Non-differentiable function proof

Question

Let $f:[0,1] \to \mathbb R$, \begin{eqnarray} f(x)= \begin{cases} 0&\quad\text {if } x \in \mathbb R - \mathbb Q \cr 1\over q&\quad\text{if } x = \frac{p}{q} \in \mathbb Q \cr \end{cases} \end{eqnarray}

Show that $f$ isnt differentiable in any point the range of $(0, 1)$, at point $x_0$

To $f(x)$: in the upper case, $x=0$ is also possible, lower case $p, q \in \mathbb N$, largest common factor is $1$. Didn't know how to edit that in.

I'm trying to look for the answer from the case of $x_0 \in \mathbb R - \mathbb Q$, but I'm not making any progres..

Answer 1

For rational $x_0$ it is not differentiable since it's not continuous.

For irrational $x_0$ the idea is to construct $p_n/q_n\rightarrow x_0$ with $p_n/q_n> x_0$. So we can take $q_n$ the nth prime, and $p_n=\lceil q_nx_0\rceil$. So for irrational $x_0$:

$$\frac{p_n-1}{q_n}

Which implies $p_n/q_n-x_0<1/q_n$ (cannot have equality since $x_0$ is irrational)

Finally:

$$\frac{f(p_n/q_n)-f(x_0)}{p_n/q_n-x_0}=\frac{1/q_n}{p_n/q_n-x_0}>\frac{1/q_n}{1/q_n}=1$$

So the derivative if it exists should be $\ge 1$.

On the other side, taking $x_n\rightarrow x_0$ irrational, we get that the derivative should be $0$, contradiction.