Without loss of generality, we only need to show that $x q \in \mathbb{I}$.
Suppose to the contrary that there are some integers $a$ and $b \neq 0$ for which
\begin{equation}
x q
= \frac{a}{b}
\end{equation}
Since $q \neq 0$ is rational, there are integers $c \neq 0$ and $d \neq 0$ such that
\begin{equation}
q
= \frac{c}{d}
\end{equation}
Therefore, we have
\begin{equation}
x
= \frac{a}{b} \frac{d}{c}
= \frac{a d}{b c}
\end{equation}
Since $a$, $b$, $d$ and $c$ are all integers, then so are $a d$ and $b c$. Since $b \neq 0$ and $c \neq 0$, then also $b c \neq 0$. But this contradicts the assumption that $x$ was irrational. Therefore, $x q$ is irrational. $\square$
