Timeline for Can $(\log\log m)/(\log\log n)$ be rational?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2010 at 4:21 | comment | added | Qiaochu Yuan | @Theo: yes. The parenthetical comment should be read as "likely to be (true but out of reach of current technology)" rather than "(likely to be true) but out of reach of current technology." | |
| Dec 21, 2010 at 3:36 | comment | added | Theo Johnson-Freyd | @asd: The overall content of Qiaochu's answer, unless I am misreading it (I am not an expert), is that the answer to your question is almost certainly "yes, it is irrational", but that a proof of this fact does not exist, and is rather a long-standing conjecture. @Qiaochu: I am not an expert. I guess your parenthetical disclaimer means that maybe OP's question is provably true even without Schanuel's conjecture? | |
| Dec 20, 2010 at 23:09 | comment | added | Wadim Zudilin | In case $\log m$ is not a rational multiple of $\log n$, it is known (from the theory of linear forms in logs) that 1, $\log m$ and $\log n$ are linearly independent over the field of algebraic numbers, and it is hypothesized (long time ago) that they are in fact algebraically independent. The latter is implied by Schanuel's conjecture. | |
| Dec 20, 2010 at 20:05 | comment | added | Qiaochu Yuan | Do you understand the statement of Schanuel's conjecture? | |
| Dec 20, 2010 at 19:26 | comment | added | asd | hi i did not convinced with the second part of answer tnx | |
| Dec 20, 2010 at 18:12 | history | answered | Qiaochu Yuan | CC BY-SA 2.5 |