(i) * is an operation as a*b = a+ b+ ab where a, b ∈ Q- {-1}. Let a = 1 and \(b=\frac{-3}{2}\)two rational numbers. 

\(a*b=1+\frac{-3}{2}+1.\frac{-3}{2}\)\(\Rightarrow \frac{2-3}{2}-\frac{3}{2}=\frac{-1-3}{2}\)\(\Rightarrow\frac{-4}{2}=-2\in Q-\{-1\}\)

So, * is a binary operation from Q - { -1 } x Q - { -1} → Q - {-1}. 

(ii) For commutative binary operation, a*b = b*a. 

\(b*a=\frac{-3}{2}+1+\frac{-3}{2}.1\)\(\Rightarrow \frac{-3+2}{2}-\frac{3}{2}\)\(=\frac{-1-3}{2}\Rightarrow \frac{-4}{2}\)\(-2\in Q-\{-1\}\)

Since a*b = b*a, hence * is a commutative binary operation. 

(iii) For associative binary operation, a*(b*c) = (a*b) *c 

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

= a+ b+ c+ bc+ ab+ ac+ abc 

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

= a+ b+ c+ ab+ ac+ bc+ abc 

Now as a*(b*c) = (a*b) *c, hence an associative binary operation. 

(iv) For a binary operation *, e identity element exists if a*e = e*a = a. As a*b = a+ b- ab 

a*e = a+ e+ ae  (1) 

e*a = e+ a +e a  (2) 

using a*e = a 

a+ e+ ae = a 

e+ ae = 0 

e(1+a) = 0 

either e = 0 or a = -1 as operation is on Q excluding -1 so a≠-1, hence e = 0. 

So identity element e = 0. 

(v) for a binary operation * if e is identity element then it is invertible with respect to * if for an element b, a*b = e = b*a where b is called inverse of * and denoted by a^-1. 

a*b = 0 

a+ b+ ab = 0 

b(1+a) = -a

\(b=\frac{-a}{(1+a)}\)

\(a^{-1}=\frac{-a}{(a+1)}\)