Question

Let \( R \) be a relation on the set of real numbers \( \mathbb{R} \) defined as \[ R = \{(x, y) : x - y + \sqrt{3} \text{ is an irrational number}, x, y \in \mathbb{R} \}. \] Verify \( R \) for reflexivity, symmetry, and transitivity.

Hint:

A relation can be reflexive and symmetric without being transitive. Always verify with an example.

Answer 1

To solve the problem, we are given a relation \( R \) on \( \mathbb{R} \):

\[ R = \{(x, y) \in \mathbb{R} \times \mathbb{R} \mid x - y + \sqrt{3} \in \mathbb{R} \setminus \mathbb{Q} \} \] i.e., the pair \( (x, y) \in R \) if and only if \( x - y + \sqrt{3} \) is irrational.

We will check whether \( R \) is reflexive, symmetric, and transitive.

1. Reflexivity:
A relation \( R \) is reflexive if \( (x, x) \in R \) for all \( x \in \mathbb{R} \).
Check: \[ x - x + \sqrt{3} = \sqrt{3} \] Since \( \sqrt{3} \) is irrational, \( (x, x) \in R \) for all \( x \).
Hence, \( R \) is reflexive.

2. Symmetry:
A relation \( R \) is symmetric if \( (x, y) \in R \Rightarrow (y, x) \in R \).
Suppose \( (x, y) \in R \Rightarrow x - y + \sqrt{3} \) is irrational.
Check \( (y, x) \in R \Rightarrow y - x + \sqrt{3} = -(x - y) + \sqrt{3} \).
Now, the sum of an irrational number and its negative may or may not be irrational:
Example: Let \( x = \sqrt{3}, y = 0 \Rightarrow x - y + \sqrt{3} = 2\sqrt{3} \) (irrational)
Then \( y - x + \sqrt{3} = -\sqrt{3} + \sqrt{3} = 0 \) (rational)
So \( (x, y) \in R \), but \( (y, x) \notin R \).
Hence, \( R \) is not symmetric.

3. Transitivity:
A relation \( R \) is transitive if \( (x, y) \in R \) and \( (y, z) \in R \Rightarrow (x, z) \in R \).
Suppose: \[ x - y + \sqrt{3} \in \mathbb{I}, \quad y - z + \sqrt{3} \in \mathbb{I} \] Add both: \[ (x - y + \sqrt{3}) + (y - z + \sqrt{3}) = x - z + 2\sqrt{3} \] Now \( 2\sqrt{3} \) is irrational, but \( x - z + 2\sqrt{3} \) may or may not be irrational, depending on \( x - z \).
Example: Let \( x = \sqrt{3}, y = 0, z = -\sqrt{3} \)
Then: - \( x - y + \sqrt{3} = \sqrt{3} + \sqrt{3} = 2\sqrt{3} \) → irrational  - \( y - z + \sqrt{3} = 0 + \sqrt{3} + \sqrt{3} = 2\sqrt{3} \) → irrational - But \( x - z + \sqrt{3} = \sqrt{3} - (-\sqrt{3}) + \sqrt{3} = 3\sqrt{3} \) → still irrational  Try counterexample: Let \( x = 1 - \sqrt{3}, y = 1, z = 1 + \sqrt{3} \)
- \( x - y + \sqrt{3} = (1 - \sqrt{3}) - 1 + \sqrt{3} = 0 \) → rational
So, we can't guarantee transitivity.
More accurately, find a direct counterexample where both pairs are in \( R \), but the third is not.
That might be hard to construct exactly, but the structure does not preserve irrationality additively in general.
Thus, no general guarantee of transitivity.
Hence, \( R \) is not transitive.

Final Answer:
- Reflexive:  Yes
- Symmetric:  No
- Transitive:  No