On a Notion of Density Related to Restricted Diophantine Approximation

Introduction

One of the foundational results in theory Diophantine approximation is Dirichlet’s approximation theorem.

It is natural to wonder how the quality of approximation in Dirichlet’s theorem would be affected if the approximating integers are restricted to some subset of the natural numbers. This question has received considerable attention over the years, with much of the research focused on restrictions to arithmetically significant subsets of the natural numbers. The arithmetic structure of these sets allows for techniques from analytic number theory to be applied. The following list gives a sample of the variety of sets which have been considered in the past.

• Prime numbers (Matomäki [8]).

• Square-free numbers (Harman [5]).

• Perfect squares (Zaharescu [11]).

• Numbers formed by the digits 0 and 1 in a given base (Iyer [6]).

• Values of a quadratic form (Baier and Rahaman [2]).

To give a specific example, we present Zaharescu’s main theorem from [11].

Results such as Zaharescu’s theorem suggest studying subsets of the natural numbers according to the quality of Diophantine approximation they permit. In this paper, we take up this task by introducing the Diophantine density of a set. We show that dense sets in the Diophantine sense, also produce good Diophantine approximations. Moreover, we hope to demonstrate that aside from its connection to rational approximation, the Diophantine density possesses a number of other interesting properties.

In the next section, we formally define Diophantine density and show that it possesses the properties of an abstract density. § 3 provides two theorems giving lower and upper bounds for the Diophantine density of a set. In § 4, we construct a set with Diophantine density equal to any prescribed value in $(0, 1)$. § 5 establishes a connection back to Diophantine approximation by proving an analogue of Dirichlet’s theorem in which the denominators of the approximating rationals are restricted to a set of positive Diophantine density.

§ 6 concerns sets arising from Piatetski-Shapiro sequences. Using the results of the preceding sections together with Zaharescu’s theorem, we establish explicit lower and upper bounds for the Diophantine density of the set of perfect squares. Finally, § 7 presents several open problems.

Diophantine density

The following definition gives us our principal object of interest for the remainder of paper.

The purpose of this section is to establish some of the basic properties of the Diophantine density and in particular, show that it possesses the features that one would expect a density to possess. The question of what properties a “generic” or “abstract” density should satisfy has been raised by a number of people over the years. The definition given below was proposed by Di Nasso and Jin [3].

Our proof of Theorem 3 will be spread out over the upcoming four propositions. Each of which will establish one of Properties (i) - (iii) as well as the fact that the codomain of the Diophantine density is $[0,1]$. We begin by showing that the Diophantine density satisfies Property (i).

Proof. For every coprime pair $(p,q)\in\mathbb{Z}\times\mathbb{N}$ we can set $v=(p^{-1}\bmod q)$, since $p$ is invertible modulo $q$. In doing so, we see that $v<q$ and

This shows immediately that $\delta(\mathbb{N})\geq1$.

To see that this is the largest possible value for $\delta(\mathbb{N})$, let us assume that some larger value is achievable. Then there would be a $\delta>1$ and $C>0$ such that $q^{-1}\leq Cq^{-\delta}$. But this implies that $q^{\delta-1}\leq C$ for all sufficiently large $q$, which is impossible since $\delta-1>0$. Therefore, we see that $\delta(\mathbb{N})=1$.

Proof. Aiming for a contradiction, let us assume that $\delta(A)>0$ and consider an infinite sequence of reduced rationals $(p_n/q_n)_{n\geq0}$, such that $p_n/q_n\to\xi$ as $n\to\infty$. By definition, for sufficiently large $n$, there are real numbers $\delta$ and $C(A,\delta)$ both greater than zero, such that

Since $A$ is finite and the sequence of rationals is infinite, by Dirichlet’s pigeonhole principle, there must be a single $v\in A$ that satisfies the above inequality for infinitely many $n$.

