Prove that $f(x)=\begin{cases}x& \text{if $x$ is rational},\\0 &\text{if $x$ is irrational}\end{cases}$ does not possess a derivative anywhere

Question

We are currently in the section of differentiability and I want to prove that

$$f(x)=\begin{cases}x& \text{if $x$ is rational},\\0 &\text{if $x$ is irrational}\end{cases}$$

does not possess a derivative anywhere. Please, can anyone help me out?

Answer 1

$f(x)$ is not continous except at $x = 0$

(I will leave that for you to prove.)

If f is not continuous it not differentiable.

At $x = 0:$

$$f'(0) = \lim_\limits{x\to 0} \frac {f(x) - f(0)}{x}$$

In a neighborhood of $0, \frac {f(x) - f(0)}{x}$ could equal 1, or it could equal 0.

The limit is not defined.

If you want to be more formal.

Suppose $f'(0)$ exits and equals $L$

Choose $\epsilon = \frac 12$

For any $L,$ for any $\delta >0,$ there is an $x$ such that $|x|<\delta$ and $|\frac {f(x) - f(0)}{x} - L|<\epsilon.$

Answer 2

Hint: note that $f$ is discontinuous at every point but zero. Hence it suffices to prove that it is not differentiable at $x=0$. Now proceed straightforward with the definition and show that the limit $$ \lim_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} $$ does not exist. In particular, what happens to that limit if we "approach" zero along a sequence of rational numbers? And with irrationals?