Easy question about an equivalence relation

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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?

Answer 4

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.