Is the binary operation * defined on set N, given by $a * b=\frac{a+b}{2}$ for all $a, b \in Q$, commutative?

Question

(i) Is the binary operation * defined on set $N$, given by $a * b=\frac{a+b}{2}$ for all $a, b \in Q$, commutative?

(ii) Is the above binary operation * associative?

Answer 1

$\begin{array}{l} \text { (i) } \because a^* b=\frac{a+b}{2} \\b^* a=\frac{b+a}{2} \quad[\because a, b \in Q] \\\therefore a^* b=b^* a\end{array}$

Hence, ' $*$ ' is commutative

(ii) Let $a, b, c \in Q$

$a{ }^*\left(b^* c\right)=a *\left(\frac{b+c}{2}\right)$

$=\frac{a+\frac{b+c}{2}}{2}=\frac{2 a+b+c}{2.2}=\frac{2 a+b+c}{4}$

and $\left(a{ }^* b\right) * c=\left(\frac{a+b}{2}\right) * c$

$=\frac{\frac{a+b}{2}+c}{2}=\frac{a+b+2 c}{4}$

and $\left(a{ }^* b\right){ }^* c \neq\left(a{ }^* b\right)$ * $c$

Hence '${ }^*$' is not associative.