# Researching the irrationality of a number [closed]

I am conducting a little research on checking if a number, written in positional numeral system is irrational.

Let $h^p_n$ be the most right non-zero digit of number $n!$ written in numeral system with base $p$. For example, in system with base $10$, $8!=40320\Rightarrow h_8^{10}=2.$ The question is following: for which $p$ number $H^p=0,h^p_1h^p_2\ldots h^p_n\ldots$ is irrational?

Obviously, $H^2$ is rational. I also know that $H^{10}$ is irrational, but have no idea on how to prove it and would really appreciate any kind of hint.

• There are recursive formulas for this rightmost nonzero digit. These recursive formulas can probably be used to show that the sequence is not periodic, hence this number will be irrational. Jun 1, 2016 at 17:20
• Have you seen home.wlu.edu/~dresdeng/papers/two.pdf ? Jun 1, 2016 at 17:23
• If $p$ is prime (so $(p-1)!=-1\pmod{p}$) and $n=\sum_id_ip^i$ in base $p$ then it works out that $h^p_n=\pm\prod_id_i!\pmod{p}$, and this is not periodic even if we ignore the $\pm$ signs. Jun 1, 2016 at 17:38
• Were it periodic with period $a = p^e b$ with $(b,p)=1$, by multiplying $b$ suitably (since it divides some $p^f - 1$), wlog $b = p^f - 1$, i.e. $a = p^{e+f} - p^e$, i.e. in base $p$ $a$ is a string of $(p-1)$'s followed by a string of $0$'s. Now take $n = 2p^{e+f-1}$, which has base $p$ digits $(0,2,0,\ldots,0,0,\ldots,0)$, and compare $h_n^p$ with the same for $n+a$, which has base $p$ digits $(1,1,p-1,\ldots,p-1,0,\ldots,0)$. One gets that $2! = 2\equiv \pm 1\pmod{p}$, which is false once $p>3$. Jun 1, 2016 at 20:15
• How, Michael, can you know $H^{10}$ is irrational, if you have no idea how to prove it? Jun 1, 2016 at 23:35