Let us denote the subsequence of $n$’s for which this is true by $(n_k)_{k\geq0}$. By taking the limit as $k\to\infty$, we find that

But this implies that $\xi$ is rational, which is a contradiction. As a result, we conclude that $\delta(A)=0$.

Proof. Since $A\subseteq B$ we see that for sufficiently large $q$,

Thus, any admissible $\delta$ for the set $A$ is also admissible for the set $B$. This shows that $\delta(B)\geq\delta(A)$.

Proof. First, we note that $\lVert x\rVert\leq1/2$ for all $x\in\mathbb{R}$. Thus, for any pair $(p,q)\in\mathbb{Z}\times\mathbb{N}$ we have

As a result, $\delta(A)\geq0$.

On the other hand, by Propositions 1 and 3 we know that for any subset $A\subseteq\mathbb{N}$, $\delta(A)\leq\delta(\mathbb{N})=1$. Combining the lower and upper bounds completes the proof.

With that we have shown that the Diophantine density is an abstract density and our proof of Theorem 3 is complete.

In their paper on abstract densities, Di Nasso and Jin [3] present four additional properties which an abstract density may possess.

• (vi) Richness: For every $x \in[0,1]$, there is an $\mathcal{A} \subseteq\mathbb{N}$ such that $d(\mathcal{A}) = x$.

• (vii) Atomless-ness: For every $\mathcal{A} \subseteq\mathbb{N}$ such that $d(\mathcal{A}) > 0$ there is a $\mathcal{B} \subseteq\mathcal{A}$ such that $0 < d(\mathcal{B}) < d(\mathcal{A})$.

Here the right and left translates of $\mathcal{A}$ are defined by

At the end of this section we show that the Diophantine density is not translation invariant, and we establish richness in Section 4. Out of the four properties listed above this leaves subadditivity and atomless-ness, which we were not able to establish. But we present some evidence that it is not subadditive in Section 7.

In another study of generic densities, Leonetti and Tringali [7] define an upper density to be a function which satisfies Properties (i), (iii), (iv) and (v), plus an additional property called $(-1)$-homogeneity.

Our next result shows that the Diophantine density is not $(-1)$-homogeneous, and therefore cannot be an upper density in the sense of Leonetti and Tringali.

Proof. Let $(p,q)\in\mathbb{Z}\times\mathbb{N}$ be coprime with $q>m$, and let

Next, let $\xi=mp/q$ in Dirichlet’s approximation theorem, so that we have

Setting $v=nm$, we see that by construction, $v\in m\mathbb{N}$ and

Moreover,

which implies that

Putting this all together we see that

This shows that $\delta(m\mathbb{N})\ge1$, and in light of the upper bound derived in the Proposition 4, we have $\delta(m\mathbb{N})=1$. □

Proposition 5 provides an example of an interesting phenomenon. Namely, the Diophantine density admits sets that possess co-infinite subsets of the same density. We revisit this example in Section 7, where a related open question is posed.

In the final proposition of this section we show that the Diophantine density is not translation invariant.

Proof. Let $m \ge2$ be a positive integer and consider the set $2m\mathbb{N}$. By Proposition 5 we know that $\delta(2m\mathbb{N})=1$. We will demonstrate that a right translation by $m$ results in $\delta(2m\mathbb{N}+m)=0$.

To see this, define $p_n=n$ and $q_n=2mn+1$, so that

For any $v\in2m\mathbb{N}+m$, $v$ is of the form $2mu+m$, and therefore

Thus, for any such $v<q_n$ the distance to the nearest integer is given by

As there are infinitely many such rationals $p_n/q_n$, this rules out any positive exponent, thereby establishing that $\delta(2m\mathbb{N}+m)=0$.

As far as the author knows, the concept of Diophantine density has not been directly studied before, but related concepts have appeared in the literature. One related idea that has appeared in the literature is that of a Heilbronn set. A set $A\subseteq\mathbb{N}$ is said to be Heilbronn if, for every real number $\xi$ and $\varepsilon>0$, there exists a $v\in A$ such that $\|v\xi\|<\varepsilon$.

