# Equivalence classes on a relationship in $R$.

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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+1 for showing your thoughts. Seems reasonable to me, but how do we know there aren't separate classes $\frac 12$ and $\frac 14$? –  Ross Millikan Sep 9 '12 at 4:50
One thing to get you started is that any rational number $[q]$ represents all $\mathbb{Q}.$ –  Shankara Pailoor Sep 9 '12 at 4:54
@ShankaraPailoor: Looks like drew gets that. I don't know how to present that, which seems to be the question. –  Ross Millikan Sep 9 '12 at 4:58
Let $t$ be irrational. If $x-t$ is rational, then $x-t = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ with $\gcd(a,b)=1$, and so $b(x-t) = a$. So $x-t$ must actually be an integer, yes? What does it mean when two decimals have integer difference? –  Kirk Boyer Sep 9 '12 at 5:33
@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 $\frac12 R \frac14$ –  drew Sep 9 '12 at 20:59

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 $\mathbb{R} / \mathbb{Q}$.