Is pi Squared Rational or Irrational?
Date: 01/07/2011 at 03:29:03 From: skand Subject: pi Is pi^2 rational or irrational?
Date: 01/07/2011 at 06:50:09 From: Doctor Ali Subject: Re: pi Hi Skand! Thanks for writing to Dr. Math. For any positive integers n and x in (0,1) we can define 2n x^n (1 - x)^n 1 --- j f(x) = ---------------- = ---- ) c_j x n! n! --- j=n Here, c_j are integers. For x in (0,1) we have, 0 < f(x) < 1/n! Now, suppose that positive integers a and b exist such that 2 a pi = --- b Define n / 2n 2n-2 2n+4 (4) P(x) = b | pi f(x) - pi f"(x) + pi f (x) - ... \ n (2n) \ ... + (-1) f (x) | / Now, we know that f(0) = 0 We also know that (j) f (0) = 0 for all j < n or j > 2n But for n <= j <= 2n, we have ... (j) j! f (0) = ---- c_j n! ... which is an integer. Hence f(x) and all its derivatives have integer values at x = 0. Since f(1 - x) = f(x), the same is true at x = 1. Now, because P(0) and P(1) are integers, we have d / \ ----|P'(x)Sin(pi x) - pi P(x) Cos(pi x)| = Sin(pi x) (P"(x) + pi^2 P(x)) dx \ / This can be simplified to n 2n+2 2 n b pi f(x) Sin(pi x) = pi a Sin(pi x) f(x) Now, - 1 - - 1 | n | P'(x)Sin(pi x) | pi | a Sin(pi x)f(x)dx = | ---------------- - P(x)Cos(pi x) | | | pi | - 0 - - 0 = P(0) + P(1) This is an integer. But notice the inequalities: - 1 n | n pi a 0 < pi | a Sin(pi x) f(x) dx < ------- < 1 | n! - 0 For a large enough n, we reach a contradiction. So pi^2 can't be written as a/b, as we originally assumed. Therefore, it's irrational. Note that from here, you can say that since the square root of any irrational number is still irrational, pi itself is also irrational. Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/
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