Suppose pi is rational; then there exists a,b natural numbers pi = a/b.

Let f be in Q[x] f=x^n(ax-b)^n/n! (pi = a/b, remember?). Suppose we define g in Q[x] as sum(0->n)(-1)^n(d/dx)^(2n)f. It should be obvious that d^2g/(dx)^2 is f-g.

Getting trickier, we see that (d/dx)[(dg/dx)sin(x)-g*cos(x)] is [d^2g/(dx)^2+g]sin(x) = f*sin(x). Integrating f*sin(x) from zero to pi with respect to x yields g(0)+g(pi).

At zero and pi, the first n derivatives of f vanish. But then the coefficients become integers by an easy result on products of n consecutive integers being divisible by n!. So our integral is an integer.

But the maximum of f on (0,pi) is (a*pi)^n/n! and the max of sin(x) is 1. Both f and sin are bounded by zero from below. Thus the integral is bounded strictly between zero and an arbitratily small number, resulting in a contradiction. Therefore pi cannot be rational.

Log in or registerto write something here or to contact authors.