Q: Equivalence Relations and distinct equivalence classes $A=\lbrace(1,3),(2,4),(-4,-8),(3,9),(1,5),(3,6)\rbrace$. $R$ is defined on $A$ as follows: For all $(a, b)\;(c, d) \in A$, $(a, b) R (c, d) \iff ad=bc$
I know what they are asking but I cannot see how to systematically do this using the definition of equivalence classes which says $[k]= \lbrace x \in R \mid x\; R\; k \rbrace$
So I found the relation to be the following $R= \lbrace (1,3),\;(3,9),\; (2,4),\;(3,6)\;(-4,8)\rbrace$. So heres one question is the empty set in this relation as well? And I know that the distinct classes should be $\lbrace (1,3),\;(3,9) \rbrace$ $\lbrace (2,4),(3,6), (-4,-8) \rbrace$ My other question is, is there a systematic way of doing this using the definition? Since this is a cross product of the cross product it can be a little confusing.

A: This is an interesting problem when considering the general case of your given relation; so let us first analyze the relation. Notice that your set $A$ contains ordered pairs containing integers. This is important. Now, lets look at your relation. It states that for $(a,b),(c,d) \in A$, $(a,b) \mathbf{R}(c,d) \iff ad = bc$. Rather than confine ourselves to your given set, as I believe you did a fine job of picking out the relation, let us conduct a more in-depth analysis. Like we said, every element of $A$ is an ordered pair in which the entries are integers; moreover, the second entry in every ordered pair is non-zero. So, let us define a new set $B$ so we can generalize the relation and enhance our understanding. 
Let $B = \lbrace (a,b)  \vert a,b \in \mathbb{Z}, b\neq 0 \rbrace$. Now let us apply your relation $\mathbf{R}$ on our set $B$. Note that $\mathbf{R}$ has been given [rightfully] as an equivalence relation. So: $(a,b)\mathbf{R}(c,d) \iff ad=bc$. Consider $ad = bc$. With some algebra, we may manipulate our equation and obtain the following form: $\frac{a}{b} = \frac{c}{d}$. Recall our definition of $B$ allows $b$ and $d$ to assume any integer values except zero. Therefore, we can interpret our relation $\mathbf{R}$ as one which partitions rational numbers: $\mathbf{R}$ divides the rational number system into smaller groups of equivalent fractions. 
Now, let us consider the quotient set [the set containing all equivalence classes of $\mathbf{R}$]. Each equivalence class will be containing an entire collection of equivalent fractions. Let us now examine this from your definition of equivalence class: $[k] = \lbrace x \in \mathbf{R} : x \mathbf{R} k \rbrace$. Relative to your relation, we can say that the equivalence class, $[(a,b)] = \lbrace (c,d) \in \mathbf{R} : (a,b)\mathbf{R}(c,d)\rbrace$. This is equivalent to the following: $[(a,b)] = \lbrace (c,d) \in \mathbf{R}: \frac{a}{b} = \frac{c}{d} \rbrace$.
Generally, with any relation, the first step is to fully understand the relation; be able to understand its uses and limitations - never fear thinking in an unorthodox manor when given a relation of any type [like functions and operations]. Only in doing so will you be able to vector the issue at hand, and in doing so, discover further properties which otherwise may have been left alone.
Hope this helps.
