# Is $n!\alpha \bmod 1$ dense in $[0,1]$?

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?

• Can you clarify if $\alpha$ or $n$ is fixed or both are allowed to vary? Jul 28, 2020 at 19:04
• Use \bmod to get proper spacing for the binary operation. Jul 28, 2020 at 19:05
• 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. Jul 28, 2020 at 19:07
• You can't just adapt the other proof, because you don't know if $k(i! - j!)$ will be a factorial. Jul 28, 2020 at 19:10
• @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. Jul 29, 2020 at 2:36

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 $$[0,1]$$.
• Of course, $n!/e$ works similarly. Jul 28, 2020 at 19:08
• 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. Jul 28, 2020 at 20:33