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:

Updated On: Nov 24, 2024
  • Only (II) is correct.
  • Only (I) is correct.
  • Both (I) and (II) are correct.
  • Neither (I) nor (II) is correct.
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The Correct Option is B

Solution and Explanation

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.

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