In honour of Pi Day, here’s a proof that Pi is irrational, i.e. it can’t be written in the form a/b where a and b are integers. It’s not too complicated so long as you choose functions with advantageous properties to help you do it.
First, we assume Pi is rational and define a polynomial of degree 2n that depends on constants a and b:
f(x) has some notable properties. First, f(0)=0. Interestingly, f(x)=f(Pi-x). Finally, the kth derivative of f(x) evaluated at 0 will always be an integer. Here are justifications for these properties:
Next we want to define g(x), which is a function of the even derivatives of f(x) up to 2n. Importantly, g(0) and g(Pi) will be integers, and g(x) + g’‘(x) will be equal to f(x):
Now we consider a quantity that seems weird, but it will become relevant I promise. We want to consider g’(x)sin(x) - g(x)cos(x), which is the anti-derivative of f(x)sin(x). It’s easiest to show this by computing the derivative of g’(x)sin(x) - g(x)cos(x):
Sooo what the fuck are we gonna do with the antiderivative of our random function f(x)sin(x)? Well, computing the definite integral of f(x)sin(x) from 0 to Pi shows us that it must be an integer, which tells us that if Pi is indeed irrational then the integral of f(x)sin(x) evaluated from 0 to Pi will be an integer as well. This fact gives us a good basis for a proof by contradiction, which we’ll try next:
Note that sin(x) is always between 0 and 1 when x is between 0 and Pi, which implies that f(x)sin(x) < f(x). Also note that in the expression f(x)=(x^n)((a-bx)^n)/n!, (a-bx) < a when x is between 0 and pi. Also, the term x^n will be between 0 and Pi^n when x is between 0 and Pi. So we create the following inequality and integrate to create a bound for the integral of f(x)sin(x) evaluated from 0 to PI:
The right side of the inequality tends towards 0 as n grows arbitrarily large:
Therefore, for large n the integral will be between 0 and 1. But we already concluded earlier that the integral of f(x)sin(x) from 0 to pi must be an integer! Remember:
So this is a contradiction. The integral of f(x)sin(x) from 0 to Pi cannot simultaneously be an integer and between 0 and 1. Therefore, we must conclude that Pi is irrational!
