Let R,= {(x,y) : x,y in N, and, x^2 -4xy +3y^ | vedclass

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Question

$R = \left\{ {\left( {x,y} \right):x,y \in N\,\,and\,\,{x^2} - 4xy + 3{y^2} = 0} \right\}$

Now, ${{x^2} - 4xy + 3{y^2} = 0}$

$ \Rightarrow

Vedclass · Accepted Answer

reflexive and transitive

Vedclass · Answer

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

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

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