Question

Prove that the relation $R=\left\{(x, y): x^2=x y\right\}$ is reflexive and transitive but not symmetric

Answer 1

$x R y \Rightarrow x^2=x y$

(i) $\quad x R x \Rightarrow x^2=x \cdot x$ (always true)

so $R$ is reflexive

(ii) $\quad(x, y) \in R, \Rightarrow x^2=x y$

But $y^2 \neq y x \quad$ eg. $(0,5) \in R$ but $(5,0) \notin R$

so $R$ is not symmetric

(iii) $\quad(x, y) \in R,(y, z) \in R \quad \Rightarrow x^2=x y$ and $y^2=y z$

Case-I $y \neq 0$

So $y^2=y z \quad \Rightarrow y=z$

$\therefore x^2=x y \quad \Rightarrow x^2=x z \quad \therefore R$ is transitive.

Case-II $y=0 \quad \Rightarrow x=0 \quad \therefore x^2=x z$

$\Rightarrow R$ is transitive