Mathematics - Sets and Relations Question with Solution | TestHub

$$T=\{(x,y):x-y\in l\}$$ as $0\in \mathrm{IT}$ is a reflexive relation. If $x-y\in 1\Rightarrow y-x\in I$ $\therefore \mathrm{T}$ is symmetrical also If $x-y=l_1$ and $y-z=l_2$ Then $x-z=(x-y)+(y-z)=l_1+I_2\in I$ $\therefore \mathrm{T}$ is also transitive. Hence $T$ is an equivalence relation. Clearly $x\ne x+1\Rightarrow (x,x)\notin S$ $\therefore S$ is not reflexive.