Proof that e is irrational

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In mathematics, the series expansion of the number e

[e = \sum_^ \frac]

can be used to prove that e is irrational.

Summary of the proof:

This will be a proof by contradiction. Initially e will be assumed to be rational. The proof is constructed to show that this assumption leads to a logical impossibility. This logical impossibility, or contradiction, implies that the underlying assumption is false, meaning that e must not be rational. Since any number that is not rational is by definition irrational, the proof is complete.

Proof:

Suppose e = a/b, for some positive integers a and b. Construct the number

[x = b\,!\left(e - \sum_^ \frac\right)]

We will first show that x is an integer, then show that x is less than 1 and positive. The contradiction will establish the irrationality of e.

To see that x is an integer, note that

[x\, ]	[= b\,!\left(e - \sum_^ \frac\right) ]
	[= b\,!\left(\frac - \sum_^ \frac\right)]
	[= a(b - 1)! - \sum_^ \frac]
	[= a(b - 1)! - \sum_^ \frac]
	[= a(b - 1)! - \sum_^(n+1)(n+2)\cdots(b-1)(b)]

The last term in the final sum is [b!/b! = 1] (i.e. it can to be interpreted as an empty product). Clearly, however, every term is an integer.

To see that x is a positive number less than 1, note that

[x\,]	[ = b\,!\sum_^ \frac] and so [0 < x]. But:
[x]	[= \frac + \frac + \frac + \cdots ]
	[< \frac + \frac + \frac + \cdots]
	[= \frac ]
	[\le 1]

Here, the last sum is a geometric series.

Since there does not exist a positive integer less than 1, we have reached a contradiction, and so e must be irrational. Q.E.D.

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