Recently, Iyer [6] defined two characterizations of sets that are more closely related to our notion of Diophantine density. Firstly, a set $A\subseteq\mathbb{N}$ is said to be an $\mathcal{H}$-set, if there exists a function $f(x)$ decreasing to $0$, such that for every $\xi\in\mathbb{R}$ and $N\in\mathbb{N}$, one has

Iyer also defines a set $A\subseteq\mathbb{N}$ to be a strong approximating set, if for all $\theta<1$ and $\xi\in\mathbb{R}$ one has

where $C_\theta$ is a constant depending only on $\theta$.

If $A$ is a strong approximating set, then by taking $\xi=p/q$ and $N=q-1$ in the definition, we see that

for all $\theta<1$ and rational $p/q$. This implies that any strong approximating set has a Diophantine density of $1$.

Bounding the Diophantine density of a set

The goal of this section to prove two theorems which provide an upper and lower bound for the Diophantine density of a given set and its complement, respectively. We begin by stating the following definition for the counting function of a subset of the natural numbers.

Remark 1. We would like to point out that our convention for the counting function $A(n)$ differs from the usual definition, in which $A(n)$ is defined as the number of elements less than or equal to $n$.

The first goal of this section is to establish a lower bound for the density of a complement $\mathbb{N} \setminus\mathcal{A}$. To accomplish this, we need the following lemma.

Proof. Fix a coprime $(p,q) \in\mathbb{Z} \times\mathbb{N}$. Our goal is to bound the smallest element left in the multiset

after removing those elements for which $k \in\mathcal{A}$.

We begin by noting that since $p$ permutes the nonzero residue classes modulo $q$ and

we see that

as multisets. Moreover, for $k<q$ we have

and thus

where each distinct element occurs twice, except for the last which occurs only once for even $q$.

Now consider what happens when we remove the elements of $D(p,q)$ which correspond to the $A(q)$ elements of $\mathcal{A}$ less than $q$. Since each distinct element of $D(p,q)$ occurs at least once, the smallest remaining element may be roughly bounded by

With that our proof is complete. □

Proof. Let us begin by defining

Since $\mathcal{A}$ is non-empty we have $0 \leq m(\mathcal{A}) \leq1$. If $m(\mathcal{A})=1$, then the lower bound $\delta(\mathbb{N}\setminus\mathcal{A}) \geq0$ follows by Proposition 4. Therefore, let us assume that $m(\mathcal{A})<1$. By Lemma 1, we see that for any coprime pair $(p,q)\in\mathbb{Z}\times\mathbb{N}$, and $\tau>m(\mathcal{A})$,

Consequently, $\delta(\mathbb{N}\setminus\mathcal{A})\geq1-\tau$ for any $\tau>m(\mathcal{A})$. This implies that

Our goal now is to show that

To achieve this, let $L$ denote the limit supremum in the previous line. Then we see that for large enough $q$,

for every $\varepsilon>0$. Thus, $A(q) \le q^{L+\varepsilon}$ and consequently, $m(A) \le L+\varepsilon$. Since this holds for every $\varepsilon$ we obtain $m(A) \le L$.

For the reverse inequality, suppose that $A(q) \ll q^\tau$. Then there exists a constant $C>0$ such that, for all sufficiently large $q$,

Taking the limit supremum gives $L \le\tau$. Since this holds for every $\tau$ for which $A(q) \ll q^\tau$, we obtain $L \le m(A)$.

Combining the upper and lower bounds for $m(A)$, we obtain $m(A)=L$, which completes the proof. □

Proof. The first step in our proof is to show that for every $0 \le\delta< \delta(A)$ there exists a $c>0$ such that $A(q) \ge cq^\delta$ for all $q$ sufficiently large. To accomplish this task, we will essentially examine the set of numerator classes modulo $q$, which are covered by the elements of $A$ less than $q$.

