Let $S=\{x_0, x_1, \cdots, x_n\} \subset [0,1]$ be a finite set of real numbers with $x_{0}=0$ and $x_{1}=1$, such that every distance between pairs of elements occurs at least twice, except for the distance $1$. Prove that all of the $x_i$ are rational.
Problem
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Tags: rational numbers
Peter
30.08.2007 01:09
Solution in first reply at http://www.mathlinks.ro/viewtopic.php?p=15303 .
hellomath010118
07.01.2021 21:18
We consider the basis of $x_i$. Let that be $\mathbb{Q}[r_1,r_2,\ldots,r_m]$. Then choose any $r$ let us say $r_1$. Since $r_1$ is in the basis thus there must exist two $x_i$ such that there difference contains $r_1$ (This ensures that the component of $r_1$ in the difference we'll define isn't $0$) . Let $X_1$ = Those $x_i$ who have the maximum component of $r_1$ and $X_2$ = Those $x_i$ having minimum component of $r_1$. Then consider the difference of "rightmost" (i.e. greatest) member of $X_1$ and "leftmost" ( i.e. smallest ) member of $X_2$. If this has to have a friend then one vertex must be in $X_1$ and other in $X_2$. But if you choose any such pair (other than this one) then its length will be strictly less than the earlier chosen pair, the most preposterous statement! So all lengths must be rational (as we must dustbin any $r_i's$ if they pop up just like that quantum fluctuations anyone?) $\square$. IMPORTANT:This stunt has been performed under the supervision of biomathematics. Do not try at home. Patreon supporters: 1. (to be disclosed) 2. (to be disclosed)