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There is a relationship on $\Bbb R$ defined aRb if a-b is a rational number. I already proved its an equivalence relation in $\Bbb R$. My question is how to describe the equivalence classes? Here is my attempt at the answer:

[0]=$\{x\in \Bbb R : xR0\}$ = $\{x\in \Bbb R : x-0 $ is rational $\}$

[a]=$\{x\in \Bbb R : xRa\}$ = $\{x\in \Bbb R : x-a $ is rational$\}$

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    @KirkBoyer Lets see if I understand this now. The reals can be partioned into the rationals and irrationals. Where in this case the rationals are [0]. The irrationals are [a] becasue like you said for $b(x-t)=a$$x-t$must have the same decimals. To anser Ross Millikan why there are not seperate classes is because if $\frac12 R 0 $ and $\frac14 R 0 $ then by symmetry and transitivity $\frac12 R \frac14 $2012-09-09

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Let $x \in \mathbb{R}$, then

$[x] = \{x + q : q \in \mathbb{Q}\}$

Thus $[0] = \{0 + q : q \in \mathbb{Q}\} = \{q : q \in \mathbb{Q}\} = \mathbb{Q}$.

To see this, note that by your definition, $[x]$ is the set of all $a$ such that $a - x = q$ for some $q \in \mathbb{Q}$. Hence $a = x + q$ for $q \in \mathbb{Q}$.


If you know some group theory, an alternatively (but essentially equivalent) way of thinking of the equivalent using coset. $(\mathbb{R}, +)$ is an abelian group. $(\mathbb{Q}, +)$ is a normal subgroup. Hence the equivalence classes are the elements of the group $\mathbb{R} / \mathbb{Q}$.