Trying to understand $\mathbb{Q} / \mathbb{Z}$ If I'm not mistaken, $\mathbb{Q} / \mathbb{Z}$ are equivalence classes of rationals where $q \sim q^\prime$ iff they differ by an integer. So I could equally imagine this set as $\mathbb{Q} \cap [0,1]$ (mod 1) (if you know what I mean?).
Is that the right way to think about this group? I'm asking because I would like to see why elements in $\mathbb{Q} / \mathbb{Z}$ have finite order.
Many thanks for your help.
 A: Yes: the elements of $[0,1)\cap\mathbb{Q}$ form a complete set of coset representatives for $\mathbb{Z}$ in $\mathbb{Q}$. That is:


*

*Every element of $\mathbb{Q}$ is congruent modulo $\mathbb{Z}$ to at least one element of $[0,1)\cap\mathbb{Q}$; and

*No two distinct elements of $[0,1)\cap\mathbb{Q}$ are congruent modulo $\mathbb{Z}$.


That means that each element of $\mathbb{Q}/\mathbb{Z}$ corresponds to one and only one element of $[0,1)\cap \mathbb{Q}$. Since the sum of classes in the quotient is equal to the class of the sum, we have that if $q_1,q_2\in\mathbb{Q}$ correspond to the rationals $r_1$ and $r_2$ in $[0,1)\cap\mathbb{Q}$, respectively, then
$$\begin{align*}
(q_1+\mathbb{Z}) + (q_2+\mathbb{Z}) &= (r_1+\mathbb{Z}) + (r_2+\mathbb{Z})\\
&= (r_1+r_2)+\mathbb{Z}\\
&=\left\{\begin{array}{ll}
(r_1+r_2)+\mathbb{Z} & \text{if }0\leq r_1+r_2\lt 1\\
(r_1+r_2-1 + \mathbb{Z} & \text{if }1\leq r_1+r_2
\end{array}\right.
\end{align*}$$
which shows you that this group is isomorphic to $\mathbb{Q}\cap[0,1]$ mod $1$. 
A: Yes, $\mathbb{Q}\cap [0,1)$ forms a set of representatives for $\mathbb{Q}/\mathbb{Z}$, and "mod 1" is a good way of thinking about it. To see why an element $\left[\frac{a}{b}\right]\in\mathbb{Q}/\mathbb{Z}$ has finite order (I am using the brackets to mean the equivalence class of $\frac{a}{b}$), consider that this is equivalent to the statement that there is a finite number $n$ such that when I add $\frac{a}{b}$ to itself $n$ times, I get an integer. Do you see an $n$ that works?
