Definitions with examples. When is a relation an equivalence relation?

Types of Relations — Reflexive, Symmetric, Transitive

Question

A relation R on the set A = {1, 2, 3} is defined as:

R = \{(1,1),\ (2,2),\ (3,3),\ (1,2),\ (2,1),\ (2,3),\ (3,2)\}

Check whether R is reflexive, symmetric, and transitive. Is R an equivalence relation?

Solution — Step by Step

A relation is reflexive if every element is related to itself — meaning $(a, a) \in R$ for all $a \in A$ .

We need $(1,1)$ , $(2,2)$ , $(3,3)$ all in R. Check: all three are present. R is reflexive.

R is symmetric if whenever $(a, b) \in R$ , we also have $(b, a) \in R$ .

Go through every pair: $(1,2) \in R$ and $(2,1) \in R$ ✓. $(2,3) \in R$ and $(3,2) \in R$ ✓. The diagonal pairs $(1,1)$ etc. are trivially symmetric. R is symmetric.

R is transitive if whenever $(a,b) \in R$ and $(b,c) \in R$ , then $(a,c) \in R$ .

We have $(1,2) \in R$ and $(2,3) \in R$ — so we need $(1,3) \in R$ . But $(1,3) \notin R$ . R is NOT transitive.

For R to be an equivalence relation, it must be reflexive, symmetric, AND transitive — all three together.

Since R fails transitivity, R is not an equivalence relation.

Why This Works

Think of these three properties as three different “fairness tests” for a relation.

Reflexive asks: does every element acknowledge itself? Like every student being their own classmate. Symmetric asks: if A is related to B, is B related to A? Like friendship — if Ravi is friends with Arjun, Arjun must be friends with Ravi.

Transitive is the trickiest. It says: if A connects to B, and B connects to C, then A must connect directly to C. Think of it as “no shortcuts missing.” In our problem, 1 connects to 2, and 2 connects to 3 — but 1 doesn’t connect to 3. The chain breaks, so transitivity fails.

An equivalence relation packages all three properties. When a relation is an equivalence relation, it partitions the set into disjoint groups called equivalence classes — this is the deeper reason why the concept matters in JEE Mains.

Alternative Method

For checking transitivity on small sets, make a relation matrix (also called Boolean matrix):

M_R = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}

Rows and columns represent {1, 2, 3}. Entry $(i,j) = 1$ if $(i,j) \in R$ .

For transitivity, compute $M_R^2$ (Boolean multiplication, where + means OR, × means AND). If every 1 in $M_R^2$ is also a 1 in $M_R$ , the relation is transitive.

In $M_R^2$ , position (1,3) becomes 1 (because row 1 of $M_R$ times column 3 of $M_R$ : $1 \cdot 0 + 1 \cdot 1 + 0 \cdot 1 = 1$ ). But $M_R$ has 0 at position (1,3). Transitivity fails — same conclusion, faster for exam conditions.

In CBSE board exams, listing all “failing pairs” earns full marks. Write: “Since $(1,2) \in R$ and $(2,3) \in R$ but $(1,3) \notin R$ , R is not transitive.” One clean counterexample is enough — no need to check every case.

Common Mistake

Checking only “obvious” pairs for transitivity. Most students verify $(1,2)$ and $(2,1)$ and call it done. The dangerous pair here is $(1,2)$ paired with $(2,3)$ — crossing from one “chain” into another. Always systematically list all pairs $(a,b)$ and $(b,c)$ and check if $(a,c)$ is present. In this problem, the pair $(1,3)$ is absent, but students miss it because they stop after checking pairs that “feel symmetric.”

Property	Condition	One counterexample breaks it
Reflexive	$(a,a) \in R$ for all $a \in A$	Any missing $(a,a)$
Symmetric	$(a,b) \in R \Rightarrow (b,a) \in R$	Find $(a,b)$ without $(b,a)$
Transitive	$(a,b),(b,c) \in R \Rightarrow (a,c) \in R$	Find chain with missing endpoint
Equivalence	All three above	Fails if any one fails

This exact question — checking all three properties for a given explicit set — appeared in CBSE Class 11 board exams multiple times. The marking scheme gives 1 mark per property check + 1 mark for the equivalence conclusion, so writing a clear counterexample for the failing property is the fastest way to secure full marks.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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