R is an equivalence relation defined as $xRy \Leftrightarrow a - b$ is an integer.
What is the cardinality of the quotient of $\mathbb R$ in respect to R? How would you prove it?
I thought about a bijection between $[0,1)$ to the quotient of $\mathbb R$ in respect to R, but can't figure out to prove it formally.
On
You should understand that this is just an instance of congruence modulo 1.
You have the equivalence class, you should be able to establish a bijection between these and the reals easily enough, yes?
On
For any $\;r\in \Bbb R\;$ there exists a unique $\;x_r\in[0,1)\;$ s.t. $\;r-x_r\in\Bbb Z\;$. We can see this as follows:
$$\text{Let us write}\;\;r=\lfloor r\rfloor+\{r\}$$
with $\;\lfloor r\rfloor =\,$ the floor function and $\;\{r\}=\;$ the fractional part function. Here, we assume that $\;\{r\}\;$ is always non-negative, and thus for example $\;\{-1.4\}=0.6=r-\lfloor r\rfloor\;$, and in any case
$$\{x\}=\left|\,\lfloor x\rfloor - x\,\right|$$
Thus, for any $\;r\in\Bbb R\;$ , we have that $\;x_r=\{r\}\;\implies r\in [\{r\}]\;$ , meaning: $\;x-\{r\}\in\Bbb Z\;$. Clearly, the element in $\;\{r\}\;$ in $\;[0,1)\;$ is unique, and thus thus $\;\left|\Bbb R/R\right|=|[0,1)|=2^{\aleph_0}\;$
Your intuition is good. The set $[0,1)$ contains a single representative of each equivalence class.
Note that your equivalence classes are sets of the form $a+\mathbb Z, a\in[0,1)$. To see why, first note that if $a\in[0,1)$, then $a+1,a-1,a+2,...$ are all in $[a]$, since they differ from $a$ by an integer. Moreover, if $x-a=n\in \mathbb Z$, then $x=a+n$, and so $x\in a+\mathbb Z$.
Also note that if $x,y\in[0,1)$ and $x\neq y$, then they are not in the same equivalence class, since their difference is not an integer.
So the collection {$a+\mathbb Z| a\in [0,1)$} is a subset of the quotient set, and the quotient set is a subset of the collection {$a+\mathbb Z|a\in [0,1)$}, which means that they are equal.