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.

> [!TLDR] > To make $R$ reflexive we must add $(b,b)$ and $(c,c)$; to make it transitive we must add $(a,c)$ (from $aRb$ and $bRc$). No symmetry is required. So the minimum number of ordered pairs to add is **3**. 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: $$ (b,b) \quad \text{and} \quad (c,c) $$ 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 $(a,c)$. > [!WARNING] > 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. > [!TIP] > Always check whether the problem asks for reflexivity + transitivity only, or also symmetry. Many students lose marks by adding symmetric pairs unnecessarily. > [!ANSWER] > The minimum number of ordered pairs to be added is $\boxed{3}$.

Let A = \a, b, c\ and the relation R be defined on A as follows: R = \(a, a), (b, c), (a, b)\. Then, write min

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A = \{a, b, c\}

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Concept understanding — Reflexive Transitive Not Symmetric

Reflexive, Transitive, Not Symmetric — A First Look

What kind of relation is reflexive and transitive but not symmetric? "Same grade as" won't do — it's symmetric. We need a relation that only goes one way. Let's build it.

The Intuition: A One-Way Street

The classic example is "divides" on the positive integers:

Reflexive: every number divides itself. $5 \mid 5$ . ✓
Transitive: if $a \mid b$ and $b \mid c$ , then $a \mid c$ . E.g. $2 \mid 4$ and $4 \mid 12$ gives $2 \mid 12$ . ✓
Not symmetric: $2 \mid 4$ is true, but $4 \mid 2$ is false. The relation goes only one way. ✓

The key insight: the relation can go from smaller to larger (or equal), but not back.

The Precise Statement

Let $R$ be a relation on a set $S$ . Then:

\text{Reflexive: } \forall a \in S,\; a\,R\,a

"Reflexive transitive not symmetric" is just a checklist of three properties — not a standard name like "equivalence relation". A relation with these (plus antisymmetry) is a partial order.

Why This Matters

Exams often ask: "Is this relation reflexive? Symmetric? Transitive?" Test each property independently — a relation can be reflexive and transitive but fail symmetry, and that's perfectly fine. For example, on the reals define $x\,R\,y$ if $x \leq y$ : reflexive yes, transitive yes, symmetric no ( $3 \leq 5$ but $5 \not\leq 3$ ).

Watch out

A common mistake: assuming that reflexive + transitive forces symmetry. False — both "divides" and " $\le$ " disprove it. Always test each property separately.

A Quick Table for Clarity

Property	Meaning	Example: "divides" on $\mathbb{N}$
Reflexive	Every element relates to itself	$3 \mid 3$ ✓
Transitive	Chain of relations implies direct relation	$2 \mid 4$ and $4 \mid 12$ ⇒ $2 \mid 12$ ✓
Not symmetric	At least one pair works one way but not the other	$2 \mid 4$ but $4 \nmid 2$ ✓

The Takeaway

A relation that is reflexive, transitive, and not symmetric is like a one-way ladder: you can stand on your own rung (reflexive), climb from rung to rung through intermediate ones (transitive), but not climb back down (not symmetric). The classic examples are "divides" and " $\le$ ".

A relation that is reflexive and transitive but not symmetric — such as "divides" or ≤ — is a classic example type in the CBSE Class 12 Relations and Functions chapter, and "relation reflexive transitive not symmetric example" is a frequently searched exam question format. Recognising that these three properties are independent of one another is a key idea tested every year in board exams.

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