Q.If a relation $R$ on the set $\{1, 2, 3\}$ be defined by $R = \{(1, 2)\}$, then $R$ is
(A) reflexive
(B) transitive
(C) symmetric
(D) none of these

The key idea is that a relation can satisfy some properties without satisfying others — here, $R$ is reflexive only if every element relates to itself, symmetric only if every pair has its reverse, and transitive only if chains close.

• Reflexive? No — $(1,1), (2,2), (3,3)$ are all missing.
• Symmetric? No — $(1,2)$ is present but $(2,1)$ is absent. …

Check each property for $R=\{(1,2)\}$ on the set $\{1,2,3\}$.

Reflexive? Needs $(1,1),(2,2),(3,3)\in R$. None are present, so $R$ is not reflexive.

Symmetric? Since $(1,2)\in R$, symmetry would require $(2,1)\in R$. It is absent, so $R$ is not symmetric. …

Method: Testing a small listed relation for R, S, T

Use this when the relation is given as an explicit short list of pairs on a small set and you must decide reflexive / symmetric / transitive.

Steps

Step 1: Reflexive — check every diagonal pair is present.

For a set $\{1,2,3\}$ the relation must contain $(1,1),(2,2),(3,3)$. If even one is missing, it is not reflexive.

Step 2: Symmetric — for each listed $(a,b)$ look for $(b,a)$.

Scan the non-diagonal pairs; if any $(a,b)$ has its reverse $(b,a)$ absent, symmetry fails immediately.

Step 3: Transitive — chase every existing chain, and apply vacuous truth. …

Common Mistakes

Mistake 1: Declaring $R=\{(1,2)\}$ 'not transitive' because it has just one pair.

Why it's wrong: transitivity is violated only when a chain $(a,b),(b,c)$ exists without $(a,c)$; here nothing starts with $2$, so no chain exists and transitivity holds vacuously. Correct approach: check whether any chain can even form before deciding.

Mistake 2: Thinking a single pair makes the relation symmetric or reflexive. …