For fixed $v$ and $\delta$, define the set of numerator classes modulo $q$ by

then the definition of Diophantine density gives

Now let us suppose that $q$ is prime. Because $v<q$ and consequently $\gcd(v,q)=1$, the map $p\mapsto vp\pmod q$ is a bijection. Therefore, the number of elements in $P_\delta(v,q)$ is equal to the nonzero residues $r$ modulo $q$ which satisfy $\lVert r/q\rVert\le Cq^{-\delta}$. In other words,

This fact allows us to easily bound the order of $P_\delta(v,q)$. To this end, observe that if $r$ is such that $\lVert r/q\rVert\le Cq^{-\delta}$, then by definition

This implies that either

and thus, $r$ must satisfy

So for fixed $v$ there are at most $2Cq^{1-\delta}$ such numerator classes that $v$ can cover. That is, $\lvert P_\delta(v,q)\rvert\le2Cq^{1-\delta}$.

By taking the union of all $v\in A$ such that $v<q$, we find that

Consolidating constants, this implies that for some $c > 0$,

for all sufficiently large prime $q$.

Next, by Bertrand’s postulate, for every sufficiently large integer $n$, there exists a prime $q$ such that $n/2 \le q \le n$. Since the counting function is non-decreasing and $\delta\ge0$,

Thus, after adjusting the constant, the bound $A(q) \ge cq^{\delta}$ holds for all sufficiently large integers $q$, not only primes.

Finally, by taking logarithms, we obtain

Therefore,

Since this holds for every $0 \le\delta< \delta(A)$ we are able to say that

as we had sought to show. □

Remark 2. By combining Theorems 4 and 5 we obtain the bounds

where we have set $A^{c}:=\mathbb{N}\setminus A$. Unfortunately, this statement cannot produce a non-trivial value for both the lower and upper bound simultaneously. That is, a lower bound that is greater than 0 and an upper bound that is less than 1.

To see this, we first note that $A^{c}(q)+A(q)=q-1$. Then suppose that we have a non-trivial lower bound, so that

It must be the case that $A^{c}(q)=o(q)$ as $q\to\infty$, and therefore that $A(q)=q-1-o(q)$. Plugging this in to the upper bound we see that

Similarly, when the upper bound is less than 1, the lower bound is seen to be 0 by a similar argument.

Remark 3. The quantities

are often called the lower and upper logarithmic densities of the set $A$ (see for instance [4]). Thus, the bound for $\delta(A)$ given in the previous remark may be written succinctly as

We point out in passing that upper logarithmic density is connected to the theory of Dirichlet series via the identity

Here the quantity on the right is referred to as the abscissa of convergence. A proof of a more general statement can be found in Apostol’s book [1].

Constructing a set with prescribed density

So far we have seen examples of sets which have a Diophantine density of 0 as well as sets which have a Diophantine density of 1. In this section, we present a construction that shows that for any $x\in(0,1)$, there exists a set whose Diophantine density is $x$. In other words, using the language of Section 2, we show that the Diophantine density is rich.

Our plan is to split the sum into two parts, one over indices less than or equal to $L$ and one over those greater than $L$. For $k \le L$ we simply use the fact that $|\mathcal{A}_{x,k}| = N_k \le2^{xk}$ and $x > 0$ to establish the bound

For those indices greater than $L$ we take advantage of the spacing $d_k$ between elements of $\mathcal{A}_{x,k}$. Since $N_k \le2^{xk}$, we have

Then because $\lfloor y \rfloor\ge y/2$ for any $y > 1$, we see that

Thus, we obtain

Using the fact that $x < 1$, we find that

Since $n < 2^{L+1}$, we find that $2^{-(1-x)(L+1)} < n^{-(1-x)}$, which shows that

Combining the two halves of the sum gives the expected bound, namely

Finally, with $C_x$ denoting the implied constant in the bound above, an application of Theorem 5 shows that

By combining this with the lower bound derived above we obtain $\delta(\mathcal{A}_x) = x$, completing the proof. □

Example 1. As an example, let $x = 1/2$. Then $\delta(\mathcal{A}_{1/2}) = 1/2$ and

$\mathcal{A}_{1/2} = \{1, 2, 4, 6, 8, 11, 12, 16, 18, 22, 23, 24, 30, 32, 33, 40, 44, 45, 46, 48, 55, 56, 64, \ldots\}$.

