Q: An integer $n$, such that $nx = 0$, where $x$ belongs to the quotient group $\Bbb Q/\Bbb Z$ Well, first, let $x$ be an element of the factor group $\mathbb Q/\mathbb Z$ ($\mathbb Q$ and $\mathbb Z$ are the additive groups, by the way).
Now, how do we find an integer n, such that, $nx = 0$, $n \ge 0$?
I managed to figure out that each x can be uniquely represented as $x= \dfrac{a}{b} + \mathbb Z$ for $0 \le a< b < 1 , \; \text{gcd}(a,b)=1$.
Now, I've considered that perhaps, I'm not understanding the question.
When $nx = 0$, I assume it means, essentially the same thing as $nx =\mathbb Z$, which is the identity element of Q/Z, right?
The obvious choice for $n$ would be a multiple of $b$.
Trying $n=b$, we have, $nx = a + b \mathbb Z $, but this is just a single coset of $b \mathbb Z$. A particular equivalence class $(a \mod b)$.
Which brings me to another question:
Consider $x+x = \dfrac{a}{b}+ \mathbb Z + \dfrac{a}{b} + \mathbb Z = 2 \dfrac{a}{b}+ \mathbb Z $,
But, $2x = \frac{2a}{b} + 2 \mathbb Z $.
I know, I must have slipped with the permissible algebra, but somehow, I just seem to be unable to reconcile the $2$, or choose between them.
Which of these is correct? (Perhaps the first one, given that addition is actually the operation of $\mathbb Q/\mathbb Z$?)
Also, how would I go about finding the integer $n$?
I would really appreciate it if, instead of actually finding $n$, someone could hand me a pointer in the right direction, and let me wrangle it myself.
Thank You!

A: As you've correctly noted, an element $x$ of $\mathbb{Q}/\mathbb{Z}$ is an equivalence class
$$x=\tfrac{a}{b}+\mathbb{Z}$$
where $\frac{a}{b}$ is a rational number. The operation $+$ on this group is defined by
$$(\tfrac{a}{b}+\mathbb{Z})+(\tfrac{c}{d}+\mathbb{Z})\overset{\text{def}}{=}(\tfrac{a}{b}+\tfrac{c}{d})+\mathbb{Z}.$$
Given an integer $n$, to multiply an element of $\mathbb{Q}/\mathbb{Z}$ by $n$ simply means to add it to itself $n$ times. Lastly, note that for any integer $a$, the equivalence class $a+\mathbb{Z}$ is the same as the equivalence class $0+\mathbb{Z}$, because
$$a+\mathbb{Z}=\{a+n\mid n\in\mathbb{Z}\}=\mathbb{Z}=0+\mathbb{Z}.$$
Therefore, 
$$b(\tfrac{a}{b}+\mathbb{Z})=\underbrace{\tfrac{a}{b}+\cdots+\tfrac{a}{b}}_{b\text{ times}}+\mathbb{Z}=a+\mathbb{Z}=0+\mathbb{Z}.$$
