Question:medium

Let \( X = {R} \times {R} \). Define a relation \( R \) on \( X \) as: \[ (a_1, b_1) \, R \, (a_2, b_2) \iff b_1 = b_2. \] Statement-I: \( R \) is an equivalence relation. Statement-II: For some \( (a, b) \in X \), the set \( S = \{(x, y) \in X : (x, y) R (a, b)\ \) represents a line parallel to \( y = x \).}

Show Hint

To check if a relation is an equivalence relation, verify reflexivity, symmetry, and transitivity. Also, when interpreting geometric shapes, note the difference between lines parallel to \( y = x \) and horizontal lines.
Updated On: Jan 14, 2026
  • Both Statement-I and Statement-II are false.
  • Statement-I is true but Statement-II is false.
  • Both Statement-I and Statement-II are true.
  • Statement-I is false but Statement-II is true.
Show Solution

The Correct Option is B

Solution and Explanation

To verify the assertions about the relation \( R \) defined on \( X = \mathbb{R} \times \mathbb{R} \), we examine each statement:

  1. Statement-I: \( R \) is an equivalence relation.
    • Reflexivity: For any \((a, b) \in X\), the condition \((a, b) \, R \, (a, b)\) implies \(b = b\), which is universally true. Hence, \( R \) is reflexive.
    • Symmetry: If \((a_1, b_1) \, R \, (a_2, b_2)\) for \((a_1, b_1), (a_2, b_2) \in X\), then \(b_1 = b_2\) by definition. For symmetry, we require \((a_2, b_2) \, R \, (a_1, b_1)\), which means \(b_2 = b_1\). Since equality is symmetric, \(R\) is symmetric.
    • Transitivity: Given \((a_1, b_1) \, R \, (a_2, b_2)\) and \((a_2, b_2) \, R \, (a_3, b_3)\) for \((a_1, b_1), (a_2, b_2), (a_3, b_3) \in X\), we have \(b_1 = b_2\) and \(b_2 = b_3\). This implies \(b_1 = b_3\), satisfying the condition for \((a_1, b_1) \, R \, (a_3, b_3)\). Thus, \(R\) is transitive.
  2. Statement-II: There exists \((a, b) \in X\) such that the set \(S = \{(x, y) \in X : (x, y) \, R \, (a, b)\}\) describes a line parallel to \(y = x\).

Conclusion: Based on the preceding analysis, Statement-I is accurate, while Statement-II is incorrect.

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