Timeline for $a^b=b^a$ and algebraicity [closed]
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 19, 2019 at 21:20 | history | closed |
abx Wojowu YCor Felipe Voloch user1073 |
Not suitable for this site | |
| May 19, 2019 at 20:19 | comment | added | Sylvain JULIEN | Thank you Todd for these words in my mother tongue :-) | |
| May 19, 2019 at 20:08 | comment | added | Todd Trimble | If you say so, Wojowu. I mean, thanks for the additional explanation, but I said easier for me, and that might still be true even after your addition. Chacun a son gout, or however it goes. | |
| May 19, 2019 at 19:56 | vote | accept | Sylvain JULIEN | ||
| May 19, 2019 at 19:55 | answer | added | Wojowu | timeline score: 8 | |
| May 19, 2019 at 19:51 | comment | added | Sylvain JULIEN | @Wojowu : thank you for your comment. Maybe you can post it as an answer so that I can accept it. | |
| May 19, 2019 at 19:31 | comment | added | Wojowu | @ToddTrimble I think it's actually (slightly) easier to keep the problem as stated. If $3^b=b^3$ and $b\neq 3$, it's clear that $b$ is not an integer, and hence that $b$ is irrational (as else $b^3$ is rational and $3^b$ isn't) and hence transcendental (Gelfond-Schneider). (also, it's Wojowu, not Wojuwo :) ) | |
| May 19, 2019 at 18:15 | comment | added | Todd Trimble | Picking up on Wojuwo's comment: it's easier for me to contemplate $a^{1/a} = b^{1/b}$. By Gelfond-Schneider, if $b$ is algebraic and irrational, then $b^{1/b}$ will be transcendental. So in the case $a = 3$, we would need a rational $b \neq 3$ to satisfy $b^{1/b} = 3^{1/3}$, and then it's just a matter of the fundamental theorem of arithmetic to rule out this possibility. | |
| May 19, 2019 at 17:20 | review | Close votes | |||
| May 19, 2019 at 21:25 | |||||
| May 19, 2019 at 16:49 | comment | added | Wojowu | No, by Gelfond-Schneider. Take $a=3$ and suitable $b$ for instance. | |
| May 19, 2019 at 16:39 | history | asked | Sylvain JULIEN | CC BY-SA 4.0 |