# Miskatonic University Press

## 6/π²

I really enjoyed the latest episode of the My Favourite Theorem podcast, with Jayadev Athreya.

JA: So when trees are planted by a paper company, they’re planted in a fairly regular grid. So imagine you have the plane, two number lines meeting at a 90 degree angle, and you have a grid, and you plant a tree at each grid point. So from a mathematician’s perspective, we’re just talking about the integer lattice, points with integer coordinates. So let’s say where I’m standing there’s a center point where maybe there’s no tree, and we call that the origin. That’s maybe the only place where we don’t plant a tree. And I stand there and I look out. Now there are a lot of trees around me. Let’s say I look around, and I can see maybe distance R in any direction, and I say, hm, I wonder how many trees there are? And of course you can do kind of a rough estimate.

Now I’m going to switch analogies and I’ll be working in flooring. I’m going to be tiling a floor. So if you think about the space between the trees as a tile and say that has area 1, you look out a distance R and say, well, the area of the region that you can see is about πR², it’s the area of the circle, and each of these tiles has size 1, so maybe you might guess that there are roughly πR² trees. That’s what’s called the Gauss circle problem or the lattice point counting problem.

But you can’t see every tree in all directions, because some are behind each other in your line of vision.

JA: A point (m,k) is visible if the greatest common divisor of the numbers m and k is 1. That’s an elementary exercise because, well maybe we’ll just talk a little bit about it, if you had m and k and they didn’t have greatest common divisor 1, you could divide them by their greatest common divisor and you’d get a tree that blocks (m,k) from where you’re sitting…. We call these lattice points, they’re called visible points, or sometimes they’re called primitive points, and a much trickier question is how many primitive points are there in the ball of radius R, or in any kind of increasingly large sequence of sets…. And miraculously enough, we agreed that in the ball of radius R, the total number of trees was roughly the area of the ball, πR². Now if you look at the proportion of these that are primitive, it’s actually 6/π².

This is another way of talking about the “probability of coprimality” (see coprime integers on Wikipedia): pick any two integers at random, and the probability that they are coprime (have no prime factors in common) is 6/π². Primes and pi coming together! Beautiful.