# Complete set of equivalence class representative

Let $\sim$ be a relation on $\mathbb{R}$ and $x\sim y$ if and only if $x-y\in \mathbb{Z}$.

(a) Show that $\sim$ is an equivalence relation
(b) Give a complete set of equivalence class representatives.

(a) is easy to show, but I really don’t understand (b).

I know that:

$$[a]_{\sim}:=\{y\in \mathbb{R} \mid a\sim y\}$$

so

$$[0]_{\sim}=\{\dots,-2,-1,0,1,2,\dots\}$$

and that

$$\dots=[-2]_{\sim}=[-1]_{\sim}=[0]_{\sim}=[1]_{\sim}=[2]_{\sim}=\,\dots$$

because if $x\sim y$ then $[x]=[y]$.

But how can I find a complete set of equivalence class representatives?

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Can you find even one other class than $[0]_\sim$? – Alexander Gruber Oct 5 '12 at 3:20

By a complete set of equivalence class representatives, I think the question wants a set $X \subset \mathbb{R}$ such that for each equivalence class, there is exactly one representative in the set - this is the sense in which it is complete (there is a $1-1$ correspondence between $X$ and $\mathbb{R}/\sim$, the set of equivalence classes).

So what you need to do is for each equivalence class, you need to pick one representative and let $X$ be the set of all your choices. As you point out, there are lots of different representatives you can choose for any given equivalence class. To ensure that your set $X$ contains only one representative for each equivalence class and that each equivalence class does have a representative in $X$, it would be good if you were able to make some sort of consistent choice of representative. For example, there is a reason I would choose $\frac{1}{2}$ to be the element in $X$ which represents $[\frac{5}{2}]_{\sim}$, and I would choose all the other representatives of equivalence classes for the same reason.

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Ok, let $\mathbb{R}/ \sim :=\left\{ \left[0 \right], \left[ \frac{1}{2} \right]_{\sim},\left[\frac{1}{3}\right]_{\sim},\left[\frac{1}{4}\right]_{\sim},.‌​....\right\}$ be the quotient set, then $X=\left\{0,\frac{1}{2},\frac{1}{3},\frac{1}{4},....\right\}\subset \mathbb{R}$ is the complete set of equivalence class representatives. Is this correct? – user43523 Oct 3 '12 at 22:38
Instead of starting to write a list, you need to be a little bit more specific. For example, I am not sure which representative you would choose for the equivalence class $[\pi]_{\sim}$. – Michael Albanese Oct 3 '12 at 22:40
I know now that we have an equivalence class for each $x \in [0, 1)$, but I have no idea how to write it down. – user43523 Oct 3 '12 at 22:50
What's wrong with writing $[0, 1)$? All you have to do now is explain: 1. why no two elements of $[0, 1)$ are representatives for the same equivalence class; and 2. why each equivalence class has a representative in $[0, 1)$. – Michael Albanese Oct 3 '12 at 22:52
It's getting late in the Netherlands, but I'll give it a try: 1. assume that $x,y \in [0,1)$ and $x \not\sim y$ (proof by contradiction) suppose that $x,y$ are in the same equivalence class, $x,y\in [a]_{\sim}$ then by def. of relation we know that $a\sim x$ and $a \sim y$ we see then that $x\sim y$(symmetric+transitive) which is a contradiction?? – user43523 Oct 3 '12 at 23:15

Hint: two numbers are equivalent under this relation iff they have the same fractional part.

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so ,correct me if i'm wrong, $\left[ \frac{1}{2} \right]_{\sim}=\left\{ ....,-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2},....\right\}$, but how can I list them all? – user43523 Oct 3 '12 at 21:15
Or can I just say that $\left\{ 0 \right\} \cup \left\{ x \in \mathbb{R} |\,0<x<1 \right\}$ is the complete set of equivalence class representatives, because for each x in the set, we can find an equivalence class? – user43523 Oct 3 '12 at 21:27
You can say that. Is there a shorter way of writing that set? And then, how would you prove that you've got a complete set of equivalence class representatives? (You'll need to prove that no two elements of your set are in the same equivalence class, and also that everything in $\Bbb R$ is related to something in the set). – user22805 Oct 3 '12 at 21:41