# Existence of Rational Powers of Irrational Numbers/Proof 3

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## Contents

## Theorem

There exist irrational numbers $a$ and $b$ such that $a^b$ is rational.

## Proof

Consider the equation:

- $\left({\sqrt 2}\right)^{\log_{\sqrt 2} 3} = 3$

We have that $\sqrt 2$ is irrational and $3$ is (trivially) rational.

It remains to be proved $\log_{\sqrt 2} 3$ is irrational.

We have:

\(\displaystyle \log_{\sqrt 2} 3\) | \(=\) | \(\displaystyle \frac{\log_2 3}{\log_2 \sqrt 2}\) | Change of Base of Logarithm | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac{\log_2 3}{1/2}\) | Logarithms of Powers: $\sqrt 2 = 2^{1/2}$ and $\log_2 2 = 1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 \, \log_2 3\) |

From Irrationality of Logarithm, $\log_2 3$ is irrational.

Therefore $2 \, \log_2 3$ is irrational.

Thus there exist two irrational numbers $a = \sqrt 2$ and $b = \log_{\sqrt 2} 3$ such that $a^b$ is rational.

$\blacksquare$

## Also see

- Positive Rational Number as Power of Number with Power of Itself for an interesting related result.

## Sources

- November 2008: Nick Lord:
*Maths bite: irrational powers of irrational numbers can be rational*(*The Mathematical Gazette***Vol. 92**: p. 534) www.jstor.org/stable/27821850