Equivalence relation help

Question

If $|A| = 30$ and the equivalence relation $R$ on $A$ partitions $A$ into (disjoint) equivalence classes $A_1$, $A_2$, and $A_3$, where $|A_1| = |A_2| = |A_3|$, then what is $|R|$?

score 1 · Accepted Answer · answered Jun 16 '13 at 23:58

1

We have $R = A_1 \times A_1 \cup A_2 \times A_2 \cup A_3 \times A_3$, so $|R|=|A_1\times A_1|+|A_2\times A_2|+|A_3\times A_3|$.

answered Jun 16 '13 at 23:58

dfeuer

9,069

Thanks.. you've helped a lot. – Dale Bryant Jun 17 '13 at 00:22
@DaleBryant: You should vote up (up-arrow) answers you like. If one of them really does it for you, click the check mark to accept it. – dfeuer Jun 17 '13 at 00:26
It says I need more reputation to vote up.. – Dale Bryant Jun 17 '13 at 00:34
@Dale: You will need $15$ reputation to be able to vote up, but then you'll be able to vote up to $40$ times per day. Welcome to the StackExchange! – Cameron Buie Jun 17 '13 at 00:41

score 0 · Answer 2 · edited Mar 24 '14 at 18:32

0

Hint: $$ R = \{(x_1,y_1):x_1,y_1\in A_1\} \cup \{(x_2,y_2):x_2,y_2\in A_2\} \cup \{(x_3,y_3):x_3,y_3\in A_3\}. $$

Count each of these separately, then add.

edited Mar 24 '14 at 18:32

Evgeny Zolin

148
8

answered Jun 16 '13 at 23:58

vadim123

82,796

score 0 · Answer 3 · answered Jun 17 '13 at 00:00

0

Hint: Note that for $j=1,2,3$, and any $x,y\in A_j,$ we have $(x,y)\in R.$ That means that $A_j\times A_j\subseteq R$ for $j=1,2,3$. In fact (why?), $$R=\bigcup_{j=1}^3A_j\times A_j,$$ and this union is disjoint. Can you take it from here?

answered Jun 17 '13 at 00:00

Cameron Buie

102,994

Yup.. thanks so much. – Dale Bryant Jun 17 '13 at 00:21

Equivalence relation help

3 Answers3