1 Introduction

The ratio set (or quotient set) of a set of positive integers A is defined as

$$\begin{aligned} R(A) := \{a / b : a,b \in A\}. \end{aligned}$$

The study of the denseness of R(A) in the set of positive real numbers \(\mathbb {R}_+\) is a classical topic. For example, Strauch and Tóth [10] (see also [11]) showed that R(A) is dense in \(\mathbb {R}_+\) whenever A has lower asymptotic density at least equal to 1 / 2. Furthermore, Bukor et al. [3] proved that if \(\mathbb {N} = A \cup B\) for two disjoint sets A and B, then at least one of R(A) or R(B) is dense in \(\mathbb {R}_+\). On the other hand, Brown et al. [1] showed that there exist pairwise disjoint sets \(A,B,C \subseteq \mathbb {N}\) such that \(\mathbb {N} = A \cup B \cup C\) and none of R(A), R(B), R(C) is dense in \(\mathbb {R}_+\). See also [2, 4, 7, 8] for other related results.

More recently, the study of when R(A) is dense in the p-adic numbers \(\mathbb {Q}_p\), for some prime number p, has been initiated. Garcia and Luca [6] proved that the ratio set of the set of Fibonacci numbers is dense in \(\mathbb {Q}_p\), for all prime numbers p. Their result has been generalized by Sanna [9], who proved that the ratio set of the k-generalized Fibonacci numbers is dense in \(\mathbb {Q}_p\), for all integers \(k \ge 2\) and prime numbers p. Furthermore, Garcia et al. [5] gave several results on the denseness of R(A) in \(\mathbb {Q}_p\). In particular, they studied R(A) when A is the set of values of Lucas sequences, the set of positive integers which are sum of k squares, respectively, k cubes, or the union of two geometric progressions.

In this paper, we continued the study of the denseness of R(A) in \(\mathbb {Q}_p\).

2 Denseness of Members of Partitions of \(\mathbb {N}\)

Motivated by the results on partitions of \(\mathbb {N}\) mentioned in Introduction, the authors of [5] showed that for each prime number p there exists a partition of \(\mathbb {N}\) into two sets A and B such that neither R(A) nor R(B) is dense in \(\mathbb {Q}_p\) [5, Example 3.6]. Then, they asked the following question [5, Problem 3.7]:

Question 2.1

Is there a partition of \(\mathbb {N}\) into two sets A and B such that R(A) and R(B) are dense in no \(\mathbb {Q}_p\)?Footnote 1

We show that the answer to Question 2.1 is negative. In fact, we will prove even more. Our first result is the following:

Theorem 2.1

Let \(A_1, \ldots , A_k\) be a partition of \(\mathbb {N}\) into k sets. Then, for all prime numbers p but at most \(k - 1\) exceptions, at least one of \(A_1, \ldots , A_k\) is dense in \(\mathbb {Z}_p\).

Then, from Theorem 2.1 it follows the next corollary, which gives a strong negative answer to Question 2.1.

Corollary 2.1

Let \(A_1, \ldots , A_k\) be a partition of \(\mathbb {N}\) into k sets. Then, for all prime numbers p but at most \(k - 1\) exceptions, at least one of \(R(A_1), \ldots , R(A_k)\) is dense in \(\mathbb {Q}_p\).

Proof

It is easy to prove that if \(A_j\) is dense in \(\mathbb {Z}_p\) then \(R(A_j)\) is dense in \(\mathbb {Q}_p\). Hence, the claim follows from Theorem 2.1. \(\square \)

The proof of Theorem 2.1 requires just a couple of easy preliminary lemmas. For positive integers a and b, define \(a + b\mathbb {N} := \{a + bk : k \in \mathbb {N}\}\).

Lemma 2.2

Suppose that \((a + b\mathbb {N}) \subseteq A \cup B\) for some positive integers a, b and some disjoint sets \(A, B \subseteq \mathbb {N}\). If p is a prime number such that \(p \not \mid b\) and A is not dense in \(\mathbb {Z}_p\), then there exist positive integers c and j such that \((c + bp^j\mathbb {N}) \subseteq B\).

