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.

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

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

We will show that x is a positive integer less than 1, and this contradiction will establish the irrationality of e.

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

$= a(b - 1)! - \sum_{n = 0}^{b}b(b-1)\cdots(n+1)$

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

$x = b\,!\sum_{n = b+1}^{\infty} \frac{1}{n!} \mbox{ and so}$

$0 < x = \frac{1}{b+1} + \frac{1}{(b+1)(b+2)} + \frac{1}{(b+1)(b+2)(b+3)} + \cdots$

$< \frac{1}{b+1} + \frac{1}{(b+1)^2} + \frac{1}{(b+1)^3} + \cdots = \frac{1}{b} \le 1$

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.

See also: Proof that e is irrational, E (mathematical constant), Empty product, Geometric series, Irrational number, Mathematics, Q.E.D.