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Let $$R_3= \{(a,b)\mid a,b \in \mathbb{Z}\text{ and there exists }k \in \mathbb{Z} \text{ such that }a-b=3k\}.$$

I know there is an equivalence relation but I'm not 100% on what it means to be an equivalence class for this problem. In class we got 3: $\{0,3,6,9,\ldots\}$ and $\{1,4,7,10,-2,-5,\ldots\}$ and $\{2, 5, 8, 11, -1, -4,\ldots\}$.

I don't understand where these cells came from. Help?

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up vote 2 down vote accepted

I'll try to put it this way:

Define a relation $\sim$ on $\mathbb Z$, such that $a \sim b \iff \exists k \in \mathbb Z ~~ \text{such that}~~~~a-b=3k$

What does this say?

Integers $a$ and $b$ are related if and only if on their difference is a multiple of $3$. Since, the remainder when $a-b$ is divided by $3$ is the difference of the remainders when $a$ and $b$ are divided by $3$, taken(all taken$\mod 3$). So, integers $a$ and $b$ are related if and only if they leave the same remainder when divided by $3$.

Now try to put all those numbers that are related to each other in the same "cell" and those that are not related in different "cells".

But, now notice that the number of distinct cells you'll need for the purpose is no more than $3$ and no less! (Why?)

Construct these "cells" to see how they coincide with what you have written down in your class.

And, now call these cells "equivalence classes".

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This question is old, but I'd like to share my view of it. I hope this offends no one.

What you are dealing with are equivalence classes modulo an integer. That is, you are dealing with equivalence classes defined by $[a]=\{b \in \mathbb{Z}:b \cong a \pmod 3\}$ for an element $a$.

In the case of the relation of congruence modulo $n$, there are $n$ equivalence classes: $[0], [1], [2], \dots, [n-1]$. You can see that supposing $[n]$ or $[n+1]$ is silly, since $b \cong n \pmod n$ reduces to $b \cong 0\pmod 0$ and $b \cong n+1 \pmod n$ reduces to $b \cong 1 \pmod n$ (and, hence $[n]=[0]$ and $[n+1]=1$). In other words, $[0], [1], [2],\dots, [n-1]$ are all of the equivalence classes.

For congruence modulo $3$, we thusly have $[0], [1], $ and $[2]$. That is, \begin{align} [0]&=\{b \in \mathbb{Z}: b\cong 0 \pmod 3\}\\ &=\{3,6,9,\dots\}.\\ [1]&=\{b \in \mathbb{Z}: b\cong 1 \pmod 3\}\\ &=\{4,7,10,\dots\}.\\ [2]&=\{b \in \mathbb{Z}: b\cong 2 \pmod 3\}\\ &=\{5,8,11,\dots\}. \end{align}

In other words, the three equivalence classes are numbers that divide $3$ with remainder $0$, all numbers that divide $3$ with remainder $1$, and all numbers that divide $3$ with remainder 2.

This is the beauty of equivalence classes: They partition the set they are defined on into disjoint sets which as a whole form the entire set. i.e. For any given equivalence relation $E$ on set $S$ and the equivalence classes $[a]=\{b \in S: aEb\}$, $$\bigcup_{a \in S}[a]=S.$$

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An exercise in set theory (if $k$ runs through $\mathbb Z$, then so does $-k$):

$[a] = \\ \{b \in \mathbb{Z} \text{ such that there exists }k \in \mathbb{Z} \text{ such that }a-b=3k\}=\\ \{b \in \mathbb{Z} \text{ such that there exists }k \in \mathbb{Z} \text{ such that }b=a-3k\}=\\ \{b \in \mathbb{Z} \text{ such that there exists }k \in \mathbb{Z} \text{ such that }b=a+3(-k)\}=\\ \{a+3k | k \in \mathbb{Z}\}=\\ \{\dots,a-6,a-3,a,a+3,a+6,\dots\}$

You should have written $ \{\ldots, -5, -2, 1, 4, 7, 10,\ldots\} $ instead of $ \{1,4,7,10,-2,-5,\ldots\} $

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I agree with you. A matter of notation, changing which we would have lost some info. +1. (May be, I'll add it to my answer later =)) – user21436 Jan 31 '12 at 21:08

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