Let N be the set of natural numbers and a rel | vedclass

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(B) The given equation is $x^{3}-3x^{2}y-xy^{2}+3y^{3}=0$.Factoring the expression:$x^{2}(x-3y) - y^{2}(x-3y) = 0$$(x^{2}-y^{2})(x-3y) = 0$$(x-y)(x+y)

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reflexive but neither symmetric nor transitive

Vedclass · Answer

(B) The given equation is $x^{3}-3x^{2}y-xy^{2}+3y^{3}=0$.Factoring the expression:$x^{2}(x-3y) - y^{2}(x-3y) = 0$$(x^{2}-y^{2})(x-3y) = 0$$(x-y)(x+y)(x-3y) = 0$This implies $x=y$ or $x=-y$ or $x=3y$.Since $x, y \in N$,$x=-y$ is impossible as $x, y > 0$.Thus,the relation is $R = \{(x, y) \in N 	imes N : x=y 	ext{ or } x=3y\}$.$1$. Reflexivity: For any $x \in N$,$(x, x) \in R$ because $x=x$ is true. Thus,$R$ is reflexive.$2$. Symmetry: Consider $(3, 1) \in R$ because $3=3(1)$. However,$(1, 3) 
otin R$ because $1 
eq 3$ and $1 
eq 3(3)=9$. Thus,$R$ is not symmetric.$3$. Transitivity: Consider $(9, 3) \in R$ (since $9=3(3)$) and $(3, 1) \in R$ (since $3=3(1)$). For $R$ to be transitive,$(9, 1)$ must be in $R$. But $9 
eq 1$ and $9 
eq 3(1)=3$. Thus,$R$ is not transitive.Therefore,$R$ is reflexive but neither symmetric nor transitive.

Let $N$ be the set of natural numbers and a relation $R$ on $N$ be defined by $R = \{(x, y) \in N \times N : x^{3}-3x^{2}y-xy^{2}+3y^{3}=0\}$. Then the relation $R$ is:

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