Q.Let $A = \{1, 2, 3\}$ and consider the relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$. Then $R$ is
(A) reflexive but not symmetric
(B) reflexive but not transitive
(C) symmetric and transitive
(D) neither symmetric, nor transitive

The key idea is that $R$ contains all three diagonal pairs, so it is reflexive. It is not symmetric because $(1,2) \in R$ but $(2,1) \notin R$. However, whenever $(a,b)$ and $(b,c)$ are in $R$, the pair $(a,c)$ is also present — for example, $(1,2)$ and $(2,3)$ give $(1,3)$, wh …

We need to check three properties: reflexivity, symmetry, and transitivity. The set is $A = \{1,2,3\}$, and $R$ is given explicitly. Let’s go through each property one by one.

1. Reflexivity — A relation on $A$ is reflexive if every element of $A$ is related to itself. That means $(1,1)$, $(2,2)$, and $(3,3)$ must all be in $R$.

   Looking at $R$, we see all three are present: $(1,1)$, $(2,2)$, $(3,3)$. So $R$ is reflexive.

2. Symmetry — A relation is symmetric if whenever $(a,b)$ is in $R$, then $(b,a)$ must also be in $R$.

   Check the non-diagonal pairs:

   • $(1,2)$ is in $R$, but $(2,1)$ is not in $R$. That single violation is enough to break symmetry. So $R$ is not symmetric.
   Watch out

   A common mistake is to check only a few pairs and miss the missing reverse. Here, $(1,2)$ is present but $(2,1)$ is absent — that’s a clear counterexample. No need to check further.

3. Transitivity — A relation is transitive if whenever $(a,b)$ and $(b,c)$ are in $R$, then $(a,c)$ must also be in $R$.

   List all pairs that could serve as the “middle” step:

   • $(1,2)$ and $(2,3)$ are both in $R$. Their “composite” is $(1,3)$, which is indeed in $R$.
   • $(1,1)$ and $(1,2)$ give $(1,2)$ — present.
   • $(1,2)$ and $(2,2)$ give $(1,2)$ — present.
   • $(2,2)$ and $(2,3)$ give $(2,3)$ — present.
   • $(1,1)$ and $(1,3)$ give $(1,3)$ — present. …

Method: Testing a listed relation and reading its structure

Use this when a relation on a small set is given explicitly and you must classify it — the same R-S-T scan, plus a shortcut for spotting a familiar order.

Steps

Step 1: Reflexive — confirm all diagonal pairs appear.

Check $(1,1),(2,2),(3,3)$ are all present. If yes, reflexive.

Step 2: Symmetric — hunt for a missing reverse.

For each off-diagonal $(a,b)$, see whether $(b,a)$ is listed. Here $(1,2)$ is present but $(2,1)$ is not, so symmetry fails on the first counterexample.

Step 3: Transitive — verify every real chain closes. …

Common Mistakes

Mistake 1: Overlooking the missing reverse and calling $R$ symmetric.

Why it's wrong: $(1,2)\in R$ but $(2,1)\notin R$, so symmetry fails; the presence of many pairs can hide the one missing reverse. Correct approach: scan every off-diagonal pair for its reverse.

Mistake 2: Wrongly deciding $R$ is not transitive.

Why it's wrong: every existing chain closes — $(1,2),(2,3)$ gives $(1,3)$, which is present, and all others hold too. Correct approach: check only the chains that actually occur; here they all succeed. …