Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the se

Q: Wait, why is the relation not symmetric? I see $(6, 12)$ and $(3, 6)$, isn't that enough?

Great question! Symmetry is a strict condition: for every pair $(a, b)$ in $R$, the pair $(b, a)$ must also be in $R$. While we have $(6, 12)$, we do not have $(12, 6)$. Because $(12, 6)$ is missing, the symmetry condition is violated. It is not enough to have some pairs; we need the reverse for every single non-reflexive pair.

Q: Do I have to check every single combination for transitivity? That seems like it would take forever!

You don't have to check every combination! You only need to check the 'chains' where a connection exists. If you have $(a, b)$ and $(b, c)$, then you check for $(a, c)$. If you don't have a chain (i.e., no $b$ such that $a \to b$ and $b \to c$), then that specific case is vacuously transitive. Focus on the paths!

Q: Is the empty relation reflexive?

No, an empty relation on a non-empty set is never reflexive. Reflexivity requires that for every element $a$ in set $A$, the pair $(a, a)$ must be in the relation. If the relation is empty, it cannot contain any $(a, a)$ pairs, so it fails the reflexivity test immediately.

Question

Let $R = \{(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)\}$ be a relation on the set $A = \{3, 6, 9, 12\}$. The relation is

Options:

(A) reflexive and transitive only

(B) reflexive only

(C) an equivalence relation

(D) reflexive and symmetric only

Sigma Percentile · Accepted Answer

### The Architecture of Connections: Understanding Relations

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to peel back the layers of a fundamental concept in discrete mathematics: the relation.

A relation is not just a set of ordered pairs; it is a map of how elements in a set interact with one another. When we look at the set $A = \{3, 6, 9, 12\}$ and the relation $R = \{(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)\}$, we are looking at a specific structure.

Let us break this down into the three pillars of relations: Reflexivity, Symmetry, and Transitivity.

### Phase 1: The Mirror of Reflexivity

Imagine our set $A$ as four distinct nodes on a graph. A relation is reflexive if every element 'loves itself'—that is, for every element $a \in A$, the pair $(a, a)$ must exist in $R$.

Think of this as a self-loop on each node. We scan our set $R$ and look for the identity pairs: $(3, 3), (6, 6), (9, 9),$ and $(12, 12)$.

They are all there! Because every single element in $A$ has a corresponding self-loop, we can confidently declare that our relation $R$ is **reflexive**. It is the foundation of our structure.

### Phase 2: The Symmetry Trap

Now, let us test for symmetry. Symmetry is about balance. It demands that if you can travel from $a$ to $b$, you must be able to travel back from $b$ to $a$.

Mathematically, this is defined as:
$$(a, b) \in R \implies (b, a) \in R$$

Let us examine our pairs. We have $(6, 12)$ in our relation. For the relation to be symmetric, we absolutely require the pair $(12, 6)$ to be present.

We search through $R$... and it is not there. The absence of this 'return ticket' is fatal to the symmetry property. Even if all other pairs were symmetric, this single missing link is enough to disqualify the entire relation. Thus, $R$ is **not symmetric**.

### Phase 3: The Transitive Journey

Finally, we arrive at transitivity, the most subtle of the three. Transitivity is the logic of shortcuts.

If you can go from $a$ to $b$, and from $b$ to $c$, then you must be able to take a direct flight from $a$ to $c$. We look for these chains.

We see $(3, 6)$ and $(6, 12)$. This forms a chain $3 	o 6 	o 12$. Does the direct pair $(3, 12)$ exist in $R$? Yes, it does!

We check other combinations, such as $(3, 9)$ and $(9, 9)$, which leads to $(3, 9)$, already present. Every chain we find has its corresponding shortcut. The logic holds firm. Therefore, the relation is **transitive**.

### The Synthesis

By systematically applying these definitions, we have peeled back the mystery. We found that the relation is reflexive because of the self-loops, not symmetric because of the missing return paths, and transitive because all the shortcuts are accounted for.

This is the beauty of set theory—it is a game of rules, and once you learn to play by them, the answers reveal themselves with absolute clarity. Keep practicing this graphical visualization; it is a powerful tool that will serve you well in your JEE journey. You are doing great!

Animated Solution for Mathematics - Sets and Relations: Let $R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)}$ be a relation on the set $A = {3, 6, 9, 12}$ . The relation is

Select Answer:

Visualized Solution

Defining Set $A$ and Relation $R$

Visualizing the Elements

Checking Reflexivity

Conclusion on Reflexivity

Plotting the Remaining Pairs

Checking Symmetry

The Missing Symmetric Link

Checking Transitivity

The Transitive Shortcut

Final Conclusion

The Sigma Insight: Types of Relations

The Architecture of Connections

Phase 1

Phase 2

Phase 3

The Synthesis

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Animated Solution for Mathematics - Sets and Relations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R={(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)} be a relation on the set A={3,6,9,12}. The relation is