If $x$ is a positive rational number, but not an integer, then can $x^{x^{x^x}}$ be a rational number ?

We can prove that if $x$ is a positive rational number but not an integer, then $x^x$ can not be rational:

Denote $x=\dfrac{b}{a},(a,b)=1,x^x=\dfrac{d}{c},(c,d)=1,$ $$\left(\dfrac{b}{a}\right)^\dfrac{b}{a}=\dfrac{d}{c} \hspace{12pt}\Rightarrow \hspace{12pt}\left(\dfrac{b}{a}\right)^b=\left(\dfrac{d}{c}\right)^a \hspace{12pt}\Rightarrow \hspace{12pt}b^b c^a=d^a a^b$$ Since $(a,b)=1,(c,d)=1,$ we have $c^a\mid a^b$ and $a^b\mid c^a$, hence $a^b=c^a.$ Since $(a,b)=1$, $a^b$ must be an $ab$-th power of an integer, assume that $a^b=t^{ab},$ then $a=t^a,$ where $t$ is a positive integer, this is impossible if $t>1,$ so we get $t=1,a=1$, hence $x$ is an integer.

Then from Gelfond–Schneider theorem , we can prove that if $x$ is a positive rational number but not an integer, then $x^{x^x}$ can not be rational. In fact, it can not be an algebraic number, because both $x$ and $x^x$ are algebraic numbers and $x^x$ is not a rational number.

  • Can we prove that $x^{x^{x^x}}$ is irrational?
  • Can $x^{x^{\dots (n-th)^{\dots x}}}~(n>1)$ be rational?
  • 2
    $\begingroup$ Closely related: This question and Schanuel's conjecture. $\endgroup$ Jun 27, 2013 at 14:34
  • $\begingroup$ Sorry but I find writing "can not" confusing. For me, "can not V"="having the possibility of not being V" whereas "cannot V"="not having the possibility of V". Am I wrong? I don't edit since I am not a native English speaker. Sorry for not being helpful. $\endgroup$
    – Taladris
    Jul 9, 2013 at 0:50
  • $\begingroup$ @Taladris Can $x^{x^{x^x}}$ be rational ? = Is there an $x$ s.t. $x^{x^{x^x}}$ is rational ? $\endgroup$
    – lsr314
    Jul 9, 2013 at 1:33
  • 4
    $\begingroup$ I can't post an answer here for lack of reputation, but I've written one on Reddit assuming Schanuel's conjecture: see reddit.com/r/math/comments/1hz56o/… $\endgroup$
    – Gro-Tsen
    Jul 10, 2013 at 11:38
  • 2
    $\begingroup$ @Gro-Tsen Please post your answer here, I have lifted the protection. $\endgroup$ Jul 10, 2013 at 15:54

1 Answer 1


Let $x_1 = x$ and by induction $x_{n+1} = x^{x_n}$: so $x_1 = x$ is rational by hypothesis, $x_2 = x^x$ is algebraic irrational, $x_3 = x^{x^x}$ is transcendental by the Gelfond-Schneider theorem, and the question is to prove that $x_4, x_5,\ldots$ are transcendental (or at least, irrational).

I will assume Schanuel's conjecture and use it to prove by induction on $n\geq 2$ that $x_3,x_4,\ldots,x_n$ are algebraically independent (and, in particular, transcendental). For $n=2$ there is nothing to prove: so let me assume the statement true for $n$ and prove it for $n+1$.

Since $x_2$ is irrational, $x_1$ and $x_2$ are linearly independent over $\mathbb{Q}$. The induction hypothesis implies that $1,x_3,\ldots,x_n$ are linearly independent over $\mathbb{Q}^{\mathrm{alg}}$ (the algebraics), so in particular $x_1,\ldots,x_n$ are linearly independent over $\mathbb{Q}$, and, of course, this implies that $x_1\cdot\log(x),\ldots,x_n\cdot\log(x)$ are also such.

Now Schanuel's conjecture then implies that among the $2n$ quantities $x_1 \log(x),\ldots,x_n \log(x), x_2,\ldots,x_{n+1}$ at least $n$ are algebraically independent. Of course, we can remove $x_2$ from that list since it is algebraic, we can similarly replace both $x_1 \log(x)$ and $x_2 \log(x)$ by simply $\log(x)$: so among $\log(x),x_3,\ldots,x_{n+1},x_3 \log(x),\ldots,x_n \log(x)$ at least $n$ are algebraically independent. But (for any $i$) this independent set cannot contain all three of $x_i$, $\log(x)$ and $x_i\log(x)$, and if it contains two of them then we can choose any two (namely, $x_i$ and $\log(x)$): so that, in fact, the $n$ quantities $\log(x),x_3,\ldots,x_n,x_{n+1}$ are algebraically independent, which concludes the induction step (and moreover shows that $\log(x)$ is also independent with the rest).


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