Let $\Bbb Q$ be the set of rationals and $f:\Bbb Q\to \Bbb R$ be a continuous function. Then $f$ is bounded on some interval? If not, what happen if in addition $f$ satisfies $f(xy)=f(x)f(y)$ for all $x, y\in \Bbb Q$?
Let $x\in \mathbb Q$. Then as $f$ is continuous, there is $\delta>0$ such that
$$|f(y) - f(x)|<1$$
whenever $|y-x| < \delta $ and $y\in \mathbb Q$. Then
$$|f(y)| < |f(x)| +1$$
for all $y \in [x-\delta/2, x + \delta/2] \cap \mathbb Q$.