Trend in equivalence classes of $ (x,y) \in R \iff x-y \in \mathbb{Z}$

Question

Question:

Define the following equivalence relation on the real numbers:

$ (x,y) \in R \iff x-y \in \mathbb{Z}$

Compute $ \ [-2], [-1], [0], [1], [2] $. Do you see a trend?

My attempt:

$ [-2] = \{ { y \in \mathbb{R} : -2-y \in \mathbb{Z} }\}$

$ [-1] = \{ { y \in \mathbb{R} : -1-y \in \mathbb{Z} }\}$

$ [0] = \{ { y \in \mathbb{R} : 0-y \in \mathbb{Z} }\}$

$ [1] = \{ { y \in \mathbb{R} : 1-y \in \mathbb{Z} }\}$

$ [2] = \{ { y \in \mathbb{R} : 2-y \in \mathbb{Z} }\}$

Will all of the above equivalence classes be equal to $ \mathbb{Z}$? Is that the trend?

Answer 1

They are all $\mathbb{Z}$. In fact, if $n\in\mathbb{Z}$, then $$ \{x\in\mathbb{R} \mid x+n\in\mathbb{Z}\} = \mathbb{Z}. $$ Proof: Denote $A:=\{x\in\mathbb{R} \mid x+n\in\mathbb{Z}\}$. Then $$ x\in A \iff x+n \in \mathbb{Z} \iff x+n-n\in \mathbb{Z} \iff x\in \mathbb{Z}. $$

Answer 2

It is clear that if $a,b\in \Bbb Z $ then $a-b\in \Bbb Z $ and $[a]=[b] $.

on the other hand $$[0]=\{x\in \Bbb R \;\;:\;\;x\in \Bbb Z\}=\Bbb Z $$

thus

if $a \in \Bbb Z $ then $[a]=[0]=\Bbb Z $ $$=[1]=[-1]=[2]=[-2] $$