# Famous Theorems of Mathematics/e is irrational

The series representation of Euler's number *e*

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[edit]

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 *b*th partial sum, which approximates the limiting value *e*. By choosing the magnifying factor to be *b*!, the fraction *a*/*b* and the *b*th 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[edit]

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

Define the number

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

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

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

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

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