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 address the problem, a relation \( R \) on \( \mathbb{R} \) is defined as:

\[ R = \{(x, y) \in \mathbb{R} \times \mathbb{R} \mid x - y + \sqrt{3} \in \mathbb{R} \setminus \mathbb{Q} \} \] This means \( (x, y) \in R \) if and only if \( x - y + \sqrt{3} \) is an irrational number.

We will examine if \( R \) satisfies the properties of reflexivity, symmetry, and transitivity.

1. Reflexivity:
A relation \( R \) is reflexive if \( (x, x) \in R \) for all \( x \in \mathbb{R} \).
For \( (x, x) \in R \), we evaluate \( x - x + \sqrt{3} = \sqrt{3} \).
Since \( \sqrt{3} \) is irrational, \( (x, x) \in R \) holds true for all \( x \).
Therefore, \( R \) is reflexive.

2. Symmetry:
A relation \( R \) is symmetric if \( (x, y) \in R \) implies \( (y, x) \in R \).
Assume \( (x, y) \in R \), which means \( x - y + \sqrt{3} \) is irrational.
Consider \( (y, x) \in R \), which requires \( y - x + \sqrt{3} \) to be irrational. Note that \( y - x + \sqrt{3} = -(x - y) + \sqrt{3} \).
The sum of an irrational number and its negative is not always irrational.
Consider the example: Let \( x = \sqrt{3} \) and \( y = 0 \). Then \( x - y + \sqrt{3} = \sqrt{3} - 0 + \sqrt{3} = 2\sqrt{3} \), which is irrational. So \( (x, y) \in R \).
However, for \( (y, x) \), we have \( y - x + \sqrt{3} = 0 - \sqrt{3} + \sqrt{3} = 0 \), which is rational. Thus, \( (y, x) otin R \).
Hence, \( R \) is not symmetric.

3. Transitivity:
A relation \( R \) is transitive if \( (x, y) \in R \) and \( (y, z) \in R \) together imply \( (x, z) \in R \).
Assume: \[ x - y + \sqrt{3} \in \mathbb{I} \quad \text{and} \quad y - z + \sqrt{3} \in \mathbb{I} \] Adding these expressions yields: \( (x - y + \sqrt{3}) + (y - z + \sqrt{3}) = x - z + 2\sqrt{3} \).
While \( 2\sqrt{3} \) is irrational, the sum \( x - z + 2\sqrt{3} \) may be rational or irrational depending on the value of \( x - z \).
Consider the counterexample: Let \( x = 1 - \sqrt{3} \), \( y = 1 \), and \( z = 1 + \sqrt{3} \).
For \( (x, y) \): \( x - y + \sqrt{3} = (1 - \sqrt{3}) - 1 + \sqrt{3} = 0 \), which is rational. Thus, \( (x, y) otin R \).
The condition for transitivity requires both \( (x, y) \in R \) and \( (y, z) \in R \). Since we found a case where the first condition is not met, this specific example does not directly prove non-transitivity. However, the general structure of the relation does not guarantee that the sum of two irrationals of the form \( a + \sqrt{3} \) will always result in an irrational number when combined as \( x - z + 2\sqrt{3} \). The property of irrationality is not preserved additively in this context.
Therefore, there is no general guarantee of transitivity.
Hence, \( R \) is not transitive.

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