# Positive Rational Number as Power of Number with Power of Itself

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## Contents

## Theorem

Every positive rational number can be written either as:

- $a^{a^a}$ for some irrational number $a$

or as:

- $n^{n^n}$ for some natural number $n$.

## Proof

\(\displaystyle \map {\frac \d {\d x} } {x^{x^x} }\) | \(=\) | \(\displaystyle \map {\frac \d {\d x} } {e^{x^x \ln x} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e^{x^x \ln x} \map {\frac \d {\d x} } {x^x \ln x}\) | Chain Rule for Derivatives | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{x^x} \paren {x^x \map {\frac \d {\d x} } {\ln x} + \map {\frac \d {\d x} } {x^x} \ln x}\) | Product Rule for Derivatives | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{x^x} \paren {x^x \frac 1 x + x^x \ln x \paren {\ln x + 1} }\) | Derivative of Natural Logarithm Function, Derivative of $x^x$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{x^x + x} \paren {\frac 1 x + \ln x \paren {\ln x + 1} }\) |

For $\ln x \le 0$ and $\ln x \ge -1$, the above is positive.

For $-1 < \ln x < 0$, we have $e^{-1} < x < 1$.

\(\displaystyle x^{x^x + x} \paren {\frac 1 x + \ln x \paren {\ln x + 1} }\) | \(>\) | \(\displaystyle x^{x^x + x} \paren {\frac 1 1 + \paren {-1} \paren {0 + 1} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) |

so we have $\dfrac \d {\d x} x^{x^x} > 0$ for every $x > 0$.

We also have:

- $\displaystyle \lim_{x \mathop \to 0^+} x^{x^x} = \lim_{x \mathop \to 0^+} x^{\lim_{x \mathop \to 0^+} x^x} = 0^1 = 0$

Thus $x^{x^x}: \R_{>0} \to \R_{>0}$ is bijective.

For each $y > 0$, we can find some $x > 0$ such that $y = x^{x^x}$.

The result above is therefore equivalent to:

- $a^{a^a}$ is irrational if $a$ is rational and not a natural number.

By Rational Number as Power of Number with Itself, as long as $a$ is not a natural number, $a^a$ is irrational.

Therefore if $a$ is not a natural number, by Gelfond-Schneider Theorem, $a^{a^a}$ is transcendental, so it is also irrational.

This proves the result.

$\blacksquare$

## Also see

## Sources

- March 2012: J. Marshall Ash and Yiren Tan:
*A rational number of the form $a^a$ with $a$ irrational*(*The Mathematical Gazette***Vol. 96**: pp. 106 – 109)