Q.Let $A = \{a, b, c\}$ and the relation $R$ be defined on $A$ as follows: $R = \{(a, a), (b, c), (a, b)\}$. Then, write minimum number of ordered pairs to be added in $R$ to make $R$ reflexive and transitive.

Concept: Reflexive and transitive closure — we add the minimum pairs needed to satisfy reflexivity and transitivity without forcing symmetry.

Step 1 — Reflexive condition

For reflexivity on $A = \{a,b,c\}$, we need $(a,a), (b,b), (c,c)$.

We already have $(a,a)$, so we must add $(b,b)$ and $(c,c)$.

Step 2 — Transitive condition

Currently $R$ has $(a,b)$ and $(b,c)$. For transitivity, if $(a,b)$ and $(b,c)$ are present, we must have $(a,c)$.

No other chain exists, so only $(a,c)$ is needed.

Step 3 — Check for further transitivity

After adding $(a,c)$, check new chains: $(a,b)$ and $(b,c)$ already gave $(a,c)$; no new pairs are forced. So we stop.

Pairs to add: $(b,b), (c,c), (a,c)$ — that’s 3 ordered pairs.

We are given $A = \{a, b, c\}$ and $R = \{(a, a), (b, c), (a, b)\}$. The task: add the minimum number of ordered pairs so that the resulting relation is both reflexive and transitive. Notice that symmetry is not required — a common trap is to assume we need to make $R$ an equivalence relation, but the problem only asks for reflexivity and transitivity.

Let’s break it down.

1. Make it reflexive A relation on $A$ is reflexive if every element of $A$ is related to itself. Currently, $R$ contains $(a,a)$ but is missing $(b,b)$ and $(c,c)$. So we must add:

That’s 2 pairs so far.

2. Make it transitive

   Transitivity means: whenever $(x,y) \in R$ and $(y,z) \in R$, then $(x,z)$ must also be in $R$.

   Look at the existing pairs (including the ones we’ve already added for reflexivity).

   • We have $(a,b)$ and $(b,c)$. Since $aRb$ and $bRc$, we need $(a,c)$.
   • Check other combinations: $(a,a)$ with $(a,b)$ already gives $(a,b)$ (present). $(b,b)$ with $(b,c)$ gives $(b,c)$ (present). $(c,c)$ with anything? No outgoing pair from $c$ exists, so no new requirement. So the only missing transitive pair is $(a,c)$.
   Watch out

   Do not add $(b,a)$ or $(c,a)$ — those would be needed for symmetry, which is not asked. Adding them would be unnecessary and would violate the “minimum” condition.

3. Count the total

   We added $(b,b)$, $(c,c)$, and $(a,c)$. That’s 3 ordered pairs.

Method: Reflexive-transitive closure by minimal augmentation

For "minimum ordered pairs to add so $R$ becomes reflexive and transitive," add only what each property forces — nothing more.

Steps

Step 1: Force reflexivity

For a set with elements $a,b,c,\dots$, reflexivity needs every diagonal pair $(a,a)$. List the diagonal pairs already present and add exactly the missing ones.

Step 2: Force transitivity from existing chains

Scan for chains $(x,y),(y,z)$ in the current relation and add the required $(x,z)$ if absent.

Step 3: Re-scan for new chains created by your additions

Every pair you add can create a new chain, so repeat Step 2 until no new pair is forced (this is the "closure" idea).

Step 4: Do NOT add symmetric pairs

Symmetry is not requested; adding reverse pairs would violate "minimum." Count only the pairs you were forced to add.

The trap is scope: reflexive + transitive is not an equivalence relation, so resist the urge to make it symmetric.

Common Mistakes

Mistake 1: Adding symmetric pairs like $(b,a)$ or $(c,a)$

Why it's wrong: the question asks only for reflexive and transitive — symmetry is not required, so those pairs violate "minimum." Correct approach: add only diagonal and forced transitive pairs.

Mistake 2: Forgetting a diagonal pair

Why it's wrong: reflexivity needs $(a,a),(b,b),(c,c)$; here $(a,a)$ is present but $(b,b),(c,c)$ are missing. Correct approach: check the diagonal element by element.

Mistake 3: Missing the forced transitive pair $(a,c)$

Why it's wrong: since $(a,b)$ and $(b,c)$ are both present, transitivity forces $(a,c)$. Correct approach: after listing pairs, test every chain and add the resulting pair.