# Curious function problem (EDIT: Not so curious, but didnt see it at the time of writing)

This one is directly from my head and although it could be something trivial I do not see the way to attack it but the problem looks interesting and I want to share it with you, here it is:

Let us define function $f$ as $f(x)=\displaystyle\frac{\pi}{x}$.

Now, I wonder is there an easy way (or any way?) to prove (or disprove) that for every interval $[x_1,x_2]\subset \mathbb R\setminus\{0\}$, $x_1\neq x_2$ there exist irrational number $x_0\in [x_1,x_2]\setminus \{\displaystyle\frac{a\pi}{b}|\displaystyle\frac{a}{b}\in\mathbb Q\}$, such that $f(x_0)$ is rational number, in other words, that every interval $[x_1,x_2]\subset \mathbb R\setminus\{0\}$ contains at least one irrational number $x_0$ such that $f(x_0)$ is rational and $x_0$ is not rational multiple of $\pi$.

Any ideas?

• "Any ideas?" Read what you are writing and think a bit before asking. Aug 17, 2013 at 12:37
• @fedja I see now, I really do not know why I did not see it at the moment of writing. So trivial. Trivializingly trivial.
– user90628
Aug 17, 2013 at 12:47
• @fedja that sounds a bit angry... :( people sometimes do ask questions which have trivial answers, why should all questions sound like problems from scientific seminars? People shoudn't be afraid of asking (and your comment isn't encouraging), no matter how trivial the question is, if they're really trying to do something. Sometimes you just can't notice something really trivial (i know it from my own experience), and need someone's hint, there's nothing bad in it... IMHO. And i'm sorry if this comment is a kind of off topic...
– W_D
Aug 17, 2013 at 13:33
• @Alex No anger, irritation, offense, etc. was implied. Those were just the best ideas I could offer to the OP. Do you disagree that they would help him in this case and in the future? Aug 17, 2013 at 13:35
• @fedja Yes, in some cases discouraging comments can cause damage. Honestly, i've been in situations like this one, that's the reason i'm writing. Well, your comment sounded a kind of, say, discouraging. Perhaps i'm exaggerating the whole thing, but that comes only from my own experience. I'm just saying, sometimes you would not notice the most-most-most trivial thing in the whole world unless someone shows you the answer. We all need communication. Did the OP ask the question connected with his homework? No, it seems. They just asked, no cheating, or something like that, it seems to me...
– W_D
Aug 17, 2013 at 15:00

$f(x_0)$ is rational if and only if $x_0$ is a rational multiple of $\pi$. Note that $x=\frac1{f(x)}\cdot \pi$.