# Irrationality of Logarithm

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## Theorem

Let $a, b \in \N_{>0}$ such that both $\nexists m, n \in \N_{>0}: a^m = b^n$.

Then $\log_b a$ is irrational.

## Proof

Aiming for a contradiction, suppose $\log_b a$ is rational.

Then:

- $\exists p, q \in \N_{>0} : \log_b a = \dfrac p q$

where $p \perp q$.

Then:

\(\displaystyle \log_b a\) | \(=\) | \(\displaystyle \dfrac p q\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle b^{\frac p q}\) | \(=\) | \(\displaystyle a\) | Definition of Real General Logarithm | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sqrt [q] {\paren {b^p} }\) | \(=\) | \(\displaystyle a\) | Definition of Rational Power | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle b^p\) | \(=\) | \(\displaystyle a^q\) | Definition of Root |

which contradicts the initial assumption:

- $\nexists m, n \in \N: a^m = b^n$

Hence the result by Proof by Contradiction.

$\blacksquare$