1
$\begingroup$

Hello, maybe this is a naive question, but so far I did not found anything related to the subject.

I would like to consider a subset of integers, say E, such that the set $\{ \frac{x}{y}, x \in E, y \in E, y \neq 0 \}$ is $\mathbb{Q}$.

Do such sets have a particular name? Is anyone known for having studied them? And is it possible to define such a set for which any (positive or non-zero) rational is uniquely represented as a ratio of elements in $E$?

Thanks by advance for your comments!

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ The representation cannot be unique for all (non-zero) rational numbers, since $-1 = \frac{x}{y}$ implies $-1 = \frac{y}{x}$. As we must have $x \neq y$, we therefore have (at least) two different representations of $-1$. $\endgroup$
    – felix
    Commented May 9, 2012 at 11:42
  • $\begingroup$ This might be a trivial comment, but if you insist on $E\subset \mathbb{Z}$ (it might be better for $E\subset \mathbb{Z}^2$), uniqueness is out of the question: clearly $E$ contains more than one nonzero element, and then $1=x/x = y/y$ for $0\neq x,y\in E$. $\endgroup$
    – Philip van Reeuwijk
    Commented May 9, 2012 at 11:43
  • $\begingroup$ Right, I forgot to add these facts, thanks! Anyway the question of uniqueness is not my first concern, mainly I would like to know if for a given set $E$, there are known methods to determine whether the set of quotients is $\mathbb{Q}$. $\endgroup$
    – Nekochan
    Commented May 9, 2012 at 11:48
  • $\begingroup$ (or $\mathbb{Q}_+$, or $\mathbb{Q}_+^*$, depends if you choose $E$ as a subset of positive integers or integers). $\endgroup$
    – Nekochan
    Commented May 9, 2012 at 11:50
  • $\begingroup$ Is there an example where E doesn't contain all multiples of some natural number? $\endgroup$
    – Gjergji Zaimi
    Commented May 9, 2012 at 15:17

3 Answers 3

Reset to default
6
$\begingroup$

The non-uniqueness of $1 = x/x = y/y$ pointed out by Philip is the only obstruction to uniqueness of representations of positive rationals. To see this, enumerate the rationals in $(0,1)$ as $a_1, a_2, \ldots$. (If we have uniqueness here, then taking reciprocals gives us uniqueness on $(1,\infty)$ as well.) We build up the set $E$ two elements at a time. At any stage we will have achieved unique representation of finitely many rationals in $(0,1)$. Say $a_n$ is the first rational in our list that is not yet represented; we just have to add two natural numbers $x$ and $y$ such that $x/y = a_n$ and the ratios of $x$ and $y$ with previous elements of $E$ don't duplicate any rationals that were already attained. But it's obvious that we can do this by taking $x = ma_n$ and $y = m$ for a large value of $m$; we can ensure that ratios of $x$ and $y$ with previous elements of $E$ are smaller than any rationals that were already attained.

Improve this answer
$\endgroup$
0
4
$\begingroup$

This is more of a long comment than an answer.

Let $\{ (p_n,q_n) :n \in \mathbb{N} \}$ be an enumeration of all pairs of integers and let $\{a_n :n \in \mathbb{N} \}$ be any sequence of non-zero integers. Then it is clear that $$E:= \{a_np_n : n \in \mathbb{N} \} \cup \{a_nq_n : n \in \mathbb{N} \}$$ satisfies what you want. The point is that you can inductively define your sequence $\{a_n :n \in \mathbb{N} \}$ in order to make $E$ as scattered as you wish.

For instance (as a reply to a comment of Gjergji Zaimi), you can make $E$ to avoid some multiple of each natural number.

Improve this answer
$\endgroup$
0
0
$\begingroup$

This should be a comment as it does not answer the question, but I do not have enough points.

A sufficient, and necessary, condition for $E$ to generate $\mathbb{Q}$ is that $0 \in E$ and for every pair of numbers $(p,q)\in \mathbb Z \times \mathbb N^\ast $ such that $gcd(p,q) = 1$, there exists $m \in \mathbb Z^\ast$ such that $pm \in E$ and $qm \in E$. The multiplicities of such $m$ corresponding to the multiplicities of representations, as stated above $p=q=1$ gives non unique solutions, actually an infinite number of them (as E needs to be bigger then the set of prime numbers).

So apart from checking for this, a faster method would depend on how E is defined.

I'm wondering what are the smallests such sets so that every quotient $\frac p q$ admits at least n representations, or what are the smallests sets, if they exist, such that every quotient (apart from the pathological cases 0, 1, -1...) admits exactly n representations?

Improve this answer
$\endgroup$
3
  • $\begingroup$ One key word in this context is "difference set" en.wikipedia.org/wiki/Difference_set ; yet these investigations are typically for finite structures. $\endgroup$
    – user9072
    Commented May 9, 2012 at 16:27
  • $\begingroup$ I don't think $p\wedge q$ is standard notation in number theory. $\endgroup$
    – Greg Martin
    Commented May 9, 2012 at 17:21
  • $\begingroup$ Duly noted, $\wedge$ is the notation I grew up with for g.c.d., but I changed it in the post. Difference sets look quite interesting, thank you quid. $\endgroup$
    – Marc Chapuis
    Commented May 10, 2012 at 9:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .