Equivalence Relation on $\mathbb{R}$ and its partition $\mathbb{R}/\sim$

Question

Define the equivalence relation $\sim$ on $\mathbb{R}$ as follows:

$$\forall a,b\in\mathbb{R},\ a\sim b\ \Leftrightarrow\ b-a\in\mathbb{Z}$$

I can prove that this is an equivalence relation, but I saw a claim that the partition $\mathbb{R}/\sim$ can be thought of as the interval $[0,1)$. To be clear, is that the set of equivalence classes $\{[a]_{\sim}:a\in[0,1)\}$?

Additionally, given a partition $\mathscr{P}_{\sim}$ on a set $S$, could I define the equivalence relation $\sim$ on $S$ by taking the cartesian product of each piece of the partition with itself?

Answer 1

The map $i:(\mathbb{R}/\sim) \to [0,1)$ defined (awkwardly) by $ \{ i([x]) \} = [x] \cap [0,1)$ is a bijection.

Notation: $[x] = \{ y \mid x \sim y \}$.

A partition defines an equivalence class in the manner you described, that is, if $P_1,...,P_n$ is the partition, then the equivalence class is $\cup_k (P_k \times P_k)$.

Answer 2

$[0,1)$ is what we call a set (or system) of distinct representatives - that is, for every $x$ there is exactly one $y\in[0,1)$ such that $x\sim y$.

Answer 3

Any real number can be written as $n+f$ uniquely with $n$ an integer $f$ the positive fractional part. (for negative numbers such as $-8.25$ you take fractional part to be $+0.75$ and integer part to be $-9$, rather than as $-.25 $ and $-8$.

Two real numbers $a,b$ are equivalent if their difference is an integer which is the same as saying their $f$ parts are equal. So this fractional part determines the equivalence class completely.