Proof that e is irrational

219,204pages on
this wiki
Add New Page
Discuss this page0 Share

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.


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<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.

See Also

Ad blocker interference detected!

Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.

Also on Fandom

Random wikia