Pi Irrational English

π is irrational Arne Smeets In this short note, I will give a more or less elementary proof of the fact that π is irrational. A little calculus will be used, but nothing more. Let r be a real number and let n be a nonnegative integer. Define Z 1
n An (r) = 1 − x2 cos (rx) dx. −1

Lemma 1 If n ≥ 2, then r2 An (r) = 2n(2n − 1)An−1 (r) − 4n(n − 1)An−2 (r). Proof Partial integration gives   Z 1
sin (rx) 0 2 n An (r) = 1−x dx r −1 1 Z

2n 1 sin (rx) n 2 2 n−1 = 1−x + x 1 − x sin (rx) dx r −1 r −1  0 Z
2n 1 cos (rx) 2 n−1 x 1−x = − dx r −1 r 1 Z

0 2n 1  2n 2 n−1 cos (rx) 2 n−1 + = x 1−x x 1 − x cos (rx) dx r r r2 −1 −1 Z Z

n−2 2n 1 4n(n − 1) 1 2 2 n−1 = 1−x x 1 − x2 cos (rx) dx. cos (rx) dx − 2 2 r −1 r −1 Using the fact that Z 1 Z
2 2 n−2 x 1−x cos (rx) dx = −1

1



1−x


2 n−2

− 1−x


2 n−1



−1

we get that An (r) =

4n(n − 1) 2n An−1 (r) − (An−2 (r) − An−1 (r)) . 2 r r2

This proves Lemma 1. 

We continue the proof with a new lemma: 1

cos (rx) dx

Lemma 2 For any nonnegative integer n, we have that n! An (r) = 2n+1 (Pn (r) sin r − Qn (r) cos r) , r where Pn (r) and Qn (r) are polynomials having integer coefficients. Proof We will prove Lemma 2 using mathematical induction. The base case is trivial: we have P0 (r) = 2, Q0 (r) = 0, P1 (r) = 4 and Q1 (r) = 4r. Suppose that the existence of Pk (r) and Qk (r) has been proven for k = n−2 and k = n−1, where n ≥ 2. We will prove the existence of Pk (r) and Qk (r) for k = n. Define Pn (r) = 2(2n − 1)Pn−1 (r) − 4r2 Pn−2 (r), Qn (r) = 2(2n − 1)Qn−1 (r) − 4r2 Qn−2 (r). From the recursion in Lemma 1, it follows that Pn (r) and Qn (r) satisfy the conditions.  An easy induction shows that deg Pn (r) ≤ n for all nonnegative integers n, since
deg Pn (r) = deg 2(2n − 1)Pn−1 (r) − 4r2 Pn−2 (r)
≤ max deg Pn−1 (r), deg r2 Pn−2 (r) = max (deg Pn−1 , 2 + deg Pn−2 ) . Suppose now that π is rational, then π/2 = a/b, where a and b are positive integers. Lemma 2 gives  2n+1   a b a = n! . Pn An b a b P Write Pn (r) = 0≤j≤dn cj,n rj where dn = deg Pn and cj,n ∈ Z. Then d

d

n n a a  a j X X a2n+1 cj,n aj b2n+1−j . An = b2n+1 Pn = b2n+1 cj = n! b b b j=0 j=0

It is obvious that the right-hand side is an integer, since 2n + 1 − j > n + 1 for 0 ≤ j ≤ dn . n Also, since (1 − x2 ) ≥ 0 and 0 ≤ cos(rx) ≤ 1 for −1 ≤ x ≤ 1 and r = 21 π, we get a An > 0. b We deduce that a n! An ≥ 2n+1 . b a Notice that X a2n+1 2 aea = . n! n≥0 Since the series converges, we have that n!

= +∞. a2n+1 n However, since (1 − x2 ) ≤ 1 for −1 ≤ x ≤ 1, we get Z 1 a Z 1 a  a  π  4
2 n An = 1 − x cos x dx ≤ cos x = A0 = . b b b 2 π −1 −1 lim

n→∞

This is an obvious contradiction. Hence π is irrational.  2