Let $X =\{1,2,3,4,5,6,7,8,9\} .$ Let $R _{1}$ be a relation in $X$ given by $R _{1}=\{(x, y): x-y$ is divisible by $3\}$ and $R _{2}$ be another relation on $X$ given by ${R_2} = \{ (x,y):\{ x,y\} \subset \{ 1,4,7\} \} $ or $\{x, y\} \subset\{2,5,8\} $ or $\{x, y\} \subset\{3,6,9\}\} .$ Show that $R _{1}= R _{2}$.
Let $X =\{1,2,3,4,5,6,7,8,9\} .$ Let $R _{1}$ be a relation in $X$ given by $R _{1}=\{(x, y): x-y$ is divisible by $3\}$ and $R _{2}$ be another relation on $X$ given by ${R_2} = \{ (x,y):\{ x,y\} \subset \{ 1,4,7\} \} $ or $\{x, y\} \subset\{2,5,8\} $ or $\{x, y\} \subset\{3,6,9\}\} .$ Show that $R _{1}= R _{2}$.
Note that the characteristic of sets $\{1,4,7\}$, $\{2,5,8\} $ and $\{3,6,9\}$ is that difference between any two elements of these sets is a multiple of $3 .$ Therefore, $(x, y) \in R _{1} \Rightarrow x-y$ is a multiple of $3 \Rightarrow\{x, y\} \subset\{1,4,7\}$ or $\{x, y\} $ $\subset\{2,5,8\}$ or $\{x, y\} \subset\{3,6,9\} \Rightarrow(x, y) $ $\in R ,$ Hence, $R _{1} \subset R _{2} .$ Similarly, $\{x, y\} \in $ $R _{2} \Rightarrow\{x, y\}$ $\subseteq\{1,4,7\}$ or $\{x, y\} \subset\{2,5,8\}$ or $\{x, y\} \subset\{3,6,9\} $ $\Rightarrow x-y$ is divisible by $3 \Rightarrow\{x, y\} \in R _{1} .$ This shows that $R _{2} \subset R _{1} .$ Hence, $R _{1}= R _{2}$
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