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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!

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    This is a great example of why the word "obvious" is dangerous in mathematics – 2017-01-17
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    Strongly related: https://math.stackexchange.com/questions/2039063/ – 2017-01-17
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    Moreover, we should notice that since $f : \Bbb Q \to \Bbb R$ is uniformly continuous, [it extends to](http://math.stackexchange.com/questions/814402) a unique continuous $g : \Bbb R \to \Bbb R$. However, I don't know why $g$ should be differentiable if $f$ is assumed to have a "rational derivative" at any point, i.e. the limit $\dfrac{f(x+h)-f(x)}{h}$ exists for all rational $x$, when the _rational number_ $h$ tends to $0$. – 2017-01-17
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    There exist functions whose derivative vanishes on a dense $G_\delta$ set – 2017-02-08

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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.

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    Wauw! That is super interesting! Can I bother you with a follow up question? Is there some way to strengthen the assumptions so that it does become true (without making the conditional trivial)? – 2017-01-17
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    Little question: why is Minkowski's question mark function uniformly continuous on $\Bbb R$? It is on every compact set, obviously, but on the whole real line I'm not sure. Thank you! – 2017-01-17
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    @Watson One can just take it on $[0,1]$ and then extend to $[0,2]$ by $f(x) = f(2-x)$ and then on whole $\mathbb{R}$ by $f(x) = f(x+2)$. Then for any given $\varepsilon>0$ $\delta$ you get for $[0,1]$ is enough everywhere else. – 2017-01-17
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    @Abstraction : you don't even have to extend it to the real line : Minkowski's question mark function is already defined on $\mathbb{R}$ and $1$-periodic. – 2017-01-17
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    @Casper One interesting statement is that if $f$ is analytic *complex* function, then it's constant on whole $\mathbb{C}$, not just real line. But of course this assumption is too strong. I'm not sure if it's enough to require $f'$ being defined everywhere - I have a feeling there still would be a counterexample, but can't build it (Minkowski function derivative isn't defined everywhere, for example see http://link.springer.com/article/10.1007/s10958-012-0750-2 ). – 2017-01-17
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    @Abstraction: Except for Wikipedia and other answers on this site, I have been unable to find a source that confirms what you say. I need to be able to cite such a source. Do you (or someone else) know of one? (I have asked this as a separate question here: http://math.stackexchange.com/questions/2126705/academic-reference-concerning-minkowskis-question-mark-function) – 2017-02-03
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    @Casper Statements should be obvious enough. Wiki even contains the reference for continuity statement; $?'(x) = 0 | x \in \mathbb{Q}, x = [a_0; a_1, ..., a_k]$ follows from that and $?(x_n) = ?(x) \pm 2^{-(s+n)}$ where $x_n = [a_0; a_1, ..., a_k, n]$ and thus $|{?(x_n) - ?(x) \over x_n - x}| \sim n2^{-n} \to 0$. I can edit my answer to elaborate on that if you think it's necessary. – 2017-02-06
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    @Abstraction: I would actually prefer to refer to proofs in the literature rather than put them in my own text, if possible. The citation on Wikipedia to Finch does not give me what I need. Finch only states that the derivative is zero almost everywhere, and I need sources for (1) that the derivative is zero on all rational numbers and (2) that on irrational numbers it is either 0 or undefined (or if you like $+\infty$). – 2017-02-06
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    @Abstraction: I have looked at some of the other papers listed on Wikipedia. They do not state these things explicitly. I suspect that the problem is that they are trivial corollaries of what they do state explicitly - trivial enough that they don't find them worth stating, but not trivial enough that I see it. – 2017-02-06
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    Update: I now have a reference for (2). But I still need one for (1). – 2017-02-07
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    @Casper elaborated on my comment in the answer. Can't help with a reference, sorry (even if I were to search for it, I would likely find a book in Russian which probably won't help you much). – 2017-02-08
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    @Abstraction: Thank you very much for your help! – 2017-02-08