Let X = {1,2,3,4,5,6,7} and R = {x,y/x–y is divisible by 3} in x. Show that R is an equivalence relation.

Let $X = \{1,2,3,4,5,6,7\}$ and $R = \{(x,y)|x-y\text{ is divisible by }3\}$ in $X$. Let us show that $R$ is an equivalence relation. Since $x-x=0$ is divisible by 3 for any $x\in X$, we conclude that $(x,x)\in R$ for any $x\in X$, and hence $R$ is a reflexive relation. If $x,y\in X$ and $(x,y)\in R,$ then $x-y$ is divisible by 3. It follows that $y-x=-(x-y)$ is also divisible by 3, and hence $(y,x)\in R.$

We conclude that the relation $R$ is symmetric. If $x,y,z\in X$ and $(x,y)\in R,\ (y,z)\in R,$ then $x-y$ is divisible by 3 and $y-z$ is divisible by 3. It follows that $x−z=(x−y)+(y−z)$ is also divisible by 3, and hence $(x,z)\in R$. We conclude that the relation $R$ is transitive. Consequently, $R$ is an equivalence relation.