Proof of Irrationality of e using Diophantine Equations I was trying to prove that e is irrational without using the typical series expansion, so starting off  $e = a/b $  Take the natural log so  $1 = \ln(a/b)$ Then $1 = \ln(a)-\ln(b)$ So unless I did something horribly wrong showing the irrationality of $e$ is the same as showing that the equation $c = \ln(a)-\ln(b)$ or $1 = \ln(a) - \ln(b)$ (whichever one is easiest) has no solutions amongst the natural numbers. I feel like this would probably be easiest with infinite descent, but I'm in high school so my understanding of infinite descent is pretty hazy. If any of you can provide a proof of that, that would be awesome.
EDIT: What I mean by "typical series expansion" is Fourier's proof http://en.wikipedia.org/wiki/Proof_that_e_is_irrational#Proof

There is the classic proof by Fourier: For the sake of contradiction, suppose $e = \frac{a}{b}$ for $a, b \in \mathbb{N}$ in lowest terms. As $e$ isn't an integer, $b > 1$. Now:
$$
e = \sum_{n \ge 0} \frac{1}{n!} \\
b! e = \sum_{n \ge 0} \frac{b!}{n!} = \sum_{0 \le n \le b} \frac{b!}{n!} + \sum_{n \ge b} \frac{b!}{n!}
$$
The first sum is an integer, we'll show that the second sum, call it $S$, satisfies $0 < S < 1$, and so can't be an integer. Firstly, $S$ is a sum of positive numbers, so $0 < S$. Now:
$$
S = \sum_{n > b} \frac{1}{(b + 1) (b + 2) \ldots n}
  = \sum_{k \ge 1} \frac{1}{(b + 1)(b + 2) \ldots (b + k)}
  < \sum_{k \ge 1} b^{-k}
$$
The inequality follows because each term in the last sum is larger than the corresponding term of the second one; the second one is just restating the first.
But:
$$
\sum_{k \ge 1} b^{-k} = b^{-1} \sum_{k \ge 0} b^{-k} = b^{-1} \cdot \frac{1}{1 - b^{-1}} = \frac{1}{b - 1} \le 1
$$
This is just an infinite geometric series, the last inequality follows from $b > 1$. So $0 < S < 1$, as promised; and $e$ isn't rational.
