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 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.

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




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


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


 * Here, the last sum is a geometric series.

If $$x=1$$, then $$b=1$$, which would imply $$e=\frac{a}{b}=a$$ is an integer. So $$0<x<1$$ and since there does not exist a positive integer less than 1, we have reached a contradiction, and so e must be irrational. Q.E.D.