Let $N$ be the set of natural numbers and a relation $R$ on $N$ be defined by $R = {(x, y) ∈ N × N: x^3 - 3x^2y - xy^2 + 3y^3 = 0}$. Then the relation $R$ is :
Correct Answer:
B
Explanation
Given the relation $x^3 - 3x^2y - xy^2 + 3y^3 = 0$. We can factor this as follows:
$x(x^2 - y^2) - 3y(x^2 - y^2) = 0$
$(x - 3y)(x^2 - y^2) = 0$
$(x - 3y)(x - y)(x + y) = 0$
Thus, the relation holds if $x = 3y$, $x = y$, or $x = -y$.
Reflexivity: For $R$ to be reflexive, $(x, x) \in R$ for all $x \in N$. If $x = y$, then $(x, x) \in R$. So, the relation is reflexive.
Symmetry: For $R$ to be symmetric, if $(x, y) \in R$, then $(y, x) \in R$. Let's consider $x = 3y$. If $(3, 1) \in R$ since $3 = 3(1)$, but $(1, 3) \notin R$ since $1 \neq 3(3)$. So, the relation is not symmetric.
Transitivity: For $R$ to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. Let $x = 3y$ and $y = z$. Then $(x, y) = (3y, y)$ and $(y, z) = (y, y)$. For transitivity, we need $(3y, y) \in R$ and $(y, y) \in R$ implying $(3y, y) \in R$. However, if we choose $x=3$, $y=1$ and $z=1$, then $x = 3y$. But if $x = 3$ and $y = 1$, we don't have $x=z$. So the relation is not transitive.
Since $R$ is reflexive but neither symmetric nor transitive, the correct option is B.