# Irrationality of Logarithm

## 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

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