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  4. Let R be the set of real numbers Statement-1: A= (x, y) ∈ R × R: y-x is an integer is an equivalence relafion on R Statement-2: B= (x, y) ∈ R × R: x=α y for some rational number α is an equivalence relation on R.

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

AIEEEAIEEE 2011Relations and Functions - Part 2

Solution:

Statement- 1 is true
We observe that
$\underline{\text { Reflexivity }}$
$x R x$ as $x-x=0$ is an integer, $\forall x \in A$
Symmetric
Let $(x, y) \in A$
$\Rightarrow y-x$ is an integer
$\Rightarrow x-y$ is also an integer
$\underline{\text {Transitivity }}$
Let $(x, y) \in A$ and $(y, z) \in A$
$\Rightarrow y-x$ is an integer and $z-y$ is an integer
$\Rightarrow y-x+z-y$ is also an integer
$\Rightarrow z-x$ is an integer
$\Rightarrow (x, z) \in A$
Because of the above properties $A$ is an equivalence relation over $R$
Statement 2 is false as $^{\prime} B^{\prime}$ is not symmetric on $R$ We observe that
$O B x$ as $0=0 x \forall x \in R$ but $(x, 0) \notin B$