Question

Let $R$ be the set of real numbers,

Statement $1:$ $\mathrm{A}=\left\{(\mathrm{x},\mathrm{y})\in \mathrm{R}\times \mathrm{R}:\mathrm{y}–\mathrm{x}\mathrm{is}\mathrm{an}\mathrm{integer}\right\}$ is an equivalence relation on $R$.

Statement $2:$ $\mathrm{B}=\left\{(\mathrm{x},\mathrm{y})\in \mathrm{R}\times \mathrm{R}:\mathrm{x}=\mathrm{\alpha y}\mathrm{for}\mathrm{some}\mathrm{rational}\mathrm{number}\mathrm{\alpha }\right\}$ is an equivalence relation on $R$ .

• A. Statement $1$ is true, Statement $2$ is true; statement $2$ is the correct explanation for statement $1$
• B. Statement $1$ is true, Statement $2$ is true; Statement $2$ is not a correct explanation for Statement $1$
• C. Statement $1$ is true, Statement $2$ is false
• D. Statement $1$ is false, Statement $2$ is true

Answer 1

The correct option is C

Statement $1$ is true, Statement $2$ is false

Determining the correct statement :

The given statement $1$

$\mathrm{A}=\left\{(\mathrm{x},\mathrm{y})\in \mathrm{R}\times \mathrm{R}:\mathrm{y}–\mathrm{x}\mathrm{is}\mathrm{an}\mathrm{integer}\right\}$ is an equivalence relation on $R$.

Checking reflexivity:

Let $x=y$

$x-x=0$is an integer.

$(x,x)\in A$

Thus $A$ is reflexive.

Checking symmetricity:

If $(x,y)\in A\Rightarrow (x–y)$ is an integer

$y–x=-\left(x-y\right)$ is also an integer

Therefore, $(y,x)\in A$

Thus $A$ is symmetric

Checking transitivity:

If $(x,y)\&(y,z)\in A$

$$\begin{gathered} x–y\in Z \\ y–z\in Z \\ (x–y)+(y–z)=x–z \end{gathered}$$

$\left(x–z\right)$ is also an integer.

Therefore, $(x,z)\in A$

So, $A$ is a transitive relation.

As $A$ is reflexive, symmetric and transitive so it is an equivalence relation.

Thus statement $1$ is true.

The given statement $2$

$\mathrm{B}=\left\{(\mathrm{x},\mathrm{y})\in \mathrm{R}\times \mathrm{R}:\mathrm{x}=\mathrm{\alpha y}\mathrm{for}\mathrm{some}\mathrm{rational}\mathrm{number}\mathrm{\alpha }\right\}$ is an equivalence relation on $R$ .

For $(x,y)=(0,x)$

$\Rightarrow 0=0x=0$ where $\alpha =0$

But for $(x,y)=(x,0)$

$\Rightarrow x=\alpha 0$

$\alpha$ become undefined .

Therefore, $(x,0)\notin B$

So, $B$ is not a symmetric relation.

Thus, it is not an equivalence relation.

Hence, statement $2$is false.

Hence, option (B) is the correct answer.