Mathematics - Sets and Relations Question with Solution | TestHub

$\mathrm{R}=\{(x,y):x,y\in \mathrm{N}$ and $x^2-4xy+3y^2=0\}$ Now, $x^2-4xy+3y^2=0$ $\Rightarrow (x-y)(x-3y)=0$ $\therefore x=y$ or $x=3y$ $\therefore \mathrm{R}=\{(1,1),(3,1),(2,2),(6,2),(3,3)$, $(9,3),\dots \dots \}$ Since $(1,1),(2,2),(3,3),\dots \dots$ 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.