Q.Let $A = \{1, 2, 3\}$. Then show that the number of relations containing $(1, 2)$ and $(2, 3)$ which are reflexive and transitive but not symmetric is three.

Setup. On $A=\{1,2,3\}$ a reflexive relation must contain $(1,1),(2,2),(3,3)$. We are told it contains $(1,2)$ and $(2,3)$, so transitivity forces $(1,3)$ as well. The base relation is

$R_0$ is already reflexive, transitive, and not symmetric ($(2,1)$ is absent), so it is one valid relation.

Which extra pairs may we add? The candidates are $(2,1),(3,2),(3,1)$.

• Add $(2,1)$: no new pair is forced, and the relation stays non-symmetric ($(3,2)$ still absent). Valid.
• Add $(3,2)$: no new pair is forced, and the relation stays non-symmetric ($(2,1)$ still absent). Valid. …

What is forced

On $A=\{1,2,3\}$ the relation must be:

• reflexive — so it contains $(1,1),(2,2),(3,3)$;
• containing $(1,2)$ and $(2,3)$ (given);
• transitive — so from $(1,2)$ and $(2,3)$ it must also contain $(1,3)$.

Collecting these, every admissible relation contains the base set

$R_0$ is transitive (check: the only chains are $(1,2)\&(2,3)\to(1,3)$, present) and it is not symmetric because $(2,1)$ is missing. So $R_0$ itself is one relation that qualifies.

What may still be added

The pairs not yet in $R_0$ are $(2,1),(3,2),(3,1)$. We test each addition, keeping the relation transitive and non-symmetric.

Add $(2,1)$. Chains created: $(2,1)\&(1,2)\to(2,2)$ ✓, $(2,1)\&(1,3)\to(2,3)$ ✓, $(1,2)\&(2,1)\to(1,1)$ ✓. Nothing new is forced. The relation is still not symmetric, since $(3,2)$ is absent. Valid.

Add $(3,2)$. Chains created: $(3,2)\&(2,3)\to(3,3)$ ✓, $(3,2)\&(2,1)$? — $(2,1)$ is not present, so no chain; $(1,3)\&(3,2)\to(1,2)$ ✓. Nothing new is forced. Still not symmetric, since $(2,1)$ is absent. Valid. …

Method: Counting Relations Under Property Constraints (Forced-Then-Optional)

When a question asks how many relations on a small set satisfy several properties plus some required pairs, don't list all $2^{n^2}$ relations. Instead find what the constraints force, then test each remaining pair as an optional add.

Steps

Step 1: Write down every pair that is forced.

Apply the mandatory properties to the given pairs. Reflexivity forces all diagonal pairs $(a,a)$; transitivity forces closures of chains (e.g. $(1,2)$ and $(2,3)$ force $(1,3)$). Collect these into a minimal base relation $R_0$.

Step 2: Verify the base relation itself satisfies all conditions.

Check $R_0$ is transitive and (here) not symmetric. If it qualifies, it is your first valid relation.

Step 3: List the still-optional pairs and test each addition. …

Common Mistakes

Mistake 1: Forgetting transitivity forces extra pairs.

Why it's wrong: with $(1,2)$ and $(2,3)$ present, transitivity requires $(1,3)$; omitting it gives a non-transitive relation that should not be counted. Correct approach: close the given pairs under transitivity before counting, building the base relation $R_0$.

Mistake 2: Missing that adding both $(2,1)$ and $(3,2)$ makes the relation symmetric.

Why it's wrong: together they force $(3,1)$, after which every pair has its reverse, so the relation becomes symmetric and is disallowed. Correct approach: test each optional addition and their combinations, discarding any that trigger full symmetry. …