Proof

Since A is not dense in \(\mathbb {Z}_p\), there exist positive integers d, j such that \((d + p^j\mathbb {N}) \cap A = \varnothing \). Hence, \((a + b\mathbb {N}) \cap (d + p^j\mathbb {N}) \subseteq B\). The claim follows by the Chinese remainder theorem, which implies that \((a + b\mathbb {N}) \cap (d + p^j\mathbb {N}) = c + bp^j\mathbb {N}\), for some positive integer c. \(\square \)

Lemma 2.3

Let a and b be positive integers. Then, \(a + b \mathbb {N}\) is dense in \(\mathbb {Z}_p\) for all prime numbers p such that \(p \not \mid b\).

Proof

It follows from the Chinese remainder theorem and the fact that \(\mathbb {N}\) is dense in \(\mathbb {Z}_p\). \(\square \)

We are now ready for the proof of Theorem 2.1.

Proof of Theorem 2.1

For the sake of contradiction, suppose that \(p_1, \ldots , p_k\) are k pairwise distinct prime numbers such that none of \(A_1, \ldots , A_k\) is dense in \(\mathbb {Z}_{p_i}\) for \(i = 1,\ldots ,k\). Since \(A_1\) is not dense in \(\mathbb {Z}_{p_1}\), there exist positive integers \(c_1\) and \(j_1\) such that \((c_1 + p_1^{j_1}\mathbb {N}) \cap A_1 = \varnothing \). Hence, \((c_1 + p_1^{j_1}\mathbb {N}) \subseteq A_2 \cup \cdots \cup A_k\) and, thanks to Lemma 2.2, there exist positive integers \(c_2\) and \(j_2\) such that \((c_2 + p_1^{j_1} p_2^{j_2}\mathbb {N}) \subseteq A_3 \cup \cdots \cup A_k\). Continuing this process, we get that \((c_{k-1} + p_1^{j_1} \cdots p_{k-1}^{j_{k-1}}\mathbb {N}) \subseteq A_k\), for some positive integers \(c_{k-1}, j_1, \ldots , j_{k-1}\). By Lemma 2.3, this last inclusion implies that \(A_k\) is dense in \(\mathbb {Z}_{p_k}\), but this contradicts the hypotheses. \(\square \)

Remark 2.1

In fact, Theorem 2.1 can be strengthened in the following way: For each partition \(A_1, \ldots , A_k\) of \(\mathbb {N}\), there exists a member \(A_j\) of this partition which is dense in \(\mathbb {Z}_p\) for all but at most \(k-1\) prime numbers p.

Indeed, for the sake of contradiction, suppose that each member \(A_j\) of the partition \(A_1, \ldots , A_k\) of \(\mathbb {N}\) has at least k prime numbers p such that \(A_j\) is not dense in \(\mathbb {Z}_p\). Then, we can choose prime numbers \(p_1, \ldots , p_k\) such that for each \(j\in \{1,\ldots ,k\}\) the set \(A_j\) is not dense in \(\mathbb {Z}_{p_j}\). Next, we provide the reasoning from the proof of Theorem 2.1 to reach a contradiction.

The next result shows that the quantity \(k - 1\) in Theorem 2.1 cannot be improved.

Theorem 2.4

Let \(k \ge 2\) be an integer, and let \(p_1, \ldots , p_{k-1}\) be pairwise distinct prime numbers. Then, there exists a partition \(A_1, \ldots , A_k\) of \(\mathbb {N}\) such that none of \(A_1, \ldots , A_k\) is dense in \(\mathbb {Z}_{p_i}\) for \(i=1, \ldots , k - 1\).

Proof

Let \(e_1, \ldots , e_{k-1}\) be positive integers such that \(p_i^{e_i} \ge k\) for \(i=1,\ldots ,k-1\), and put

$$\begin{aligned} V := \left\{ 0, \ldots , p_1^{e_1} - 1\right\} \times \cdots \times \left\{ 0, \ldots , p_{k-1}^{e_{k-1}} - 1\right\} . \end{aligned}$$

We shall construct a partition \(R_0, \ldots , R_{k-1}\) of V (note that the indices of \(R_i\) start from 0) such that if \((r_1, \ldots , r_{k-1}) \in R_j\) then none of the components \(r_1, \ldots , r_{k-1}\) is equal to j. Then, we define

$$\begin{aligned} A_j := \left\{ n \in \mathbb {N}: \exists (r_1, \ldots , r_{k-1}) \in R_{j-1}, \; \forall i = 1, \ldots , k - 1, \quad n \equiv r_i \pmod {p_i^{e_i}} \right\} , \end{aligned}$$

for \(j = 1, \ldots , k\). At this point, it follows easily that \(A_1, \ldots , A_k\) is a partition of \(\mathbb {N}\) and that none of \(A_1, \ldots , A_k\) is dense in \(\mathbb {Z}_{p_i}\), since \(A_{j+1}\) misses the residue class \(j \pmod {p_i^{e_i}}\).

