# Proof by contradiction

I want to prove that, for all algebraic $n\neq0\text{ or }1$, $\ln n$ is transcendental. Here's how I tried to do it:

$n$ is an algebraic number, $n\neq0\text{ or }1$.

Assume $x$ is algebraic.

$\ln n=x$

$e^x=n$

By the Lindemann-Weierstrass theorem, $n$ must be transcendental. However, it was established at the beginning that $n$ is algebraic.

Our assumption must then be wrong. $\ln n=x$ is therefore transcendental.

I've never done proof by contradiction before, or any proofs for that matter. Did I do this right, and is there anything that needs changing?

• Wow. This is really not a good topic for the first proof that you ever look at. How do you even know what the words in your post mean if you've never done a proof? – DanielV Mar 16 '15 at 0:03
• @DanielV Well, I know what algebraic and transcendental mean, and I looked up the Lindemann-Weierstrass theorem on Wikipedia and read what it said (I don't understand how it works, but I understand what it says at the end). – user3932000 Mar 16 '15 at 0:16
• I'm honestly fascinated. How did you come to wanting to prove the irrationality of $\ln n$ but never have proven anything before? It would be like me saying "I'd like to learn to dance like Michael Jackson, but I never learned to walk". I'm not trying to be rude, but it's quite a remarkable thing to say. – DanielV Mar 16 '15 at 0:22
• @DanielV I asked this question before whether $\ln n$ was transcendental for all rational $n>1$, and the answer said it was by the Lindemann-Weierstrass theorem. I read the Wikipedia page on it but didn't find anything mentioning $\ln$, so I wanted to see if the answer was right myself. – user3932000 Mar 16 '15 at 0:25

Let $n$ be an algebraic number, $n\neq 0$ or 1. Let $x=\ln n$. Suppose $x$ were algebraic. Then by the Lindemann-Weierstrass theorem, $n=e^x$ would be transcendental. But this would contradict that $n$ is algebraic. Therefore, $x=\ln n$ must be transcendental.