Let $^*$ be a binary operation on the set $Q$ of rational numbers defined as $a \,^*\, b = a b^{2}$. Determine whether the operation is commutative and associative.

On the set $Q$,the operation $^*$ is defined as $a \,^*\, b = a b^{2}$.
To check for commutativity,we compare $a \,^*\, b$ and $b \,^*\, a$.
Consider $a = \frac{1}{2}$ and $b = \frac{1}{3}$ in $Q$.
$\frac{1}{2} \,^*\, \frac{1}{3} = \frac{1}{2} \times (\frac{1}{3})^{2} = \frac{1}{2} \times \frac{1}{9} = \frac{1}{18}$.
$\frac{1}{3} \,^*\, \frac{1}{2} = \frac{1}{3} \times (\frac{1}{2})^{2} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$.
Since $\frac{1}{18} \neq \frac{1}{12}$,the operation $^*$ is not commutative.
To check for associativity,we compare $(a \,^*\, b) \,^*\, c$ and $a \,^*\, (b \,^*\, c)$.
Consider $a = \frac{1}{2}, b = \frac{1}{3}, c = \frac{1}{4}$ in $Q$.
$(\frac{1}{2} \,^*\, \frac{1}{3}) \,^*\, \frac{1}{4} = (\frac{1}{2} \times (\frac{1}{3})^{2}) \,^*\, \frac{1}{4} = \frac{1}{18} \,^*\, \frac{1}{4} = \frac{1}{18} \times (\frac{1}{4})^{2} = \frac{1}{18 \times 16} = \frac{1}{288}$.
$\frac{1}{2} \,^*\, (\frac{1}{3} \,^*\, \frac{1}{4}) = \frac{1}{2} \,^*\, (\frac{1}{3} \times (\frac{1}{4})^{2}) = \frac{1}{2} \,^*\, \frac{1}{48} = \frac{1}{2} \times (\frac{1}{48})^{2} = \frac{1}{2 \times 2304} = \frac{1}{4608}$.
Since $\frac{1}{288} \neq \frac{1}{4608}$,the operation $^*$ is not associative.