Timeline for Researching the irrationality of a number [closed]

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15 events

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Jun 2, 2016 at 9:02	history	closed	Greg Martin Wolfgang Myshkin Alexey Ustinov Stefan Kohl♦		Not suitable for this site
Jun 2, 2016 at 7:48	comment	added	Michael Freimann		@GerryMyerson I have encountered this fact somewhere.
Jun 1, 2016 at 23:35	comment	added	Gerry Myerson		How, Michael, can you know $H^{10}$ is irrational, if you have no idea how to prove it?
Jun 1, 2016 at 20:23	comment	added	alpoge		As for the formula given for $h_n^p$, use the multinomial Lucas theorem on the decomposition of $n = \underbrace{1 + \cdots + 1}_{d_0} + \cdots + \underbrace{p^i + \cdots + p^i}_{d_i} + \cdots$ to get $h_n^p\equiv \prod d_i! \cdot \prod (h_{p^i}^p)^{n_i} \pmod{p}$. Since $h_{p^i}^p = h_{p^i - 1}^p$ it follows by induction on $i$ that $h_{p^i}^p = \pm 1$! Lemme know if I've made a mistake in the above!
Jun 1, 2016 at 20:15	comment	added	alpoge		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 18:06	comment	added	Michael Freimann		@NeilStrickland Very interesting, but can you give some explanation on how you figured out that $h_n^p=\pm\prod_i d_i! \pmod p$ and why is it not periodic?
Jun 1, 2016 at 17:42	review	Close votes
Jun 2, 2016 at 9:02
Jun 1, 2016 at 17:38	comment	added	Neil Strickland		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 17:33	comment	added	Michael Freimann		@so-calledfriendDon, that's a great thing, thank you. At first look, this proof heavily uses that $p=10$.
Jun 1, 2016 at 17:28	comment	added	Michael Freimann		These formulas don't seem really helpful, because the use "greatest integer function", which complicates the analysis. But thanks a lot, I will examine it to see, if anything can be done here.
Jun 1, 2016 at 17:23	comment	added	so-called friend Don		Have you seen home.wlu.edu/~dresdeng/papers/two.pdf ?
Jun 1, 2016 at 17:20	comment	added	Greg Martin		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 16:57	history	edited	Michael Freimann		tags added
Jun 1, 2016 at 16:43	review	First posts
Jun 1, 2016 at 16:49
Jun 1, 2016 at 16:38	history	asked	Michael Freimann	CC BY-SA 3.0