Let R be the set of real numbers. 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 R.

Question

Let $\mathrm{R}$ be the set of real numbers.

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 R.

(1) Statement- 1 is true, Statement-2 is true; Statement- 2 is a correct explanation for Statement- 1.

(2) Statement- 1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement1.

(3) Statement- 1 is true, Statement-2 is false.

(4) Statement- 1 is false, Statement-2 is true.

Answer 1

Statement - 1 :

(i) $x-x$ is an integer $\forall x \in R$ so $A$ is reflexive relation.

(ii) $y-x \in I \Rightarrow x-y \in I$ so $A$ is symmetric relation.

(iii) $y-x \in I$ and $z-y \in I \Rightarrow y-x+z-y \in I$

$\Rightarrow z-x \in I$ so $A$ is transitive relation.

Therefore $A$ is equivalence relation.

Statement - 2 :

(i) $x=\alpha x$ when $\alpha=1 \Rightarrow B$ is reflexive relation

(ii) for $x=0$ and $y=2$, we have $0=\alpha(2)$ for $\alpha=0$

But $2=\alpha(0)$ for no $\alpha$

so $B$ is not symmetric so not equivalence.