The construction of \(R_0, \ldots , R_{k-1}\) is algorithmic. We start with \(R_0, \ldots , R_{k-1}\) all empty. Then, we pick a vector \(\mathbf {x} \in V\) which is not already in \(R_0 \cup \cdots \cup R_{k-1}\). It is easy to see that there exists some \(j\in \{0,\ldots ,k-1\}\) such that j does not appear as a component of \(\mathbf {x}\). We thus throw \(\mathbf {x}\) into \(R_j\). We continue this process until all the vectors in V have been picked.

Now, by the construction it is clear that \(R_0, \ldots , R_{k-1}\) is a partition of V satisfying the desired property. \(\square \)

3 Denseness of Ratio Sets of Members of Partitions of \(\mathbb {N}\)

The result in Corollary 2.1 is not optimal. Let \(\lfloor x \rfloor \) denote the greatest integer not exceeding x, and write \(\log _2\) for the base 2 logarithm. Our next result is the following:

Theorem 3.1

Let \(A_1, \ldots , A_k\) be a partition of \(\mathbb {N}\) into k sets. Then, for all prime numbers p but at most \(\lfloor \log _2 k\rfloor \) exceptions, at least one of \(R(A_1), \ldots , R(A_k)\) is dense in \(\mathbb {Q}_p\).

Before proving Theorem 3.1, we need to introduce some notations. For a prime number p and a positive integer w, we identify the group \((\mathbb {Z}/p^w\mathbb {Z})^*\) with \(\{a \in \{1, \ldots , p^w\} : p \not \mid a\}\). Moreover, for each \(a \in (\mathbb {Z}/p^w\mathbb {Z})^*\) we define

$$\begin{aligned} (a)_{p^w} := \left\{ x \in \mathbb {Q}_p^* : x / p^{\nu _p(x)} \equiv a \bmod p^w \right\} , \end{aligned}$$

where, as usual, \(\nu _p\) denotes the p-adic valuation. Note that the family of sets

$$\begin{aligned} (a)_{p^w} \cap \nu _p^{-1}(s) = \left\{ \left( a + rp^w\right) p^s : r \in \mathbb {Z}_p\right\} \end{aligned}$$

where w is a positive integer, \(a \in (\mathbb {Z}/p^w\mathbb {Z})^*\), and \(s \in \mathbb {Z}\), is a basis of the topology of \(\mathbb {Q}_p^*\). Finally, for all integers \(t \le m\) and for each set \(X \subseteq \mathbb {N}\), we define

$$\begin{aligned} V_{p^w, t, m} := \left\{ (a)_{p^w} \cap \nu _p^{-1}(s) : a \in \left( \mathbb {Z}/p^w\mathbb {Z}\right) ^*,\; s \in \mathbb {Z} \cap {[t, m - 1]} \right\} \end{aligned}$$

and

$$\begin{aligned} V_{p^w, t, m}(X) := \left\{ I \in V_{p^w, t, m} : X \cap I \ne \varnothing \right\} . \end{aligned}$$

Note that the following trivial upper bound holds

$$\begin{aligned} \#V_{p^w, t, m}(X) \le \#V_{p^w, t, m} = (m - t) \varphi \left( p^w\right) , \end{aligned}$$

where \(\varphi \) is the Euler’s totient function.

Now, we are ready to state a lemma that will be crucial in the proof of Theorem 3.1.

Lemma 3.2

Fix a prime number p, two positive integers w, t, a real number \(c > 1/2\), and a set \(X \subseteq \mathbb {N}\). Suppose that \(\#V_{p^w, 0, m}(X) \ge c m \,\varphi (p^w)\) for some positive integer \(m > t / (2c - 1)\). Then, the ratio set R(X) intersects nontrivially with each set in \(V_{p^w, 0, t}\).

