Question

Let $R=\left\{\left(x,y\right): x, y\,\in\,N \,x ^{2}-4xy + 3y^{2} = 0\right\}$ , where $N$ is the set of all natural numbers. Then the relation $R$ is :

• A. reflexive but neither symmetric nor transitive
• B. symmetric and transitive
• C. reflexive and symmetric
• D. reflexive and transitive

Answer 1

Answer: (A)

$R=\left\{\left(x,y\right): x, y\,\in\,N \,x ^{2}-4xy + 3y^{2} = 0\right\}$
Now, $x^{2} -4xy + 3y^{2} = 0$
$\Rightarrow \left(x - y \right) \left(x - 3y\right) = 0$
$\therefore x =y or x = 3y$
$\therefore R=\left\{\left(1,1\right), \left(3,1\right), \left(2, 2\right), \left(6, 2\right), \left(3, 3\right),\left(9,3\right),.....\right\}$
Since $\left(1,1\right), \left(2,2\right), \left(3,3\right),.......$ are present inthe relation, therefore $R$ is reflexive.
Since $\left(3, 1\right)$ is an element of $R$ but $\left(1, 3\right)$ is not the element of $R$, therefore R is not symmetric
Here $\left(3,1\right) \in R\,and\left(1, 1\right) \in\, R \Rightarrow \left(3,1 \right) \in \, R$
$\left(6,2\right) \in R\,and\left(2, 2\right) \in \, R \Rightarrow \left(6,2 \right) \in \, R$
For all such $\left(a, b\right) \in \, R$ and $\left(b, c\right) \in \, R$
$\Rightarrow \left(a,c\right)\in \, R$
Hence $R$ is transitive.