Elements of $\mathbb{R}/\mathbb{Z}$ I read that $\mathbb{R}/\mathbb{Z}=[0,1)$ and I can't understand why. I mean, I understand the proof, but I am confused since I knew that quotient groups are sets of equivalence classes and an equivalence class is a set. How is it possible for this to be equal to $[0,1)$? I expected something like $\mathbb{R}/\mathbb{Z}=\{\hat{x}|x\in [0,1)\}$.
EDIT: After the discussion in the comments I came up with the following idea : if $\cdot$ is the operation with which $\mathbb{R}/\mathbb{Z}$ is a group, then I believe that ($\mathbb{R}/\mathbb{Z}, \cdot$) $\approx ([0,1), *)$, where $*$ is defined as $x*y=\{x+y\}$ for $x, y \in [0,1)$ ($\{a\}$ denotes the fractional part of $a$). Am I right?
 A: We claim that $[0, 1)$ is a group with respect to the operation $x * y = \{x + y \},$ as you suggest. I assume throughout that $\{x \} = x - \lfloor x \rfloor$ in order to conclude that $\{r + s \} = \{\{r \} + \{s \} \} ;$ however, there are differing opinions about that. Unfortunately, if you subscribe to Knuth's definition of $\{x \},$ then you might need to come up with a different operation $*.$
1.) Clearly, $[0, 1)$ is closed under $*$ by definition of $\{ \cdot \}.$
2.) Considering that addition is associative, $*$ is likewise associative, as  we have that $$(x * y) * z = \{x + y \} * z = \{\{x + y \} + z \} = \{x + \{y + z \} \} = x * \{y + z \} = x * (y * z).$$
3.) Further, we have that $x * 0 = \{x + 0 \} = \{x \} = x = 0 * x$ by hypothesis that $x$ is in $[0, 1),$ hence $0$ is the identity element in $[0, 1)$ with respect to $*.$
4.) Last, we have that $x * (-x) = \{x + (-x) \} = \{0 \} = 0 = (-x) * x,$ hence every element in $[0, 1)$ has an inverse with respect to $*.$ We conclude that $[0, 1)$ is a group with respect to $*.$
Consider the map $\varphi : (\mathbb R, +) \to ([0, 1), *)$ defined by $\varphi(r) = \{r \}.$ We have that $$\varphi(r + s) = \{r + s \} = \{\{r \} + \{s \} \} = \{r \} * \{s \} = \varphi(r) * \varphi(s),$$ hence $\varphi$ is a group homomorphism. Clearly, $\varphi$ is surjective. By the First Isomorphism Theorem, therefore, we have that $(\mathbb R, +) / \ker \varphi \cong ([0, 1), *).$ Observe that we have $r \in \ker \varphi$ if and only if $\varphi(r) = 0$ if and only if $\{r \} = 0$ if and only if $r \in \mathbb Z,$ as desired. QED.
