Question

Let a relation \( R \) on \( \mathbb{N} \times \mathbb{N} \) be defined as: $$(x_1, y_1) \, R \, (x_2, y_2) \text{ if and only if } x_1 \leq x_2 \text{ or } y_1 \leq y_2.$$ 
Consider the two statements:
[(I)] \( R \) is reflexive but not symmetric. 
[(II)] \( R \) is transitive.
Then which one of the following is true:

• A. Only (II) is correct.
• B. Only (I) is correct.
• C. Both (I) and (II) are correct.
• D. Neither (I) nor (II) is correct.

Answer 1

The Correct Option is B

To verify the properties of \( R \), consider all \( (x_1, y_1), (x_2, y_2) \in R \) where \( x_1, y_1 \in \mathbb{N} \).

1. \( R \) is reflexive: For all \( (x_1, y_1) \in \mathbb{N} \times \mathbb{N} \), \[ x_1 \leq x_1 \, \text{or} \, y_1 \leq y_1 \] is always true.
   Hence, \( R \) is reflexive.
2. \( R \) is not symmetric: For example, consider \( (1, 2) R (2, 3) \) because \( 1 \leq 2 \). However, \( (2, 3) \notin R(1, 2) \) because neither \( 2 \leq 1 \) nor \( 3 \leq 2 \). Hence, \( R \) is not symmetric.
3. \( R \) is not transitive: For example, consider \( (2, 4)R(3, 3) \) and \( (3, 3)R(1, 3) \). However, \( (2, 4) \notin R(1, 3) \), so \( R \) is not transitive.

Thus, only statement (I) is correct.

Answer 2

Step 1: Understand the relation \( R \).
The relation \( R \) on \( \mathbb{N} \times \mathbb{N} \) is defined as:
\[ (x_1, y_1) \, R \, (x_2, y_2) \text{ if and only if } x_1 \leq x_2 \text{ or } y_1 \leq y_2. \] This means that for any two pairs \((x_1, y_1)\) and \((x_2, y_2)\), the relation holds if either \(x_1 \leq x_2\) or \(y_1 \leq y_2\).

Step 2: Check if \( R \) is reflexive.
For \( R \) to be reflexive, it must satisfy:
\[ (x_1, y_1) \, R \, (x_1, y_1) \text{ for all } (x_1, y_1) \in \mathbb{N} \times \mathbb{N}. \] This is true because:
- \( x_1 \leq x_1 \) (so the first condition holds),
- \( y_1 \leq y_1 \) (so the second condition holds).
Thus, \( R \) is reflexive.

Step 3: Check if \( R \) is symmetric.
For \( R \) to be symmetric, it must satisfy:
\[ (x_1, y_1) \, R \, (x_2, y_2) \implies (x_2, y_2) \, R \, (x_1, y_1). \] However, this is not true in general. For example, if \( (x_1, y_1) = (1, 2) \) and \( (x_2, y_2) = (2, 3) \), we have \( 1 \leq 2 \) (so \( (1, 2) \, R \, (2, 3) \)), but \( 2 \not\leq 1 \) and \( 3 \not\leq 2 \), so \( (2, 3) \not\ R \ (1, 2) \). Therefore, \( R \) is not symmetric.

Step 4: Check if \( R \) is transitive.
For \( R \) to be transitive, it must satisfy:
\[ \text{If } (x_1, y_1) \, R \, (x_2, y_2) \text{ and } (x_2, y_2) \, R \, (x_3, y_3), \text{ then } (x_1, y_1) \, R \, (x_3, y_3). \] Let’s examine whether this holds:
- If \( x_1 \leq x_2 \) or \( y_1 \leq y_2 \), and \( x_2 \leq x_3 \) or \( y_2 \leq y_3 \), it’s not guaranteed that \( x_1 \leq x_3 \) or \( y_1 \leq y_3 \). For instance, consider:
- \( (x_1, y_1) = (1, 2) \),
- \( (x_2, y_2) = (2, 1) \),
- \( (x_3, y_3) = (3, 2) \).
Here, \( (1, 2) \, R \, (2, 1) \) because \( x_1 \leq x_2 \), and \( (2, 1) \, R \, (3, 2) \) because \( y_2 \leq y_3 \), but \( (1, 2) \not\ R \ (3, 2) \) because neither \( x_1 \leq x_3 \) nor \( y_1 \leq y_3 \). Therefore, \( R \) is not transitive.

Step 5: Conclusion.
- Statement (I): \( R \) is reflexive but not symmetric. This is correct.
- Statement (II): \( R \) is transitive. This is incorrect.
Thus, the correct answer is:
Only (I) is correct.