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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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homework should not be used as a standalone tag; see tag-wiki and meta. –  Martin Sleziak Sep 9 '12 at 4:48
+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

1 Answer 1

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

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