Is there a differentiable function such that $f(\mathbb Q) \subseteq \mathbb Q$ but $f'(\mathbb Q) \not \subseteq \mathbb Q$? Is there a differentiable function $f:\mathbb R \rightarrow \mathbb R$ such that $f(\mathbb Q) \subseteq \mathbb Q$, but $f'(\mathbb Q) \not \subseteq \mathbb Q$? A friend of mine asserted this without giving any examples. I seriously doubt it, but I had hard time trying to disprove it since analysis isn't really my thing. I can't even think of any class of differentiable functions with $f(\mathbb Q) \subseteq \mathbb Q$ other than the rational functions.
 A: Here is another answer. Define $h:[0,+\infty)\to \mathbb{R}$ as follows:
$$h(x):= \frac{1}{(n+1)!}(n+1-n(x-n)) \quad \text{for } x\in [n,n+1)$$
The function $h$ is continuous and satisfies $h(k)=1/k!$ for every $k$ integer in its domain. Moreover:
$$ \int _0^{q}h(t)dt \in \mathbb{Q} \quad \forall q \in \mathbb{Q}$$
since on every interval of the type $[n,n+1)$ the function $h$ is  an affine function with rational coefficients. A direct computation shows:
$$ \int _0^{\infty}h(t)dt = \sum_{k=0}^{+\infty} \frac{1}{2}\big( \frac{1}{k!} + \frac{1}{(k+1)!}\big) = e - \frac{1}{2}$$
Define now $f:\mathbb{R}\to \mathbb{R}$ as follows:
$$f(x):=
\begin{cases}
0 & x=0\\
x \int _0^{1/|x|}h(t)dt & x\neq 0
\end{cases}
$$
Then $f$ satisfies $f(\mathbb{Q})\subseteq \mathbb{Q}$ and it is differentiable in $x\neq 0$ for the fundamental theorem of integral calculus. What about $x=0$? We have:
$$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{x \int _0^{1/|x|}h(t)dt}{x}= \int _0^{+\infty}h(t)dt = e - \frac{1}{2}$$
Hence $f$ is differentiable on the whole real line and $f'(0)= e-1/2 \notin \mathbb{Q}$
A: Nice question.
Here is my  solution.  It would be nice to know if there is a simpler example.
This is basically a "spline" example.  First, if $A,B,C,D,r,s$ are all rational numbers, and $r < s$, then there exist rational numbers $a,b,c,d$ so that the cubic polynomial $f(x) = ax^3+bx^2+cx+d$ satisfies $f(r)=A,f'(r)=B, f(s)=C, f'(s)=D$.  The proof is: write down the system of equations and solve for $a,b,c,d$.  As long as $r \ne s$ the solution is rational in $A,B,C,D,r,s$.
Remark 1. Computation shows that if $B=D=0$, then the graph of $f$ on $[r,s]$ lies inside
the rectangle with opposite corners $(r,A)$ and $(s,C)$.  [Differentiate the cubic, then factor to find that the derivative vanishes at the two endpoints $r,s$.]

The construction First, choose your favorite irrational number, let's say $\pi$.  Consider the two curves $\phi_1(x)=\pi x$ and $\phi_2(x) = \pi x + x^3$.

Of course, any function $f$ with graph between these two has $f'(0)=\pi$.  Define such a function by using a sequence of rational "knots" $r_n \searrow 0$ and $A_n$ so that
$$
\phi_1(r_n) < A_n < \phi_2(r_n)
$$
Then fill in with the splines as explained, to get a function $f$ with $f(r_n) = A_n$ and $f'(r_n) = 0$ for all $n$.  Do the same thing on the negative side.  Finallly let $f(0)=0$.
This gives us $f : \mathbb R \to \mathbb R$ such that $f(\mathbb Q) \subseteq \mathbb Q$, $f$ is differentiable except possibly at $0$, and
$$
\lim_n \frac{f(r_n)-f(0)}{r_n-0} = \pi
$$
is irrational.
What remains: we need to choose the sequences $r_n$ and $A_n$ so that
the rectangles with opposite vertices $(r_n,A_n)$ and $(r_{n+1},A_{n+1})$ lie entirely between $\phi_1$ and $\phi_2$.  Then by Remark 1, $f$ remains between $\phi_1$ and $\phi_2$, so $f'(0)=\pi$.
A: This answer shares the use of cubic splines with that of GEdgar.
Consider the following polynomial function:
$$ p(x)=2(x-1)^3-3(x-1)^2+1$$
$p$ satisfies $p'(1)=p'(2)=0$. Studying $p$ in the interval $[1,2]$  we see that in this interval $p$ is non negative and strictly decreasing, with $p(1)=1$ its maximum and $p(2)=0$ its minimum.
Therefore the function:
$$
h(x)=
\begin{cases}
1& |x|\leq 1\\
p(|x|)&  1<|x| \leq 2\\
0 &|x|>2
\end{cases}
$$
is in $C^1(\mathbb{R})$.
Now set:
$$ g_n(x)=h(nx)/n!$$
and consider $g(x):=\sum_{n=0}^{+\infty} g_n(x)$.
What we are going to prove is that $f(x):=xg(x)$ is a differentiable function and satisfies $f(\mathbb{Q})\subseteq \mathbb{Q}$ and $f'(0)=e$.
Let's first discuss the values taken by $f$ on the rationals.
The function $g_n$ takes only rational values in every rational point and for $n>0$ we have $g_n$ is zero outside $[-2/n,2/n]$. This means that for every $q \in \mathbb{Q}-\{0\}$ the sequence $(g_n(q))$ is eventually zero, so that $g(q)$ is in fact a finite sum of rationals and $f(q)=qg(q)$ is also rational. We also have $g(0)=e$ but clearly $f(0)=0\cdot e=0 \in \mathbb{Q}$.
To show $f$ is differentiable, it is enough to show $g$ is. Since $g'_n(x)= h'(nx)/(n-1)!$ for $n>1$, we have $\sum g'_n$ converges uniformly on $\mathbb{R}$ and since $\sum g_n(0)$ converges we deduce that also $g$ is differentiable, with $g'=\sum g_n'$.
Finally, $f'(0)= g(0)=e$. This concludes the proof.
