Timeline for Elementary results with p-adic numbers
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24, 2011 at 16:39 | comment | added | KConrad | My previous comment about needing $2\pi$ instead of $\pi$ when $p=2$ was incorrect. One only has to consider $p=2$ if no odd $p$ appears in the numerator of $\pi$. Then the numerator of $\pi$ is a power of 2. The numerator can't be 2 since $\pi > 3$, so the numerator of $\pi$ would be divisible by 4 and we can use the argument for $p=2$ directly on $\pi$ instead of $2\pi$. | |
| Nov 23, 2011 at 19:55 | comment | added | KConrad | Laurent: the series $\sin(a/b)$ converges $p$-adically if $p|a$ when $p$ is odd, but the 2-adic disc of convergence of $\sin(x)$ in ${\mathbf Q}_2$ is $4{\mathbf Z}_2$ rather than $2{\mathbf Z}_2$. So if $\pi$ has an even numerator then one can gets a (fake) contradiction with $p=2$ by running through the argument with $2\pi$ in place of $\pi$. | |
| Nov 22, 2011 at 3:09 | comment | added | KConrad | One can write down an example of this very easily. Fixing a prime number $p$, as $n \rightarrow \infty$ the sequence $p^n/(p^n+1)$ converges to 1 in the reals and to 0 in the $p$-adics while $1/(p^n+1)$ converges to 0 in the reals and to 1 in the $p$-adics. So for your favorite rationals (or just integers) $r$ and $s$, $rp^n/(p^n+1)+s/(p^n+1)$ converges to $r$ in the reals and to $s$ in the $p$-adics. | |
| Nov 20, 2011 at 8:23 | comment | added | Laurent Berger | If you're not careful with this, you can give a very short proof that $\pi$ is irrational. Say it equals $a/b$, write $\sin(a/b) = 0$ and expand the series : it converges $p$-adically if $p$ divides $a$, and you get a contradiction ... except that it does not necessarily converge to $\sin(\pi)=0$! | |
| Nov 19, 2011 at 23:31 | comment | added | Lubin | All the suggestions have been good, but this simple fact, a commonplace to us old hands, is just the sort of thing that can raise an outsider's eyebrows. | |
| Nov 19, 2011 at 21:22 | history | answered | Aeryk | CC BY-SA 3.0 |