Question

Let $S$ be a relation on $R^{+}$ defined by $xSy\iff x^2-y^2=2(y-x)$, then $S$ is

• A. Only reflecxive
• B. Only symmetric
• C. Only Trasitive
• D. Equivalence

Answer 1

The correct option is D Equivalence

Let $S$ be the relation on $R^{+}$ defined by $xSy\iff x^2-y^2=2(y-x)$.

Now, $xSx$ holds as $x^2-x^2=2(x-x)$ as $0=0$. So, $S$ is reflexive.

Also, $xSy\iff ySx$ holds as $x^2-y^2=2(y-x)\iff y^2-x^2=2(x-y)$. So, $S$ is symmetric.

Let, $xSy,ySz$ holds.

Then, $x^2-y^2=2(y-x)$.........(1) and $y^2-z^2=2(z-y)$.......(2).

Now, adding (1) and (2) we get, $x^2-z^2=2(z-x)\Rightarrow xSz$ holds. So, $S$ is also transitive.

Moreover $S$ is an equivalence relation.