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Suppose that I am given an irrational $\alpha$, then the set $$\{\{n\alpha\}:n\in \mathbb N\}$$

with $\{x\}=x-\lfloor x\rfloor$ is dense in $[0,1]$ (see [1]).

Now given $\epsilon>0$ and some target number $0<p<1$, I would like to find an upper bound on $n\in\mathbb N$ that verifies $$\left|p-\{n\alpha\}\right|<\epsilon$$

Any ideas appreciated. There are many related questions. For instance:

[1] Density of positive multiples of an irrational number

[2] Multiples of a given irrational number can be arbitrarily close to a natural number

[3] Dense set in the unit circle- reference needed

Euclean
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  • S'pose $p$ is near $1$, and $\alpha$ is really tiny. Then $n$ will have to be enormous, on the order of $1/\alpha$. Your big-oh has to allow for dependence on $\alpha$, doesn't it? – Gerry Myerson Oct 15 '20 at 12:42
  • indeed, what I conjecture more precisely is $\mathcal O(f(\alpha)\log(1/\epsilon)$, but with the big O notation as a function of epsilon the function on $\alpha$ is just a prefactor. Of course, I might be wrong. – Euclean Oct 15 '20 at 12:47
  • You'd also have to let it depend on $p$. Given $N$, the points ${n\alpha}$ for $1 \leqslant n \leqslant N$ partition the unit interval into $N+1$ intervals, the longest of which must be longer than $\frac{1}{N+1}$, so there is a $p \in (0,1)$ whose distance to the nearest point ${n\alpha}$, $n \leqslant N$, is larger than $\frac{1}{2(N+1)}$. Hence the supremum over all $p$ is at least $\frac{1}{2\epsilon} - 1$. For fixed $p$ you may get a smaller bound, but I don't think $O(\log 1/\epsilon)$ is realistic. – Daniel Fischer Oct 15 '20 at 14:04
  • @DanielFischer, fair enough, thanks for the comment, I will remove the conjecture from the question statement. Maybe you can rephrase the comment into an answer? – Euclean Oct 16 '20 at 20:10

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