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I was told the following in class: If we define an equivalence relation on $[0,1)$ by declaring that $x \sim y$ iff $x-y$ is rational, then there are uncountably many equivalences classes. Why is that? I think I may not understand the definition of equivalence class. Thanks.

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  • $\begingroup$ Irrationals are singleton. Right? $\endgroup$ – Ergin Suer Sep 9 '14 at 14:50
  • $\begingroup$ @ErginSuer You mean with not equivalent to any other number? No, they are not. $\endgroup$ – almagest Sep 9 '14 at 14:53
  • $\begingroup$ Not really, @ErginSuer. For example, $$\frac14+\frac1{\sqrt2}\;,\;\;\frac1{\sqrt2}\;$$ are two irrational numbers in the same equivalence lcass. $\endgroup$ – Timbuc Sep 9 '14 at 14:53
  • $\begingroup$ Sorry, you are right! $\endgroup$ – Ergin Suer Sep 9 '14 at 14:56
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If there were only countably many equivalence classes, since each class has a countable number of elements, this would mean that $[0,1)$ is countable, which is false.

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  • $\begingroup$ messages do not require email $\endgroup$ – robjohn Sep 15 '14 at 12:48
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One equivalence class here consists of a translated "copy" of $\mathbb{Q}$, for example the number $t\in\mathbb{R}$ belongs to the equivalence class $t+\mathbb{Q}$. Hence one equivalence class contains only a countable number of points.

What whould be the conclusion if there were only a countable number of equivalence classes?

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Arguing for a contradiction, pick a representative $x_n$ from each class. Then $$ [0,1)=\bigcup_{n\in\mathbb N} (x_n+\mathbb Q) $$ But this is a countable union of countable sets, making $[0.1)$ countable. There is your contradiction.

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All of the rational numbers are in one equivalence class. Other equivalence classes include $r+\sqrt{n}$ where $r$ is rational and $n$ is a fixed prime number. You can imagine others.

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