Let `R` be a relation on the set `N` be defined by `{(x,y)|x,yepsilonN,2x+y=41}`. Then prove that the `R` is neither reflexive nor symmetric and nor transitive.

Explanation

To prove that the relation \( R \) defined on the set of natural numbers \( N \) by the condition \( R = \{(x,y) | x,y \in N, 2x + y = 41\} \) is neither reflexive, nor symmetric, nor transitive, we will analyze each property step by step.

### Step 1: Check for Reflexivity
A relation \( R \) is reflexive if for every element \( a \in N \), the pair \( (a, a) \) is in \( R \). This means we need to check if \( 2a + a = 41 \) holds true for any natural number \( a \).

1. Set up the equation:
\[ 2a + a = 41 \]
This simplifies to:
\[ 3a = 41 \]
Therefore:
\[ a = \frac{41}{3} \]

2. Since \( \frac{41}{3} \) is not a natural number, there is no \( a \in N \) such that \( (a, a) \in R \).

**Conclusion**: \( R \) is not reflexive.

### Step 2: Check for Symmetry
A relation \( R \) is symmetric if for every pair \( (x, y) \in R \), the pair \( (y, x) \) is also in \( R \).

1. Suppose \( (x, y) \in R \), which means:
\[ 2x + y = 41 \]
We need to check if \( (y, x) \in R \), meaning:
\[ 2y + x = 41 \]

2. From \( 2x + y = 41 \), we can express \( y \):
\[ y = 41 - 2x \]

3. Substitute \( y \) into \( 2y + x = 41 \):
\[ 2(41 - 2x) + x = 41 \]
Simplifying this gives:
\[ 82 - 4x + x = 41 \\ 82 - 3x = 41 \\ 3x = 41 \\ x = \frac{41}{3} \]

4. Since \( \frac{41}{3} \) is not a natural number, we can find a specific example to demonstrate non-symmetry. Let’s take \( x = 20 \) and \( y = 1 \):
\[ 2(20) + 1 = 41 \quad \text{(True)} \]
Now check \( (1, 20) \):
\[ 2(1) + 20 = 2 + 20 = 22 \quad \text{(Not equal to 41)} \]

**Conclusion**: \( R \) is not symmetric.

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

1. Assume \( (x, y) \in R \) and \( (y, z) \in R \):
\[ 2x + y = 41 \quad (1) \\ 2y + z = 41 \quad (2) \]

2. We need to check if \( (x, z) \in R \), which means:
\[ 2x + z = 41 \]

3. From equation (1), express \( y \):
\[ y = 41 - 2x \]

4. Substitute \( y \) into equation (2):
\[ 2(41 - 2x) + z = 41 \\ 82 - 4x + z = 41 \\ z = 41 - 82 + 4x \\ z = 4x - 41 \]

5. Now check if \( 2x + z = 41 \):
\[ 2x + (4x - 41) = 41 \\ 6x - 41 = 41 \\ 6x = 82 \\ x = \frac{82}{6} = \frac{41}{3} \]

6. Again, \( \frac{41}{3} \) is not a natural number. Let’s take specific values, e.g., \( x = 20, y = 1, z = 39 \):
- Check \( (20, 1) \):
\[ 2(20) + 1 = 41 \quad \text{(True)} \]
- Check \( (1, 39) \):
\[ 2(1) + 39 = 41 \quad \text{(True)} \]
- Check \( (20, 39) \):
\[ 2(39) + 20 = 78 + 20 = 98 \quad \text{(Not equal to 41)} \]

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

### Final Conclusion
The relation \( R \) is neither reflexive, nor symmetric, nor transitive.

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