Question

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

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

Answer 1

The correct option is C reflexive but neither symmetric nor transitive

$x^3-3x^{2}y-x{y}^2+3y^3=0$
$\Rightarrow x^2(x-3y)-y^2(x-3y)=0\Rightarrow (x-y)(x+y)(x-3y)=0\dots \left(1\right)$

$\because \left(1\right)$ holds for all $(x,x)$
$\therefore R$ is reflexive
If $(x,y)$ holds, then $(y,x)$ may or may not hold for factor $(x-3y)$
For example $(3,1)$
$\therefore R$ is not symmetric

Similarly factor $(x-3y)$ doesn’t hold for transitive
For exapmle $(9,3)\in R$ and $(3,1)\in R$ but $(9,1)\notin R$

Hence, relation $R$ is reflexive but neither symmetric nor transitive