Basic understanding of quotients of "things"? My modern algebra needs some work. Am I right in thinking that $\mathbb{Z}/2\mathbb{Z}$ refers to the two sets $$\{\pm0, \pm2, \pm4, \pm6, \ldots\}$$ and $$\{\pm1, \pm3, \pm5, \pm7\}~~?$$ What about $\mathbb{R}/2\mathbb{Z}$ if that makes sense to write? Would that mean $$\{,\ldots,[0,1),[2,3),[4,5),\ldots\}$$ and $$\{\ldots,[1,2),[3,4),[5,6),\ldots\}~~?$$ I haven't got round to looking at these "quotients" as I think they're called. It's on my list of things to do. I believe they're to do with equivalence classes? Are there some more "exotic" examples with concrete examples of the sets produced as above? Just asking to see if I'm thinking on the right track. So my original understanding of $\mathbb{R}/2\mathbb{Z}$ was incorrect (as pointed out in answer(s) below).
EDIT
Not sure if this is a good way to visualise what's going on with e.g. $\mathbb{R}/2\mathbb{Z}$, but imagine the Cartesian plane with $x$ and $y$ axes. For $\mathbb{R}/2\mathbb{Z}$ I can see the left hand side ($\mathbb{R}$) corresponding to the $y$-axis and the right hand side ($2\mathbb{Z}$) corresponding to the integers on the $x$-axis. If I imagine $2\mathbb{Z}$ on the $x$-axis slicing the plane vertically then what I'm left with is an infinite number of slices each of width $2$. The quotient kind of takes all these slices and stacks them on top of each other so that the only information available to me belongs to $[0,2)$. Positioning along the $x$-axis is lost.
 A: Taking quotients means considering equality up to something you are not interested in. This always corresponds to building equivalence classes, as you correctly noted. However, in most practical uses, one is interested in cases where the mapping that sends elements to their equivalence classes furthermore respects the underlying structure.
For example, the fact that you can only take quotients of groups by normal subgroups, not just any subgroup, is related to this: 
Indeed, if you have a subgroup $N$ of a group $G$, then $a\sim b:\Leftrightarrow ab^{-1}\in N$ always provides a partition of $G$ into disjoint subsets (this is not trivial but also not too difficult to show). Let's call the set of these disjoint subsets $H$. 
You may now define $\phi:G\to H$ by $a\mapsto \phi(a):=aN$ and this identifies elements $a,b\in G$ for which $a=nb$ for some $n\in N$, i.e. elements that are equal up to some element of $N$.
However, $\phi$ is NOT always a group homomorphism. Let's find out some necessary condition for $\phi$ to be a homomorphism by assuming it is: Clearly, the image $\phi(e)=N\in H$ of the identity element $e\in G$ must be the identity element of $H$ no matter which group structure you use on $H$. Since $\phi^{-1}(N\in H)=N\subset G$ (show this yourself) this shows that $N=\ker\phi$. But any subgroup that is a kernel of some homomorphism satisfies $a(\ker\phi)=(\ker\phi) a$ for any $a\in G$, which is the characterization of normal subgroups. Thus, we have seen that the map that quotients out $N$ can only be a homomorphism when $N$ is a normal subgroup.  (It is easy to show that normality is not only necessary but also sufficient for $\phi$ to be a homomorphism if $H$ is equipped with the obvious structure $aN\times bN:= (a\times b)N$.)
Summary: you can take quotients w.r.t. any subgroup, and this quotient taking will identify elements up to something you are not interested in. However, if your subgroup is not normal, this partitioning is quite useless because you cannot do any further algebra with it. 
A: The elements of $\Bbb R/2\Bbb Z$ are of the form:
$$\{x,x\pm2,x\pm4,\ldots,x\pm 2n,\ldots\},$$
where $0\le x<2$.
A: A good way to think about quotients is to pretend that nothing has changed except your concept of equality. 
You can think of $\mathbb{Z}/2\mathbb{Z}$ as just the integers (under addition) but multiples of 2 (i.e. elements of $2\mathbb{Z}$) are eaten up like they're zero. So in this quotient world $1=3=-5=41$ etc. and $0=6=104=-58$ etc.
What is $1+1$? Well, $1+1=2$. But in this quotient group $2=0$ so $1+1=0$. Notice that $1=-3=99$ so $1+1 =-3+99=96$ (which also $=0$). Equivalent "representatives" give equivalent answers.
Formally, yes, $\mathbb{Z}/2\mathbb{Z} = \{0+2\mathbb{Z}, 1+2\mathbb{Z}\}$ where $0+2\mathbb{Z}=2\mathbb{Z}=$ even integers and $1+2\mathbb{Z}=$ odd integers.
A more formal version of my previous calculation: $(1+2\mathbb{Z})+(1+2\mathbb{Z}) = (-3+2\mathbb{Z})+(99+2\mathbb{Z}) = (-3+99)+2\mathbb{Z} = 96+2\mathbb{Z}=0+2\mathbb{Z}$.
If we move to $\mathbb{R}/2\mathbb{Z}$, then elements are equivalence classes: $x+2\mathbb{Z} = \{x+2k \;|\; k \in \mathbb{Z}\} = \{\dots,x-4,x-2,x,x+2,x+4,\dots\}$. Addition works exactly the same as it does in $\mathbb{R}$ (except we have enlarged what "equals" means). So $((3+\sqrt{2})+2\mathbb{Z})+((-10+\pi)+2\mathbb{Z}) = (7+\sqrt{2}+\pi)+2\mathbb{Z}$. Of course, here, $7+\sqrt{2}+\pi$ could be replaced by something like $-3+\sqrt{2}+\pi$.
In fact, every $x+2\mathbb{Z}$ is equal to $x'+2\mathbb{Z}$ where $x' \in [0,2)$ (add an appropriate even integer to $x$ to get within the interval $[0,2)$). So as a set $\mathbb{R}/2\mathbb{Z}$ is essentially $[0,2)$ (each equivalence class in the quotient can be uniquely represented by a real number in $[0,2)$).
Alternatively, think of this group like $[0,2]$ with $0=2$. Take the interval $[0,2]$ and glue the ends together. It's a circle group. Basically $\mathbb{R}/2\mathbb{Z}$ as a group is just like adding angles (but $2=0$ not $2\pi=0$). :)
