every element of A is invertible.

#### Solution

We have a*b = a + b + ab for all a, b ∈ A, where A = R – { – 1}

Commutativity: For any a, b ∈ R – { – 1}

To prove: a*b = b*a

Now, a*b = a + b + ab .....(1)

b*a = b + a + ab .....(2)

From (1) and (2), we get

a*b = b*a

Hence, * is commutative.

Associative: For any a, b, c ∈ R – { – 1}

To prove: a*(b*c) = (a*b)*c

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

= a + (b + c + bc) + a(b + c + bc)

= a + b + c + ab + ac + bc + abc .....(3)

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

= a + b + ab + c + (a + b + ab)c

= a + b + c + ab + bc + ca + abc .....(4)

From (3) and (4), we have

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

Hence, a*b is associative.

Identity element:

Let e be the identity element. Then,

a*e = e*a = a

a*e = a + e + ae = a

e(1 + a) = 0

Therefore, e = 0 [∵ a ≠ – 1]

Hence, the identity element for* is e = 0.

Existence of inverse: Let a = R – { – 1} and b be the inverse of a.

Then, a * b = e = b * a

⇒a*b = e⇒a + b + ab = 0⇒b= `−a/(a+1)`

Since,

a∈R−(−1)

∴a≠−1

⇒a+1≠0

`⇒b=−a/(a+1)∈R`

Also, `−a/(a+1)` =−1⇒−a = −a−1⇒−1=0, which is not possible.

Hence, ` −a/(a+1)∈R−(−1)`

So, every element of R – { – 1} is invertible and the inverse of an element a is `−a/(a+1).`