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We know that positive integer times a irrational number modulo $1$ generate a dense set in $[0,1]$. According the answer of this post:Multiples of an irrational number forming a dense subset. I see no reason why the proof cannot be extended to $n!\alpha$ for $\alpha$ be an irrational number. We can just replace $i$ and $j$ with $i!$ and $j!$ and the argument still holds. Is that true?

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  • $\begingroup$ Can you clarify if $\alpha$ or $n$ is fixed or both are allowed to vary? $\endgroup$ Jul 28, 2020 at 19:04
  • $\begingroup$ Use \bmod to get proper spacing for the binary operation. $\endgroup$ Jul 28, 2020 at 19:05
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    $\begingroup$ Neither $n!e$ nor $n!/e$ are dense modulo $1.$ Use that $e=\sum_{k=0 }^{\infty} \frac 1{k!}$ for the first and $e^{-1}= \sum_{k=0 }^{\infty} \frac {(-1)^k}{k!}$ for the second. $\endgroup$ Jul 28, 2020 at 19:07
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    $\begingroup$ You can't just adapt the other proof, because you don't know if $k(i! - j!)$ will be a factorial. $\endgroup$ Jul 28, 2020 at 19:10
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    $\begingroup$ @IzaakvanDongen: your comment is the most important here. While studying a proof one must try to grasp where and how each hypotheses is related to some part of a proof. $\endgroup$
    – Paramanand Singh
    Jul 29, 2020 at 2:36

1 Answer 1

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I presume you mean $n!\alpha$ (not $n!/\alpha$).

Try $\alpha=e$. (Yes, that $e$.) Then $$n!e=\text{integer}+\frac1{n+1}+\frac1{(n+1)(n+2)}+\cdots$$ so modulo $1$, $n!e$ is between $1/(n+1)$ and $1/n$, so the $n!e$ are certainly not dense modulo $1$ in $[0,1]$.

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  • $\begingroup$ Of course, $n!/e$ works similarly. $\endgroup$ Jul 28, 2020 at 19:08
  • $\begingroup$ Great answer! I like it! $\endgroup$ Jul 28, 2020 at 19:16
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    $\begingroup$ And the same is true of all numbers of the form $x = \sum_k c_k/n!$ where $\{c_k\}$ is a bounded set of nonnegative integers. And the cardinality of these is the continuum. So even if you didn't know $e$ is transcendental, uncountably many of these are transcendental. $\endgroup$ Jul 28, 2020 at 20:33

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