Let R = {(x,y) : x,y in N text{ and } x^2 - 4 | vedclass

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(A) Given $R = \{(x,y) : x,y \in N 	ext{ and } x^2 - 4xy + 3y^2 = 0\}$.Factorizing the equation: $x^2 - 4xy + 3y^2 = (x - y)(x - 3y) = 0$.This implie

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reflexive but neither symmetric nor transitive

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(A) Given $R = \{(x,y) : x,y \in N 	ext{ and } x^2 - 4xy + 3y^2 = 0\}$.Factorizing the equation: $x^2 - 4xy + 3y^2 = (x - y)(x - 3y) = 0$.This implies $x = y$ or $x = 3y$.$1$. Reflexivity: For any $x \in N$,$x = x$ is true,so $(x,x) \in R$. Thus,$R$ is reflexive.$2$. Symmetry: $(3,1) \in R$ because $3 = 3(1)$. However,$(1,3) 
otin R$ because $1 
eq 3$ and $1 
eq 3(3)$. Thus,$R$ is not symmetric.$3$. Transitivity: Let $(a,b) \in R$ and $(b,c) \in R$. Then $(a=b 	ext{ or } a=3b)$ and $(b=c 	ext{ or } b=3c)$.- If $a=b$ and $b=c$,then $a=c$,so $(a,c) \in R$.- If $a=b$ and $b=3c$,then $a=3c$,so $(a,c) \in R$.- If $a=3b$ and $b=c$,then $a=3c$,so $(a,c) \in R$.- If $a=3b$ and $b=3c$,then $a=9c$,which is not necessarily $c$ or $3c$. Wait,checking $(9,3) \in R$ and $(3,1) \in R$: $9=3(3)$ and $3=3(1)$. Here $a=9, b=3, c=1$. $a=9c$. Since $9 
eq 1$ and $9 
eq 3(1)$,$(9,1) 
otin R$. Therefore,$R$ is not transitive.Correction: $R$ is reflexive only.

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

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