Solutions

We infer that $f$ must be constant on $[a,b]$. Indeed, for contradiction, suppose it were not so. This means that there exist $c,d\in[a,b]$ such that $f(c)\ne f(d)$ and $c\ne d$. Without loss of generality, suppose $f(c)<f(d)$. By Theorem 3, for any irrational number $e$ such that $f(c)<e<f(d)$, there exists $x\in[c,d]$ such that $f(x)=e$. But this contradicts the assumption that $f$ is always rational.