Diophantine density and restricted approximation

In this section we turn to the original motivation for this paper, namely, restricted Diophantine approximation. The next theorem gives us an idea of how well an arbitrary irrational number can be approximated by a subset of $\mathbb{N}$.

Proof. Let $p_n/q_n$ be the $n$-th convergent of $\xi$. By a standard result from the theory of continued fractions (see e.g. [10]) we know that

Next, fix a real number $\delta'$, such that $\delta<\delta'<\delta(\mathcal{A})$. Since $\delta'<\delta(\mathcal{A})$, there is a $v_n\in\mathcal{A}$ and constant $C:=C(\mathcal{A},\delta')$ such that for large enough $n$, we have

where $N_n$ denotes the nearest integer to $v_np_n/q_n$.

Thus, by the triangle inequality and the fact that $v_n<q_n$, we find that

By Proposition 4, we know that $\delta(\mathcal{A})\leq1$, so that $1+\delta'<2$. Thus, we obtain the bound

All that we have left to show is that $v_n\to\infty$ as $n\to\infty$. If this were not the case, and there were only finitely many distinct denominators, then by the pigeonhole principle there must be a single $v$ for which $v_n=v$, for infinitely many $n$. Denote the subsequence of $n$’s for which this is true by $(n_k)_{k\geq0}$. Since $\delta'>0$, we find that

This is clearly a contradiction, as $\xi$ is irrational.

Therefore, we conclude that there are infinitely many distinct $v_n\in\mathcal{A}$, such that

Finally, since $\delta'-\delta>0$, we have $C+1<v_n^{\delta'-\delta}$ for all sufficiently large $n$. Consequently, we may write

With that our proof is complete. □

Definition 6

Definition 6. The restricted irrationality exponent $\mu_{\mathcal{A}}(\xi)$, of an irrational number $\xi$ with respect to a set $\mathcal{A}\subseteq\mathbb{N}$, is defined as the supremum of all real numbers $\mu$ such that the inequality

possesses infinitely many solutions $(p,q)\in\mathbb{Z}\times\mathcal{A}$. When $\mathcal{A}=\mathbb{N}$, we write $\mu(\xi)$ for the ordinary irrationality exponent of $\xi$.

Just as Dirichlet’s theorem guarantees that $\mu(\xi)\geq2$ for any irrational $\xi$, our Theorem 7 gives us the following lower bound for the restricted irrationality exponent.

Approximations involving Piatetski-Shapiro sequences

In this section, we investigate approximations whose denominators either are elements or avoid elements of Piatetski-Shapiro sequences, which are sequences of the form $(\lfloor n^\tau\rfloor)_{n\ge1}$ where $\tau>1$. Often, $\tau$ is defined to be non-integral, but here we do not impose such a restriction. These sequences are named after Piatetski-Shapiro [9], who showed that for $\tau\in(1,12/11)$, there are infinitely many primes of the form $\lfloor n^\tau\rfloor$.

To prepare for the remainder of this section, let us define the set

consisting of the elements of a given Piatetski-Shapiro sequence. The following lemma provides an upper and lower bound for the counting function of $S_\tau$.

Proof. Using Lemma 2 along with Theorem 5 easily establishes that

:::

Proof. We will begin by establishing the lower bound. By setting $\xi= p/q$ and

in Zaharescu’s theorem, we find that

Thus, for every coprime pair $(p,q)\in\mathbb{Z}\times\mathbb{N}$,

for any $\delta<2/7$. Consequently, by taking the supremum over all such $\delta$ we find that $\delta(S_2)\ge2/7$.

