From: kovarik@mcmail.cis.mcmaster.ca (Zdislav V. Kovarik)
Newsgroups: sci.math
Subject: Re: Help! irrationals drive me crazy!
Date: 29 Jan 1995 03:36:07 -0500
Keywords: irrationals pi

In article <3g3hcd$rbf@doc.armltd.co.uk>,
David Seal <dseal@armltd.co.uk> wrote:
>djd@shell.portal.com (Duggan - Dieterly) writes:
>
>>	Why is pi, or any other irrational number, not expressable as a ratio 
>> 	of two rational numbers?

The following proof of the irrationality of pi was suggested in the book
IV of the series "Elements of Mathematics", called "Functions of one real
variable"  Chapter III, Sec. 2.4, Exercise 8. Author: Nicolas Bourbaki

For any positive number q and for every nonnegative integer n, 
consider the integrals

                         n   / pi                       
                        q    |              n           
                 A(n) = --   |  (x (pi - x))   sin x dx 
                        n!  /                           
                             0                          

These numbers are positive, and they  converge to 0 as n goes to 
infinity because their sum is finite.
(The sum is
                   pi
                  /
                  |     qx(pi-x)
                  |    e          sin x dx
                 /
                   0
since the series of integrands converges uniformly on [0, pi].)
We calculate directly for n = 0, 1, ...
(integrating by parts where needed)
                                                        2
A(0) = 2,   A(1) = 4q,   A(n+2) = (4n+6)qA(n+1) - (q*pi)  A(n)

Now we are ready to argue by contradiction: If  q  were a positive
integer such that q*pi would also be an integer, then all A(n)
would be positive integers (by induction from the recursion formula), 
but then they would not converge to 0, being always >=1.

Hence no such integer q exists, and pi is irrational.

Z. Kovarik.