# Extending the Definition of Asymptotic Density to rationals

$\require{enclose}$

Edit: Replace rationals with $\enclose{horizontalstrike}{\mathbb{Z}\times\mathbb{Z}}$. And view fractions as $\enclose{horizontalstrike}{(\text{Numerator},\text{Denominator})}$ instead of $\enclose{horizontalstrike}{\frac{\text{Numerator}}{\text{Denominator}}}$. For more information look at Hurkyl's answer

I want to extend the definition of asymptotic density to countably dense sets. The Asymptotic density of a subset of $\mathbb{N}$ gives the ratio of the number of elements from the subset, compared to the number of elements from $\mathbb{N}$, between $[0,n]$ as $n\to\infty$.

I want to apply a similar concept to the subsets of rationals which gives a ratio of number of elements from the subset of $\mathbb{Q}$, compared to number of elements from $\mathbb{Q}$, based on restricted intervals. Note this is not the same as extending the definition of asymptotic density to $\mathbb{Z}\times\mathbb{Z}$, as this takes the density of the numerator and denominator separately and counts the same element more than once.

This "new density" should act as an informal measure. If such a density exists and that density of set $X$ is $D(X)$, then the density for sets $A$ and $B$ should be meet the following requirements such that

• If set $A=B$ then $D(A)=D(B)$
• If set $A\subset{B}$ then $D(A)\le D(B)$

However, I am not sure how to answer this question. So far I determined the following.

The rationals or $\left\{\left.\frac{m}{n}\right|m,n\in\mathbb{Z}\right\}$ can be divided into groups of sets that contain eachother.

\begin{equation} \left\{\left.\frac{2^{k}m}{2n+1}\right|m,n\in\mathbb{Z}\right\}\subset...\subset\left\{\left.\frac{2m}{2n+1}\right|m,n\in\mathbb{Z}\right\}\subset\left\{\left.\frac{m}{2n+1}\right|m,n\in\mathbb{Z}\right\}\subset\left\{\left.\frac{m}{4n+2}\right|m,n\in\mathbb{Z}\right\}\subset...\subset\left\{\left.\frac{m}{2^{k}(2n+1)}\right|m,n\in\mathbb{Z}\right\}=\left\{\left.\frac{m}{n}\right|m,n\in\mathbb{Z}\right\} \qquad (1) \end{equation}

However, each set can be permuted diffferently. For example $\left\{\left.\frac{m}{2n+1}\right|m,n\in\mathbb{Z}\right\}=\left\{\left.\frac{m}{3(2n+1)}\right|m,n\in\mathbb{Z}\right\}=\left\{\left.\frac{m}{5(2n+1)}\right|m,n\in\mathbb{Z}\right\}$

Hence we need identical sets with different permutations to be permuted in the same permutation before taking their density.

I believe all sets should be rearranged to have permutations similar to the sets in (1) for two reasons. One, the union of the numerator and denominator of all the set covers every integer that could be in the numeator and denominator. Second, due to their permutations, the sets can easily be shown as the subsets of one another. For example, we can convert $(1)$ into

\begin{equation} \left\{\left.\frac{2^{2k}m}{2^k(2n+1)}\right|m,n\in\mathbb{Z}\right\}\subset...\subset\left\{\left.\frac{2^{k+1}m}{2n+1}\right|m,n\in\mathbb{Z}\right\}\subset\left\{\left.\frac{2^{k}m}{2n+1}\right|m,n\in\mathbb{Z}\right\}\subset\left\{\left.\frac{2^{k-1}m}{4n+2}\right|m,n\in\mathbb{Z}\right\}\subset...\subset\left\{\left.\frac{m}{2^{k}(2n+1)}\right|m,n\in\mathbb{Z}\right\}=\left\{\left.\frac{m}{n}\right|m,n\in\mathbb{Z}\right\} \qquad (1) \end{equation} then we can compare the density of the numerators.

From here, I attempted an answer below this post but I am not sure if its correct. If I'm wrong could there still be way of extending asymptotic density to subsets of rationals?

