If $m, n$ are two different positive integers, is it true that the ratio $\displaystyle \frac{\log\log m}{\log\log n}$ is necessarily irrational? By $\log$ I mean the logarithm in base $e$ (not base $10$). I guess it is irrational, but I don't know why?
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$\begingroup$ What about (10^(10^1))/(10^(10^2))? $\endgroup$– Eric TresslerCommented Dec 20, 2010 at 17:49
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1$\begingroup$ Here I meant : log n = log_e n $\endgroup$– asdCommented Dec 20, 2010 at 17:53
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1$\begingroup$ I have made some typographical and copyediting changes. If I have changed the meaning of anything or otherwise introduced any errors, I apologize, and you can roll back the edits in the edit window. $\endgroup$– Theo Johnson-FreydCommented Dec 21, 2010 at 3:31
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I'll assume you mean the natural logarithm. The condition that $\log \log m = q \log \log n$ where $q = \frac{a}{b}$ is rational is equivalent to the condition $(\log m)^b = (\log n)^a$. If $m \neq n$ then $a \neq b$ and this condition cannot be satisfied if $\log m$ is a rational multiple of $\log n$ because $\log n$ is transcendental by Lindemann-Weierstrass.
If $\log m$ is not a rational multiple of $\log n$, then this condition cannot be satisfied if one assumes Schanuel's conjecture, since then $\mathbb{Q}(n, m \log n, \log m)$ would have transcendence degree $1$. So this is likely to be true but out of reach of current technology (although I don't know enough about transcendence theory to say this with any authority).
answered Dec 20, 2010 at 18:12
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$\begingroup$ hi i did not convinced with the second part of answer tnx $\endgroup$– asdCommented Dec 20, 2010 at 19:26
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2$\begingroup$ Do you understand the statement of Schanuel's conjecture? $\endgroup$ Commented Dec 20, 2010 at 20:05
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3$\begingroup$ 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. $\endgroup$ Commented Dec 20, 2010 at 23:09
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$\begingroup$ @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? $\endgroup$ Commented Dec 21, 2010 at 3:36
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2$\begingroup$ @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." $\endgroup$ Commented Dec 21, 2010 at 4:21