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At first, I thought that proving the irrationality of $\ln(2)$ was so easy that it was trivial. However, someone in the comments of the $115$ vote answer at Can an irrational number raised to an irrational power be rational? said that "OTOH, proving that e and log2 are irrational is not trivial." (Ihf said this). Thus, I decided to give it a so to see how difficult it is.

First, we assume that $\ln(2)$ is rational and can thus be written as the fraction $\frac{m}{n}$ with integers $m,n$ and $m$ and $n$ in lowest terms, meaning that $\gcd(m,n)$ $=1$. So we have that $$\ln(2) = \frac{m}{n} \implies$$ $$2 = e^\frac{m}{n} \implies$$ $$2^n = e^m $$

Since $n$ is an integer $2^n$ is an integer as well. For $\frac{m}{n}$ to be a valid fraction, we need $m$ to be an integer as well. However, since $m$ must be an integer, $e^m$ cannot be an integer because it is only an integer when we have $e^{\ln(a)}$ for some natural number $a$, but we know that $\ln(a)$ can't be an integer, with exception of $\ln(1) = 0$, but $m$ isn't $0$ because that would make our fraction $0$. Thus $2^n$ is an integer, while $e^m$ isn't, which is a contradiction. Thus our initial assumption that $\ln(2)$ is rational and can be written in lowest terms as $\frac{m}{n}$ is false, meaning that $\ln(2)$ is irrational. This proof seemed too easy, what did I do wrong?

Some Guy
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The proof can be streamlined in the following way. Suppose that $\log 2=m/n$ for two coprime integers $m$ and $n$. Then, $$ e^{m/n}=2 \implies e^m=2^n \, . $$ Hence, $e$ would be a root of the polynomial $$ x^m-2^n=0 \, . $$ But this would imply that $e$ is algebraic, which is untrue. Hence, $\log 2$ is irrational.

Joe
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  • Thats circular reasoning, you used the fact that $e$ is transcedental to prove that it is irrational, but for $e$ to be transcedental, it had to be proven irrational first, otherwise if it were rational it would be the root of an equation $px = q$ – Some Guy Mar 01 '21 at 19:12
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    @SomeGuy: I don't understand what part of my proof you think is circular. I'm using the fact that $e$ is transcendental to show that $\log 2$ is irrational, not that $e$ is irrational. – Joe Mar 01 '21 at 19:15
  • Oh nevermind, you're good @Joe – Some Guy Mar 01 '21 at 19:20