Example 19: $(x,y)\,R\,(u,v) \iff xv=yu$

Basics you need

Solution

1. Try the rule on a few pairs

   $(x,y)$ and $(u,v)$	$xv$	$yu$	Related?
   $(1,2)$ and $(3,6)$	$1\cdot 6 = 6$	$2\cdot 3 = 6$	yes
   $(3,6)$ and $(5,10)$	$3\cdot 10 = 30$	$6\cdot 5 = 30$	yes
   $(2,3)$ and $(4,6)$	$2\cdot 6 = 12$	$3\cdot 4 = 12$	yes
   $(1,2)$ and $(2,1)$	$1\cdot 1 = 1$	$2\cdot 2 = 4$	no

   The related pairs are those with the same ratio: $\frac12 = \frac36 = \frac{5}{10}$⁠, $\frac23 = \frac46$⁠, but $\frac12 \ne \frac21$⁠.

   The condition $xv = yu$ is the cross-multiplied form of $\frac{x}{y} = \frac{u}{v}$ (Basic 4). Seeing this first tells us what to expect, an equivalence relation, and what the classes will look like. The proof itself will use only the integer equation $xv = yu$⁠.
   Cross-multiplication: $\frac{x}{y} = \frac{u}{v} \iff xv = yu$⁠, valid when $y \ne 0$ and $v \ne 0$⁠.

2. Reflexive: $(x,y)\,R\,(x,y)$

   Take any $(x,y)\in A$⁠. To compare it with itself, put $u = x$ and $v = y$ in the rule. The condition becomes

   $$xy = yx,$$

   which is true because multiplication of integers is commutative. So $(x,y)\,R\,(x,y)$ for every $(x,y)\in A$⁠, and $R$ is reflexive.

   The rule compares two pairs, $(x,y)$ and $(u,v)$⁠. Reflexivity compares a pair with itself, so the second pair is the same as the first: $u = x$⁠, $v = y$⁠. The resulting equation $xy = yx$ holds for all integers, so no particular pair is needed.
   Commutativity: $xy = yx$ for all integers $x, y$ (Basic 3).

3. Symmetric: swap the two pairs

   Suppose $(x,y)\,R\,(u,v)$⁠, so $xv = yu$⁠. We must show $(u,v)\,R\,(x,y)$⁠. By the rule (with the roles of the pairs swapped), that means $uy = vx$⁠. Now

   $$xv = yu \ \Longrightarrow\ yu = xv \ \Longrightarrow\ uy = vx,$$

   reading the equation from right to left, then using commutativity on each side. So $(u,v)\,R\,(x,y)$⁠, and $R$ is symmetric.

   The main care here is getting the target right. "$(u,v)\,R\,(x,y)$" puts $(u,v)$ first, so in the rule "first-entry of the first pair times second-entry of the second pair equals second-entry of the first pair times first-entry of the second pair" we get $u\cdot y = v\cdot x$⁠. Once that is written down, it is the given equation rearranged.
   An equation can be read in either direction ($p = q \Rightarrow q = p$⁠), and $yu = uy$⁠, $xv = vx$⁠.

4. Transitive: combine and cancel

   Suppose $(x,y)\,R\,(u,v)$ and $(u,v)\,R\,(a,b)$⁠. Then

   $$xv = yu \qquad\text{and}\qquad ub = va.$$

   We want $(x,y)\,R\,(a,b)$⁠, that is $xb = ya$⁠. Multiply the two equations side by side (left times left, right times right):

   $$(xv)(ub) = (yu)(va).$$

   Rearrange each side, using commutativity and associativity, so that the factor $uv$ stands out:

   $$(xb)(uv) = (ya)(uv).$$

   Since $u$ and $v$ are positive integers, $uv > 0$⁠, so $uv \ne 0$ and we may cancel it:

   $$xb = ya.$$

   Hence $(x,y)\,R\,(a,b)$⁠, and $R$ is transitive.

   We need a relation between $x, y$ and $a, b$ only, so the middle pair's entries $u, v$ must be removed. Multiplying the two equations makes both $u$ and $v$ appear on each side, as the common factor $uv$⁠; cancelling it removes them. Cancelling is only allowed for a non-zero factor, and this is exactly where the question's word positive is used. The textbook does the same thing in another way: from $ub = va$ it gets $\frac{a}{u} = \frac{b}{v}$⁠, multiplies $xv = yu$ by $\frac{a}{u}$⁠, replaces $\frac{a}{u}$ on the left by $\frac{b}{v}$⁠, and simplifies $xv\cdot\frac{b}{v} = yu\cdot\frac{a}{u}$ to $xb = ya$⁠. Both routes divide by $u$ and $v$⁠.
   Cancellation: if $pk = qk$ and $k \ne 0$⁠, then $p = q$ (Basic 3). Multiplying equal quantities by equal quantities gives equal results.

5. Conclude, and describe the classes

   $R$ is reflexive, symmetric and transitive, so $R$ is an equivalence relation on $A$.

   By Step 1, $(x,y)\,R\,(u,v)$ means $\frac{x}{y} = \frac{u}{v}$⁠. So each equivalence class is the set of all pairs that represent one positive fraction. For example

   $$[(1,2)] = \{(1,2), (2,4), (3,6), \dots\} = \{(k, 2k) : k \in \N\}.$$
   This is the real meaning of the relation: it treats $(1,2)$ and $(2,4)$ as "the same" because they describe the same fraction $\frac12$⁠. In fact, this is one standard way of constructing the positive rational numbers from pairs of integers. The picture also shows the partition: every dot lies on exactly one line through the origin.
   An equivalence relation splits the set into disjoint classes. Basics: equivalence classes

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