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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 $\mathbf{R}$ Statement-2 : $B=\{(x, y) \in R \times R: x=\alpha y$ for
some rational number $\alpha\}$ is an equivalence relation of $\mathbf{R}$.

1) Statement- 1 is true, Statement- 2 is false
2) Statement-1 is false, Statement- 2 is true
3) Statement-1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement-1
4) Statement- 1 is true, Statement 2 is true; Statement- 2 is not a correct explanation for Statement-1
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(1)

Explanation

Since $x-x=0 \in Z,(x, x) \in A$
$\Rightarrow A$ is reflexive
$(x, y) \in A \Rightarrow x-y \in z \Rightarrow y-x \in Z$
$\Rightarrow(y, x) \in A \Rightarrow A$ is symmetric
$(x, y) \in A,(y, z) \in A \Rightarrow x-y \in z, y-z \in Z$
$\Rightarrow x-z \in Z \Rightarrow(x, z) \in A$
$\Rightarrow A$ is equivalence relation $(0,1) \in B \quad \because 0=(0)(1), 0 \in \ldots \mathrm{Q}$
But $(1,0) \notin B \Rightarrow B$ is not symmetric
$\Rightarrow B$ is not equivalence.

by Diamond (75,948 points)

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