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