Proof

Given \((a_0)_{p^w} \cap \nu _p^{-1}(s_0) \in V_{p^w, 0, t}\), we have to prove that \(R(X) \cap (a_0)_{p^w} \cap \nu _p^{-1}(s_0) \ne \varnothing \). For the sake of convenience, define \(A := V_{p^w, t, m}(X)\) and

$$\begin{aligned} B := \left\{ (a_0a)_{p^w} \cap \nu _p^{-1}(s_0 + s) : (a)_{p^w} \cap \nu _p^{-1}(s) \in V_{p^w, t - s_0, m - s_0}(X) \right\} . \end{aligned}$$

We have

$$\begin{aligned} \#A = \#V_{p^w, 0, m}(X) - \#V_{p^w, 0, t}(X) \ge (c m - t) \varphi \left( p^w\right) > \frac{1}{2} (m - t) \varphi \left( p^w\right) , \end{aligned}$$
(1)

where we used the inequality \(m > t / (2c - 1)\). Similarly,

$$\begin{aligned} \#B&= \#V_{p^w, 0, m}(X) - \#V_{p^w, 0, t - s_0}(X) - \#V_{p^w, m - s_0, m}(X) \nonumber \\&\ge (cm - (t-s_0) - s_0)\varphi \left( p^w\right) > \frac{1}{2} (m - t) \varphi \left( p^w\right) . \end{aligned}$$
(2)

Now, A and B are both subsets of \(V_{p^w, t, m}\), while \(\#V_{p^w, t, m} = (m - t) \varphi (p^w)\). Therefore, (1) and (2) imply that \(A \cap B \ne \varnothing \). That is, there exist \((a_1)_{p^w} \cap \nu _p^{-1}(s_1) \in A\) and \((a_2)_{p^w} \cap \nu _p^{-1}(s_2) \in V_{p^w, t - s_0, m - s_0}(X)\) such that \(a_1 / a_2 \equiv a_0 \pmod {p^w}\) and \(s_1 - s_2 = s_0\), so that \(R(X) \cap (a_0)_{p^w} \cap \nu _p^{-1}(s_0) \ne \varnothing \), as claimed. \(\square \)

Proof of Theorem 3.1

For the sake of contradiction, put \(\ell := \lfloor \log _2 k \rfloor + 1\) and suppose that \(p_1, \ldots , p_\ell \) are \(\ell \) pairwise distinct prime numbers such that none of \(R(A_1), \ldots , R(A_k)\) is dense in \(\mathbb {Q}_{p_i}\) for \(i = 1, \ldots , \ell \). Hence, there exist positive integers w and t such that for each \(i \in \{1, \ldots , k\}\) and each \(j \in \{1, \ldots , \ell \}\) we have \(R(A_i) \cap (a_{i,j})_{p_j^w} \cap \nu _{p_j}^{-1}(s_{i,j}) = \varnothing \), for some \(a_{i,j} \in (\mathbb {Z}/p_j^w\mathbb {Z})^*\) and some \(s_{i,j} \in \{-(t-1), \ldots , t-1\}\). Clearly, since ratio sets are closed under taking reciprocals, we can assume \(s_{i,j} \ge 0\). Put \(c := 1 / \root \ell \of {k}\), so that \(c > 1/2\), and pick a positive integer \(m > t / (2c - 1)\). There are

$$\begin{aligned} N := m^\ell \prod _{j=1}^\ell \varphi \left( p_j^w\right) \end{aligned}$$

sets of the form

$$\begin{aligned} \bigcap _{j = 1}^\ell \left( (a_j)_{p_j^w} \cap \nu _{p_j}^{-1}(s_j)\right) , \end{aligned}$$
(3)

