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Step-by-Step Explanation
1. Understand the Relation
The relation $R$ on the set of natural numbers $N$ is defined by:
$$(x, y) \in R \iff x^3 - 3x^2y - xy^2 + 3y^3 = 0.$$
We first factor the expression on the right-hand side:
$$x^3 - 3x^2y - xy^2 + 3y^3 = (x - 3y)(x - y)(x + y).$$
For $(x, y) \in R$, we need
$$(x - 3y)(x - y)(x + y) = 0.$$
2. Possible Solutions for Natural Numbers
Since $x, y \in N$ (the set of natural numbers), the factor $(x + y) = 0$ is impossible because $x + y > 0$ for all natural $x, y$. Hence the solutions come from either:
$x = y$, or
$x = 3y.$
Thus, $(x, y) \in R$ if and only if $x = y$ or $x = 3y$ (with $x, y \in N$).
3. Check Reflexivity
A relation $R$ on $N$ is reflexive if $(a, a) \in R$ for every $a \in N$.
Clearly, for every $a \in N$, if we take $x = a$ and $y = a$, we get $x = y$, which satisfies our factorization. Hence $(a, a) \in R$ for all $a$, making $R$ reflexive.
4. Check Symmetry
A relation $R$ on $N$ is symmetric if whenever $(x, y) \in R$, then we also have $(y, x) \in R$.
Suppose $(x, y) \in R$ because $x = 3y$. Then to have $(y, x) \in R$, we would need either $y = x$ or $y = 3x$. But if $x = 3y$, then $y = x$ means $y = 3y$ which forces $y = 0$ (not in natural numbers), or $y = 3x$ would mean $y = 3(3y) = 9y$ which forces $y = 0$. So $(y, x)$ does not hold for $x = 3y \neq y.$
Thus, symmetry fails. For example, $(3, 1) \in R$ because $3 = 3(1)$, but $(1, 3) \notin R.$
5. Check Transitivity
A relation $R$ on $N$ is transitive if whenever $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$ must also hold.
Take $x = 3$, $y = 1$ so that $(3, 1) \in R$ (since $3 = 3\times1$). Now choose $z$ such that $(1, z) \in R$. This would mean either $1 = z$ or $1 = 3z$. The second equation $1 = 3z$ is impossible in the naturals, so $z = 1$. Thus $(1, 1) \in R.$
Then $(3, 1) \in R$ and $(1, 1) \in R$ but do these guarantee $(3, 1) \in R \implies (3, 1)$ again? This example does not necessarily break transitivity, so let's look at a more general mismatch where $(y,z)$ might involve $y = 3z$, etc. A more precise argument is that if $x = 3y$ and $y = 3z$, we would need $x = 3z$ or $x = z$ to stay in $R$. But $x = 3y$ and $y = 3z$ imply $x = 9z$, which is neither $3z$ nor $z$ unless $z = 0$, not allowed in naturals. Hence transitivity fails for certain combinations.
Therefore, $R$ is not transitive.
6. Conclusion
Since $R$ is reflexive but fails to be symmetric or transitive, the correct classification is:
“Reflexive but neither symmetric nor transitive.”
