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.

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
 $x\,$ $= b\,!\left(e - \sum_{n = 0}^{b} \frac{1}{n!}\right)$ $= b\,!\left(\frac{a}{b} - \sum_{n = 0}^{b} \frac{1}{n!}\right)$ $= a(b - 1)! - \sum_{n = 0}^{b} \frac{b!}{n!}$ $= a(b - 1)! - \sum_{n = 0}^{b} \frac{1\cdot2\cdot3\cdots(n-1)(n)(n+1)\cdots(b-1)(b)}{1\cdot2\cdot3\cdots(n-1)(n)}$ $= a(b - 1)! - \sum_{n = 0}^{b}(n+1)(n+2)\cdots(b-1)(b).$
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: $x$ $= \frac{1}{b+1} + \frac{1}{(b+1)(b+2)} + \frac{1}{(b+1)(b+2)(b+3)} + \cdots$ $< \frac{1}{b+1} + \frac{1}{(b+1)^2} + \frac{1}{(b+1)^3} + \cdots$ $= \frac{1}{b}$ $\le 1.$
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 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.