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. $\square$