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We have the relation $\sim$ in $\mathbb{R}^n$: $x\sim y \leftrightarrow d(x,y)\in \mathbb{Q}$, where $d(x,y)=\sqrt{\sum^n_{i=1}(x_i-y_i)^2}$. How do you prove that this isn't an equivalence relation for $n>1$? I know for sure that it's reflexive, so it must either not be symmetric or not transitive.

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    $\begingroup$ $\sqrt{2}=\sqrt{1+1}\notin \Bbb Q,\sqrt{4}=2\in \Bbb Q,$ $\endgroup$ – Surb Dec 18 '14 at 15:23
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$d(0,e_1)\in\mathbb{Q}$, $d(0,e_2)\in\mathbb{Q}$, yet $d(e_1,e_2)\notin\mathbb{Q}$.

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    $\begingroup$ For completeness, $d(e_1,e_2) = \sqrt2$, $d(0,e_i) = 1$ $\endgroup$ – AlexR Dec 18 '14 at 15:24
  • $\begingroup$ @vadim123 what is $e_1$? Is that $(1,0)$? $\endgroup$ – Mathematicxcz Dec 18 '14 at 15:25
  • $\begingroup$ @Mathematicxcz, yes, if $n=2$. If $n=3$ then $e_1=(1,0,0)$, and so on. $\endgroup$ – vadim123 Dec 18 '14 at 15:25
  • $\begingroup$ Nevrmind, thanks @AlexR $\endgroup$ – Mathematicxcz Dec 18 '14 at 15:25

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