Famous Theorems of Mathematics/e is irrational

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The series representation of Euler's number e

e = \sum_{n = 0}^{\infty} \frac{1}{n!}\!

can be used to prove that e is irrational. Of the many representations of e, this is the Taylor series for the exponential function ey evaluated at y = 1.

Summary of the proofEdit

This is a proof by contradiction. Initially e is assumed to be a rational number of the form a/b. We then analyze a blown-up difference x of the series representing e and its strictly smaller bth partial sum, which approximates the limiting value e. By choosing the magnifying factor to be b!, the fraction a/b and the bth partial sum are turned into integers, hence x must be a positive integer. However, the fast convergence of the series representation implies that the magnified approximation error x is still strictly smaller than 1. From this contradiction we deduce that e is irrational.


Suppose that e is a rational number. Then there exist positive integers a and b such that e = a/b.

Define the number

\ x = b!\,\biggl(e - \sum_{n = 0}^{b} \frac{1}{n!}\biggr)\!

To see that x is an integer, substitute e = a/b into this definition to obtain

x = b!\,\biggl(\frac{a}{b} - \sum_{n = 0}^{b} \frac{1}{n!}\biggr)
= a(b - 1)! - \sum_{n = 0}^{b} \frac{b!}{n!}\,.

The first term is an integer, and every fraction in the sum is an integer since nb for each term. Therefore x is an integer.

We now prove that 0 < x < 1. First, insert the above series representation of e into the definition of x to obtain

x = \sum_{n = b+1}^{\infty} \frac{b!}{n!}>0\,.\!

For all terms with nb + 1 we have the upper estimate


which is even strict for every nb + 2. Changing the index of summation to k = nb and using the formula for the infinite geometric series, we obtain

=\sum_{n = b+1}^{\infty} \frac{b!}{n!}
< \sum_{k=1}^\infty\frac1{(b+1)^k}
= \frac{1}{b}
\le 1.

Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so e must be irrational.