Over $\mathbb{Q}$ there is a nonconstant differentiable function with identically zero derivative, so Rolle and the mean value theorem both fail

The notion of derivative used here is stated in full, and is not imported. Let $F$ be an ordered field, $D\subseteq F$, $\phi :D\to F$ and $c\in D$ a point that is not isolated in $D$, meaning that for every $\delta >0$ in $F$ there is $x\in D$ with $0<\mathrm{\mid }x-c\mathrm{\mid }<\delta$. Say $\phi$ is differentiable at $c$ with derivative $L\in F$ when

$\text{for every }\epsilon >0\text{ in }F\text{ there is }\delta >0\text{ in }F\text{ with }\mid \frac{\phi (x)-\phi (c)}{x-c}-L\mid <\epsilon \text{ for all }x\in D\text{ with }0<\mathrm{\mid }x-c\mathrm{\mid }<\delta ,$

and write ${\phi }^{\mathrm{\prime }}(c)=L$. This is the ordinary difference-quotient condition, read entirely inside $F$. Nothing below cites a definition of the derivative from elsewhere in this library, because there is none yet.

Refuted claim: over every ordered field $F$, if $\phi :[a,b{]}_F\to F$ with $a<b$ is differentiable at every point of $[a,b{]}_F$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field), then

1. (Rolle) $\phi (a)=\phi (b)$ implies ${\phi }^{\mathrm{\prime }}(c)=0$ for some $c\in [a,b{]}_F$, and
2. (Mean value) $\phi (b)-\phi (a)={\phi }^{\mathrm{\prime }}(c)(b-a)$ for some $c\in [a,b{]}_F$.

The witness is $F=\mathbb{Q}$ and $D=[0,2{]}_{\mathbb{Q}}$ with

$\phi (x)=\{\begin{array}{ll}0& x^2<2,\\ 1& x^2>2,\end{array}\text{and}\rho (x)=\phi (x)-\frac{x}{2}\mathrm{.}$

$\phi$ is well defined on $D$ because no rational squares to $2$ (FALSE: some rational number squares to 2). It is locally constant, hence differentiable everywhere on $D$ with ${\phi }^{\mathrm{\prime }}\equiv 0$, and it is not constant, since $\phi (0)=0$ and $\phi (2)=1$; that refutes clause 2. And $\rho$ satisfies $\rho (0)=\rho (2)=0$ while ${\rho }^{\mathrm{\prime }}(c)=-1\mathrm{/}2$ for every $c$; that refutes clause 1.

• Where the classical proof breaks. Rolle's theorem is proved by taking a point where the function attains its maximum and showing the derivative vanishes there. Over $\mathbb{Q}$ the maximum need not exist: that is On a closed interval of $\mathbb{Q}$ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property, proved on the same interval and by the same missing $\sqrt{2}$. So this counterexample is not independent of that one, it is its consequence for the differential calculus.

• A locally constant function need not be constant when the domain is disconnected, and $D$ is disconnected in exactly the way $\mathbb{Q}$ is: the sets $\{x\in D:x^2<2\}$ and $\{x\in D:x^2>2\}$ are disjoint, nonempty, cover $D$, and each is open in the $\epsilon$-$\delta$ sense. Over $\mathbb{R}$ no such split of an interval exists, and that is the connectedness that the mean value theorem really rests on.

• The derivative here is genuinely a derivative, not a degenerate reading: the difference quotient is not merely small near $c$, it is exactly $0$ for $\phi$ and exactly $-1\mathrm{/}2$ for $\rho$ on a whole punctured neighbourhood, so the limit exists in the strongest possible sense.

• This item does not use, and does not need, a general theory of differentiation. The difference-quotient condition is stated in the Statement refuted and every use of it above is a direct verification, so the item is self-contained and nothing here waits on a later page. As with On a closed interval of $\mathbb{Q}$ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property, that is deliberate: the claim refuted is a claim about an arbitrary ordered field and is refuted over $\mathbb{Q}$, so a derivative defined for real functions on subsets of $\mathbb{R}$ would not apply to it. The condition above is the ordinary difference-quotient one read inside $F$, and it specialises to the real-variable definition at $F=\mathbb{R}$.