Irrationality of Logarithm

From ProofWiki
Jump to navigation Jump to search


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.


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


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

where $p \perp q$.


\(\displaystyle \log_b a\) \(=\) \(\displaystyle \dfrac p q\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle b^{\frac p q}\) \(=\) \(\displaystyle a\) 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.