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:

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

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, $b\ast a=b{a}^2$

Since $a\ast b\ne b\ast a$

$\ast$ is not commutative.

$\text{Checking associative..}$

$\ast$ is associative if,

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

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

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

$=a{b}^2{c}^2$

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

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

$=a{b}^2{c}^4$

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

$\ast$ is not an associative binary operation.