
Suppose π is rational. Then π² is rational, so π²=a/b, where a and b are integers (b not zero).
Since the limit as n approaches infinity of a^{n}/n! = 0, it follows
that
there exists an M > 0 such that if n ³ M, a^{n}/n!
< 1/π, or πa^{n}/n! < 1.
Choose an integer N ³ M.
It follows that πa^{N}/N! < 1.
Define a polynomial
f(x) = (x^{N}(1x)^{N})/N!
Expanding the polynomial,
f(x) = (1/N!)(c_{N}x^{N} + c_{N+1}x^{N+1} + c_{N+2}x^{N+22} + ... + c_{2N}x^{2N}), where each c_{N} is an integer.
For integers k such that N ≤ k ≤ 2N, the k^{th} derivative f^{(k)}(x) is given by
f^{(k)}(x) = (1/N!)
_{2N} ∑ ^{n=k} (n!/(nk)!) c_{N}x^{Nk}
Now I will show by induction that f^{(k)}(x) = (1)^{(k)}f^{(k)}(1x) for all integers, k.
It is true when k=0; f^{(0)}(x) = (1)^{(0)}f^{(0)}(1x), because (x^{N}(1x)^{N})/N! = ((1x)^{N}x^{N})/N!
Suppose f^{(k)}(x) = (1)^{(k)}f^{(k)}(1x)
f^{(k+1)}(x) = d(f^{(k)}(x))/dx
f^{(k+1)}(x) = d[ (1)^{(k)} f^{(k)}(1x) ]/dx
f^{(k+1)}(x) = (1)^{(k) }d[ f^{(k)}(1x) ]/dx
f^{(k+1)}(x) = (1)^{(k) }(1)f^{(k+1)}(1x)
f^{(k+1)}(x) = (1)^{(k+1)} f^{(k+1)}(1x)
For integer values of k such that 0 ≤ k ≤ N, each term in the expansion of f^{(k)}(0)
is 0,
and f^{(k)}(1) = (1)f^{(k)}(0) = 0
For integer values of k such that N ≤ k ≤ 2N, the only term that does not
contain a positive integer power of x is the first (when n=k), so
f(k)(0) = k!c_{k}/N!, which is an integer, since k ³
N. f^{(k)}(1) is also an integer because f^{(k)}(1) =
(1)f^{(k)}(0).
Now define another function,
F(x) = (b^{N}) 

(1)j π^{2N2j} f^{(2j)}(x) 
F(x) = (b^{N}) 

(1)j (a/b)^{(Nj)} f^{(2j)}(x) 
F(x) = (b^{N}) 

(1)j a^{(Nj)}b^{(jN)} f^{(2j)}(x) 
F(x) = 

(1)j a^{(Nj)}b^{j} f^{(2j)}(x) 
It follows that F(0) and F(1) are both integers, and thus F(0)+F(1) is an integer. Now define
g(x) = F'(x) sin(πx)  πF(x) cos(πx)
It follows that
g'(x) = F''(x) sin(πx) + πF'(x) cos(πx)  πF'(x) cos(πx) + π²F(x) sin(πx)
g'(x) = [ F''(x) + π²F(x) ] sin(πx)
Now
F(x) = b^{N} [ π^{2N} f(x)  π^{2N2} f''(x) + π^{2N4} f^{(4)}(x)  ... + (1)^{j} f^{(2N)}(x) ], and so
F''(x) = b^{N} [ π^{2N} f''(x)  π^{2N2} f^{(4)}(x) + π^{2N4} f^{(6)}(x)  ... + (1)^{j} π^{2} f^{(2N+2)}(x) ]. Also,
π²F(x) = b^{N} [ π^{2N+2} f(x)  π^{2N} f''(x) + π^{2N2} f^{(4)}(x)  ... + (1)^{j} π^{2} f^{(2N)}(x) ]
Notice that f(x) is a polynomial of degree 2N, so f^{(2N+2)}(x) = 0 for all x. Thus
F''(x) + π^{2} F(x) = b^{N} π^{(2n+2)} f(x). So
g'(x) = b^{N} π^{(2N+2)} f(x) sin(πx)
g'(x) = (a^{N}/π^{(2N)}) π^{(2N+2)} f(x) sin(πx)
g'(x) = π^{2} a^{N} f(x) sin(πx)
Since g(x) is continuous on [0,1], and g' exists on (0,1), by the
Mean Value Theorem, there exists a c in (0,1) such that
g(1)g(0)=g'(c).
Now g(1) = F'(1) sin(π)  πF(1) cos(π) = πF(1), and
g(0) = F'(0) sin(0)  πF(0) cos(0) = πF(0).
So g(1)g(0) = πF(1) + πF(0) = π [F(1) + F(0)], and thus
π [F(1) + F(0)] = π^{2} a^{N} f(c) sin(πc)
F(1) + F(0) = π a^{N} f(c) sin(πc)
Since 0 < c < 1, 0 < sin(πc) < 1, and also 0 < 1c < 1.
Since f(c) = (c^{N} (1c)^{N}) / N! < 1/N!,
and remember that we selected N in the first place so that πa^{N}/N! <
1,
so it follows that
0 < π a^{N} f(c) < π a^{N}/N! < 1
It now follows that 0 < π a^{N} f(c) sin(πc) < 1, and so 0 < F(1) + F(0) < 1 is an integer. But this is a contradiction because F(1)+F(0) is an integer, and there can be no integer between 0 and 1.
Proof that cos a is irrational, when a is rational and not zero
Rumor has it that this proof that pi is irrational can be adapted to one that shows cos a irrational whenever a is rational and not zero. Mathworld: Irrational Number says cos a is irrational for every rational number a not equal to 0, and credits Niven 1956, and Stevens 1999 for proving this.
http://www.meikleriggs1.freeonline.co.uk/pi/index.htm
Mathworld: Irrational Number
Perfect Squares  proof that sqrt(n) is irrational, as long as n isn't a perfect square.
The webmaster and author of this Math Help site is Graeme McRae.