I am looking for a set of positive integers $(\mathbb{S} \space\subset\space \mathbb{N} := \{1,2,3,...\})$ such that its "2-element quotients" are dense over the full set of positive rational numbers (which would also make it dense over the full set of positive real numbers).

One obvious satisfying set is $\{An+B \space\space|\space\space n\in\mathbb{N}\}$ where $A$ and $B$ are any given positive integers.

What about higher powers ($C$ is a given integer greater than 1) or square-free numbers?

$$\{An^C+B \space\space|\space\space n\in\mathbb{N}\}$$ $$\{\prod _{primes} {p^{b_p} \space\space|\space\space b_p \in \{0,1}\}\}$$

By the way, I believe I could eliminate any finite number of elements from any satisfying set and still have a satisfying set, so I am really thinking about the set's "asymptotic form".


4 Answers 4


Say $\mathbb{S}\subset\mathbb{N}$ has the dense rational property if the image of $\div:\mathbb{S}\times\mathbb{S} \rightarrow\mathbb{R}^+$ is dense. Let $s_n$ be the enumeration of $\mathbb{S}$, in increasing order. Then:

Claim: $\mathbb{S}$ has the dense rational property if $$ \lim\limits_{n\rightarrow\infty} \frac{s_n}{s_{n+1}}=1 $$ In particular, any set with a natural density has the dense rational property, including the square-free numbers. Also, the images of polynomials have the dense rational property, as do the prime numbers.

Conversely, $\mathbb{S}$ does not have the dense rational property if $$ \limsup\limits_{n\rightarrow\infty} \frac{s_n}{s_{n+1}} < 1 $$

Proof: Suppose $\limsup\limits_{n\rightarrow\infty} \frac{s_n}{s_{n+1}} < 1$. Then, there exists $R<1$ and $N\in\mathbb{N}$ such that $n\ge N$ implies $$ \frac{s_n}{s_{n+1}} < R $$ Thus for any $n>m\ge N$, we have $$ \frac{s_m}{s_n} < R^{n-m} \leq R $$ and $$ \frac{s_n}{s_m} > R^{m-n} \geq R^{-1} $$ hence the intersection of $\mathbb{S}\div \mathbb{S}=\left\{\frac{s}{t}\mid s,t\in\mathbb{S}\right\}$ has a finite intersection with the interval $[R,R^{-1}]$, and therefore cannot be dense.

Conversely, suppose $$ \lim\limits_{n\rightarrow\infty} \frac{s_n}{s_{n+1}} = 1 $$ It suffices to show that $\mathbb{S}\div \mathbb{S}$ is dense in $(1,\infty)$, since it is closed under taking reciprocals, and thus it suffices to show that for any $(a,b)\subset(1,\infty)$, there exist integers $n_1,n_2$ such that $$ \frac{s_{n_1 + n_2}}{s_{n_1}} \in (a,b) $$ Let $n_1$ be such that $n\ge n_1$ implies $\frac{s_{n + 1}}{s_{n}} < \frac{b}{a}$. Then, we observe that $$ \lim_{n\rightarrow\infty} \frac{s_{n_1+n}}{s_{n_1}} = \infty $$ Then we can define $$n_2 = \max\left\{n \text{ }:\text{ } \frac{s_{n_1+n}}{s_{n_1}} < b\right\} $$ Trivially, $\frac{s_{n_1+n_2}}{s_{n_1}} < b$ and $\frac{s_{n_1+n_2+1}}{s_{n_1}} \ge b$. We also have \begin{eqnarray} b&\le& \frac{s_{n_1+n_2+1}}{s_{n_1}} = \frac{s_{n_1+n_2}}{s_{n_1}} \cdot \frac{s_{n_1 + n_2+1}}{s_{n_1 +n_2}} < \frac{s_{n_1+n_2}}{s_{n_1}}\cdot \frac b a\\ a &<& \frac{s_{n_1+n_2}}{s_{n_1}} \end{eqnarray} as desired.

  • $\begingroup$ This is by far the most general answer (yet), solving the problem for vast classes of sets! $\endgroup$
    – Ingix
    Jul 27, 2022 at 21:38
  • $\begingroup$ After "implies", I think you want $n$ (instead of $n_1$). All is very easy to follow then and gives an easy general check, thank you. Pushing further with Dirichlet's theorem, this shows primes of the form p=An+B (for any coprime A and B) have the dense rational property too. And, with the Generalized Bunyakovsky conjecture, perhaps primes satisfying any "coprime" system of polynomials have the dense rational property. $\endgroup$
    – bobuhito
    Jul 28, 2022 at 13:46
  • $\begingroup$ Thanks for the interesting question! It isn't fully general, as there are sequences where $\limsup \frac{s_n}{s_{n+1}} = 1$ but $\liminf \frac{s_n}{s_{n+1}} < 1$. I will update if I find any finer tests for such sets. $\endgroup$ Aug 2, 2022 at 11:43

One set that qualifies is the products of powers of any two distinct prime numbers

$$\lbrace p^mq^n | m, n \in \mathbb{N}; p, q \in \mathbb{P}; p \ne q\rbrace$$

Then any quotient between them has the form $p^mq^n$ with $m, n \in \mathbb{Z}$. And in any real interval $(a, b)$, you can choose $n > \frac{1}{\log_q \frac{b}{a}}$ and then $m \in (\log_p{\frac{a}{q^n}}, \log_p{\frac{b}{q^n}})$, so that $p^mq^n \in (a, b)$, thus satisfying your density requirement.

