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, $x^2-4xy+3y^2=0$
$\Rightarrow \left(x-y\right)\left(x-3y\right)=0$
$\therefore x=yorx=3y$
$\therefore R=\left\{\left(1,1\right),\left(3,1\right),\left(2,2\right),\left(6,2\right),\left(3,3\right),\left(9,3\right),.....\right\}$
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 $\left(3,1\right)\in Rand\left(1,1\right)\in R\Rightarrow \left(3,1\right)\in R$
$\left(6,2\right)\in Rand\left(2,2\right)\in R\Rightarrow \left(6,2\right)\in R$
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.