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:
\(\ds \log_b a\) | \(=\) | \(\ds \dfrac p q\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds b^{\frac p q}\) | \(=\) | \(\ds a\) | Definition of Real General Logarithm | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sqrt [q] {\paren {b^p} }\) | \(=\) | \(\ds a\) | Definition of Rational Power | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds b^p\) | \(=\) | \(\ds a^q\) | Definition of Root of Number |
which contradicts the initial assumption:
- $\nexists m, n \in \N: a^m = b^n$
Hence the result by Proof by Contradiction.
$\blacksquare$