To determine the properties of the relation \( R \) defined on the set of natural numbers \( N \) by the equation \( 2x + y = 41 \), we will analyze whether the relation is reflexive, symmetric, transitive, or none of these.
### Step 1: Identify the Domain and Range
The relation is defined by the equation \( 2x + y = 41 \). Here, \( x \) and \( y \) are natural numbers.
- **Domain of \( R \)**: The values of \( x \) that satisfy the equation when \( y \) is a natural number.
- **Range of \( R \)**: The corresponding values of \( y \) for each \( x \).
Rearranging the equation gives:
\[ y = 41 - 2x \]
To ensure \( y \) is a natural number, \( 41 - 2x \) must be greater than 0:
\[ 41 - 2x > 0 \]
\[ 41 > 2x \]
\[ x < 20.5 \]
Since \( x \) must be a natural number, the possible values of \( x \) are \( 1, 2, 3, \ldots, 20 \).
Calculating \( y \) for these values of \( x \):
- If \( x = 1 \), \( y = 41 - 2(1) = 39 \)
- If \( x = 2 \), \( y = 41 - 2(2) = 37 \)
- If \( x = 3 \), \( y = 41 - 2(3) = 35 \)
- ...
- If \( x = 20 \), \( y = 41 - 2(20) = 1 \)
Thus, the pairs \( (x, y) \) in \( R \) are:
\[
(1, 39), (2, 37), (3, 35), (4, 33), (5, 31), (6, 29), (7, 27), (8, 25), (9, 23), (10, 21), (11, 19), (12, 17), (13, 15), (14, 13), (15, 11), (16, 9), (17, 7), (18, 5), (19, 3), (20, 1)
\]
### Step 2: Check for Reflexivity
A relation \( R \) is reflexive if for every \( x \) in the domain, \( (x, x) \) is in \( R \).
For \( (x, x) \) to be in \( R \):
\[ 2x + x = 41 \]
\[ 3x = 41 \]
This equation does not yield a natural number solution, as \( x \) would not be an integer. Therefore, \( R \) is **not reflexive**.
### Step 3: Check for Symmetry
A relation \( R \) is symmetric if whenever \( (x, y) \) is in \( R \), then \( (y, x) \) is also in \( R \).
From our relation:
If \( (x, y) \) is in \( R \), then:
\[ 2x + y = 41 \]
For symmetry, we need to check if \( (y, x) \) is in \( R \):
\[ 2y + x = 41 \]
This is generally not true. For example, if \( x = 1 \) and \( y = 39 \), \( (1, 39) \) is in \( R \), but \( (39, 1) \) is not in \( R \) since \( 2(39) + 1 \neq 41 \). Therefore, \( R \) is **not symmetric**.
### Step 4: Check for Transitivity
A relation \( R \) is transitive if whenever \( (x, y) \) and \( (y, z) \) are in \( R \), then \( (x, z) \) must also be in \( R \).
Assume \( (x, y) \) and \( (y, z) \) are in \( R \):
1. \( 2x + y = 41 \)
2. \( 2y + z = 41 \)
From the first equation, we can express \( y \):
\[ y = 41 - 2x \]
Substituting \( y \) into the second equation:
\[ 2(41 - 2x) + z = 41 \]
\[ 82 - 4x + z = 41 \]
\[ z = 41 - 82 + 4x \]
\[ z = 4x - 41 \]
For \( z \) to be a natural number, \( 4x - 41 \) must be greater than 0:
\[ 4x > 41 \]
\[ x > 10.25 \]
Thus, \( x \) must be at least 11. However, the values of \( z \) calculated from \( 4x - 41 \) do not guarantee that \( (x, z) \) will be in \( R \) for all \( x \) in the domain. Therefore, \( R \) is **not transitive**.
### Conclusion
Since \( R \) is neither reflexive, symmetric, nor transitive, we conclude that the relation \( R \) is **none of these**.
