Question

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

• A. reflexive and symmetric, but not transitive
• B. reflexive but neither symmetric nor transitive
• C. symmetric but neither reflexive nor transitive
• D. an equivalence relation

Hint:

When given a relation defined by a polynomial equation, the first step is always to try and factor it. This simplifies the condition and makes it much easier to test for reflexivity, symmetry, and transitivity.

Answer 1

The Correct Option is B

Given: \[ x^3 - 3x^2y - xy^2 + 3y^3 = 0 \] Factorising, \[ x^2(x-3y) - y^2(x-3y) = 0 \] \[ (x^2-y^2)(x-3y)=0 \] \[ (x-y)(x+y)(x-3y)=0 \] Since $x,y\in\mathbb{N}$, $x+y\neq0$. Hence, \[ (x,y)\in R \iff x=y \ \text{or}\ x=3y \] Reflexive: For every $x\in\mathbb{N}$, $x=x$, hence $(x,x)\in R$. So, $R$ is reflexive. Symmetric: $(3,1)\in R$ since $3=3(1)$, but $(1,3)\notin R$. Hence, $R$ is not symmetric. Transitive: $(9,3)\in R$ and $(3,1)\in R$, but $(9,1)\notin R$. Hence, $R$ is not transitive. \[ \boxed{\text{R is reflexive but neither symmetric nor transitive}} \]