Timeline for Two algebraically independent irrational numbers $\alpha,\beta$ s.t. $\alpha^\beta$ is a rational number
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 31 at 13:27 | comment | added | Oleg Eroshkin | @GeraldEdgar One of Gel'fond's results implies that at least one of $2^{\ln 2}$, $2^{\ln^2 2}$, $2^{\ln^3 2}$ is algebraically independent with $\ln 2$. So you can take $\alpha=2^{\ln^k 2}$ and $\beta=\ln^{-k} 2$ for one of $k=1,2,3$. I don't know if that is sufficiently explicit for you. | |
| May 30 at 15:54 | comment | added | Gerald Edgar | So, for almost all $t$ we have $e^t$ and $\ln 2/t$ are algebraically independent. Great. Now exhibit one. | |
| May 30 at 15:19 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
a typo in the title
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| May 30 at 14:57 | history | edited | Oleg Eroshkin | CC BY-SA 4.0 |
deleted 4 characters in body; edited title
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| May 30 at 14:55 | comment | added | Ali Taghavi | @EmilJeřábek I edit to "algebraic independent" | |
| May 30 at 14:54 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
edited title
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| May 30 at 14:53 | comment | added | Ali Taghavi | @OlegEroshkin Thank you for your helpful comment | |
| May 30 at 14:53 | comment | added | Ali Taghavi | @EmilJeřábek Thank you for your helpful comments | |
| May 30 at 14:52 | history | undeleted | Ali Taghavi | ||
| May 30 at 14:42 | history | deleted | Ali Taghavi | via Vote | |
| May 30 at 14:34 | comment | added | Oleg Eroshkin | It's true unconditionally. An algebraic curve $P(x,y)=0$ and the analytic curve $(e^t,\frac{ln 2}{t})$ with $t>0$ has only countably many intersections. So there is a value of $t$ that is not on any algebraic curve $P(x,y)=0$ with $P$ in $\mathbb{Z}[X,Y]$. Then $\alpha=e^t$ and $\beta=\frac{ln 2}{t}$ are algebraically independent with $\alpha^\beta=2$. | |
| May 30 at 14:21 | comment | added | Emil Jeřábek | I also suppose you want this to be unconditionally provable, as it is trivial to construct such examples using Schanuel’s conjecture (e.g., $\alpha=e$ and $\beta=\log2$). | |
| May 30 at 14:06 | comment | added | Emil Jeřábek | I suppose you mean nonzero polynomial. Then this is normally called “algebraically independent”. | |
| May 30 at 14:04 | comment | added | Ali Taghavi | @EmilJeřábek There is no polynomial $F(x,y)$ with integer coefficients with $F(\alpha,\beta)=0$ | |
| May 30 at 14:02 | comment | added | Emil Jeřábek | What does “transcendentally independent” mean? | |
| May 30 at 13:52 | history | asked | Ali Taghavi | CC BY-SA 4.0 |