Famous Theorems of Mathematics/e is irrational

The series representation of Euler's number e

$e=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}$

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

Summary of the proof

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.

Proof

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

Define the number

$x=b!(e-\sum _{n=0}^b\frac{1}{n!})$

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

$x=b!(\frac{a}{b}-\sum _{n=0}^b\frac{1}{n!})=a(b-1)!-\sum _{n=0}^b\frac{b!}{n!}.$

The first term is an integer, and every fraction in the sum is an integer since n≤b 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

$x=\sum _{n=b+1}^{\mathrm{\infty }}\frac{b!}{n!}>0.$

For all terms with n ≥ b + 1 we have the upper estimate

$\frac{b!}{n!}=\frac{1}{(b+1)(b+2)\cdots (b+(n-b))}\le \frac{1}{(b+1{)}^{n-b}},$

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

$x=\sum _{n=b+1}^{\mathrm{\infty }}\frac{b!}{n!}<\sum _{k=1}^{\mathrm{\infty }}\frac{1}{(b+1{)}^k}=\frac{1}{b+1}\left(\frac{1}{1-\frac{1}{b+1}}\right)=\frac{1}{b}\le 1.$

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