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

\(a*b=\frac{1}{2}*2\Rightarrow 1\in Q^+\)

So, * is a binary operation from Q⁺ x Q⁺ → Q. 

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

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

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

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

\(a*(b*c)=a*\frac{bc}{2}\)\(\Rightarrow \frac{a\frac{bc}{2}}{2}=\frac{abc}{4}\)

\((a*b)*c=\frac{ab}{2}*c\)\(\Rightarrow \frac{\frac{ab}{2}c}{2}=\frac{abc}{4}\)

As a*(b*c) = (a*b) *c, hence * is an associative binary operation. For a binary operation *, e identity element exists if a*e = e*a = a. 

\(a*e=\frac{ae}{2}\) (1) 

\(a*e=\frac{ea}{2}\) (2) 

using a*e = a 

\(\frac{ae}{2}=a\Rightarrow \frac{ae}{2}-a=0\)\(\Rightarrow \frac{a}{2}(e-2)=0\)

Either a = 0 or e = 2 as given a≠0, so e = 2. 

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 = 2

\(\frac{ab}{2}=2\Rightarrow b=\frac{4}{a}\)

\(a^{-1}=\frac{4}{a}\)