Let R be the relation defined on the set N of natural numbers by the rule xRy iff x + 2y = 8, then domain of R is

Options

A. {2, 4, 8}
B. {2, 4, 6}
C. {2, 4, 6, 8}
D. {1, 2, 3, 4}

Answer

To find the domain of the relation R defined on the set of natural numbers N by the rule \( xRy \) if \( x + 2y = 8 \), we need to determine the values of \( x \) for which there exists a corresponding natural number \( y \).

### Step-by-Step Solution:

1. **Start with the equation**:
\[ x + 2y = 8 \]

2. **Rearrange the equation to solve for \( y \)**:
\[ 2y = 8 - x \]
\[ y = \frac{8 - x}{2} \]

3. **Determine the values of \( x \)**: Since \( y \) must be a natural number, \( 8 - x \) must be a non-negative even number (because it is divided by 2). This means \( 8 - x \geq 0 \) and \( 8 - x \) must be even.

4. **Find possible values of \( x \)**:
- \( 8 - x \geq 0 \) implies \( x \leq 8 \).
- The even numbers less than or equal to 8 are \( 0, 2, 4, 6, 8 \). However, since we are considering natural numbers, we only consider \( 2, 4, 6, 8 \).

5. **Check each value of \( x \)**:
- For \( x = 2 \):
\[ y = \frac{8 - 2}{2} = \frac{6}{2} = 3 \quad (\text{natural number}) \]
- For \( x = 4 \):
\[ y = \frac{8 - 4}{2} = \frac{4}{2} = 2 \quad (\text{natural number}) \]
- For \( x = 6 \):
\[ y = \frac{8 - 6}{2} = \frac{2}{2} = 1 \quad (\text{natural number}) \]
- For \( x = 8 \):
\[ y = \frac{8 - 8}{2} = \frac{0}{2} = 0 \quad (\text{not a natural number}) \]

6. **Conclusion**: The values of \( x \) that yield a natural number \( y \) are \( 2, 4, \) and \( 6 \). Therefore, the domain of the relation \( R \) is:
\[ \{2, 4, 6\} \]

### Final Answer:
The domain of \( R \) is \( \{2, 4, 6\} \).