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

On $Q,$ the operation $^*$ is defined as $a ^* b = a - b.$
$1$. Commutativity:
For the operation to be commutative,$a ^* b = b ^* a$ must hold for all $a, b \in Q.$
Consider $a = \frac{1}{2}$ and $b = \frac{1}{3}.$
$\frac{1}{2} ^* \frac{1}{3} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6}$
$\frac{1}{3} ^* \frac{1}{2} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6}$
Since $\frac{1}{6} \neq -\frac{1}{6},$ the operation $^*$ is not commutative.
$2$. Associativity:
For the operation to be associative,$(a ^* b) ^* c = a ^* (b ^* c)$ must hold for all $a, b, c \in Q.$
Consider $a = \frac{1}{2}, b = \frac{1}{3}, c = \frac{1}{4}.$
$(a ^* b) ^* c = (\frac{1}{2} - \frac{1}{3}) ^* \frac{1}{4} = \frac{1}{6} ^* \frac{1}{4} = \frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = -\frac{1}{12}$
$a ^* (b ^* c) = \frac{1}{2} ^* (\frac{1}{3} - \frac{1}{4}) = \frac{1}{2} ^* \frac{1}{12} = \frac{1}{2} - \frac{1}{12} = \frac{6 - 1}{12} = \frac{5}{12}$
Since $-\frac{1}{12} \neq \frac{5}{12},$ the operation $^*$ is not associative.