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
$\log_b a$ is rational.

Then:


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

where $p \perp q$.

Then:

which contradicts the initial assumption:
 * $\nexists m, n \in \N: a^m = b^n$

Hence the result by Proof by Contradiction.