Famous Theorems of Mathematics/e is irrational

The series representation of Euler's number e

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

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.

ProofEdit

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

Define the number

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

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

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

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

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

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

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

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

${\displaystyle x=\sum _{n=b+1}^{\infty }{\frac {b!}{n!}}<\sum _{k=1}^{\infty }{\frac {1}{(b+1)^{k}}}={\frac {1}{b+1}}{\biggl (}{\frac {1}{1-{\frac {1}{b+1}}}}{\biggr )}={\frac {1}{b}}\leq 1.}$

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