Q.Let R be the relation in the set $\{1, 2, 3, 4\}$ given by $R = \{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}$. Choose the correct answer. (A) R is reflexive and symmetric but not transitive. (B) R is reflexive and transitive but not symmetric. (C) R is symmetric and transitive but not reflexive. (D) R is an equivalence relation.

Test the three properties of $R=\{(1,1),(1,2),(1,3),(2,2),(3,2),(3,3),(4,4)\}$ on the set $\{1,2,3,4\}$.

Reflexive: every element has its self-pair — $(1,1),(2,2),(3,3),(4,4)$ are all present. So $R$ is reflexive.

Symmetric: $(1,2)\in R$ but $(2,1)\notin R$. One missing reverse is enough — $R$ is not symmetric. …

We check reflexivity, symmetry and transitivity of

on $S=\{1,2,3,4\}$, one property at a time.

1. Reflexive?

Reflexive means $(a,a)\in R$ for every $a\in S$. Checking: $(1,1),(2,2),(3,3),(4,4)$ are all present. So $R$ is reflexive.

2. Symmetric?

Symmetric means whenever $(a,b)\in R$ we also have $(b,a)\in R$. Look at $(1,2)$: its reverse $(2,1)$ is not in $R$. A single missing reverse breaks the property, so $R$ is not symmetric.

3. Transitive?

Transitive means whenever $(a,b)\in R$ and $(b,c)\in R$, the pair $(a,c)\in R$. We only need the chains whose middle element matches:

• $(1,2)$ and $(2,2)$ $\Rightarrow$ need $(1,2)$ — present.
• $(1,3)$ and $(3,2)$ $\Rightarrow$ need $(1,2)$ — present. …

Method: Testing reflexive / symmetric / transitive on a listed relation

For a relation given as an explicit finite set of pairs on a set $S$, check the three properties mechanically, in order of ease.

Steps

Step 1: Reflexive — confirm $(a,a)\in R$ for every $a\in S$. Scan for the self-pair of each element; one missing self-pair kills reflexivity.

Step 2: Symmetric — for every pair $(a,b)\in R$ with $a\neq b$, check that $(b,a)\in R$ too. A single missing reverse pair means "not symmetric". …

Common Mistakes

Mistake 1: Declaring $R$ symmetric without checking every reverse pair.

Why it's wrong: $(1,2)\in R$ but $(2,1)\notin R$, so symmetry fails — a single missing reverse is enough. Correct approach: for each pair with $a\neq b$, verify its reverse is present before concluding symmetric.

Mistake 2: Calling $R$ non-transitive because it "looks incomplete". …