# Does my proof of the irrationality of $\ln(2)$ hold up?

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?

• A lot of silent work is being done by "$e^m$ cannot be an integer . . ." and thereabouts. I agree that this is a nontrivial result. Commented Mar 1, 2021 at 18:23
• How do you know $log(a)$ is only an integer for $a=1$? Commented Mar 1, 2021 at 18:25
• How do you know that $\ln(a)$ is not an integer for $a \neq 1$? Commented Mar 1, 2021 at 18:26
• You want to use the theorem that $e$ is transcendental. Commented Mar 1, 2021 at 18:34
• You said it, but you didn't justify that claim. Commented Mar 1, 2021 at 18:35

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.
• 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 $px = q$ Commented Mar 1, 2021 at 19:12
• @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.