Question

Let $\ast$ be a binary operation on the set $Q$ of rational numbers as follows. Find the binary operation given below is associative or commutative:

(iv) $a\ast b=(a-b{)}^2$

Answer 1

$\text{Checking commutative.}$

Given: $a\ast b=(a-b{)}^2$

$\forall a,b\in Q,(a-b{)}^2\in Q.$ So, it is binary

$\ast$ commutative if,

$a\ast b=b\ast a$

Now, $a\ast b=(a-b{)}^2$

And, $b\ast a=(b-a{)}^2$

$=(a-b{)}^2$

Since $a\ast b\ne b\ast a\forall a,b\in Q$

$\ast$ is not commutative.

$\text{Checking associative.}$

$\ast$ is associative if,

$(a\ast b)\ast c=a\ast (b\ast c)$

Now, $(a\ast b)\ast c=(a-b{)}^2\ast c$

$=\left(\right(a-b)-c{)}^2$

And, $a\ast (b\ast c)=a\ast (b-c{)}^2$

$=(a-(b-c{)}^2{)}^2$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not an associative binary operation.