• Asymptotic density is a property of subsets of the positive integers; it's not about cardinality at all really, but rather how the subset interacts with the fixed total ordering on $\Bbb N$. You are now considering sets of rational numbers, and so a completely different concept of "size" is necessary. Cardinality and Lebesgue measure are both possibilities, but you might complain (reasonably) that they don't distinguish between any of your sets. But unless you have some fixed underlying set, you're not going to be able to define anything like density—and even then it's not trivial. – Greg Martin Mar 26 '17 at 23:53
• A rather well-known example of this "asymptotic density" is loosely described as the probability of two random integers being coprime. The exact computation of $\frac{6}{\pi^2}$ can be rigorously justified as the limit over sets $[-n,n]\times[-n,n]$ as $n$ tends to $\infty$. – hardmath Apr 17 '17 at 4:13
• @hardmath Can you show how this asymptotic density can be applied? I want to see if my version is similar to the one you mentioned. – Arbuja Apr 17 '17 at 13:28
• @GregMartin Your comment is incorrect. Asymptotic density can be extended into $\mathbb{Z}\times\mathbb{Z}$. – Arbuja Apr 17 '17 at 15:15
• I think you've identified the inconsistency in your method because it does not "count" a rational number only in its lowest (reduced) form, i.e. coprime numerator and denominator. Do you want me to expand on why using a criterion that is not "well-defined" leads to inconsistencies? – hardmath Apr 17 '17 at 16:09

## 2 Answers

Based on your methods, I strongly suspect what you really want to do is study the asymptotic density of subsets of $\mathbb{Z}^2$ and are too focused on the interpretation of $(p,q)$ as the rational number $p/q$.

• What does $\mathbb{Z}^2$ mean? Is it the same as rationals? – Arbuja Apr 16 '17 at 17:23
• @Arbuja: ... The set of pairs of integers. – Hurkyl Apr 16 '17 at 17:25
• I made minor edits. Hopefully, understanding the post is easier. – Arbuja Apr 16 '17 at 22:01
• Also do you think I correctly found an extension of asymptotic density to $\mathbb{Z}^2$? – Arbuja Apr 17 '17 at 1:17
• @Arbuja I am not sure to which extent this might be useful for you, but I have collected a few references to asymptotic density for subsets of $\mathbb N\times\mathbb N$ here. – Martin Sleziak Apr 17 '17 at 4:47

EDIT: In steps (5), (6), (7) and (8) I replaced $n$ with $2n+1$ and in step (4) I changed $R$ to $L$

My initial answer to the question was incorrect but I think I have the right idea.

Consider the following group of sets

$T_1=\left\{\left.\frac{M_1(c_1)}{R_1(q_1)}\right|c_1,q_1\in\mathbb{Z}\right\}$ $,T_2=\left\{\left.\frac{M_2(c_2)}{R_2(q_2)}\right|c_2,q_2\in\mathbb{Z}\right\},...,$ $T_p=\left\{\left.\frac{M_p(c_p)}{R_p(q_p)}\right|c_p,q_p\in\mathbb{Z}\right\}$

Where $M_1,M_2,..M_p$ and $R_1,R_2,..R_p$ are functions that allow $T_1,T_2,T_p,..$ to be subsets of $\mathbb{Q}$

Inorder to define a density for $T_1,T_2,..T_p$ in $[a,b]$, we must arrange the elements of each set into "packets" based on their numerator and denominator when simplified. "Packets" are infinite unions of finite disjoint sets which result by dividing up $\mathbb{Q}$.

Inorder, to create "packets" we begin by noting.

\begin{equation} \mathbb{Q}=\bigcup_{k\in\mathbb{Z}}\left\{\left.\frac{2m+1}{2^k(2n+1)}\right|m,n\in\mathbb{Z}\right\}\qquad (1) \end{equation}

I chose the sets in the RHS of $(1)$ because union of denominator values and numerators values combined covers any rational.

