Proof that e is irrational

In mathematics, the series expansion of the number e

$e=\sum _{n=0}^{\infty }{\frac {1}{n!}}$

can be used to prove that e is irrational.

Summary of the proof:

This will be a proof by contradiction. Initially e will be assumed to be rational. The proof is constructed to show that this assumption leads to a logical impossibility. This logical impossibility, or contradition, implies that the underlying assumption is false, meaning that e must not be rational. Since any number that is not rational is by definition irrational, the proof is complete.

Proof:

Suppose e = a/b, for some positive integers a and b. Construct the number

$x=b\,!\left(e-\sum _{n=0}^{b}{\frac {1}{n!}}\right)$

We will first show that x is an integer, then show that x is less than 1 and positive. The contradiction will establish the irrationality of e.

• To see that x is an integer, note that

$x\,$	$=b\,!\left(e-\sum _{n=0}^{b}{\frac {1}{n!}}\right)$
	$=b\,!\left({\frac {a}{b}}-\sum _{n=0}^{b}{\frac {1}{n!}}\right)$
	$=a(b-1)!-\sum _{n=0}^{b}{\frac {b!}{n!}}$
	$=a(b-1)!-\sum _{n=0}^{b}b(b-1)\cdots (n+1)$

Here, the last term in the final sum is to be interpreted as an empty product.

• To see that x is a positive number less than 1, note that

$x\,$	$=b\,!\sum _{n=b+1}^{\infty }{\frac {1}{n!}}{\mbox{ and so }}0<x.{\mbox{ But }}x$
	$={\frac {1}{b+1}}+{\frac {1}{(b+1)(b+2)}}+{\frac {1}{(b+1)(b+2)(b+3)}}+\cdots$
	$<{\frac {1}{b+1}}+{\frac {1}{(b+1)^{2}}}+{\frac {1}{(b+1)^{3}}}+\cdots$
	$={\frac {1}{b}}$
	$\leq 1$

Here, the last sum is a geometric series.

Since there does not exist a positive integer less than 1, we have reached a contradiction, and so e must be irrational. This completes the proof.

Q.E.D.