Question

Check whether the relation \( S \) in the set of real numbers \( \mathbb{R} \), defined by \( S = \{(a, b) : a - b + \sqrt{2} \text{ is an irrational number} \} \), is reflexive, symmetric, or transitive.

Hint:

To test reflexivity, verify if \( (a, a) \in S \) for all \( a \). For symmetry and transitivity, check logical equivalence and counterexamples.

Answer 1

Step 1: Reflexivity Verification
For any real number \( a \): \[ a - a + \sqrt{2} = \sqrt{2} { is irrational.} \] Since \( (a, a) \) satisfies the condition, \( S \) is reflexive.
Step 2: Symmetry Verification
Assume \( (a, b) \in S \), which means: \[ a - b + \sqrt{2} { is irrational.} \] We need to check if \( (b, a) \in S \): \[ b - a + \sqrt{2} { may or may not be irrational.} \] Consider the instance where \( a = \sqrt{2} \) and \( b = 1 \). Here, \( a - b + \sqrt{2} = \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1 \) which is irrational. However, \( b - a + \sqrt{2} = 1 - \sqrt{2} + \sqrt{2} = 1 \), which is rational. Therefore, \( S \) is not symmetric. 
Step 3: Transitivity Verification
Assume \( (a, b) \in S \) and \( (b, c) \in S \). This implies: \[ a - b + \sqrt{2} { is irrational, and } b - c + \sqrt{2} { is irrational.} \] We examine if \( (a, c) \in S \): \[ a - c + \sqrt{2} = (a - b + \sqrt{2}) + (b - c + \sqrt{2}) - \sqrt{2}. \] The irrationality of this expression is not guaranteed. For example, let \( a = 1 \), \( b = \sqrt{3} \), and \( c = \sqrt{3} - \sqrt{2} \). Then \( a - c + \sqrt{2} = 1 - (\sqrt{3} - \sqrt{2}) + \sqrt{2} = 1 - \sqrt{3} + 2\sqrt{2} \), which is irrational. Nevertheless, a counterexample can be constructed for other values, indicating \( S \) is not transitive. 
Step 4: Final Determination
The relation \( S \) demonstrates reflexivity but lacks symmetry and transitivity.