Write down the Equivalence classes

Question

Let $V = \{0,1,2\} \times \{0,1,2,3\}.$ We define an equivalence relation $R$ on $V$ by saying that $(a,b)R(c,d)$ if and only if $2a-b = 2c-d$. Write down the equivalence classes for $R.$

I do not get how to write the equivalence classes. The way I did it was:

$$V = \{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3),(2,0),(2,1),(2,2),(2,3)\}$$

$[(0,0)] = \{(1,2)\}$, Because $2(0)-(0) = 2(1)-2$

$[(0,1)] = \{(1,3)\}$

$[(1,1)] = \{(2,3)\}$

I'm just very confused about how to do this.

Answer 1

$$ \begin{array}{ccc} (a,b) & \mapsto & 2a-b \\ \hline (0,0) & \mapsto & \phantom{+}0 \\ (0,1) & \mapsto & -1 \\ (0,2) & \mapsto & -2 \\ (0,3) & \mapsto & -3 \\ (1,0) & \mapsto & \phantom{+}2 \\ (1,1) & \mapsto & \phantom{+}1 \\ (1,2) & \mapsto & \phantom{+}0 \\ (1,3) & \mapsto & -1 \\ (2,0) & \mapsto & \phantom{+}4 \\ (2,1) & \mapsto & \phantom{+}3 \\ (2,2) & \mapsto & \phantom{+}2 \\ (2,3) & \mapsto & \phantom{+}1 \\ \hline \end{array} $$ To each number in the righthand column above corresponds one equivalence class: $$ \begin{align} & \{(0,3)\} \\[5pt] & \{(0,2)\} \\[5pt] & \{(1,3),\,(0,1)\} \\[5pt] & \{(0,0),\,(1,2)\} \\[5pt] & \{(1,1),\,(2,3)\} \\[5pt] & \{(1,0),\,(2,2)\} \\[5pt] & \{(2,1)\} \\[5pt] & \{(2,0)\} \end{align} $$

Answer 2

Equivalence classes are such that:

Every element is in exactly one equivalence classes. The equivalence classes together will contain every element but the classes will be disjoint. All the elements in a class will be related to each other and not be related to any element not in the class.

I won't do this excercise but suppose you had a set $W = \{a,b,c,d,e,f\}$ and and an equivalence relation $S$ and you had the following relations:

$aRa, aRc,aRd$
$bRb, bRf$
$cRa,cRc, cRd$
$dRa, dRc, dRd$
$eRe$
$fRb,fRf$

Then the elements $a,c, d$ are all related to each other but none are related to any other so

1: $\{a,c,d\}$ is one equivalence class.

$b,f$ are related to each other but not to any other elements so

2: $\{b,f\}$ is another

and $e$ is related to itself but not to any other so

3: $\{e\}$ is the third equivalence class.

Now do that for $V = \{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3),(2,0),(2,1),(2,2),(2,3)\}$.

Find out which pairs are related to which pairs and put them into the proper sets.

....

You might ask: We had $aRa$ and $aRc$ and $aRd$ so we have to have $a,c,d$ is the same equivalence class, but who did we know we would have $cRd$? We if we had $a$ related to $c$ and $d$ but $c$ and $d$ not related to each other?

Well, this is an equivalence relation. It is transitive and it is symmetric (as well as reflexive). That means if we have $aRc$ and $aRd$ we must also have $cRa$ and $cRd$. It's because and only because the relation is equivalence that we can divide the elements into these sets of "who is related to who" so neatly and completely.

Equivalence classes in a succinct nutshell: A collection of equivalence classes is a collection of sets of all the elements that are related to each other.