Proof that e is irrational

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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.

[edit] 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 contradiction, 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.

[edit] Proof

Suppose e = a/b, the definition of a rational number, for some positive integers a and b (WLOG, b>1; the expression is not assumed to be reduced). 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

$x\,$	$= b\,!\left(e - \sum_{n = 0}^{b} \frac{1}{n!}\right)$
	$= b\,!\left(\frac{a}{b} - \sum_{n = 0}^{b} \frac{1}{n!}\right)$
	$= a(b - 1)! - \sum_{n = 0}^{b} \frac{b!}{n!}$
	$= a(b - 1)! - \sum_{n = 0}^{b} \frac{1\cdot2\cdot3\cdots(n-1)(n)(n+1)\cdots(b-1)(b)}{1\cdot2\cdot3\cdots(n-1)(n)}$
	$= a(b - 1)! - \sum_{n = 0}^{b}(n+1)(n+2)\cdots(b-1)(b).$

The last term in the final sum is

b! / b! = 1

(i.e. it can be interpreted as an empty product). Clearly, however, every term is an integer.

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

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