Proving Transcendence

Date: 05/15/99 at 20:47:04
From: Tony Johnson
Subject: How does one prove transcendence in a number?

I've been asked by my mathematics teacher to present a talk about 
transcendental numbers in class. I've been to nearly every site that 
is listed by my search engine regarding transcendental numbers. They 
all seem to be telling me which numbers are transcendental; however, 
none has shown how to prove transcendence in a number. Can you help 
me please?

Date: 05/17/99 at 13:14:22
From: Doctor Wilkinson
Subject: Re: How does one prove transcendence in a number?

Hi, Tony!  In general it is a very difficult task to prove any 
particular number is transcendental. In fact, not a single example was 
known until 1851, when Liouville exhibited a whole class of them, and 
it was not until 1873, when Hermite proved the transcendence of e, 
that a number not specifically constructed for the purpose was proved 
to be transcendental.

Although the details are very difficult, all transcendence proofs that 
I know of are based an a very simple idea. If you want to show that a 
number is transcendental, you first assume it isn't (indirect proof or 
proof by contradiction). You then use this assumption to construct a 
number that

   (a) is a whole number;

and

   (b) is greater than 0 and less than 1.

Since there is no such number, this is a contradiction and the 
original assumption must have been false.

For an example of how such a proof goes, let's prove that e is 
irrational. We have the formula for e:

   e = 1 + 1/1! + 1/2! + 1/3! +...

Let's suppose e is rational, say e = a/b, where a and b are whole 
numbers. Multiply the equation for e by b!.  This gives

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

Now the left side is an integer, and so are the terms on the right 
side up to b!/b!. So the difference beween the left side and those 
terms on the right side is an integer, and it's greater than zero, 
because it's just the sum of the remaining terms on the right. The sum 
of those terms, however, is

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

and this is less than

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

which is a geometric series with sum 1/b, which is less than 1. So 
here we have a number that is an integer, and is greater than 0 and 
less than 1, which is impossible.

You might want to fill in all the details of this proof and present it 
as part of your talk.  Transcendence proofs are much more difficult, 
but they use the same basic idea.

Good luck with your talk!

- Doctor Wilkinson, The Math Forum
  http://mathforum.org/dr.math/