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

Question

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? — Adina Goldberg, Jul 28 '20 at 19:04
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. — Thomas Andrews, Jul 28 '20 at 19:07
You can't just adapt the other proof, because you don't know if $k(i! - j!)$ will be a factorial. — Izaak van Dongen, Jul 28 '20 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. — Paramanand Singh, Jul 29 '20 at 02:36

score 12 · Answer 1 · answered Jul 28 '20 at 19:06

12

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 $[0,1]$.

answered Jul 28 '20 at 19:06

Angina Seng

158,341

Of course, $n!/e$ works similarly. – Thomas Andrews Jul 28 '20 at 19:08
Great answer! I like it! – Marktmeister Jul 28 '20 at 19:16
1

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. – Robert Israel Jul 28 '20 at 20:33

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

1 Answers1