# Is $f$ differentiable at $0$, where $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise?

This is the function: $$f(x)=\begin{cases}x& \text{if x is rational}\\0 &\text{if x is irrational}\end{cases}$$

My attempt:

It's easy to verify that $$f$$ is continuous at $$x=0$$ using the sequential definition of continuity. I claim that $$f$$ is not differentiable at $$x=0$$. Assume the contrary and let $$f'(0)=L$$. Now, we pick an $$\varepsilon$$ such that $$0<\varepsilon < |L|$$. For this choice of $$\varepsilon$$ there is a $$\delta >0$$ such that if $$0<|x-0|<\delta$$ then we have $$\left| \frac{f(x)-f(0)}{x-0} -L\right| < \varepsilon$$. Now, pick $$x' \in \mathbb{R}\setminus\mathbb{Q}$$ with $$0<|x'| <\delta$$. Then we have $$\left| \frac{f(x')-f(0)}{x'-0} -L\right| = |L| > \varepsilon$$. A contradiction!

Is this proof correct?

• The proof looks correct to me. – b00n heT Sep 30 '18 at 7:17
• en.wikipedia.org/wiki/Nowhere_continuous_function – georg Sep 30 '18 at 7:27
• @b00nheT. The $<$ in the last line should have been $>$ (a typo) so I fixed it. But the flaw is assuming $L\ne 0$. The proposer has only shown that if $f'(0)=L$ exists then $L=0.$ But $f'(0$) does not exist – DanielWainfleet Nov 17 '18 at 7:05
• – nmasanta Jun 9 '19 at 16:38
• @Ashish K if you look at the answer given there, you can get your answer. – nmasanta Jun 10 '19 at 9:40

EDIT: As @DanielWainfleet pointed out in the comments, your proof does not apply in the case $$L=0$$. The proof basically goes exactly like the one you have written: first you choose an $$0<\varepsilon<1$$ small enough, then you take $$\delta$$ as you please and look at the difference quotient for any $$|x|<\delta$$. If $$f$$ was differentiable, then it should hold that $$\left|\frac{f(x)-f(0)}{x-0}\right|<\varepsilon$$ but by density of the irrationals, you can find an irrational $$x$$ such that $$|x|<\delta$$ and for that one $$\left|\frac{f(x)-f(0)}{x-0}\right|=\left|\frac{x-0}{x-0}\right|=1>\varepsilon$$ contradicting the assumption.
Another working approach is showing that approaching $$0$$ via irrationals and via rationals leads to different difference quotients:
• choose a sequence of irrational numbers ($$a_n=\sqrt{2}/n$$) tending to zero. Then $$\frac{f(a_n)-0}{a_n-0}=0$$ for all $$n$$
• choose a sequence of rational numbers ($$a_n=1/n$$) tending to zero. Then $$\frac{f(a_n)-0}{a_n-0}=1$$ for all $$n$$
• The flaw in the proposer's proof is assuming that if $L$ exists then $L\ne 0.$ If $L=0$ then we cannot take any $\epsilon$ such that $0<\epsilon <|L|.$ – DanielWainfleet Nov 17 '18 at 7:07