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

Step 1: Analyze Relation R1

Relation $R_1$ states that $x-y$ must be divisible by 3. This means that $x$ and $y$ must have the same remainder when divided by 3. We can categorize the elements of set $X$ based on their remainders when divided by 3.

Step 2: Categorize Elements by Remainder (R1)

We divide the set $X$ into three disjoint subsets based on their remainders when divided by 3. $C_0$ contains numbers divisible by 3, $C_1$ contains numbers with a remainder of 1, and $C_2$ contains numbers with a remainder of 2.

Step 3: Express R1 using Categories

For $x-y$ to be divisible by 3, both $x$ and $y$ must belong to the same category ($C_0$, $C_1$, or $C_2$). This means that if $(x,y) \in R_1$, then $x$ and $y$ must come from the same remainder class.

Step 4: Analyze Relation R2

Relation $R_2$ is defined such that both $x$ and $y$ must belong to one of the specified subsets: $\{1{,}4{,}7\}$, $\{2{,}5{,}8\}$, or $\{3{,}6{,}9\}$.

Step 5: Compare R1 and R2

By comparing the definitions, we see that the subsets specified in $R_2$ are exactly the remainder classes we identified for $R_1$. Specifically, $\{1{,}4{,}7\} = C_1$, $\{2{,}5{,}8\} = C_2$, and $\{3{,}6{,}9\} = C_0$.

Step 6: Conclusion

Since both relations define the same condition for $(x,y)$ to be related (i.e., $x$ and $y$ must belong to the same remainder class modulo 3), we can conclude that $R_1 = R_2$.

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