This, for example, allows Pythagorean tuning ($p = 2, q = 3$) to approximate any possible musical interval.

  • 1
    $\begingroup$ Good qualifying set, thank you. Any comment on whether my two sets qualify or not? Even if the answer for those two is not known, I hope to see more answers like yours just to better understand this general problem. $\endgroup$
    – bobuhito
    Jul 26, 2022 at 16:33

$$\{An^C+B \space\space|\space\space n\in\mathbb{N}\}$$

works for any real $C>0$.

$0$ can obviously easily approximated to any desired accurary wth this set, so let $T>0$ be the target.

The arguments that prove $$\{An+B \space\space|\space\space n\in\mathbb{N}\}$$ works make sure that you can get arbitrarily close to $\frac{n^C}{m^C}$ for any positve intergers $n,m$. By choosing $n,m$ such that $\frac{n}{m}$ is close to $T^\frac1C$, we can thus get as close as desired to T, since $x^C$ is continuous.

  • $\begingroup$ Doh, I guess I should have seen that! So, in some sense, the integers need to go up at least exponentially to not have dense rationals. $\endgroup$
    – bobuhito
    Jul 27, 2022 at 5:19

The square free numbers work. By https://en.wikipedia.org/wiki/Square-free_integer, the estimate on $Q(x)$, the number of square-free numbers up to $x$, $$Q(x) = \frac{6}{\pi^2}x+O(\sqrt{x})$$ implies that for some constant $c$, for any real number $x$ there exists a square-free number between $x$ and $x+c\sqrt{x}$. From the aforementioned estimate, there exists some constant $C$ such that $$Q(x) \in \left[\frac{6}{\pi^2}x - C\sqrt{x}, \frac{6}{\pi^2}x + C\sqrt{x}\right].$$ Pick $c$ to be determined later, and note that $$Q(x+c\sqrt{x}) - Q(x) \geq \frac{6}{\pi^2}c\sqrt{x} - C\sqrt{x} - C\sqrt{x + c\sqrt{x}}.$$ Since sqrt is concave and thus subadditive, we see that $$Q(x+c\sqrt{x}) - Q(x) \geq \frac{6}{\pi^2}c\sqrt{x} - C\sqrt{x} - C\sqrt{x} -C \sqrt{c}\sqrt{\sqrt{x}}.$$ We can find some $c$ that for all $x$ greater than 10, this difference is at least 1. More generally, this shows that sets which increase roughly linearly have a sort of bound like this.

To leverage this, pick a number $p$ greater than 1. Pick $N$ large to be determined later, and approximate $p^{N+1}$ and $p^N$ by square-free numbers $a,b$ in the intervals $[p^{N+1}, p^{N+1} + c\sqrt{p^{N+1}}], [p^{N+1}, p^{N+1} + c\sqrt{p^{N+1}}].$ Now note that $$\left|\frac{a}{b} - p\right| \geq cp^{-N/2+1}.$$

The case for when $0 < p < 1$ is solved by approximating $1/p$.

The other comments established that sets of the form $\{An^B+C : n \in \mathbb{N}\}$ work. The simplest counterexample I could come up with for a set that doesn't work is indeed "exponential" in a sense: $$A = \{2^n : n \in \mathbb{N}\}.$$ For this set, the counting function $A(x)$ that computes the number of elements of $A$ up to $x$ is basically $\log_2(x) \pm 1.$ This basically says that for any number $x$, we can find an element of $A$ in the interval $(x,2x)$, and if we try to repeat the reasoning I used before to approximate a number $p > 1$, it doesn't work.

In contrast, if for a set $B$ we have a counting function $B(x)$ that grows any faster than $\log$ (i.e., suppose $B(x)=x^b$ for $0 < b < 1$,) we can find some power $d < b$ such that $$B(x) \sim x^b + O(x^d).$$ A similar argument as before would show that this set satisfies your hypotheses, though the onus would be on you to prove enough about the counting function of your set so that you could use the arguments above.

  • 2
    $\begingroup$ There are some big flaws in this proof (e.g., to find the gaps between square-free numbers, you will need to find an upper bound on C), but your conclusion is true. To take this further, I don't even need products of primes: I can just use individual primes (which are trivially square-free) for my set (math.stackexchange.com/questions/1013414). I can probably even pare it down further (e.g., using only primes of the form 4n+1). $\endgroup$
    – bobuhito
    Jul 27, 2022 at 12:56

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