Proof That PI is Irrational

The first proof of the irrationality of PI was found by Lambert in 
1770 and published by Legendre in his "Elements de Geometrie".  A
simpler proof, essentially due to Niven, goes like this:

For any integer n and real number r we can define a quantity A[n] by
the definite integral
        
                   / 1
          A[n]  =  |    (1 - x^2)^n cos(rx) dx
                   /
                  x=-1

If you integrate this by parts you find that the quantities A[n] for
n=2,3,4,...etc satisfy the recurrence relation

                    2n(2n-1) A[n-1]  -  4n(n-1) A[n-2]
          A[n]  =   ----------------------------------
                                   r^2

We can obviously express A[n-1] and A[n-2] in terms of lower members
of the recurrence, and so on, all the way back to A[0] and A[1].  The
result is
                       n!
           A[n]  =  -------- [ P(r) sin(r) - Q(r) cos(r) ]
                    r^(2n+1)
                       
where P(r) and Q(r) are polynomials in r of degree less than 2n-1 with
integer coefficients.  (See below if you're interested.)

Now let's assume that PI=a/b, where a and b are integers, and let's
set r=PI/2.  Of course, this means that r=a/(2b).  If we substitute 
this value into the preceding equation, and remember that sin(PI/2)=1 
and cos(PI/2)=0, we have

               (a/2b)^(2n+1) A[n]  =  n! P(a/2b)

Multiplying both sides by (2b)^(2n+1) and dividing by n! gives

             a^(2n+1) A[n] 
             -------------    =  (2b)^(2n+1) P(a/2b)
                  n!

Remember that P(a/2b) is a polynomial in a/2b with integer coefficients, 
and its degree is less than 2n+1.  Therefore, when we multiply through 
by the factor (2b)^(2n+1) we clear out all the 2b's in the denominators, 
so the quantity on the right hand side of the preceding equation is 
clearly an integer.  It follows that the left side is also an integer.

Now, recall that A[n] was defined as the integral of

                        (1 - x^2)^n cos(rx) dx

from x=-1 to +1.  Clearly the leading factor of this integrand,
(1 - x^2)^n, is always between 0 and 1, so an upper bound on the 
value of A[n] is given by

                          / 1
            A[n]    <     |    cos(rx) dx
                    -     /
                         x=-1

This integral has some constant value C (doesn't matter what it is),
which is the upper bound on the value of A[n] for ANY n.  But now we 
see the contradiction, because we showed previously that the quantity

                    a^(2n+1) A[n]
                    -------------
                         n!

is always an integer (assuming PI is rational).  But since A[n] is less 
than C, it follows that the integer given by the above expression is 
less than
                     a^(2n+1) C 
                     ----------
                         n!

But that's impossible, because n! increases faster than a^(2n+1), 
so there is some value of n beyond which this ratio will be less 
than 1.  This proves the quantity can't be an integer, so we have a
contradiction.  We are forced to conclude that PI is not a rational
number.  QED.


By the way, the actual expressions for the first few A[n] and the 
polynomials P[r] and Q[r] are

            2   /        \
  A[0]  =  --- (  sin(r)  )
            r   \        /


            4   /                         \
  A[1]  =  --- (  [1] sin(r) - [r] cos(r)  )
           r^3  \                         /


            16   /                              \
  A[2]  =  ---- ( [3 - r^2] sin(r) - [3r] cos(r) )
           r^5   \                              /


            96   /                                       \
  A[3]  =  ---- ( [15 - 6r^2] sin(r) - [15r - r^3] cos(r) )
           r^7   \                                       /


           768   /                                                  \
  A[4]  =  ---- ( [105 - 45r^2 + r^4] sin(r) - [105r - 10r^3] cos(r) )
           r^9   \                                                  /


and so on.  If we let double curly brackets {{*}} denote the "two-step 
factorial function, e.g., {{5}} = 5*3*1, {{8}} = 8*6*4*2, and 
{{-1}}={{0}}=1, then the general expression for A[n] is


              2^(n+1) n!  /                              \
     A[n]  =  ---------- (  P_n[r] sin(r) - Q_n[r] cos(r) )
               a^(2n+1)   \                              /


where the coefficient of r^2k in the polynomial P_n is

                        / n-k \  {{2n-2k-1}}
              (-1)^k   (       ) -----------
                        \  k  /   {{2k-1}}

For example, to determine the coefficient of r^2 in A[4] we have
k=1 and n=4, which gives

                       / 3 \  {{5}}          
              (-1)^1  (     ) -----    =   - 45
                       \ 1 /  {{1}} 

Of course, the coefficients of odd powers of r in P_n are all 
zero, as are the coefficients of all even powers of r in Q_n.  The 
coefficient of r^(2k+1) in Q_n is

                       / n-k-1 \  {{2n-2k-1}}
              (-1)^k  (         ) -----------
                       \   k   /   {{2k+1}}

For example, the coefficient of r^3 in A[4] is found by setting
k=1 and n=4 to give

                       / 2 \  {{5}}
              (-1)^1  (     ) -----   =   -10
                       \ 1 /  {{3}}