JEE Main 2008 — Sets And Relations Question with Solution

Given $$S=\{\left(x,y\right):y=x+1$$

and $$0<x<2\}$$

As $$x\ne x+1$$

for any $$x\in \left(0,2\right)\Rightarrow \left(x,x\right)\notin S$$

$$\therefore$$ $$S$$ is not reflexive.

Hence $$S$$ in not an equivalence relation.

Also $$T=\{x,y):x-y$$ is an integer $$\}$$

as $$x-x=0$$ is an integer $$\forall x\in R$$

$$\therefore$$ $$T$$ is reflexive.

If $$x-y$$ is an integer then $$y-x$$ is also an integer

$$\therefore$$ $$T$$ is symmetric

If $$x-y$$ is an integer and $$y-z$$ is an integer then

$$(x-y)+(y-z)=x-z$$ is also an integer.

$$\therefore$$ $$T$$ is transitive

Hence $$T$$ is an equivalence relation