However, the sets are still countably infinite. We can make the sets finite by dividing them by values of $n$ and restriciting them to elements between $[a,b]$. Then we make the sets disjoint by ensuring the greatest common factor of the numerator and denominator is one.

\begin{equation} \bigcup_{k\in\mathbb{Z}}\bigcup_{n\in\mathbb{Z}}\left\{\left.\frac{2m+1}{2^k(2n+1)}\in[a,b]\right|m\in\mathbb{Z}\ \land\ \text{gcf}(2m+1,2^k(2n+1))=1\right\}\qquad (2) \end{equation}

Finally we can restrict $k$ and $n$ using integers $r$ and $t$ to get a union of "packets".

\begin{equation} \bigcup_{\left\{k\in\mathbb{Z}\right\}\cap[-r,r]}\bigcup_{\left\{p\in 2\mathbb{N}+1\right\}\cap[-t,t]}\left\{\left.\frac{2m+1}{2^k n}\in[a,b]\right|m\in\mathbb{Z}\ \land\ \text{gcf}(2m+1,2^k n)=1\right\}\qquad (3) \end{equation}

Where $r,t\in\mathbb{Z}$

I will take part of $(3)$ for later use.

$$V=\left\{\left.\frac{2m+1}{2^k n}\in[a,b]\right|m\in\mathbb{Z}\ \land\ \text{gcf}(2m+1,2^k n)=1\right\}\qquad (4)$$

Next we take the elements of sets $T_1, T_2,..T_n$ in the elements of the packets by taking the common elements between the sets and packets. We sum the number of common elements based on the number of packets from the restrictions of $r$ and $k$.

$$L_1(r,t)=\sum_{\left\{k\in\mathbb{Z}\right\}\cap[-r,r]}\sum_{\left\{p\in 2\mathbb{N}+1\right\}\cap[-t,t]}\left|T_1\bigcap V\right|\qquad (5)$$

$$L_2(r,t)=\sum_{\left\{k\in\mathbb{Z}\right\}\cap[-r,r]}\sum_{\left\{p\in 2\mathbb{N}+1\right\}\cap[-t,t]}\left|T_2\bigcap V\right|\qquad (6)$$

$$L_p(r,t)=\sum_{\left\{k\in\mathbb{Z}\right\}\cap[-r,r]}\sum_{\left\{p\in 2\mathbb{N}+1\right\}\cap[-t,t]}\left|T_p\bigcap V\right|\qquad (7)$$

Then we set

$$O(r,t)=\sum_{\left\{k\in\mathbb{Z}\right\}\cap[-r,r]}\sum_{\left\{p\in 2\mathbb{N}+1\right\}\cap[-t,t]}\left|\mathbb{Q}\bigcap V \right|\qquad (8)$$

Finally, we define this "extended asymptotic density" (I call it rational density) of $T_1,T_2,..T_p$ in $[a,b]$ which will be denoted as $\underset{[a,b]}{\text{D}}$

$$\underset{[a,b]}{\text{D}} (T_1)=\lim_{(r,t)\to\infty}\frac{L_1(r,t)}{O(r,t)}\qquad (9)$$

$$\underset{[a,b]}{\text{D}} (T_2)=\lim_{(r,t)\to\infty}\frac{L_2(r,t)}{O(r,t)}\qquad (10)$$

$$\underset{[a,b]}{\text{D}} (T_p)=\lim_{(r,t)\to\infty}\frac{L_p(r,t)}{O(r,t)}\qquad (11)$$

We could go further and compare the densities of different subsets. For example, if we want to compare the rational density of $T_1$ and $T_2$, we get

$$\frac{\underset{[a,b]}{\text{D}} (T_1)}{\underset{[a,b]}{\text{D}} (T_2)}=\lim_{(r,t)\to\infty} \frac{L_1(r,t)}{L_2(r,t)}\qquad (12)$$

However, with asymptotic density, we take the interval of natural numbers from $[0,z]$ as $z\to\infty$. We can extend the definition of rational density by setting $a=-z$ and $b=z$ as $z\to\infty$. We will call this "total rational density".

For example with $(12)$, we can modify the original definition to get

$$\lim_{z\to\infty}\frac{\underset{[-z,z]}{\text{D}} (T_1)}{\underset{[-z,z]}{\text{D}} (T_2)}= \lim_{z\to\infty}\left(\lim_{(r,t)\to\infty} \frac{L_1(r,t)}{L_2(r,t)}\right)\qquad (13)$$

In $(13)$, if we set $T_1=A$ and $T_2=\mathbb{N}$ where $A\subseteq\mathbb{N}$ then we get the definition of natural density.

• I wished someone could state what is not clear about my answer. – Arbuja Apr 17 '17 at 17:54