Q.Let $R$ be a relation on the set $A$ of ordered pairs of positive integers defined by $(x, y) \, R \, (u, v)$ if and only if $xv = yu$. Show that $R$ is an equivalence relation.

Concept: Equivalence Relation Proof — we check reflexivity, symmetry, and transitivity.

1. Reflexive: For any $(x,y) \in A$, $xy = yx$ holds, so $(x,y) R (x,y)$.

2. Symmetric: If $(x,y) R (u,v)$, then $xv = yu$. This implies $uy = vx$, so $(u,v) R (x,y)$.

3. Transitive: Suppose $(x,y) R (u,v)$ and $(u,v) R (w,z)$. Then $xv = yu$ and $uz = vw$. Multiply the first by $z$ and the second by $y$: $xvz = yuz$ and $yuz = yvw$. Hence $xvz = yvw$, so $xz = yw$ (since $v>0$ cancels). Thus $(x,y) R (w,z)$.

All three properties hold, so $R$ is an equivalence relation.

The key insight here is that the condition $xv = yu$ is exactly the cross-multiplication test for equality of two fractions: $\frac{x}{y} = \frac{u}{v}$. So $R$ is really saying "these two ordered pairs represent the same rational number." Once you see that, proving it's an equivalence relation becomes natural — you're just checking that equality of fractions behaves properly.

Let's work through the three properties systematically.

1. Reflexivity: We need to show $(x,y)R(x,y)$ for any positive integers $x,y$.

   The condition becomes $x \cdot y = y \cdot x$, which is $xy = yx$. Since multiplication of integers is commutative, this is always true. So every pair is related to itself.

2. Symmetry: If $(x,y)R(u,v)$, then $xv = yu$. We need to show $(u,v)R(x,y)$.

   The condition for the reverse is $uy = vx$. But $uy = yu$ and $vx = xv$, so $uy = vx$ is exactly the same equation as $xv = yu$, just written differently. Since equality is symmetric, if $xv = yu$ holds, then $uy = vx$ holds automatically. So symmetry follows.

3. Transitivity: This is the trickiest one. Suppose $(x,y)R(u,v)$ and $(u,v)R(w,z)$. That means:

   • $xv = yu$
   • $uz = vw$

   We need to show $(x,y)R(w,z)$, i.e., $xz = yw$.

   Multiply the first equation by $z$: $xvz = yuz$

   Multiply the second equation by $y$: $yuz = yvw$

   So $xvz = yvw$.

   Now, $v$ is a positive integer, so we can cancel it from both sides (since we're working with integers and $v \neq 0$). This gives $xz = yw$, which is exactly what we needed.

Since all three properties — reflexivity, symmetry, and transitivity — hold, $R$ is an equivalence relation on $A$.

Method: The Three-Property Check for an Equivalence Relation

Use this whenever a relation is given by an algebraic condition (here $(x,y)R(u,v) \iff xv = yu$) and you are asked to prove it is an equivalence relation. You verify reflexivity, symmetry, and transitivity in that order, each time substituting the specific pairs into the defining equation.

Steps

Step 1: Reflexivity — plug the same pair into itself.

Write the defining condition with both slots equal and check it is an identity. For a cross-multiplication rule this reduces to $xy = yx$, true by commutativity.

Step 2: Symmetry — rewrite the same equation with sides swapped.

Assume the condition for $(a)R(b)$, then algebraically rearrange it into the condition for $(b)R(a)$. Often the two equations are literally the same equality written in a different order, so symmetry is immediate.

Step 3: Transitivity — combine two given equations to eliminate the middle pair.

This is the step that needs real algebra. Given the equations for $(a)R(b)$ and $(b)R(c)$, multiply each by a well-chosen factor so a common term appears, chain the equalities, then cancel the shared factor:

Step 4: Justify every cancellation.

Before cancelling a factor, confirm it is non-zero — here the domain is positive integers, so $v \ge 1$ and cancellation is legal. State this explicitly; it is where marks are won or lost.

Common Mistakes

Mistake 1: Cancelling a factor without checking it is non-zero.

Why it's wrong: the transitivity step divides both sides by $v$ to get $xz = yw$; this is only valid because the domain is positive integers, so $v \ge 1$. Cancelling blindly is illegal if the factor could be $0$. Correct approach: state explicitly that $v > 0$ (positive integers) before cancelling.

Mistake 2: Writing the wrong condition for symmetry or transitivity.

Why it's wrong: the relation is $xv = yu$, so the condition for $(u,v)R(x,y)$ is $uy = vx$, not $xu = yv$ or a random rearrangement. Getting the cross-multiplication slots wrong invalidates the proof. Correct approach: substitute the pairs carefully into $(\text{first-first})(\text{second-second}) = (\text{first-second})(\text{second-first})$ each time.

Mistake 3: Confusing this with an order relation and checking antisymmetry.

Why it's wrong: antisymmetry is a property of partial orders, not equivalence relations. Correct approach: verify exactly reflexive, symmetric, and transitive — nothing else.