(i) For all a , b ∊ N , a * b = $\frac{\mathit{a}+\mathit{b}}{2}$
Now , b * a = $\frac{\mathit{b}+\mathit{a}}{2}$ = $\frac{\mathit{a}+\mathit{b}}{2}$ = a * b
Thus, the binary operation * is commutative.
(ii) Let a , b , c ∊ N
a * b * c = a * $\left(\frac{\mathit{b}+\mathit{c}}{2}\right)$ = $\frac{\mathit{a}+\frac{\mathit{b}+\mathit{c}}{2}}{2}$ = $\frac{2\mathit{a}+\mathit{b}+\mathit{c}}{4}$
a * b * c = $\left(\frac{\mathit{a}+\mathit{b}}{2}\right)$ * c = $\frac{\frac{\mathit{a}+\mathit{b}}{2}+\mathit{c}}{2}$ = $\frac{\mathit{a}+\mathit{b}+2\mathit{c}}{4}$
∴ a * b * c ≠ a * b * c
Thus, the binary operation * is not associative.