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### The Irrationality of e

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Date: 09/04/98 at 00:17:10
Subject: How is it possible to prove a number is transcendental?

Hi, I was wondering if you could help me.

numbers. I know that a transcendental number is a number such that it
can't be put in normal algeabraic forms, such as pi. However, I was
wondering how this is proven. Thanks in advance for any response.

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Date: 09/04/98 at 11:57:49
From: Doctor Wilkinson
Subject: Re: How is it possible to prove a number is transcendental?

In spite of the fact that most numbers are transcendental (a fact
proved by Cantor), proving that any particular number is transcendental
is usually quite difficult. There is a large class of such numbers
known as Liouville numbers, after the French mathematician Liouville
who defined them and showed they were transcendental. A typical
Liouville number is:

0.10100100000010000000000000000000000001...

Here the successive groups of zeros are of length 1!, 2!, 3!, 4!, and
so on.

As you mentioned, we know that pi is transcendental, but this was
proved only in 1882 after being studied for more than 2000 years. The
number e is also transcendental, and so is a^b, if a and b are
algebraic and b is irrational (like 2 to the square root of 2 power,
for example). But all these theorems are very hard.

The basic idea, however, in all these proofs is the same: you assume
that the number in question is algebraic, and you arrive at a
the assumption implies the existence of an integer between 0 and 1.
I do not know any of these proofs, although I have studied the proof
for e, but I can give you some of the idea by showing that e is
irrational.

e is defined as 1 + 1/1! + 1/2! + 1/3! + 1/4! + ....

Suppose e is rational, say e = a/b, where a and b are integers. Then
b! e is certainly an integer. So multiply the equation for e by b!.
This gives us:

b! e = b! + b!/1! +... + b!/b! + b!/(b+1)! + b!/(b+2)! + ...

Now all the terms up to b!/b! on the right are integers, and the left
side is an integer, so the sum of the rest of the terms on the right
must be an integer. This is:

1/(b+1) + 1/(b+1)(b+2) + 1/(b+1)(b+2)(b+3) + ...

Now if we replace b+2, b+3, ... by b+1 we make the terms on the right
bigger, so the right side is less than:

1/(b+1) + 1/(b+1)^2 + 1/(b+1)^3 + ...

which is a geomtric series. Its sum is:

1/(b+1) (1 - 1/(b+1)) = 1/b.

Note that b is certainly bigger than 1, since e is not an integer, and
we have arrived at the promised contradiction: an integer between 0
and 1.

- Doctor Wilkinson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Associated Topics:
High School Transcendental Numbers

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