Let $X = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Let $R_{1}$ be a relation in $X$ given by $R_{1} = \{(x, y) : x - y \text{ is divisible by } 3\}$ and $R_{2}$ be another relation on $X$ given by $R_{2} = \{(x, y) : \{x, y\} \subset \{1, 4, 7\} \text{ or } \{x, y\} \subset \{2, 5, 8\} \text{ or } \{x, y\} \subset \{3, 6, 9\}\}$. Show that $R_{1} = R_{2}$.

(N/A) The set $X = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is partitioned into three subsets based on remainders when divided by $3$: $S_{1} = \{1, 4, 7\}$,$S_{2} = \{2, 5, 8\}$,and $S_{3} = \{3, 6, 9\}$.
For any $x, y \in X$,$x - y$ is divisible by $3$ if and only if $x$ and $y$ belong to the same subset $S_{i}$ (where $i \in \{1, 2, 3\}$).
If $(x, y) \in R_{1}$,then $x - y$ is a multiple of $3$,which implies $x$ and $y$ have the same remainder when divided by $3$. Thus,$\{x, y\} \subset S_{1}$ or $\{x, y\} \subset S_{2}$ or $\{x, y\} \subset S_{3}$,which means $(x, y) \in R_{2}$. Hence,$R_{1} \subset R_{2}$.
Conversely,if $(x, y) \in R_{2}$,then $\{x, y\}$ is a subset of $S_{1}$,$S_{2}$,or $S_{3}$. In any of these cases,the difference $x - y$ is a multiple of $3$,so $(x, y) \in R_{1}$. Thus,$R_{2} \subset R_{1}$.
Since $R_{1} \subset R_{2}$ and $R_{2} \subset R_{1}$,we conclude that $R_{1} = R_{2}$.