# a question on subsets of rational number and analysis. [duplicate]

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Subset of $\mathbb{Q}$

Let $S= \{x_0,\dots,x_n\}$ be a finite subset of $[0,1]$ , $x_0=0$ and $x_1=1$ such that every distance between pair of elements of $S$ occurs at least twice, except for the distance $1$, then we are to show that $S$ is a subset of $\mathbb{Q}$.

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## marked as duplicate by Martin Sleziak, MJD, Henry, martini, Asaf KaragilaMay 31 '12 at 22:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

So the idea is that for each $0≤i<j≤n$ (except $(i,j) = (0, 1)$), we must be able to find $\{p,q\} \ne \{i, j\}$ with $|a_i - a_j| = |a_p - a_q|$? Is that right? –  MJD May 31 '12 at 20:08
This question seems to be an identical duplicate from the above one. –  Thomas E. May 31 '12 at 20:16

## 1 Answer

I would use induction. First assume WLOG all the elements of the set are distinct. Let $x_0=0, x_1=1$. Suppose there is another element $x_2$. Then $x_2-x_0=x_1-x_2\implies x_2=\frac{1}{2}$. So the statement is true if there are 3 elements...I think this can work.

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I don't think this is going to work. I think you'll end up having to show that if $S\subset\Bbb R$ fails the distance criterion, but $S\cup\{x\}$ satisfies the criterion, then $x\in\Bbb Q$, even if $S\not\subset\Bbb Q$. But this seems unlikely. –  MJD May 31 '12 at 20:33