e is irrational

From the Taylor series for $e^x$ we know the following equation:

$$e=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}.$$ (1)

Now let us assume that $e$ is rational. This would there are two natural numbers $a$ and $b$, such that:

$$e=\frac{a}{b}.$$

This yields:

$$b!e\in \mathbb{N}.$$

Now we can write $e$ using (1):

$$b!e=b!\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}.$$

This can also be written:

$$b!e=\sum _{k=0}^b\frac{b!}{k!}+\sum _{k=b+1}^{\mathrm{\infty }}\frac{b!}{k!}.$$

The first sum is obviously a natural number, and thus

$$\sum _{k=b+1}^{\mathrm{\infty }}\frac{b!}{k!}$$

must also be . Now we see:

Since $\frac{1}{b}\le 1$ we conclude:

We have also seen that this is an integer, but there is no integer between 0 and 1. So there cannot exist two natural numbers $a$ and $b$ such that $e=\frac{a}{b}$, so $e$ is irrational.

Title	e is irrational
Canonical name	EIsIrrational
Date of creation	2013-03-22 12:33:02
Last modified on	2013-03-22 12:33:02
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	13
Author	mathwizard (128)
Entry type	Theorem
Classification	msc 11J82
Classification	msc 11J72
Related topic	ErIsIrrationalForRinmathbbQsetminus0
Related topic	NaturalLogBase