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Let we have a Dedekind cut set $\alpha$ and $w$ be a positive rational number. How to prove that there exists an integer $a$ such that $aw \in \alpha$ ? I am able to prove using Archemedian property that there exists a natural number $b$ such that $bw$ does not belong to $\alpha$. Please don't tell the solution. Only a hint is enough.

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HINT: $a$ can be a negative integer.

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  • $\begingroup$ Does this proof also uses Archemedian property ?? $\endgroup$ – RIchard Williams Jul 5 '13 at 8:46
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    $\begingroup$ @prasenjit: Yes. Start with some $r\in\alpha$. If $r\ge 0$, take $w=-1$. If $r<0$, consider the positive number $-r$ and find a useful application of the Archimedean property. $\endgroup$ – Brian M. Scott Jul 5 '13 at 8:50
  • $\begingroup$ understood. Because $\alpha$ is non-empty let $y \in \alpha$. Now select any integer $a$ with $a < \frac{y}{w} $. So, $aw < y$ and so $aw \in \alpha$ using property $(2)$ of cut set. $\endgroup$ – RIchard Williams Jul 5 '13 at 8:54
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    $\begingroup$ @prasenjit: You have to be a little careful there. If $y<0$, what you’re really doing is using the Arch. prop. to get a positive integer $n$ such that $nw>-y$ and then observing that if $a=-n$, then $a<\frac{y}w$. $\endgroup$ – Brian M. Scott Jul 5 '13 at 8:57
  • $\begingroup$ understood now. $\endgroup$ – RIchard Williams Jul 5 '13 at 9:43

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