Niven’s proof for the irrationality of π

In this post I will present a beautiful proof owing to Ivan Niven on the irrationality of \pi. An analytical definition of \pi is that \pi is twice the smallest positive x for which \cos x equals 0. (This definition is better then the usual circumference/diameter  one since it is not dependent on Euclidean axioms).

Theorem (Niven): \pi is irrational.

Proof: Let, by way of contradiction \pi=a/b with a,b\in\mathbb{N}. Now let 

f(x)=\frac{x^n(a-bx)^n}{n!} and 


where we will specify n later. Now n!f(x)=a_0x^n+a_1x^{n+1}+\cdots a_nx^{2n} with a_i\in\mathbb{Z} and so f(0)=0. Not only this, f^{(j)}(0)=0 for 1\le j\le n-1. For n\le j\le 2n we have n!f^{(j)}(x)=n!b_0+b_1x+\cdots b_nx^n with b_i\in\mathbb{Z} and so f^{(j)}(0) is integral in this case as well. Also note that f(x)=f(\pi-x) and so for any j\ge 0,j\in \mathbb{Z} we have f^{(j)}(x)=(-1)^jf^{(j)}(\pi-x) following which f^{(j)}(\pi)=(-1)^jf^{(j)}(0) i.e. f^{(j)}(\pi) is also integral.

Now it follows by elementary calculus that \frac{d}{dx}(F'(x)\sin x-F(x)\cos x)=(F''(x)+F(x))\sin x=f(x)\sin x and so \int_0^\pi f(x)\sin x dx=(F'(x)\sin x-F(x)\cos x)_0^\pi=F(\pi)+F(0). Now F(\pi)+F(0) is integral because f^{(j)}(\pi) and f^{(j)}(0) are both integers for any j. It is also easy to see that for 0<x<\pi we have 0<f(x)<\frac{a^n\pi^n}{n!} and hence 0<f(x)\sin x<\frac{a^n\pi^n}{n!}. But now  0<\int_0^\pi f(x)\sin x dx\le\frac{\pi(a\pi)^n}{n!} and as n! grows faster then any exponential so this upper bound for the integral can be made arbitrarily small. Hence the value of the integral lies between 0 and 1. This is the contradiction which proves the result.\Box

One final word. Even though the proof is credited to Ivan Niven, it is very much possible that the idea was given by Hermite much earlier and all did Niven was to tweak Hermite’s idea to give this proof.


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