$S = \{(x,y)|x,y \in\mathbb{R}, (x−y)/3\in\mathbb{Z}\}$ is a relation. Prove: $S$ is an equivalence relation.

Question

Assume $S$ is a relation defined on $\mathbb{R}$ and $S = \{(x,y)|x,y \in\mathbb{R}, (x−y)/3\in\mathbb{Z}\}$. Prove that $S$ is an equivalence relation.

How do I prove this equivalence relation?

I am not sure what to use.

Answer 1

If $a$ and $b$ are related by $S$ i.e., $(a,b)\in S=\{(x,y)|x,y\in\mathbb{R},(x-y)/3\in\mathbb{Z}\}$, let them be represented by $a\sim b$.

A relation $S$ on $\mathbb{R}$ is an equivalence relation on $\mathbb{R}$ if it follows the following properties:

1. Reflexive property: $x\sim x$
2. Symmetric property: $x\sim y\Rightarrow y\sim x$
3. Transitive property: If $x\sim y$ and $y\sim z$ then, $x\sim z$

To show that $S$ is an equivalence relation, it is enough to show that two arbitrary numbers $x,y\in\mathbb{R}$ satisfy these properties.

1. $x\in\mathbb{R}$ $$\frac{x-x}{3}=\frac{0}{3}=0\in\mathbb{Z}$$
   $\therefore x\sim x$
2. Let $x\sim y$. Then, $\frac{x-y}{3}\in\mathbb{Z}.$
   $$\frac{x-y}{3}\in\mathbb{Z}\Rightarrow -\left(\frac{x-y}{3}\right)\in\mathbb{Z}\Rightarrow\frac{y-x}{3}\in\mathbb{Z}.$$ $\therefore x\sim y \Rightarrow y\sim x$
3. Let $x\sim y,y\sim z$. Then, $\frac{x-y}{3},\frac{y-z}{3}\in\mathbb{Z}.$
   $$\left(\frac{x-y}{3}\right)-\left(\frac{y-z}{3}\right)\in\mathbb{Z}\Rightarrow \left(\frac{x-y+y-z}{3}\right)\in\mathbb{Z}\Rightarrow \frac{x-z}{3}\in\mathbb{Z}$$ $\therefore$ If $x\sim y$ and $y\sim z$ then, $x\sim z$.

Hence, $S$ is an equivalence relation.