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?

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 $6$.
