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Is there $n\in \mathbb{N}$ such that $\ln(\ln(n)) \in \mathbb{Q}$? If such $n$ exists, we will get $$\ln(\ln(n)) = \frac{p}{q}, \quad p, q \in \mathbb{Z}.$$ Hence we will get $n = e^{e^{p/q}},$ where the question about the nature of $e^{e^{p/q}}$ haven't been answered yet.

Is there any other direction ? Thank you in advance.

uniquesolution
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A. PI
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    Well, we know $e^{p/q}$ is transcendental, and its hard to discern anything about transcendentals to the power of transcendentals. You could also add on the side question about different bases for the logarithms. – Simply Beautiful Art Dec 18 '16 at 16:24
  • Just a an idea. Maybe Iit could be helpful to take Taylor expansion of $ln(ln(x))$ (with $x\geq1$). In my opinion it is closely connected with having simulteniously $a^{b}$ algebraic and $b^{a}$ algebraic with both $a$ and $b$ transcendental. (Though here is asked only irrationality.) – kolobokish Dec 18 '16 at 16:27
  • I reminds me of [this](https://math.stackexchange.com/questions/1245845) question… – Watson Dec 18 '16 at 17:03
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    What do you mean by 'is there any other direction'? You've showed that it's exactly identical to the question of whether $e^{e^r}$ is an integer for some $r\in\mathbb{Q}$ and from the way you describe it suggests that you know that that problem is open. Since there's a direct equivalence, that question being open means that yours is too. – Steven Stadnicki Dec 18 '16 at 17:12
  • @StevenStadnicki I know that the question was answered for $\ln(n) = \frac{p}{q}.$ The proof depends on the fact that $e^r$ is transcendental. In this setting, we have different function $\ln\circ \ln$ and it might be another proof depending on the properties of this function itself. – A. PI Dec 18 '16 at 17:17
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    Ahh, I see. The problem is that no one knows whether $e^{e^r}$ is rational or not, so your question is still completely open (and likely to be hard). – Steven Stadnicki Dec 18 '16 at 17:32
  • @StevenStadnicki Thank you for your comment. In fact, I am about to improve my knowledge in this field. – A. PI Dec 18 '16 at 17:37
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    And integer $n$ such that $e^{e^n}$ is an integer, was called a "humdrum" number a couple decades ago. Someone posed the problem at West Coast Number Theory and so this probably ended up in Guy's "Unsolved problems in Number Theory". – B. Goddard Dec 18 '16 at 21:10
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    I am not sure, but I think, Schanuel's conjecture would even imply that $e^{e^n}$ is transcendental for rational $n$ , which would imply that $r:=\ln(\ln(q))$ is irrational for rational $q>1$ , otherwiese we would have a rational $r$ with rational $q=e^{e^r}$. Maybe someone can work this out. – Peter Dec 19 '16 at 13:49
  • Q: "Is $\ln(\ln(n))$ irrational for any integer $n>1$?" ... A: Yes. (Of course it is true. But no one knows the proof.) – GEdgar Oct 05 '17 at 00:06

1 Answers1

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@GEdgar in a comment suggested that this is not known, and @Peter suggested that someone work out how it follows from

Schanuel's Conjecture: If $z_1,\dots,z_k$ are linearly independent over $\mathbb Q$, then $\mathbb Q(z_1,\dots,z_k,e^{z_1},\dots,e^{z_k})$ has transcendence degree at least $k$ over $\mathbb Q$.

So here we go:

Proposition: Schanuel's Conjecture implies that $\ln(\ln(n))\not\in\mathbb Q$ for all $n\in\mathbb N$ with $n>1$.

Proof: Suppose $\ln(\ln(n))=r\in\mathbb Q$, so that $e^{e^r}=n$. Let $z_1=r$ and $z_2=e^r$. Suppose $$az_1+bz_2=0,\qquad a,b\in\mathbb Q.$$ Then either $b=0$ or $e^r=-az_1/b\in\mathbb Q$. But by Schanuel's Conjecture with $k=1$, we have that $e^r$ is transcendental. So it must be that $b=0$. If $z_1=0$ then $e^{e^0}=n$ which is false (simply since $2<e<3$). We conclude that $(a,b)=(0,0)$ and so we have shown that $z_1, z_2$ are linearly independent over $\mathbb Q$.

So by Schanuel's Conjecture, $$\mathbb Q(z_1,z_2,e^{z_1},e^{z_2})=\mathbb Q(r,e^r,e^r,e^{e^r})=\mathbb Q(e^r,e^{e^r})$$ has transcendence degree at least 2 over $\mathbb Q$. But this implies that both $e^r$ and $e^{e^r}$ are transcendental numbers. $\Box$

Bjørn Kjos-Hanssen
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  • So you shifted the problem to another unsaved problem ! – A. PI Oct 08 '17 at 09:33
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    @A.MONNET Unfortunately, problems like this one are, given tools available today, completely hopeless. This answer at least shows that your numbers are irrational conditionally on a widely believed conjecture. – Wojowu Oct 08 '17 at 10:14
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    Note: $k=1$ is also the Lindemann-Weierstrass theorem. – japh Oct 08 '17 at 12:59