# Prove or disprove: $\phi: \mathbb{N} \to \mathbb{N}\text{, }\phi(n) = \lfloor n \cdot | \sin( \sqrt{2} \cdot n ) | \rfloor$ is surjective

How do I go about proving or disproving that the following function is surjective? Is there some sort of standard trick for integer functions?

$$\phi: \mathbb{N} \to \mathbb{N} \\ \phi(n) = \lfloor n \cdot | \sin( \sqrt{2} \cdot n ) | \rfloor$$

where $\lfloor x \rfloor$ denotes the function that returns the greatest integer less than or equal to $x$.

• I imagine that a proof in the positive direction will make use of the fact that $\sin(n)$ is dense in $[-1, 1]$. – Duncan Ramage Dec 12 '17 at 20:09
• Replacing $\sin(\sqrt{2} n)$ by $n\frac{\sqrt{2}}{2\pi }-\lfloor n\frac{\sqrt{2}}{2\pi} \rfloor$ doesn't change the problem a lot. Then you are asking if we can make the Diophantine approximation/irrationality measure of $\omega_m = \frac{\sqrt{2}}{2\pi m}$ uniform on $m$. – reuns Dec 12 '17 at 20:20
• @reuns Thank you very much. I had never seen any of the concepts on that page, and I'm having a hard time grasping them, but I'll keep trying. Is the answer to the question more or less independent of the choice of $\sqrt{2}$ (instead of any other irrational number)? Edit: nevermind the last part, I obviously hadn't read enough of the wikipedia page. – mlaci Dec 12 '17 at 20:28