Question

Let $R$ be the real line. Consider the following subsets of the plane $R\times R$
$S=\{x,y\}:y=x+1$ and $0<x<2$
$T=\{x,y\}:x-y$ is an integer
Which one of the following is true?

• A. $S$ is an equivalence relation on $R$ but $T$ is not
• B. $T$ is an equivalence relation on $R$ but $S$ is not
• C. Neither $S$ nor $T$ is an equivalence relation on $R$
• D. Both $S$ and $T$ are equivalence relations on $R$

Answer 1

Answer: (B)

$T$ is an equivalence relation on $R$ but $S$ is not

$y=x+1$
$\therefore y<3$

Let $S$ is symmetric
So $(x,y)\in R$
$(y,x)\in R$

but $(y,x)$ is not satisfying $0<x<2$
So $S$ is not an equivalence relation.(for verification take $y=2$ and $x=1$)
$T(x,y)=x-y$ is integer
$(x,x)$ is an integer always

If $(x,y)\in R$, then $y-x$ is also an integer.
$\because (y,x)\in IR$

If $(x,y)\in IR$ & $(y,z)\in IR$, then $(x-y)$ is an integer, $(y-z)$ is an integer, sum of $x-y$ & $y-z=x-z$ is also an integer. Hence T is equivalence relation.