Most recent change of ProofThatEIsIrrational

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HEADERS_END
The number /e/ also known as Euler's Number, is irrational.
This page could be replaced by a reference to a good proof
elsewhere.
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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!}+...)

* EQN:RHS:\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 we get
from both sides and this expression has to be an integer:

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

Now on the RHS 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 RHS is bigger than 0, but less than
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 equation (*) above.
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.