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 and only if \( x + 2y = 8 \), we need to determine the values of \( x \) for which \( y \) is a natural number.

### Step-by-Step Solution:

1. **Start with the equation**:
\[ x + 2y = 8 \]
Rearranging gives:
\[ 2y = 8 - x \]
\[ y = \frac{8 - x}{2} \]

2. **Identify the conditions for \( y \)**:
Since \( y \) must be a natural number, \( \frac{8 - x}{2} \) must also be a natural number. This means that \( 8 - x \) must be a positive even number.

3. **Determine the values of \( x \)**:
- For \( y \) to be a natural number, \( 8 - x \) must be positive:
\[ 8 - x > 0 \implies x < 8 \]
- Also, \( 8 - x \) must be even. The even numbers less than 8 are 0, 2, 4, 6, and 8. However, since \( x \) is a natural number, we only consider \( 1, 2, 3, 4, 5, 6, 7 \).

4. **Check each value of \( x \)**:
- \( x = 1 \):
\[ y = \frac{8 - 1}{2} = \frac{7}{2} \quad \text{(not a natural number)} \]
- \( x = 2 \):
\[ y = \frac{8 - 2}{2} = \frac{6}{2} = 3 \quad \text{(natural number)} \]
- \( x = 3 \):
\[ y = \frac{8 - 3}{2} = \frac{5}{2} \quad \text{(not a natural number)} \]
- \( x = 4 \):
\[ y = \frac{8 - 4}{2} = \frac{4}{2} = 2 \quad \text{(natural number)} \]
- \( x = 5 \):
\[ y = \frac{8 - 5}{2} = \frac{3}{2} \quad \text{(not a natural number)} \]
- \( x = 6 \):
\[ y = \frac{8 - 6}{2} = \frac{2}{2} = 1 \quad \text{(natural number)} \]
- \( x = 7 \):
\[ y = \frac{8 - 7}{2} = \frac{1}{2} \quad \text{(not a natural number)} \]

5. **List the valid values of \( x \)**:
From the checks above, the values of \( x \) that yield a natural number \( y \) are:
\[ x = 2, 4, 6 \]

6. **Conclusion**:
The domain of the relation \( R \) is:
\[ \{2, 4, 6\} \]