Question

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

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

Answer 1

The correct option is A

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

Explanation for the correct option:

Determining the correct statement:

Step 1: Checking if $\mathrm{S}$ is an equivalence relation

Given, $S=\left\{\right(x,y):y=x+1and0<x<2\}$

Therefore, $y<3$

Now for $(y,x)\Rightarrow x=y+1$

So it is not satisfying $0<x<2$

Hence, $\mathrm{S}$ is not symmetric so not equivalence.

Step 2: Checking if $\mathrm{T}$ is an equivalence relation

Given,$\mathrm{T}=\left\{\right(\mathrm{x},\mathrm{y}):\mathrm{x}-\mathrm{y}\mathrm{is}\mathrm{an}\mathrm{integer}\}$

For $(x,x)\Rightarrow x-x=0$ is an integer

So it is reflexive.

For $(y,x)\Rightarrow y-x=-(x-y)$ is always an integer.

So it is symmetric.

If $(x,y)\in R\&(y,z)\in R$, then

$\Rightarrow (x-y)$is an integer,

$\Rightarrow (y-z)$ is an integer,

Now adding both we get

$\Rightarrow x-z$ which is also an integer.

Thus,$T$ is an equivalence relation.

Hence, option (A) is the correct answer.