Why π is irrational (Part 1 of 5)

Through the next several posts, I will be taking you through a proof of why π is irrational. The proof was first given by Ivan Niven in 1947 and requires only basic calculus to understand it.

Table of Contents
1. Outline
2. A polynomial and its derivatives
3. A new function
4. The integral is an integer (pending)
5. The size of the integral (pending)

This proof uses the method known as “proof by contradiction” (also called “indirect proof”). We assume premise P and then (rigorously) prove result Q from it. We then find that Q is in contradiction with an earlier result or an axiom; thus, something must be wrong. If our logic is sound, then the thing that must be wrong is premise P–thus, P must be false.

In this proof, we will assume that π is rational. In particular, we shall assume that there are integers a and b such that

Using these integers, we will define a function f(x) and show that

is and integer and that

Our contradiction will be that, although there are no integers strictly between 0 and 1, the integral must be an integer and, by adjustment of a parameter, can be “squeezed” into this interval.


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