What is the simplest proof that the density of coprime pairs does not go to zero?

Question

By density of coprime pairs, I mean the proportion of pairs integers between $1$ and $x$ which are coprime. This is known to be asymptotically $1/\zeta(2)$. I want something much weaker, namely that the proportion does not go to zero. I would like a proof that is short and self-contained, so it shouldn't rely on other theorems.

Answer 1

I would say that the standard proof is elementary enough, but here is an argument that avoids the Möbius function and the Riemann zeta function.

Let $A_d$ be the set of pairs $(a,b)$ where $a,b\le x$ are both divisible by $d$. Your set of pairs is $A_1\setminus \bigcup_{p\le x} A_p$ where $p$ ranges over primes. So its size is, by a union bound, $\ge |A_1|-\sum_{p\le x} |A_p|$. Since $|A_d|=\lfloor x/d \rfloor^2 \le x^2/d^2$, it suffices to show that $\sum_p 1/p^2$ is strictly less than $1$. This is the case since this series is dominated by $1/4 +\sum_{n\ge 3}1/(n(n-1))=3/4$.

Answer 2

A purely geometric argument involving no convergent sum (but a simple limit):

Suppose $x=2y$ even for simplicity.

The small square $s=[0,y]^2$ contains $(y+1)^2$ integer points and the large square $S=[0,2y]^2$ contains $(2y+1)^2$ integer points. Consider a line $\mathbb R(a,b)$ containing $(0,0)$ and an integer point $(a,b)\in S\setminus s$. Such a line intersects the large square $S$ in $K+1\geq 2$ integer points and it intersect the small square $s$ in $k+1$ integer points. $a,b$ are coprime if and only if $k=0$. We have however $2(k+1)-1\leq K+1\leq 2(k+1)$ (by elementary geometric considerations) showing $k\leq K-k\leq k+1\leq 2k$ if $k\geq 1$. Since $\mathbb R(a,b)$ contains $K-k$ integral points in $S\setminus s$, this implies that $S\setminus s$ contains at least $(2y+1)^2-(y+1)^2-2(y+1)^2$ coprime pairs. The proportion of coprime integers up to $x$ (even but this does not matter asymptotically) is thus at least equal to $1/4$.

Answer 3

I have put a longer writeup around the method described below in the Aperiodical's 2024 edition of "The Big Internet Math-Off," which can be found here!

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If you imagine coloring the points with $x,y$ such that $\gcd(x,y) = 1$ in black, then you can consider what it would look like to color those that have $\gcd(x,y) = 2$ in red: there would be, in some sense, about $1/4$ as many of them by the matching that takes a black point like $(3,7)$ and transforms it into a red point by multiplying each factor by $2$ to get the point $(6,14)$.

If you push this geometric view/interpretation through, then you can see that there should be $1/9$ as many of the black points when looking at those pairs with $\gcd(x,y) = 3$.

Denoting by $p$ the probability that $\gcd(x,y) = 1$ (again, to make this sensible we are not really quantifying over "all" positive integers but rather those that are $\leq N$ for some large $N$).

Then we have:

$$p + \frac{1}{2^2}p + \frac{1}{3^2}p + \frac{1}{4^2}p + \cdots = 1$$

where the left hand side is the sum of the various probabilities of different gcd's, and the right hand side is, therefore, one (since the two numbers must have some gcd).

But now we end up with:

$$p = \frac{1}{\sum_{n\geq1}\frac{1}{n^2}}$$

The denominator is $\zeta(2)$, so this is the full result (modulo slight hand waving at the top) and uses relatively simple techniques. If you really want only that the density does not go to zero, then it suffices to show that the denominator converges; this is a pre-Calculus or Calculus result with a "$p$-series for $p > 1$".

For a collection of problems that scaffold in this way, you can find some wonderfully assembled ones from PCMI here.