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If $x\in\mathbb{R}-\mathbb{Q}$, I can prove that $$x-\lfloor x\rfloor,\ 2x-\lfloor 2x\rfloor,\ \dots,\ nx - \lfloor nx\rfloor $$ are all distinct. If we assume otherwise then $$ (i-j)x = \lfloor ix\rfloor -\lfloor jx\rfloor$$ for $i\neq j$, and therefore $x\in\mathbb{Q}$.

Is this true if we consider $x\in \mathbb{Q}$?

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    $\begingroup$ What happens if $x=1/2$ and $n=3$? $\endgroup$ – kimchi lover Jan 20 '18 at 18:13
  • $\begingroup$ @kimchilover Thank you! $\endgroup$ – mathma Jan 20 '18 at 18:47
  • $\begingroup$ @user8734617 True, I just corrected it $\endgroup$ – mathma Jan 20 '18 at 18:56
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If $x=\frac{p}{q}$, then, for $k,n\in\mathbb Z$:

$$(k+nq)x-\lfloor (k+nq)x\rfloor=kx+np-\lfloor kx+np\rfloor=kx+np-\lfloor kx\rfloor - np = kx-\lfloor kx\rfloor$$

thus the values $kx-\lfloor kx\rfloor$ are never all distinct if $x$ is rational, and they repeat with the period at most $q$.

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