# Is it true that $\lim_{n \to \infty } nt - \lfloor nt \rfloor =0$?

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!

• Consider $t=\dfrac12$ – J. W. Tanner Nov 4 '19 at 15:29
• 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? – postmortes Nov 4 '19 at 15:30
• Okey, got it! Thanks! – MrFranzén Nov 4 '19 at 15:33

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.