Q.On the set $A = \{1, 2, 3\}$, a relation is given by $R = \{(1,1),(2,2),(3,3),(1,2),(2,1)\}$. Determine whether $R$ is reflexive, symmetric and transitive. Is it an equivalence relation?

Check each property against its definition: all self-pairs present (reflexive), every pair's reverse present (symmetric), and every two-step chain's shortcut present (transitive).

Test each property on the underlying set $A = \{1, 2, 3\}$.

Reflexive? Need $(a, a) \in R$ for every $a \in A$. The self-pairs $(1,1)$, $(2,2)$, $(3,3)$ are all present. Yes, reflexive.

Symmetric? Need $(b, a) \in R$ whenever $(a, b) \in R$. The only non-self pairs are $(1, 2)$ and $(2, 1)$, and each is the reverse of the other. The self-pairs are their own reverses. Yes, symmetric.

Transitive? Need $(a, c) \in R$ whenever $(a, b), (b, c) \in R$. Check the chains that use the off-diagonal pairs:

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

Because the only pairing of distinct elements is $1$ and $2$ (both directions present) together with all three self-pairs, $R$ is really 'equality, plus treating $1$ and $2 …

A frequent error is declaring transitivity without checking the chains through the off-diagonal pairs, or declaring reflexivity while forgetting to confirm $(3 …