Puede $x^{x^{x^x}}$ ser un número racional?

Question

Si $x$ is a positive rational number, but not an integer, then can $x^{x^{x^x}}$ ser un número racional ?

Podemos demostrar que si $x$ is a positive rational number but not an integer, then $x^x$ no puede ser racional:

Denotar $x=\dfrac{b}{a},(a,b)=1,x^x=\dfrac{d}{c},(c,d)=1,$ $$\left(\dfrac{b}{a}\right)^\dfrac{b}{a}=\dfrac{d}{c} \hspace{12pt}\Rightarrow \hspace{12pt}\left(\dfrac{b}{a}\right)^b=\left(\dfrac{d}{c}\right)^a \hspace{12pt}\Rightarrow \hspace{12pt}b^b c^a=d^a a^b$$ Desde $(a,b)=1,(c,d)=1,$ we have $c^a\mid a^b$ and $a^b\mid c^a$, hence $a^b=c^a.$ Desde $(a,b)=1$, $a^b$ 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 $x$ es un número entero.

Luego de Gelfond–Schneider teorema , podemos demostrar que si $x$ 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 $x^x$ no es un número racional.

• Podemos demostrar que $x^{x^{x^x}}$ es irracional?
• Puede $x^{x^{\dots (n-th)^{\dots x}}}~(n>1)$ ser racional?

Answer 1

Deje $x_1 = x$ 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 $x_4, x_5,\ldots$ son trascendentales (o al menos, irracional).

Voy a suponer que Schanuel de la conjetura y la usan para demostrar por inducción sobre $n\geq 2$ 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 $n+1$.

Desde $x_2$ is irrational, $x_1$ and $x_2$ are linearly independent over $\mathbb{Q}$. The induction hypothesis implies that ,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 $x_1\cdot\log(x),\ldots,x_n\cdot\log(x)$ son tales.

Ahora Schanuel la conjetura implica entonces que entre las n$ 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 $\log(x)$ es también independiente con el resto).