For the upper bound, the previous proposition gives us $\delta(S_2)\le1/2$, and our proof is complete. $\square$

Next, let us take a look at Diophantine approximation using the complement of a Piatetski-Shapiro set.

Proof. From Lemma 2 we know that $S_\tau(n)\ll n^{1/\tau}$. Then for some constant $C>0$, we have

We can then use Theorem 4 to produce the bound

Since we have required that $\tau>1$ so that $1-1/\tau>0$, Theorem 7 tells us that for any $\delta$ satisfying $0<\delta<1-1/\tau$, we have infinitely many pairs $(p,q)\in\mathbb{Z}\times(\mathbb{N}\setminus S_\tau)$, such that

As this is equivalent to the stated result, our proof is complete. $\square$

Open questions

There are a number of fundamental questions involving Diophantine density that have yet to be answered and that could provide promising avenues for further research. We present a few such problems below.

Question 1. Is the Diophantine density atomless?

Question 2. It is immediate that $\delta(\mathcal{A}\cup\mathcal{B})\ge\max\{\delta(\mathcal{A}),\delta(\mathcal{B})\}$. Are there sets $\mathcal{A}$ and $\mathcal{B}$ such that $\delta(\mathcal{A}\cup\mathcal{B})>\max\{\delta(\mathcal{A}),\delta(\mathcal{B})\}$?

Let us sketch one possible approach to this problem. The idea is to construct two sets, such that each set is missing a subset of $\mathbb{N}$ which other set possesses. So that separately, the Diophantine density of each set is limited by the missing subsets, but together, their union produces all of $\mathbb{N}$.

More specifically, let $x<1$ and define the sequence of sets $(E_n)_{n\ge0}$ by

where $(p_n/q_n)_{n\ge0}$ is an associated sequence of rational numbers. Now suppose we can construct the sequence $(p_n/q_n)_{n\ge0}$ recursively, so that the pairwise intersection of the even-indexed and odd-indexed sets of the sequence $(E_n)_{n\ge0}$ is empty. That is,

Then if we define

by construction we have $\max\{\delta(\mathcal{A}),\delta(\mathcal{B})\}\le x$, and $\delta(\mathcal{A}\cup\mathcal{B})=\delta(\mathbb{N})=1$. Since we have required that $x<1$, this construction gives a specific example of sets satisfying

Moreover, this would show that the Diophantine density is not subadditive. This is because for sufficiently small $x$, $\delta(\mathcal{A})+\delta(\mathcal{B})<1$, and therefore

If Question 1 can be answered in the affirmative, then we pose a related follow-up question. To state it, we will need the following definition.

In other words, a minimally dense set cannot be partitioned into two infinite subsets, where either subset is as dense as the whole. This behavior is in contrast to the behavior of $\mathbb{N}$, for instance, which is not minimally dense. By Proposition 5 we know that

gives a partition which violates the conditions of minimal density, since $\delta(\mathbb{N})=\delta(m\mathbb{N})$.

Question 3. Do minimally dense sets exist?

Question 4. How dependent is the Diophantine density of $\mathcal{A}$ on the growth of its counting function? By Theorem 5 we know that if $\mathcal{A}(q)\asymp q^{\alpha}$, then $0\le\delta(\mathcal{A})\le\alpha$. Can $\delta(\mathcal{A})$ take a value that is strictly less than $\alpha$?

Acknowledgments

The author would like to thank Jeff Ylvisaker and Jerry Federspiel for taking an interest in this work as well as for their questions and comments which helped shape the content of this paper.

Declaration of AI use. This work has benefited significantly from the assistance of ChatGPT 5.5, 5.6 and 6. Notably, ChatGPT was instrumental in developing the proofs of Theorems 4, 5 and 6.

Madison, Wisconsin, USA

Email address: [email protected]