Let $R$ be the set of real numbers. Statement-1: $A=\{(x, y) \in R imes R: y-x$ is an integer $\}$

Question

Let $R$ be the set of real numbers. Statement-1: $A=\{(x, y) \in R 	imes R: y-x$ is an integer $\}$ is an equivalence relation on $R$. Statement-2 : $B=\{(x, y) \in R 	imes R: x=\alpha y$ for some rational number $\alpha\}$ is an equivalence relation on $R$.

Professor PC Thomas Classes · Answer

(a) : $y-x=$ integer\ and $z-y=$ integer $\Rightarrow z-x=$ integer\ $	herefore \quad(x, y) \in A$ and $(y, z) \in A$\ $\Rightarrow \quad(x, z) \in A \Rightarrow$ Transitive\ Also $(x, x) \in A$ is true $\Rightarrow$ Reflexive\ As $(x, y) \in A \Rightarrow(y, x) \in A \Rightarrow$ Symmetric Hence $A$ is an equivalence relation but $B$ is not. Since, $(0, y)$ is in $B$ but $(y, 0)$ is not in $B$.\