4
$\begingroup$

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 $x$.

Thanks in advance.

(PS: it's not a homework question or anything, I'm just curious)

$\endgroup$
  • 1
    $\begingroup$ I imagine that a proof in the positive direction will make use of the fact that $\sin(n)$ is dense in $[-1, 1]$. $\endgroup$ – Duncan Ramage Dec 12 '17 at 20:09
  • 1
    $\begingroup$ 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 $m$. $\endgroup$ – reuns Dec 12 '17 at 20:20
  • $\begingroup$ @DuncanRamage I thought so as well, I just couldn't figure out how to how to compensate the "growth" of the n multiplier... But it's really just a silly question, if there isn't any obvious way of doing it, I'll just delete the question. $\endgroup$ – mlaci Dec 12 '17 at 20:20
  • 1
    $\begingroup$ @mlaci Please don't delete this question, I think it's very interesting, even if it's just silly. $\endgroup$ – Duncan Ramage Dec 12 '17 at 20:21
  • $\begingroup$ @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. $\endgroup$ – mlaci Dec 12 '17 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.