Let R : \(\mathbf{R} \rightarrow \mathbf{R}\), where \(\mathbf{R}\) is the set of real numbers and R be a relation. An element \((x, y) \in \mathrm{R}\) if \(x + y - \sqrt{2}\) belongs to the set of irrational numbers. Then the relation R is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Symmetric
Reflexive: we need \(x + x - \sqrt{2} = 2x - \sqrt{2}\) to be irrational for every real x. Take \(x = \frac{\sqrt{2}}{2}\):
\( \displaystyle 2 \times \frac{\sqrt{2}}{2} - \sqrt{2} = 0, \text{ which is rational} \)
So \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \notin \mathrm{R}\), and R is not reflexive.
Symmetric: if \((x, y) \in \mathrm{R}\), then \(x + y - \sqrt{2}\) is irrational. Since \(y + x - \sqrt{2}\) is the same number, \((y, x) \in \mathrm{R}\). So R is symmetric.
Transitive: take x = 1, \(y = \sqrt{3}\), \(z = \sqrt{2}\):
\(x + y - \sqrt{2} = 1 + \sqrt{3} - \sqrt{2}\), irrational, so \((1, \sqrt{3}) \in \mathrm{R}\)
\(y + z - \sqrt{2} = \sqrt{3}\), irrational, so \((\sqrt{3}, \sqrt{2}) \in \mathrm{R}\)
\(x + z - \sqrt{2} = 1\), rational, so \((1, \sqrt{2}) \notin \mathrm{R}\)
So R is not transitive, and therefore not an equivalence relation. R is only symmetric.