In a previous post we proved that is irrational. In this post we prove the irrationality of e.
A proof of the irrationality of e must start by defining e. There are some different ways to do that. We’ll take e to be the unique positive number b such that . (This property itself can be proved from other ways to define e. See, for example, Fitzpatrick’s Advanced Calculus , Section 5.2.) We’ll also make the assumptions that (1) and (2) is an increasing function.
Under this definition of e, the Lagrange Remainder Theorem says that, given and , there exists such that
Taking , we obtain, for some ,
Now, suppose e is rational. Then there exist with . Therefore,
Multiplying by , this becomes
Since the above inequality is true for all , it is true for . For this value of n, is an integer, and .
This means that there is an integer in the interval . However, we know that there is no integer in the interval . Contradiction; thus e is irrational.
 Patrick M. Fitzpatrick, Advanced Calculus, American Mathematical Society, 2006.