The ol' switcheroo function

Suppose for the sake of contradiction such a function $f$ exists. Since $f(x)$ is sometimes rational and sometimes irrational, $f$ isn't constant. So, there are values $a<b$ such that $f(a)\neq f(b)$ . Without loss of generality, suppose $f(a)<f(b)$ . By the intermediate value theorem, for each irrational number $y\in (f(a),f(b))$ , there exists an $x\in (a,b)$ such that $f(x)=y$ , where we know that $x$ is rational by definition of $f$ . Thus, $f$ defines a surjection from the rational numbers in $(a,b)$ to the irrational numbers in $(f(a),f(b))$ , which is impossible since the former set is countable and the latter set is uncountable. So, no such function $f$ exists.