Question

Define a relation `~' on R by 'a ~ b if a — b is an integer'. Show that this is an equivalence relation. Give the equivalence classes of 0 and √2 .

Answer 1

Let us show that a relation '$\sim$' on $\mathbb R$ by '$a \sim b$ if $a - b$ is an integer' is an equivalence relation, that is reflexive, symmetric and transitive.

For each $a\in\mathbb R$, $a-a=0\in\mathbb Z$ , and thus $a\sim a$ each $a\in\mathbb R$, and the relation is reflexive.

If $a\sim b$, then $a-b\in\mathbb Z$. It follows that $b-a=-(a-b)\in\mathbb Z$, and thus $b\sim a.$ Consequently, the relation is symmetric.

Let $a\sim b$ and $b\sim c$. Then $a-b\in\mathbb Z, \ b-c\in\mathbb Z$. It follows that $a-c=(a-b)+(b-c)\in\mathbb Z$, and thus $a\sim c$. Therefore, the relation is transitive.

By defenition, the equivalence class of an element $a$ is $[a]=\{b\in\mathbb R\ |\ b-a\in\mathbb Z\}$.

Let us give the equivalence classes of 0 and $\sqrt{2}$ :

$[0]=\{b\in\mathbb R\ |\ b-0\in\mathbb Z\}=\{b\in\mathbb R\ |\ b\in\mathbb Z\}=\mathbb Z,$

$[\sqrt{2}]=\{b\in\mathbb R\ |\ b-\sqrt{2}\in\mathbb Z\}=\{b\in\mathbb R\ |\ b=\sqrt{2}+n,\ n\in\mathbb Z\}=\sqrt{2}+\mathbb Z.$