Elements of $\mathbb{Z}/(n)$ Let $(n) = \{ \lambda n | \lambda \in \mathbb{Z} \}$. In my book it has shown that every element in  $\mathbb{Z}/(n)$ can be expressed uniquely in the form $r + (n)$ where $0 \leq r \leq n-1$ now I look at a practical application of this, say:
$\mathbb{Z}/6 = \{ 0,1,2,3,4,5 \}$ and when computing elements in this set we just do it in modulo $6$, so everytime I see a $6$ I write $0$, so for $\mathbb{Z}/(n)$ everytime I see $n$ I write $0$. However, I don't see how this leads on from the theorem I stated above, i.e. if we know that every element in  $\mathbb{Z}/(n)$ can be represented in the form $r + (n)$ with $ 0 \leq r \leq n-1$ why does that mean, when I see $n$ I write $0$? I'm just not comfortable with $r + (n)$ I mean, what is the $(n)$ doing there?
 A: Remember that $r + (n)$ in this context is a coset, it itself is not a single number but an infinite collection of integers (namely those elements $k \in \mathbb{Z}$ such that $r-k$ is a multiple of $n$.  You should think of the coset $r+(n)$ as the set of elements that are $n$-translates of $r$.  In your example above you had $\mathbb{Z}/(6)$ so the coset corresponding to $r=5$ is 
$$
5 + (6) \; \; = \;\; \{\ldots, -7, -1, 5, 11, 17, 23, \ldots \}.
$$
So the $6$-translates of $6$ would be the coset
$$
6 + (6) \;\; =\;\; \{\ldots, -18, -12, -6, 0, 6, 12, 18, \ldots\}
$$
We see that $0 \in [6 + (6)]$ so if we want to represent the coset uniquely in terms of nonnegative integers less than $6$ we can use $0$ to represent the coset containing $6$.
A: Notice that $r$ is the remainder of the division of any integer number by $n$. And as $0 \leq r < n$ then you may choose a class representant to the residue class $[r]$ then 
$$[r] = r + (n) = \{r + x ; x \in (n)\}$$
You may also notice that 
$$a \equiv b\mod (n) \iff b-a \in (n)$$
A: The problem is that $\mathbb Z/(6)$ is not the set $\{0,1,2,3,4,5\}$. Instead, $\mathbb Z/(6)$ is the set of equivalence classes of $\mathbb Z$ under the equivalence relation $x\sim y:\Leftrightarrow x-y\in(6)$. These equivalece classes are not single numbres, but are sets of numbers, for example
$$\{\ldots,-24,-18,-12,-6,0,6,12,18,24,\ldots\} $$
and 
$$\{\ldots,-23,-17,-11,-5,1,7,13,19,25,\ldots\} $$
are two of these equivalence classes.
The first is just $(6)$ (cf. the definition of $(n)$). Since $0$ "does nothing" we can also write it as $0+(6)$. The second equivalence class is the set $(6)$ with $1$ added to each element, which is suggestively written as $1+(6)$. There is nothing special about the $1$ for we might with the same justification cwrite this set as $-5+(6)$ or $6667+(6)$ or many other ways of the form $r+(6)$. But the theorem you quote says that we can pick $r$ such that $0\le r\le 5$ (which we actually did in our first attempt, with $r=1$).
So more precisely
$$\mathbb Z/(6)=\{(6),1+(6),2+(6),3+(6),4+(6),5+(6)\} $$
Remark: You may sometimes find the notation $6\mathbb Z$ instead of $(6)$, so with that convention $\mathbb Z/6\mathbb Z=\{6\mathbb Z,1+6\mathbb Z,2+6\mathbb Z,3+6\mathbb Z,4+6\mathbb Z,5+6\mathbb Z\}$
