Let $\mathbb{J}=\mathbb{R}\setminus\mathbb{Q}$ denote the irrationals. There is no continuous function $f\colon \mathbb{R} \to \mathbb{R}$ such that $f(\mathbb{Q})\subseteq \mathbb{J}$ and $f(\mathbb{J})\subseteq\mathbb{Q}$

Proof

Suppose $f$ is such a function. Since $\mathbb{Q}$ is countable, $f(\mathbb{Q})$ and $f(\mathbb{J})$ are also countable. Therefore the image of $f$ is countable. If $f$ is not a constant function, then by the intermediate value theorem the image of $f$ contains a nonempty interval, so the image of $f$ is uncountable. We have just shown that this isn't the case, so there must be some $c$ such that $f(x)=c$ for all $x\in\mathbb{R}$ Therefore $f(\mathbb{Q})=\{c\}\subset\mathbb{J}$ and $f(\mathbb{J})=\{c\}\subset\mathbb{Q}$ Obviously no number is both rational and irrational, so no such $f$ exists.