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