Question

Let $R$ be the real line. Consider the following subsets of the plane $R\times R,S=\left\{(x,y):y=x+1\&0<x<2\right\},T=\left\{(x,y):x–yisaninteger\right\}$ Which one of the following is true?

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

Answer 1

The correct option is C

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

Explanation For The Correct Option:

Determining the correct statement

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

Checking reflexivity:

Let $y=x$

$$\begin{gathered} \Rightarrow x=x+1 \\ \Rightarrow 0=1 \end{gathered}$$

Which is not possible.

$(x,x)\notin S$

Thus, it is not reflexive .

Hence $S$ is not an equivalence on R.

Given relation: $T=\left\{(x,y):x–yisaninteger\right\}$

Checking reflexivity:

Let $y=x$

$$\begin{gathered} \Rightarrow x-x \\ \Rightarrow 0 \end{gathered}$$

Since $0$ is an integer so $(x,x)\notin T$

Thus it is reflexive.

Checking symmetricity:

Put $(x,y)=(y,x)$

$$\begin{gathered} \Rightarrow y-x \\ \Rightarrow -\left(x-y\right)\notin T \end{gathered}$$

$\therefore (y,x)\in T$

Thus, it is symmetric relation.

Checking transitivity:

If $(x,y)\in T$, $(y,z)\in T$ then $(x,z)\in Z$

Consider $y–x={\mathrm{I}}_1\&z–y={\mathrm{I}}_2$

$$\begin{gathered} z–x={\mathrm{I}}_1+{\mathrm{I}}_2=I’\left(say\right) \\ \therefore (x,z)\in T \end{gathered}$$

Thus, $T$ is transitive.

Hence,$T$ is an equivalence relation.

Hence, option C is the correct answer.