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 by $a\ast b=(a-b{)}^2$
For $a,b\in Q$,we have
$a\ast b=(a-b{)}^{2}andb\ast a=(b-a{)}^2=[-(a-b){]}^2=(a-b{)}^2$
Therefore, $a\ast b=b\ast a$
Thus, the operation $\ast$ is commutative.

It can be observed that
$(1\ast 2)\ast 3=(1-2{)}^2\ast 3=(-1{)}^2\ast 3=1\ast 3=(1-3{)}^2=(-2{)}^2=41\ast (2\ast 3)=1\ast (2-3{)}^2=1\ast (-1{)}^2=1\ast 1=(1-1{)}^2=0$
$\therefore (1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ where $1,2,3\in Q$
Thus, the operation $\ast$ is not associative.