Irrationality of Logarithm

Theorem
Let $a,b \in \N_{>0}$ such that neither $\nexists n \in \N: a = b^n$ nor $\nexists n \in \N: a^n = b$

Then $\log_b a$ is irrational.

Proof
Let $p,q \in \N_{>0}$ such that $p \perp q$.

Suppose $\log_b a$ is rational, then $\exists p,q: \log_b a = \dfrac p q$.

Note that $\dfrac p q$ may not be $0$, as the only $a$ that satisfies $\log_b a = 0$ is $a = 1 = b^0$.

There are three cases to consider, representing the permitted relationships between $a$ and $b$:
 * 1. Neither of $a$ and $b$ is a multiple of the other
 * 2. $a$ is a power of $b$, but also multiplied by a number not a power of $b$
 * 3. $b$ is a power of $a$, but also multiplied by a number not a power of $a$

Case 1

 * $a \perp b$

But $a \perp b$, so different prime factorizations are representing the same number.

This contradicts the Fundamental Theorem of Arithmetic.

Thus, the supposition that $\log_b a$ is rational is false.

Case 2

 * $a = n b^m b$, $n \perp b$

But from case 1, $\log_b n$ is irrational.

Case 3

 * $n a^m a = b$, $n \perp a$

But from case 2, $\log_b \left({n a^m a}\right)$ is irrational.