# Is $\ln(\ln(n))$ irrational for any integer $n>1$?

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.

• 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. Dec 18, 2016 at 16:24
• 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. Dec 18, 2016 at 17:12
• 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). Dec 18, 2016 at 17:32
• 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". Dec 18, 2016 at 21:10
• 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. Dec 19, 2016 at 13:49

@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. $\Box$

• So you shifted the problem to another unsaved problem ! Oct 8, 2017 at 9:33
• @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. Oct 8, 2017 at 10:14
• Note: $k=1$ is also the Lindemann-Weierstrass theorem.
– japh
Oct 8, 2017 at 12:59