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
For \( a \in \mathbb{R} \): \[ a - a + \sqrt{2} = \sqrt{2} \text{ is irrational.} \] Thus, \( (a, a) \in S \), and \( S \) is reflexive. 

Step 2: Symmetry
Let \( (a, b) \in S \), so: \[ a - b + \sqrt{2} \text{ is irrational.} \] Now, check if \( (b, a) \in S \): \[ b - a + \sqrt{2} \text{ may or may not be irrational.} \] For example: \[ a = \sqrt{2}, \, b = 1 \implies a - b + \sqrt{2} = \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1 \text{ (irrational), but } \] \[ b - a + \sqrt{2} = 1 - \sqrt{2} + \sqrt{2} = 1 \text{ (rational).} \] Thus, \( S \) is not symmetric. 

Step 3: Transitivity
Let \( (a, b) \in S \) and \( (b, c) \in S \), so: \[ a - b + \sqrt{2} \text{ is irrational, and } b - c + \sqrt{2} \text{ is irrational.} \] Check if \( (a, c) \in S \): \[ a - c + \sqrt{2} = (a - b + \sqrt{2}) + (b - c + \sqrt{2}) - \sqrt{2} \text{ may or may not be irrational.} \] For example: \[ a = 1, b = \sqrt{3}, c = \sqrt{3} - \sqrt{2} \implies a - c + \sqrt{2} = 1 - (\sqrt{3} - \sqrt{2}) + \sqrt{2} = 1 - \sqrt{3} + 2\sqrt{2}. \] This is irrational, but a counterexample exists for other values. Thus, \( S \) is not transitive. 

Final conclusion
The relation \( S \) is reflexive but neither symmetric nor transitive.