where \(a_j \in (\mathbb {Z}/p_j^w\mathbb {Z})^*\) and \(s_j \in \{0, \ldots , m - 1\}\). Therefore, there exists \(i_0 \in \{1, \ldots , k\}\) such that \(A_{i_0}\) intersects nontrivially with at least N / k of the sets of form (3). Consequently, there exists \(j_0 \in \{1, \ldots , \ell \}\) such that \(A_{i_0}\) intersects nontrivially with at least \(c m \varphi (p_{j_0}^w)\) sets of the form \((a)_{p_{j_0}^w} \cap \nu _{p_{j_0}}^{-1}(s)\), where \(a \in (\mathbb {Z}/p_{j_0}^w\mathbb {Z})^*\) and \(s \in \{0, \ldots , m - 1\}\). In other words, \(\#V_{p_{j_0}^w, 0, m}(A_{i_0}) \ge c m \varphi (p_{j_0}^w)\). Hence, by Lemma 3.2, the set \(R(A_{i_0})\) intersects nontrivially with all the sets of the form \((a)_{p_{j_0}^w} \cap \nu _{p_{j_0}}^{-1}(s)\), where \(a \in (\mathbb {Z}/p_{j_0}^w\mathbb {Z})^*\) and \(s \in \{0, \ldots , t-1\}\), but this is in contradiction with the fact that \(R(A_{i_0}) \cap (a_{i_0, j_0})_{p_{j_0}^w} \cap \nu _{p_{j_0}}^{-1}(s_{i_0, j_0}) = \varnothing \). \(\square \)

The bound \(\lfloor \log _2 k \rfloor \) in Theorem 3.1 is sharp in the following sense:

Theorem 3.3

Let \(k \ge 2\) be an integer, and let \(p_1< \cdots < p_\ell \) be \(\ell := \lfloor \log _2 k \rfloor \) pairwise distinct prime numbers. Then, there exists a partition of \(\mathbb {N}\) into k sets \(A_1, \ldots , A_k\) such that none of \(R(A_1), \ldots , R(A_k)\) is dense in \(\mathbb {Q}_{p_i}\) for \(i = 1, \ldots , \ell \).

Proof

We give two different constructions. Put \(h := 2^\ell \) and let \(S_1, \ldots , S_h\) be all the subsets of \(\{1, \ldots , \ell \}\). For \(j = 1, \ldots , h\), define

$$\begin{aligned} B_j := \{n \in \mathbb {N} : \forall i = 1, \ldots , \ell \quad \nu _{p_i}(n) \equiv \chi _{S_j}(i) \pmod 2\}, \end{aligned}$$

where \(\chi _{S_j}\) denotes the characteristic function of \(S_j\). It follows easily that \(B_1, \ldots , B_h\) is a partition of \(\mathbb {N}\) and that none of \(R(B_1), \ldots , R(B_h)\) is dense in \(\mathbb {Q}_{p_i}\), for \(i=1, \ldots , \ell \), since each \(R(B_j)\) contains only rational numbers with even \(p_i\)-adic valuations. Finally, since \(h \le k\), the partition \(B_1, \ldots , B_h\) can be refined to obtain a partition \(A_1, \ldots , A_k\) satisfying the desired property.

The second construction is similar. For \(j=1,\ldots ,h\), define

$$\begin{aligned} C_j=\left\{ n\in \mathbb {N} : \left( \frac{n/p_i^{v_{p_i}(n)}}{p_i}\right) =(-1)^{\chi _{S_j}(i)} \text{ for } \text{ each } i\in \{1,\ldots ,\ell \}\right\} , \end{aligned}$$

where \(\left( \frac{a}{p}\right) \) means the Legendre symbol and in case of \(p_1=2\) we put \(\left( \frac{a}{2}\right) = a \pmod {4}\). It follows easily that \(C_1, \ldots , C_h\) is a partition of \(\mathbb {N}\), and that none of \(R(C_1), \ldots , R(C_h)\) is dense in \(\mathbb {Q}_{p_i}\), for \(i = 1, \ldots , \ell \), since each \(R(C_j)\) contains only products of powers of \(p_i\) and quadratic residues modulo \(p_i\) (in case of \(p_1 = 2\) we have only products of powers of 2 and numbers congruent to 1 modulo 4). Finally, since \(h \le k\), the partition \(C_1, \ldots , C_h\) can be refined to obtain a partition \(A_1, \ldots , A_k\) satisfying the desired property. \(\square \)

In light of Remark 2.1, it is worth to ask the following question.

Question 3.1

Let us fix a positive integer k. What then is the least number \(m=m(k)\) such that for each partition \(A_1, \ldots , A_k\) of \(\mathbb {N}\) there exists a member \(A_j\) of this partition such that \(R(A_j)\) is dense in \(\mathbb {Q}_p\) for all but at most m prime numbers p?

In virtue of Remark 2.1, we know that m(k) exists and \(m(k)\le k-1\). On the other hand, by Theorem 3.3 the value m(k) is not less than \(\left\lfloor \log _2 k\right\rfloor \).