Q.Let $R = \{(3, 1), (1, 3), (3, 3)\}$ be a relation defined on the set $A = \{1, 2, 3\}$. Then $R$ is symmetric, transitive but not reflexive.

The key idea is that a relation can be reflexive, symmetric, or transitive independently — here we check each property against the definition.

• Reflexive: Every element must relate to itself. $A = \{1,2,3\}$ requires $(1,1)$, $(2,2)$, and $(3,3)$. Only $(3,3)$ is present; $(1,1)$ and $(2,2)$ are missing. So $R$ is not reflexive.
• Symmetric: If $(a,b) \in R$, then $(b,a)$ must also be in $R$. We have $(3,1)$ and $(1,3)$ — both present. Also $(3,3)$ is its own pair. So $R$ is symmetric. …

Test each property of $R=\{(3,1),(1,3),(3,3)\}$ on $A=\{1,2,3\}$.

Symmetric? $(3,1)\in R$ and its reverse $(1,3)\in R$; $(3,3)$ is its own reverse. So $R$ is symmetric. ✓

Reflexive? Reflexivity needs $(1,1),(2,2),(3,3)$. Only $(3,3)$ is present, so $R$ is not reflexive. ✓

Transitive? Check the chains. From $(1,3)\in R$ and $(3,1)\in R$, transitivity requires $(1,1)\in R$ — but $(1,1)\notin R$. The condition fails, so $R$ is not transitive. …

Method: Verifying a Claim About a Relation's Properties

Use this for a true/false statement asserting that a listed relation is (say) symmetric, transitive, and not reflexive.

Steps

Step 1: Test each named property against its definition, independently.

Reflexive needs $(a,a)$ for every element of the base set; symmetric needs $(b,a)$ whenever $(a,b)$ is present; transitive needs $(a,c)$ whenever $(a,b)$ and $(b,c)$ are present.

Step 2: For transitivity, check every chain — including reversed pairs. …

Common Mistakes

Mistake 1: Declaring the relation transitive without checking all chains.

Why it's wrong: with $(3,1)$ and $(1,3)$ present, transitivity requires $(1,1)$; since $(1,1)$ is missing, the relation is not transitive. Correct approach: test every pair-of-pairs, especially those formed by a pair and its reverse.

Mistake 2: Accepting the claim wholesale without testing each property. …