Q.The relation $R$ on the set $A = \{1, 2, 3\}$ defined as $R = \{(1, 1), (1, 2), (2, 1), (3, 3)\}$ is reflexive, symmetric and transitive.

Check the three properties of $R=\{(1,1),(1,2),(2,1),(3,3)\}$ on $A=\{1,2,3\}$.

Reflexive? Need $(a,a)$ for every $a\in A$: we have $(1,1)$ and $(3,3)$ but $(2,2)$ is missing — not reflexive.

Symmetric? $(1,2)$ has its reverse $(2,1)$; the self-pairs $(1,1),(3,3)$ reverse to themselves — symmetric.

Transitive? From $(2,1)$ and $(1,2)$ transitivity would require $(2,2)$, which is missing — not transitive. …

The idea

The claim says $R=\{(1,1),(1,2),(2,1),(3,3)\}$ on $A=\{1,2,3\}$ is reflexive, symmetric and transitive — i.e. an equivalence relation. We test each property against its definition; one counterexample settles each.

Step 1 — reflexive?

Reflexivity requires $(a,a)\in R$ for every $a\in A$. We need $(1,1),(2,2),(3,3)$. Of these $(1,1)$ and $(3,3)$ are present, but $(2,2)$ is not in $R$. So $R$ is not reflexive.

Step 2 — symmetric?

Symmetry requires that whenever $(a,b)\in R$, also $(b,a)\in R$. Check each pair:

• $(1,1)$ — reverse is itself. ✓
• $(1,2)$ — reverse $(2,1)$ is present. ✓
• $(2,1)$ — reverse $(1,2)$ is present. ✓
• $(3,3)$ — reverse is itself. ✓

Every pair's reverse is present, so $R$ is symmetric.

Step 3 — transitive?

Transitivity requires: if $(a,b)\in R$ and $(b,c)\in R$ then $(a,c)\in R$. Look at the two-step chains: …

Method: Testing a Listed Relation on a Small Set

Use this when the relation is given as an explicit list of ordered pairs and you must check reflexive / symmetric / transitive.

Steps

Step 1: Reflexive — list the required self-pairs

Write down $(a,a)$ for every element of the base set, then check each is present. A single missing self-pair (like $(2,2)$) breaks reflexivity.

Step 2: Symmetric — mirror every pair

For each $(a,b)$ in the list, confirm $(b,a)$ is also present. Self-pairs mirror to themselves automatically.

Step 3: Transitive — chase every two-step chain …

Common Mistakes

Mistake 1: Overlooking a missing self-pair

Why it's wrong: having $(1,1)$ and $(3,3)$ but not $(2,2)$ still fails reflexivity, because reflexivity needs every element. Correct approach: check the self-pair of each base-set element, not just the ones that appear.

Mistake 2: Skipping the chain that forces $(2,2)$

Why it's wrong: from $(2,1)$ and $(1,2)$, transitivity demands $(2,2)$, which is absent, so the relation is not transitive. Correct approach: test both orders of a symmetric pair when checking chains. …