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I would like to prove the following fact:

$\displaystyle\lim_{n \to \infty } (nt - \lfloor nt \rfloor) =0 $, where $\lfloor x \rfloor := \max \{m \in \mathbb Z: \ m \le x \} $ denotes the floor function.


My idea to prove this was the following:

first, prove it for $t\in\mathbb Q$ and then use that the rationals are dense in the reals. $$$$ Thus,$(t=\frac {p } {q }\in\mathbb Q)\;\land\;(n=q^j,j\in\mathbb N)\implies nt\in\mathbb Z\implies\lfloor nt \rfloor =nt\iff nt - \lfloor nt \rfloor =0 $.

But what happens with the difference $ nt - \lfloor nt \rfloor$ when $q^{j-1 } < n < q^j $?

Does it decrese as $j \to \infty $?

$$$$ Secondly if $\exists\displaystyle\lim_{n\to\infty}(nt-\lfloor nt\rfloor)$ may we argue that if $|x-t|< \epsilon, \ x \in \mathbb R, \ t \in \mathbb Q $ then: $$\lim_{n \to \infty } nt - \lfloor nt \rfloor =0 \implies\lim_{n \to \infty } nx - \lfloor nx \rfloor < \epsilon ?$$


Of course, any other approach is fine if it proves or disproves the limit!

Most grateful for any help provided!

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    $\begingroup$ Consider $t=\dfrac12$ $\endgroup$ – J. W. Tanner Nov 4 '19 at 15:29
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    $\begingroup$ If we take $t=1/9=0.11111....$ then there's a subsequence for $n=10^m$ where $nt-\lfloor nt \rfloor$ remains $0.11111....$ isn't there? $\endgroup$ – postmortes Nov 4 '19 at 15:30
  • $\begingroup$ Okey, got it! Thanks! $\endgroup$ – MrFranzén Nov 4 '19 at 15:33
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It's not true. If $t=\dfrac12$, then $a_n=nt-\lfloor nt\rfloor$ alternates between $0$ and $\dfrac 12$,

so the limit as $n\to\infty$ does not exist.

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