The height of probabilistic interpretation

Girls only love men as tall as 6' and above. Socrates, ca. 2023

It is undeniable that heights strongly influence our daily lives. Be it our heights, or the height of a mountain we scale, or the height of all problems - humans. Mathematics too hasn't been able to escape its clutches, with height functions being useful in several fields, including but not limited to - Diophantine Geometry, Automorphic forms and the Weil-Mordell theorem - something you should have heard before if you attend my talks.

If you have attended school (or maybe you are a climate activist) - then try recalling the elementary school days when fractions were introduced. Albeit unknowingly, but we had as children classified fractions into proper and improper - based on whether the denominator was larger than the numerator or vice versa. Well, it seems mathematicians have stuck with this classification - giving us the crux of todays discussion - height of a rational number.

Given a rational number $x=\frac mn$ in its lowest terms, let us define

$$H(x):=\max\left\{|m|,|n|\right\}$$

Let $R(\kappa)$ be the number of rational numbers $x$ with $H(x)$ less than $\kappa$. A popular result is that $R(\kappa)$ is a finite set - and is fairly easy to prove that it is bounded by $(2\kappa+1)\kappa$.  However, we will here provide a nice intuitive way to see and prove a very neat result.

When you strictly bound $H(x)$ by $\kappa$, you automatically bound both $m$ and $n$ in the set $(-\kappa,\kappa)$. Take a tuple in the integer lattice $\mathbb{Z}^2$ - it looks like $(m,n)$, where $m,n\in\mathbb{Z}$. The product set $S_{\kappa}=(-\kappa,\kappa)\times(-\kappa,\kappa)$ denotes a square with center of mass at the origin and side of length $2\kappa$. This is a finite sized square, so the number of lattice points inside this square is finite. Since rationals are necessarily in their lowest terms, so the set $R(\kappa)\subseteq S_{\kappa}$, hence it is finite.

We are now equipped with enough mathematical hodge-podge to look at the main result (which is just a side character in the exercises section of  Andrew Tate's book),

$$\lim_{\kappa \to \infty} \frac{R(\kappa)}{\kappa^2}=\frac{12}{\pi^2}$$

where we see that our good friend $6/\pi^2$ has somehow sneaked in, just how Basel slid into Euler's DM's. But how do we connect the sum of squares of reciprocals of positive integers to rational heights?

Our not so good friend Probability Theory hints at a connection.

What is the probability that two randomly uniformly chosen positive integers are co-prime? Well, if we have some prime $p$ dividing both, then we can't proceed. So, no prime whatsoever can divide both the chosen integers. As there is one multiple of $p$ in every $p$ integers,

$$P(\text{Randomly uniformly chosen integer is divisible by $p$}) =\frac 1p$$

So, the probability that two randomly chosen integers not being divisibly by $p$ has a probability $\left(1-\frac {1}{p^2}\right)$. Since this needs to hold for all primes $p$, we have

$$P(\text{randomly chosen positive integers are coprime}) = \prod_{p}\left(1-\frac{1}{p^2}\right)$$

Imagine standing at the origin of $\mathbb Z^2$ and looking around you. What are the points that you will be able to see, and what are the ones that will hide behind some other point?

By Courtesy of Euler L., Mathologer, Wikipedia et. al., this sum is precisely $1/\zeta(2) = 6/\pi^2$. So our final ingredient is a connection between this result and $R(\kappa)$. But oi! What do we know about rationals? They are of the form $m,n\in\mathbb{Z}$ where $m,n$ are co-prime! And $n\neq 0$, which are exactly the conditions we require to retrieve $R(\kappa)$.

In other words, $R(\kappa)$ gives the number of tuples $(m,n)$ in the square $S_{\kappa}\subset \mathbb{R}^2$ where $m,n$ are coprime (which means the associated fraction is in lowest terms). We define a formalism where we only allow $m$ to take negative values - this is because we don't want to overcount fractions of the form $-m/n$ as both $(-m,n)$ and $(m,-n)$. This reduces our square to only the upper half plane - with $S_{\kappa} := (-\kappa,\kappa)\times (0,\kappa)$ being the redefinition.

The total number of lattice points in the square is $2\kappa^2$ (it won't change the result if $\kappa$ is replaced by $\lfloor\kappa\rfloor$ in case it is not an integer - but that is trivial to show), hence the probability of finding such a number is $R(\kappa)/2\kappa^2$.

I hope that you've successfully jumped the bandwagon by now, on to the never ending journey to perfecting Mathematics.

We finally state our happy-go-lucky proof

$$\lim_{\kappa\to\infty} \frac {R(\kappa)}{\kappa^2} = 2P(\text{two randomly chosen positive integers are coprime}) = 2\times\frac{6}{\pi^2} = \frac {12}{\pi^2}.$$

If you're bothered so easily by the lack of rigour in our proofs, you probably haven't taken enough math courses so far! However, since you are the nerd who has stuck so long, note that

$$R(k+1)=R(k)+\left|\left\{x \in\mathbb{Q} | H(x)=k\right\}\right|=\\R(k)+4\cdot\varphi(k)$$

where $\varphi(n)$ is the Euler's totient function.

As a result: 

$$\frac{R(k+1)}{(k+1)^2}= \frac{R(1)}{(k+1)^2}+\frac{4}{(k+1)^2}\left(\sum_{n=1}^{k}\varphi(n)\right)$$

Now, according to a result of Arnold Walfisz (search within the Wikipedia article):

$${\displaystyle \sum_{n=1}^{k}\varphi (n)={\frac {3k^{2}}{\pi ^{2}}}+O\left(k(\log k)^{\frac {2}{3}}(\log \log k)^{\frac {4}{3}}\right)\quad {\text{as }}k\rightarrow \infty ,}$$

and the result follows.