# e is irrational

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

 $e=\sum_{k=0}^{\infty}\frac{1}{k!}.$ (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):

 $b!e=b!\sum_{k=0}^{\infty}\frac{1}{k!}.$

This can also be written:

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

The first sum is obviously a natural number, and thus

 $\sum_{k=b+1}^{\infty}\frac{b!}{k!}$

must also be . Now we see:

 $\sum_{k=b+1}^{\infty}\frac{b!}{k!}=\frac{1}{b+1}+\frac{1}{(b+1)(b+2)}+...<\sum% _{k=1}^{\infty}\left(\frac{1}{b+1}\right)^{k}=\frac{1}{b}.$

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

 $0<\sum_{k=b+1}^{\infty}\frac{b!}{k!}<1.$

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 EIsIrrational 2013-03-22 12:33:02 2013-03-22 12:33:02 mathwizard (128) mathwizard (128) 13 mathwizard (128) Theorem msc 11J82 msc 11J72 ErIsIrrationalForRinmathbbQsetminus0 NaturalLogBase