public-notes/Probability of Choosing 2 Coprime Numbers.md

#Math #NT #Probability

Problem

Calculate:

P(x, y \in \mathbb{N}: gcd(x, y) = 1)

Solution

Each number has a \frac 1 p chance to be divisible by prime p, so the probability that two numbers do not share prime factor p is

1 - p^{-2}

Therefore, the probability two numbers are coprime is:

\prod _{p \in \mathbb{P}} 1 - p^{-2}

Since 1 - x = (\frac 1 {1 - x})^{-1} = (\sum _{n = 0}^{\infty} x^n)^{-1}, we can express the above as:

(\prod _{p \in \mathbb{P}} \sum _{n = 0}^{\infty} p^{-2n})^{-1}

We can choose any n for p^{2n} for each prime p, so by the Unique Factorization Theorem (any natural number can be prime factored one and only one way), we get:

(\sum _{n = 0} n^{-2})^{-1} \\
= (\frac {\pi^2} 6)^{-1} \\
= \frac 6 {\pi^2}