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Let $\rho$ be a relation defined on $N$, the set of natural numbers, as $\rho=\{(x, y) \in N \times N: 2 x+y=41\} .$ Then
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Verified Answer
The correct answer is:
$\rho$ is not transitive
We have, $\rho=\{(x, y) \in N \times N: 2 x+y=41\}$
For reflexive,
$\begin{array}{l}
x p x \Rightarrow 2 x+x=41 \\
\Rightarrow 3 x=41
\end{array}$
$\Rightarrow x=\frac{41}{3} \notin N$
So, $\rho$ is not reflexive.
For symmetric,
$x \rho y \Rightarrow 2 x+y=41$
and $y \rho x \Rightarrow 2 y+x=41$
$\Rightarrow x p y \neq y \rho x$
So, $\rho$ is not symmetric.
For transitive,
$x p y \Rightarrow 2 x+y=41$
and $y \rho z \Rightarrow 2 y+z=41$
$\Rightarrow x p z$
$\Rightarrow \rho$ is not transitive.
For reflexive,
$\begin{array}{l}
x p x \Rightarrow 2 x+x=41 \\
\Rightarrow 3 x=41
\end{array}$
$\Rightarrow x=\frac{41}{3} \notin N$
So, $\rho$ is not reflexive.
For symmetric,
$x \rho y \Rightarrow 2 x+y=41$
and $y \rho x \Rightarrow 2 y+x=41$
$\Rightarrow x p y \neq y \rho x$
So, $\rho$ is not symmetric.
For transitive,
$x p y \Rightarrow 2 x+y=41$
and $y \rho z \Rightarrow 2 y+z=41$
$\Rightarrow x p z$
$\Rightarrow \rho$ is not transitive.
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