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Rational Number Equation Summation Integer Irrational Natural Number Taylor Series

1	Summation
1	Natural Number
1	Integer
1	Rational Number
1	Irrational
1	Equation
2	Taylor Series

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E Is Irrational Proof

From the Taylor series for we know the following equation:

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

(1)

Now let us assume that

is rational. This would mean there are two natural numbers

and

, such that:

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

This yields:

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

Now we can write

using (1):

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

This can also be written:

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

The first sum is obviously a natural number, and thus

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

must also be natural. Now we see:

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

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

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

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

and

such that $e=\frac{a}{b}$ , so

is irrational.