Question

Let $\rho$ be a relation defined on $N$, the set of natural numbers, as $\rho =\left\{\right(x,y)\in N\times N:2x+y=41\}$. Then

• A. $\rho$ is an equivalence relation
• B. $\rho$ is only reflexive relation
• C. $\rho$ is only symmetric relation
• D. $\rho$ is not transitive

Answer 1

The correct option is D $\rho$ is not transitive

$\rho =\left\{\right(x,y)\in N\times N:2x+y=41\}$

For reflexive relation,
$(a,a)\in \rho ,\forall a\in \rho$
$\Rightarrow 2x+x=41\Rightarrow x=\frac{41}{3}\notin N$
$\therefore \rho$ is not reflexive.

For symmetric relation,
$(a,b)\in \rho \Rightarrow (b,a)\in \rho$
$\Rightarrow 2x+y=41\ne 2y+x$
Let $x=20,y=1$
$\Rightarrow 2x+y=41\ne 2y+x$
$\therefore \rho$ is not symmetric.

For transitive relation,
$(a,b)\in \rho ,(b,c)\in \rho \Rightarrow (a,c)\in \rho$
Let $a=12,b=17,c=7$
$(a,b)\in \rho \Rightarrow 2x+y=41$
$(b,c)\in \rho \Rightarrow 2y+z=41$
$(a,c)\notin \rho \Rightarrow 2x+z=31\ne 41$
​​​​​​​$\therefore \rho$ is not transitive.