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