Question

Let $R$ be the real line, consider the following subset of the plane $R\times R$
$S=\{(x,y):y=x+1,0<x<2\}$
$T=\left\{(x,y)\right\}$ $x-y$ is an integer} which of the following is true

• A. both $S$ and $T$ are equivalence relation on $R$
• B. $T$ is equivalence relation on $R$ but $S$ is not true
• C. neither $S$ not $T$ is an equivalence relation on $R$
• D. $S$ is an equivalence relation on $R$ but $T$ is not

Answer 1

Answer: (B)

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

$S=\left\{\right(x,y):y=x+1,0<x<2\}$

let $x=\{.1,.2,.3....1,1.1,1.2,...1.9\}$

then $y=\{1.1,1.2,...,1.9,2,2.1,\mathrm{2.2....},2.9\}$

$\therefore S=(x,y)=\left\{\right(.1,1.1),(.2,1.2),(.3,1.3)...\}$ so many ordered pairs can be taken in $S$

Now $(.1,1.1)\in R$ but $(.1,.1)\notin R$

$\Rightarrow S$ is not reflexive as $x\ne x+1$

$\Rightarrow S$ can not be equivalence relation

Again $T=\{(x,y):(x-y\left)isaninteger\right\}$

Now $T$ is reflexive $\therefore (x,x)=x-x=0$ is an interger

$T$ is symmetric as $(x,y)=x-y$ is an interger

$\Rightarrow (y,x)=y-x$ is also an interger

$T$ is transitive also

$(y,z)=y-z$ is an integer

$\Rightarrow (x,z)=x-z=(x-y)+(y-z)$= $integer$+ $integer$ =$integer$

$\therefore T$ is reflexive ,symmetric as well as transitive

also so $T$ is an equivalence relation.

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