Let a relation R on N × N be defined as: (x1, y1) R (x2, y2) if and only if x1 ≤ x2 or y1 ≤ y2. Cons

Q: Wait, why is it reflexive? I thought reflexive relations usually involve equality, like $x_1 = x_2$ and $y_1 = y_2$.

That is a common confusion! A relation is reflexive if every element is related to itself. Here, the condition is $x_1 \le x_2 \text{ or } y_1 \le y_2$. When we plug in the same point $(x_1, y_1)$ for both, we get $x_1 \le x_1 \text{ or } y_1 \le y_1$. Since $x_1 \le x_1$ is always true, the 'or' condition is satisfied. It doesn't matter if the other part is true or false; the gate is already open!

Q: Why can't we just say it's symmetric? If $1 \le 2$, then $2 \ge 1$, so it should work, right?

Ah, but the relation is defined by $\le$, not $\ge$. Symmetry requires that if $(1, 1) R (2, 3)$ is true, then $(2, 3) R (1, 1)$ must also be true. While $1 \le 2$ is true, the reverse requires $2 \le 1$ or $3 \le 1$. Both are false! The relation is strictly directional based on the $\le$ operator, making it non-symmetric.

Q: I'm still struggling with transitivity. How do I know which points to pick for a counterexample?

Great question! The trick is to pick points that 'barely' satisfy the conditions. For $A \to B$, pick a point where the first condition is true but the second is false. For $B \to C$, pick a point where the second condition is true but the first is false. Then, check if $A \to C$ fails both conditions. That is exactly how we broke the chain with $(2, 4) \to (3, 3) \to (1, 3)$!

Question

Let a relation $R$ on $N 	imes N$ be defined as: $(x_1, y_1) R (x_2, y_2)$ if and only if $x_1 \le x_2$ or $y_1 \le y_2$. Consider the two statements:
(I) $R$ is reflexive but not symmetric.
(II) $R$ is transitive
Then which one of the following is true?

Options:

(A) (1) Both (I) and (II) are correct.

(B) (2) Only (II) is correct.

(C) (3) Neither (I) nor (II) is correct.

(D) (4) Only (I) is correct.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical landscape. Today, we are not just solving a problem; we are dissecting the very architecture of a relation.

In the world of JEE Advanced, relations are not just abstract definitions; they are rules of engagement, boundaries that define how elements in a set interact. Let us dive into the relation $R$ defined on $\mathbb{N} 	imes \mathbb{N}$, where $(x_1, y_1) R (x_2, y_2)$ if and only if $x_1 \le x_2$ or $y_1 \le y_2$.

This 'or' is the heartbeat of our problem. It is a permissive, inclusive rule that tells us that as long as one of our conditions is met, the connection is established.

### Phase 1: The Mirror Test (Reflexivity)

Imagine you are standing in front of a mirror. A relation is reflexive if every point in our coordinate space can 'see' itself.

Mathematically, this means for any point $(x_1, y_1)$, the statement $(x_1, y_1) R (x_1, y_1)$ must hold true. Let us apply our rule: $(x_1, y_1) R (x_1, y_1)$ is true if $x_1 \le x_1$ or $y_1 \le y_1$.

Since any natural number is equal to itself, both $x_1 \le x_1$ and $y_1 \le y_1$ are undeniably true. Because the 'or' condition only requires one of them to be true, and here both are, the relation is perfectly **reflexive**.

### Phase 2: The Two-Way Street (Symmetry)

Now, let us test the symmetry. A relation is symmetric if the path from $A$ to $B$ implies the path from $B$ to $A$.

To disprove this, we do not need to check every point in the infinite set of natural numbers; we only need one 'rebellious' counterexample. Let us pick $A = (1, 1)$ and $B = (2, 3)$.

First, check $A R B$: Is $1 \le 2$ or $1 \le 3$? Yes, $1 \le 2$ is true. The forward path is open.

Now, check the reverse, $B R A$: Is $2 \le 1$ or $3 \le 1$? Both are false. The reverse path is blocked. Because the connection is not reciprocal, we have successfully proven that $R$ is **not symmetric**.

### Phase 3: The Chain Reaction (Transitivity)

Finally, we arrive at the most elusive property: transitivity. A relation is transitive if $A R B$ and $B R C$ implies $A R C$.

Let us hunt for a counterexample. We choose $A = (2, 4)$, $B = (3, 3)$, and $C = (1, 3)$.

1. Check $A R B$: Is $2 \le 3$ or $4 \le 3$? Since $2 \le 3$ is true, the first link is solid.

2. Check $B R C$: Is $3 \le 1$ or $3 \le 3$? Since $3 \le 3$ is true, the second link is solid.

3. Now, the moment of truth: $A R C$. Is $2 \le 1$ or $4 \le 3$? Both are false! The direct bridge from $A$ to $C$ does not exist. The chain is broken, meaning $R$ is **not transitive**.

### The Final Verdict

We have walked through the logic, tested the boundaries, and found the truth.

Statement (I) claimed $R$ is reflexive but not symmetric—our investigation confirmed this completely. Statement (II) claimed $R$ is transitive, but our counterexample proved it is not.

Thus, we conclude that **only statement (I) is correct**. This problem teaches us that in mathematics, as in life, definitions are everything. By carefully testing the boundaries of 'or', we have unraveled the nature of this relation.

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Visualized Solution

Defining the Relation $R$

Checking Reflexivity

Reflexivity Condition

Checking Symmetry - The Setup

Symmetry - Forward Relation

Symmetry - Reverse Relation Fails

Checking Transitivity - The Setup

Transitivity - First Link ( $A \to B$ )

Transitivity - Second Link ( $B \to C$ )

Transitivity - The Missing Link ( $A \to C$ )

Final Conclusion

The Sigma Insight: Types of Relations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Verdict

Similar Questions

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Visualized Solution

Defining the Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Checking Reflexivity

Reflexivity Condition

Checking Symmetry - The Setup

Symmetry - Forward Relation

Symmetry - Reverse Relation Fails

Checking Transitivity - The Setup

Transitivity - First Link ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A→B)

Transitivity - Second Link ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B→C)

Transitivity - The Missing Link ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A→C)