Most recent change of ProofThatEIsIrrational

Edit made on January 08, 2013 by ColinWright at 15:04:57

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The number /e/ also known as Euler's Number, is irrational.
This page could be replaced by a reference to a good proof
This will be a proof by contradiction.

The number /e/ can be defined as:

* EQN:e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...

Let's suppose (to get our proof by contradiction started) that /e/ is a rational number, so we write EQN:e=a/b

Now we multiply both sides by /b!/

* EQN:LHS:\quad{\quad}b!e=b!(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...) LHS: EQN:\quad{\quad}b!e=b!(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...)

* EQN:RHS:\quad{\quad}a(b-1)! RHS: EQN:\quad{\quad}a(b-1)!

Many of the terms on the LHS simplify to integers, up to term EQN:\frac{1}{(b+1)!} from which point
we start to get left over denominators. We can subtract the first /b/ terms (which are all integers)
from both sides and this expression has to be an integer:

* EQN:\frac{1}{(b+1)}+\frac{1}{(b+1)(b+2)}+\frac{1}{(b+1)(b+2)(b+3)}+...\quad\quad(*)

Now, we can replace all the /(b+c_i)/ values with /(b+1)/ which makes them all slightly bigger
(reducing the denominator increases the value) and now we see that the expression is bigger than 0,
but less than

* EQN:b^{-1}+b^{-2}+b^{-3}+b^{-4}+...

which is a geometric series that converges to EQN:\frac{1}{b-1} which is greater than 0, and less than 1.

Hence EQN:\frac{1}{(b+1)}+\frac{1}{(b+1)(b+2)}+\frac{1}{(b+1)(b+2)(b+3)}+... is strictly between 0 and 1.

This contradicts the requirement that expression (*) must be an integer.

Thus we cannot have EQN:e=a/b and so /e/ is not a rational number - It's an irrational number.