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Let $X \subseteq \mathbb{N}^2$ be the set $\{(x,y) \, : \, y \text{ is the greatest power of 2 dividing } x\}$. I'm wondering how the set gets when multipled by the non negative rationals, i.e. what is $\mathbb{Q}_{\geq 0}X$?

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"How the set gets"? – rschwieb Oct 26 '12 at 16:31
up vote 2 down vote accepted

Claim: $$\mathbb Q_{\ge 0}X = \left\{(x,y)\in\mathbb Q_{>0}^2\mid \frac xy\in 2\mathbb N-1\right\}\cup\{(0,0)\}$$


"$\supseteq$": Clearly $(0,0)= 0\cdot (1,1)\in \mathbb Q_{\ge 0}X$ becuse $(1,1)\in X$. Let $x,y,$ be positive rational numbers such that $k:= \frac xy$ is an odd natural number. Then $(k,1)\in X$ and $y\cdot(k,1)=(x,y)\in\mathbb Q_{\ge 0}X$.

"$\subseteq$": Let $(n,m)\in X$ and $q\in \mathbb Q_{\ge0}$. If $q=0$, then $q\cdot(n,m)=(0,0)$ and we are done. Otherwise, $q=\frac rs$ for suitable natural numbers $r,s$. Also, we have $m=2^k$ with $k\in\mathbb N_0$ and $n=um$ with $u$ an odd natural number. Note that $q\cdot(n.m)=(x,y)$ with $x=qn$, $y=qm$. Then $x>0$ and $y>0$ because $q$, $n$ and $m$ are positive. Moreover, $\frac xy=\frac{n}{m}=u$ is an odd natural number, i.e. an element of $2\mathbb N-1$.$_\blacksquare$

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Thank you! Do you mind giving a short explanation? Or just an intuition? – Mark T. Oct 26 '12 at 16:23

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