For real numbers x and y, define xRY if and only if `x - y + sqrt(2)` is an irrational number. Then the relation R is

Options

A. Reflexive
B. Symmetric
C. Transitie
D. None of these

Answer

To determine the properties of the relation \( R \) defined by \( x R y \) if and only if \( x - y + \sqrt{2} \) is an irrational number, we will analyze its reflexivity, symmetry, and transitivity.

### Step 1: Check Reflexivity
A relation \( R \) is reflexive if for every \( x \in \mathbb{R} \), \( x R x \) holds true.

**Calculation:**
- Let \( y = x \).
- Then, \( x R x \) means \( x - x + \sqrt{2} \) is irrational.
- This simplifies to \( \sqrt{2} \), which is indeed irrational.

**Conclusion:**
Since \( \sqrt{2} \) is irrational for all \( x \), the relation \( R \) is reflexive.

### Step 2: Check Symmetry
A relation \( R \) is symmetric if whenever \( x R y \) holds, then \( y R x \) also holds.

**Calculation:**
- Assume \( x R y \) holds, which means \( x - y + \sqrt{2} \) is irrational.
- Now, we check \( y R x \):
- This means \( y - x + \sqrt{2} \) should also be irrational.
- We can rewrite \( y - x + \sqrt{2} \) as \( -(x - y) + \sqrt{2} \).
- If \( x - y + \sqrt{2} \) is irrational, then \( -(x - y) + \sqrt{2} \) is rational (since the negation of an irrational number is irrational).

**Conclusion:**
Since \( y - x + \sqrt{2} \) is not guaranteed to be irrational, the relation \( R \) is not symmetric.

### Step 3: Check Transitivity
A relation \( R \) is transitive if whenever \( x R y \) and \( y R z \) hold, then \( x R z \) must also hold.

**Calculation:**
- Assume \( x R y \) and \( y R z \):
- This means \( x - y + \sqrt{2} \) is irrational and \( y - z + \sqrt{2} \) is irrational.
- We need to check if \( x R z \) holds, which means \( x - z + \sqrt{2} \) should be irrational.
- We can express \( x - z \) as:
\[ x - z = (x - y) + (y - z) \]
Therefore,
\[ x - z + \sqrt{2} = (x - y + \sqrt{2}) + (y - z) \]
- Since both \( x - y + \sqrt{2} \) and \( y - z + \sqrt{2} \) are irrational, their sum is not guaranteed to be irrational.

**Conclusion:**
Thus, \( R \) is not transitive.

### Final Conclusion
The relation \( R \) is:
- Reflexive: Yes
- Symmetric: No
- Transitive: No