Obvious but unprovable: If $f'$ is the zero function on $\mathbb Q$, then $f$ is constant Let $f:\mathbb Q\to\mathbb R$ be a uniformly continuous function and assume that $f'(x)=0$ for all $x\in\mathbb Q$. That $f$ is constant is obvious...and, as far as I can tell, unprovable. Please tell me that I'm wrong!
 A: It is unprovable because there are counterexamples, such as Minkowski's question mark function (Wiki).
See also: Is there any function continuous in R
and differentiable in rational numbers with zero derivative?
Upd: $?'(q) = 0 | \forall q \in \mathbb{Q}$ can be proven by taking $q_n \to q^+$ such that ${?(q_n) - ?(q) \over q_n - q} \to 0$ and $r_n \to q^-, {?(r_n) - ?(q) \over r_n - q} \to 0$ (since the function is monotonic, for any other sequence $x_n \to q^+$, ${?(x_n) - ?(q) \over x_n - q}$ can be "sandwiched" between two similar subsequences based on $q_n$; since it's continuous, pointwise limit value equals derivative value).
Let $q = [q_0; q_1, ..., q_k]$, then take $q_n = [q_0; q_1, ..., q_k, n]$ and see that $?(q_n) = ?(q) + (-1)^k2^{-(a_1+a_2+...+a_k+n-1)}$; $q_n = {an+b \over cn+d}$ where $q = {a \over c}$, $a,b,c,d$ don't depend on $n$. Then ${?(q_n) - ?(q) \over q_n - q} = (-1)^k{c(cn+d) \over (bc-ad)2^{a_1+...+a_k+n-1}} \sim n2^{-n} \to 0$. Taking $r_n = [q_0; q_1, ..., q_k -1, 1, n]$ concludes the proof.
