This arises from the standard question at the beginning of any basic number theory course: determining whether $ab$ is always irrational, given that $a$ and $b$ both are.
As far as I can tell, the counterexamples to the above statement fall into the following two categories (given $p \in \mathbb{Q}, q \in \mathbb{Q}$):
- $a = b \cdot q \text{ and } b = \sqrt{p}$
- $a = b^{-1} \cdot q$
The answers to this question appear to me to be a case of the latter.
Are there any examples beyond those listed above?
No, there aren't. If $ab=q$, with $q\in\mathbb Q$, then $a=b^{-1}q$,
Note that if $a=bq$ and $b=\sqrt p$ for some $p\in\mathbb Q$, then$$a=b^{-1}b^2q=b^{-1}pq.$$Therefore, the first case is a particular type of the second one.