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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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    +1 for showing your thoughts. Seems reasonable to me, but how do we know there aren't separate classes $\frac 12$ and $\frac 14$?2012-09-09
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    One thing to get you started is that any rational number $[q]$ represents all $\mathbb{Q}.$2012-09-09
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    @ShankaraPailoor: Looks like drew gets that. I don't know how to present that, which seems to be the question.2012-09-09
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    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?2012-09-09
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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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