Finding cardinality of a set given equivalent classes

Question

I have the following question:

If $|A| = 30$ and the equivalence relation $R$ on $A$ partitions $A$ into equivalence classes $A_1$, $A_2$, $A_3$, where:

$$|A_1| = |A_2| = |A_3|$$

what is $|R|$?

I understand why $|A_1| = |A_2| = |A_3| = 10$ because of the fact that $\frac{30}{10}=3$, but the answer says that $|R|$ is $10\times 10\times 3 = 300$, I don't understand where the $10\times 10$ and $\times 3$ comes from. Is it from the relation of $|A_1| = |A_2|$, $|A_2| = |A3|$, & $|A_1| = |A3|$ which every relation has $10\times 10$ and there are $3$ of them? Please help.

Answer 1

You are asked with the cardinality of the relation $R$ is. So what's $R$? Well, the elements of $R$ are ordered pairs $(x,y)$, where $x\sim y$ - in other words, $x,y\in A_j$ for some $j\in \{1,2,3\}$.

So how many elements does $R$ have? Well, there are $10$ choices of $x$ in $A_1$ and $10$ choices of $y$. Thus, there are $10\times 10=100$ such pairs stemming from each $A_j$. Since there are $3$ different values of $j$, we conclude that $|R|=300$.

Answer 2

There is exactly one $(x,y) \in R$ for every pair $x,y \in A_1$, for every pair $x,y \in A_2$, and for every pair $x,y \in A_3$ and no others, as $x$ and $y$ in different classes cannot be equivalent so $(x,y) \notin R$ then, and all $x,y$ in the same class must be $R$-related, so $(x,y) \in R$.

So of the $900= 30 \times 30$ pairs in $A \times A$, we have $300= 3 \times (10 \times 10)$ that are in $R$. $10$ is from the size of each class. $3$ because there three classes.