most recent 30 from http://mathoverflow.net 2013-05-24T02:47:14Z http://mathoverflow.net/feeds/question/96434 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96434/particular-subset-of-integers-generating-rational-numbers Nekochan 2012-05-09T11:19:16Z 2012-05-10T09:36:57Z

<p>Hello, maybe this is a naive question, but so far I did not found anything related to the subject.</p> <p>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 $\mathbb{Q}$.</p> <p>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 $E$? </p> <p>Thanks by advance for your comments!</p>

http://mathoverflow.net/questions/96434/particular-subset-of-integers-generating-rational-numbers/96457#96457 Marc Chapuis 2012-05-09T14:57:22Z 2012-05-10T09:36:57Z

<p>This should be a comment as it does not answer the question, but I do not have enough points.</p> <p>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).</p> <p>So apart from checking for this, a faster method would depend on how E is defined.</p> <p>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?</p>

http://mathoverflow.net/questions/96434/particular-subset-of-integers-generating-rational-numbers/96473#96473 Ramiro de la Vega 2012-05-09T16:32:54Z 2012-05-09T16:32:54Z

<p>This is more of a long comment than an answer.</p> <p>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.</p> <p>For instance (as a reply to a comment of Gjergji Zaimi), you can make $E$ to avoid some multiple of each natural number.</p>

http://mathoverflow.net/questions/96434/particular-subset-of-integers-generating-rational-numbers/96476#96476 Nik Weaver 2012-05-09T16:50:22Z 2012-05-09T16:50:22Z

<p>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.</p>