Question

This problem outlines a proof that the number π is irrational. Suppose not, Then there are relatively prime positive integers a and b for which π = a/b. If p is any polynomial let Ip= ∫0a/bp(x) sinx dx. i. Show that if p is non-negative and not identically 0 on [0,a/b] then Ip>0; ii. Show that if p and all of its derivatives are integer-valued at 0 and a/b then Ip is an integer. iii. Let N be a large integer and define p(x) = xN(a-b x)N . N! Show that p satisfies the conditions of i. and ii. Show that if N is sufficiently large, then Ip<1. This gives us a positive integer that is < 1. This is impossible, so the assumption that .π= a/b leads to a contradiction

Answer 1