Showing that $R / I$ is a ring 
Let $R$ be a ring and let $I$ be an ideal (that is, $I \le R$ and $R \cdot I \subset I)$. Define $$R / I = \{r + I : r \in R \}$$
  Show that $R/I$ is a ring under the operations $(r+I) + (s+I) = (r+s)+I$ and $(r+I) \cdot (s+I) = rs + I$.

The hint tells me to show that each operation is well-defined. 
 A: You can definitely do the laborious way that @k,stm did, and in fact, such is the most sure-fire way to be correct.  However, seeing as you have (hopefully) been exposed to groups, and quotients of groups, here is what you can do.  (So group stuff is not totally irrelevant here).
(I assume $R$ is commutative.  If not, slightly more words are required, but this will do for now)
First, note a particular definition/characterization of what a ring $R$ is:
$R$ is a ring if $(R, +)$ is an Abelian group and there exists a map $R\times R \to R$ sending $(r_1, r_2) \mapsto r_1r_2$ such that $1r = r$, $(rs)t = r(st)$, and $r(r_1 + r_2) = rr_1+rr_2$.  In short, one says that there is an action of $R$ on $R$.  This is really viewing $R$ as a $R$-module.
Well, $I$ being an ideal means that $I$ is a subgroup of $R$ (with respect to $+$), and that $R\cdot I \subset I$ (i.e. $I$ "absorbs" multiplication by $R$).
Hence, we can first view $R/I$ as a quotient group.  This automatically grants us that viewing $R/I$ as cosets of $I$, i.e. $R/I = \{a+I : a\in R\}$, addition is well-defined.  And now, we only need check that $R/I\times R/I\overset{mult}\to R/I$ is well-defined., which follows rather easily from the fact that $I$ "absorbs" multiplication by $R$.
I fear this may not have been as helpful in solving the problem, but to me viewing $R/I$ really as a quotient group with additional multiplication structure (that is what a ring is: an abelian group with mult. structure) has helped me a lot.
A: This is a very formal way of checking that $R/I$ is a ring.

Consider the maps


*

*$a \colon R × R → R/I,\, (r,s) ↦ (r+s)+I$,

*$m \colon R × R → R/I,\, (r,s) ↦ (r·s)+I$, and

*$π×π \colon R × R → R/I × R/I,\, (r,s) ↦ (r+I,s+I)$


(where $π \colon R → R/I,\, r ↦ r+I$).
Show that


*

*$∀(r,s) ∈ R × R \colon\quad (r-r',s-s') ∈ I × I ⇒ a(r,s) = a(r',s')$,

*$∀(r,s) ∈ R × R \colon\quad (r-r',s-s') ∈ I × I ⇒ m(r,s) = m(r',s')$, and

*$∀(r,s) ∈ R × R \colon\quad (r-r',s-s') ∈ I × I ⇔ (π×π)(r,s) = (π×π)(r',s')$.


Now, there’s a general theorem saying:

For any sets $A, B, F$ with a map $f \colon A → B$ and a surjective map $p \colon A → F$, if
  $$∀a,a' ∈ A \colon\quad p(a) = p(a') ⇒ f(a) = f(a'),$$
  then there is a unique map $\bar{f} \colon F → B$ such that $f = \bar{f}∘π$.

Prove and understand that once and for all. It’s a good idea to use this theorem (as well as the homomorphism theorems for any structures) whenever you need to check whether a map given by representatives is well-defined (representations of elements are usually surjective maps in disguise).
Using this on $p = π×π$ and $f = a$ or $f = m$, it follows that $\bar{a} = +$ and $\bar{m} = ·$ are well-defined maps. Then you need to check the ring axioms, but you can do this by using representatives, exploiting their validity in $R$.
