Question

Let $\ast$ be a binary operation on the set Q of rational number as follows:
$\left(i\right)a\ast b=a{b}^2$
Find which of the binary operation are commutative and which are associative?

Answer 1

On Q, the operation $\ast$ is defined as $a\ast b=a{b}^2$
It can be observed that for $2,3\in Q$
$2\ast 3={2.3}^2=18$ and $3\ast 2={3.2}^2=12$
Hence, $2\ast 3\ne 3\ast 2$
Also, $\frac{1}{2}\ast \frac{1}{3}=\frac{1}{2}(\frac{1}{3}{)}^2=\frac{1}{2}.\frac{1}{9}=\frac{1}{18}$
$\frac{1}{3}.\frac{1}{2}=\frac{1}{3}(\frac{1}{2}{)}^2=\frac{1}{3}.\frac{1}{4}=\frac{1}{12}$
$\therefore \frac{1}{2}\ast \frac{1}{3}\ne \frac{1}{3}\ast \frac{1}{2}$ where, $\frac{1}{2},\frac{1}{3}\in Q$
Thus, the operation $\ast$ is not commutative.
It can also be observed that for $1,2,3\in Q$

$(1\ast 2)\ast 3=\left({1.2}^2\right)\ast 3=4\ast 3={4.3}^2=361\ast (2\ast 3)=1\ast \left({2.3}^2\right)=1\ast 18={1.18}^2=324(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$
Also, $(\frac{1}{2}\ast \frac{1}{3})\ast \frac{1}{4}=[\frac{1}{2}.(\frac{1}{3}{)}^2]\ast \frac{1}{4}=\frac{1}{18}\ast \frac{1}{4}=\frac{1}{18}.(\frac{1}{4}{)}^2=\frac{1}{18\times 16}$
$\frac{1}{2}\ast (\frac{1}{3}\ast \frac{1}{4})=\frac{1}{2}\ast \left[\frac{1}{3}\right(\frac{1}{4}{)}^2]=\frac{1}{2}\ast \frac{1}{48}=\frac{1}{18}=\frac{1}{2}(\frac{1}{48}{)}^2=\frac{1}{4608}$
$\therefore (\frac{1}{2}\times \frac{1}{3})\ast \frac{1}{4}\ne \frac{1}{2}\ast (\frac{1}{3}\ast \frac{1}{4})where\frac{1}{2},\frac{1}{3},\frac{1}{4}\ne Q$
Thus, the operation $\ast$ is not associative.