Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am given the function

$$f(x)=\begin{cases} 0 \text{ ; when } x \text{ is irrational} \\\frac 1 q \text{ ; for } x=\frac p q \text{ irreducible fraction}\end{cases}$$

Spivak proved that for $a\in (0,1)$, we have $$\lim_{x\to a} f(x)=0$$

That is, this function is only continuous at the irrationals. I assume we consider $f$ for $x\geq 0$, since for negative $x$ it wouldn't be defined:

If $x=-\dfrac p q=\dfrac {-p }q=\dfrac {p }{-q}$ should we take $f(x)=-\dfrac 1 q $ or $f(x)=\dfrac 1q$?

Also, this function is periodic with period $1$, so we can just prove this on $(0,1)$ (as in the first case when proving continuity).

Now, Spivak wants me to prove this function is not differentiable at $a$, for any $a$. It is clear that, since it isn't continuous at the rationals, it is not differentiable there. Thus we need to prove the claim for $a$ irrational.

He gives the following hint.

Suppose $a=n,a_1a_2a_3\dots$. Consider the expression $$\frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}$$ for $h$ rational, and also for $h=-0,0\dots0a_{n+1}a_{n+2}\dots$

I have been trying to work this out for a while, but I can't. I am not sure, also, if he meant to have the $n$ in the subindices match the $n$ of $a$, or if it is only a typo. The book has $$\frac{{f\left( {a + h} \right) - f\left( h \right)}}{h}$$ instead of $$\frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}$$ which is (I guess) a typo, too.

For $h$ rational, one gets $$\frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} = 0$$ and for $h=-0,0\dots0a_{n+1}a_{n+2}\dots$, you get,assuming $a+h=m/u$ $$\frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} = \frac{{f\left( {a + h} \right)}}{h} = f\left( {\frac{m}{u}} \right)\frac{1}{h}$$

but I really don't know what to do with this. I know I have to show the limit $$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right)}}{h}$$ doesn't exist for any $a$, but I can't see how.

I'm looking for good hints rather than full solutions.

share|cite|improve this question
I'm used to the convention that a "reduced fraction" always has positive denominator. (the numerator can be either positive, negative, or zero) – Hurkyl Sep 16 '12 at 23:48
@Hurkyl Good to know. – Pedro Tamaroff Sep 17 '12 at 0:00
Without the hint: 1. Recall that another equivalent definition of differentiability of $f$ at $a$ is _$f$ is diff. at $a$ iff $$\displaystyle{\lim_{x\to a} \frac{f(x)-f(a)}{x-a}}\in\Bbb R$$._ 2. Use the fact that for any function $h$, $\lim_{x\to a}h(x)=l$ iff for each sequence $(x_n)$ with $x_n\to a$, $\lim_{n\to\infty}f(x_n)=l$. – leo Sep 17 '12 at 0:41
@leo See Brian's comment. I see what you mean now. I guess the proof of $2$ shouldn't be awfully difficult, right? – Pedro Tamaroff Sep 17 '12 at 0:51
Well it is a good exercise, just let flow the definitions of limit of a function and limit of a sequence. – leo Sep 17 '12 at 0:55
up vote 4 down vote accepted

First, I agree with Hurkyl that you should understand the denominator to be positive.

In the hint the index $n$ in the expansion of $h$ is not the same as the $n$ before the comma in the expansion of $a$.

If $h=-0,0\dots0a_{n+1}a_{n+2}\dots$, then $a+h=n,a_1\dots a_n$, which can be written as a fraction with denominator $10^n$. In lowest terms, therefore, its denominator is at most $10^n$, and $f(a+h)\ge 10^{-n}$. On the other hand, $|h|<10^{-n}$. Can you finish it from there? Consider letting $n\to\infty$.

share|cite|improve this answer
Hmm OK. Now, I should say that, for every $h$ of the form $-0,0\dots0a_{n+1}a_{n+2}\dots$ we'll have $f(a+h)/h>1$, thus, for any $\epsilon \geq 1$ no $\delta $ will work. That is, we can make $h$ as small as we want (keeping it of the desired form) but $f(a+h)/h>1$ in any case, while for $h$ rational $f(a+h)/h=0$, so the limit doesn't exist? Why would I want to let $n\to \infty$? – Pedro Tamaroff Sep 17 '12 at 0:43
@Peter: You’ve got it. There’s at least this one ‘bad’ sequence of $h$’s approaching $0$ for which the difference quotient is always greater than $1$, while there are other sequences for which it’s always $0$. – Brian M. Scott Sep 17 '12 at 0:45
So that's what @leo was saying! Great. – Pedro Tamaroff Sep 17 '12 at 0:52
@PeterTamaroff yes I was following those lines of thought – leo Sep 17 '12 at 0:58

Fix $\epsilon = 1$. We want to show that $\forall \delta > 0$, there exists $x + h = \dfrac{p}{q} \in (x - \delta, x + \delta)$ for which $\dfrac{f(x + h)}{h} > 1$.

Notice that if $x + h = \dfrac{p}{q}$, then $f(x + h) \ge \dfrac{1}{q}$.

Pick $q > \dfrac{1}{\delta}$. We can find an integer $p$ so that $p - 1 < qx < p$. Set $x + h = \dfrac{p}{q}$. We have:

$$ \dfrac{p}{q} - \dfrac{1}{q} < x \Rightarrow h < \dfrac{1}{q} $$


$$ \dfrac{f(x + h)}{h} > \frac{1}{q} \cdot q = 1 $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.