Let $R$ be the set of real numbers This question has Statement $-1$ and Statement $-2$. Of the four choices…
Let $R$ be the set of real numbers
This question has Statement $-1$ and Statement $-2$. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-1 : $A=\{(x, y) \in R \times R: y-x$ is an integer $\}$ is an equivalence relation on $R$.
Statement-2 : $B=\{(x, y) \in R \times R: x=\alpha y$ for some rational number $\alpha\}$ is an equivalence relation on $\mathrm{R}$.
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$
Statement $-1$ is true, Statement- 2 is false.
Statement $-1$ is false, Statement $-2$ is true.
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$
Solution
$\mathrm{x}-\mathrm{y}$ is an integer
$x-x=0$ is an integer $\Rightarrow A$ is Reflexive
$x-y$ is an integer $\Rightarrow y-x$ is an integer $\Rightarrow A$ is symmetric
$x-y, y-z$ are integers
As sum of two integers is an integer.
$\Rightarrow(x-y)+(y-z)=x-z$ is an integer
$\Rightarrow \mathrm{A}$ is transitive. Hence statement $-1$ is true.
Also $\frac{\mathrm{x}}{\mathrm{x}}=1$ is a rational number $\Rightarrow B$ is reflexive
$\frac{\mathrm{x}}{\mathrm{y}}=\alpha$ is rational $\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}$ need not be rational
i.e., $\frac{0}{1}$ is rational $\Rightarrow \frac{1}{0}$ is not rational
Hence B is not symmetric
$\Rightarrow B$ is not an equivalence relation.