Q.Show that the relation R defined in the set A of all polygons as $R = \{(P_1, P_2) : P_1$ and $P_2$ have same number of sides$\}$, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides $3, 4$ and $5$?

Concept: Relation Properties — We check reflexivity, symmetry, and transitivity to prove equivalence.

Step 1 — Reflexive: Any polygon $P$ has the same number of sides as itself, so $(P,P) \in R$ for all $P \in A$. Hence $R$ is reflexive.

Step 2 — Symmetric: If $P_1$ and $P_2$ have the same number of sides, then $P_2$ and $P_1$ also have the same number of sides. So $(P_1,P_2) \in R \implies (P_2,P_1) \in R$. Hence $R$ is symmetric.

Step 3 — Transitive: If $P_1$ and $P_2$ have the same number of sides, and $P_2$ and $P_3$ have the same number of sides, then $P_1$ and $P_3$ have the same number of sides. So $(P_1,P_2) \in R$ and $(P_2,P_3) \in R \implies (P_1,P_3) \in R$. Hence $R$ is transitive. …

The core idea here is that "having the same number of sides" is a natural way to classify polygons. Whenever you define a relation based on equality of some property (here, the count of sides), you almost always get an equivalence relation. The three required properties — reflexivity, symmetry, transitivity — follow directly from the fact that equality itself has those properties.

Let’s verify each property carefully.

1. Reflexive: For any polygon $P_1$ in A, does $P_1$ have the same number of sides as itself? Obviously yes. So $(P_1, P_1) \in R$ for every $P_1 \in A$. Reflexivity holds.

2. Symmetric: If $P_1$ and $P_2$ have the same number of sides, then $P_2$ and $P_1$ also have the same number of sides — it’s the same fact stated in reverse. So if $(P_1, P_2) \in R$, then $(P_2, P_1) \in R$. Symmetry holds.

3. Transitive: Suppose $P_1$ and $P_2$ have the same number of sides, and $P_2$ and $P_3$ have the same number of sides. Then $P_1$ and $P_3$ must also have that same number of sides (since the number is fixed). So $(P_1, P_2) \in R$ and $(P_2, P_3) \in R$ implies $(P_1, P_3) \in R$. Transitivity holds. …

Method: Proving an "equal-property" relation is an equivalence relation

Use this whenever a relation is defined by "$x$ and $y$ share the same value of some property" (same number of sides, same remainder, same age).

Steps

Step 1: Restate the relation as an equality of a quantity

Identify the single quantity the relation compares — here, the number of sides $s(P)$. Then $P_1\,R\,P_2 \iff s(P_1)=s(P_2)$. Once it reads as an equality, all three properties follow from properties of "$=$".

Step 2: Reflexive

Every element equals its own quantity: $s(P)=s(P)$, so $(P,P)\in R$ for all $P$.

Step 3: Symmetric

$s(P_1)=s(P_2)$ immediately gives $s(P_2)=s(P_1)$, so the relation reverses. …

Common Mistakes

Mistake 1: Thinking the side lengths $3,4,5$ decide the equivalence class of $T$.

Why it's wrong: the relation compares only the NUMBER of sides, not lengths or shape. Correct approach: note $T$ is a triangle (3 sides), so its class is all triangles, whatever their side lengths.

Mistake 2: Trying to prove transitivity by testing many triples of polygons. …