why $(r+I)(s+I) = rs + I$ in the quotient ring $R ?$ Say $R$ is a commutative ring and $I\in R$ is an ideal. Let us consider the quotient $R/I$. It is created by taking every element $a\in R$, and adding all the elements of $I$ to it. The elements of $R/I$ are of the form $a+I$ and $b+I$, $\forall a,b\in R$. 
Now $$(a+I).(b+I)=(a+i_{1}).(b+i_{2}),\forall (i_{1},i_{2})\in I \times I$$
$$(a+i_{1}).(b+i_{2})=ab+a.i_{2}+b.i_{1}+i_{1}i_{2}=ab+I$$
We know that $a.i_{2}+b.i_{1}+i_{1}i_{2}\in I$. 
If $(a+I).(b+I)=ab+I$, then taking suitable $(i_{1},i_{2})\in I\times I$, we should be able to prove $(a+i_{1}).(b+i_{2})=ab+i_{3},\forall i_{3}\in I$. However, can every element in $I$ be generated by taking suitable $i_{1},i_{2}\in I$? And if not, is that the reason why $(a+I).(b+I)$ has to be defined as equal to $ab+I$ in violation of the distributive property of ring elements? This has confused me for a long time. 
EDIT: I figured addition does not have to be defined as such because $(a+i_{1})+(b+i_{2})=a+b+i_{1}+i_{2}$, where $i_{1}+i_{2}\in I$. In fact, every element $i\in I$ can be constructed by taking $i_{1}=0$ and $i_{2}=i$. Hence, $(a+I) + (b+I)=a+b+I$ naturally. This is the logic I followed to determine that $(a+I).(b+I)$ doesn't quite work as nicey. I'm not sure if this logic is flawed or not as I haven't had the opportunity to ask anybody. 
Thanks for your help in advance! 
 A: In the case of group cosets, one could define coset addition in terms of equivalence relations or in terms of set operations, which we will call the modulo and Frobenius versions, respectively, for clarity's sake. Suppose that $G$ is a (multiplicative) group and let $C, D\subseteq G$. Then we define the (Frobenius) product of subsets as $CD := \{cd\mid c\in C, d\in D\}\subseteq G$. Hence, we can create a product on the subsets of a group by defining multiplication between sets element-wise. Notice that a coset is defined using this Frobenius product, ie, $$gN = \{g\}N = \{gn\mid n\in N\}.$$ Thus, it seems natural to define the product of cosets to be their Frobenius product as subsets of $G$. If $N$ is a normal subgroup of $G$, then $$(gN)(hN) = g(Nh)N = g(hN)N = (gh)(NN) = (gh)N,$$ where each equality holds as a (Frobenius) set equality. With this perspective there is no need to "define" a product of cosets as such and such because there already exists a product of subsets that restricts exactly to the product of cosets we need. Conversely, the modulo product of cosets, $(gN) \sharp (hN)$ as the unique coset containing $gh$, namely, $(gN)\sharp (hN):= (gh)N$.
So the confusion here is the overloading of coset multiplication. For the case of groups, these two different definitions of coset multiplication produce the exact same binary operation on the set of cosets (not just isomorphic, they are literally the same operation on the set of cosets). Therefore, for group cosets, $(gN)(hN) = (gN)\sharp (hN)=(gh)N$. Hence, we do not have to define a new operation when introducing coset multiplication on groups because the product is already well-defined and is associative. One only needs to show that the product of two cosets is a coset (which we have already shown). That is, the Frobenius product makes the power set of a group into a semigroup and the restriction to cosets of a subgroup forms a subsemigroup of the power set if and only if this subgroup is normal. This subsemigroup of the power set is in fact a group, which is identical to the quotient group created using the modular approach.
It is natural to suppose this same pattern occurs for ring cosets. Let $R$ be a ring and $I$ an ideal. It is true that $(a+I) + (b+I) = \{(a+i) + (b+j) \mid i,j\in I\}=\{(a+b)+I\mid i\in I\} = (a+b)+I$, where this is the sum of two subsets of $R$ in the Frobenius-sense. This follows from above since $(R,+)$ is an abelian group and $I$ is an additive subgroup of $(R,+)$. Unfortunately, the assumption that $I$ is an ideal is insufficient to prove that $(a+I)(b+I) = ab+I$ (in the Frobenius-sense). Even in a commutative ring, this is false. Let $R=\mathbb{Z}_2[x]/(x^3-x^2)$ and let $I=(x) = \{0, x, x^2, x^2+x\}$. Note that $$I^2 = \{0, x, x^2, x^2+x\}\{0, x, x^2, x^2+x\} = \{0, x^2\} \neq I.$$ This defect is a consequence of $(R,*)$ only being a semigroup (well, commuative monoid in this case). Without cancellation, we cannot guarantee the Frobenius product of two cosets is itself a coset.
It is easy to show from the properties of an ideal that $$(a+I)(b+I) \subseteq ab+I$$ always, but, as we saw above, equally can fail. So while the Frobenius product of subsets of a ring still applies in this context, the Frobenius product of two cosets is not necessarily a coset of the ideal. But, and this is an important but, because the system of cosets of a fixed ideal of a ring forms a partition of the ring, if $(a+I)(b+I) \subseteq ab+I$, then $ab+I$ is the only coset that contains $(a+I)(b+I)$. Therefore, we can still define the modular product of cosets and the above set containment is sufficient to prove this modular coset multiplication is well-defined.
A: The multiplication in the quotient ring is not defined by
$$
(a+I)(b+I)=
\{\,(a+i_{1})(b+i_{2}): (i_{1},i_{2})\in I \times I\,\}
$$
but by
$$
(a+I)(b+I)=ab+I.
$$
This is a definition, nothing else. Why do we define it in this way? Because it does what we want, together with
$$
(a+I)+(b+I)=(a+b)+I,
$$
namely it makes $A/I$ into a ring and
\begin{align}
\pi\colon A &\to A/I\\
a&\mapsto a+I
\end{align}
a ring homomorphism.
The only thing to show is that the definitions are “correct”: if $a_1+I=a_2+I$ and $b_1+I=b_2+I$, then we should have
$$
a_1b_1+I=a_2b_2+I
$$
that is
$$
a_1b_1-a_2b_2\in I.
$$
This is true because
$$
a_1b_1-a_2b_2=a_1b_1-a_1b_2+a_1b_2-a_2b_2=
a_1(b_1-b_2) + (a_1-a_2)b_2
$$
and, by hypothesis, $a_1-a_2\in I$ and $b_1-b_2\in I$; apply the properties of $I$ to end the proof. Similarly for the addition. The verification of the ring properties is easy.
A: You have a ring $R$ and an ideal $I\subseteq R$ (note that is is not a member of $R$, it is a subset).
Now you define an equivalence relation by $\forall a,b \in R, a\sim b \iff a-b \in I$
Then you look at $R/\sim = \left\{\overline{a}\mid a \in R\right\}$ where $\forall a \in R,\overline{a}=\left\{b \in R \mid b \sim a\right\}=a+I$ is the equivalence class of $a$.
Not you want to define $\forall a,b \in R,\overline{a}+\overline{b}=\overline{a+b}$ and $\overline{a}\times\overline{b}=\overline{a\times b}$ but to be able to do that, you need $\forall a,b,c,d \in R,a\sim c \land b \sim d \implies a+b \sim c+d$ and the same property with $\times$.  Because otherwise, you would have $\exists a,b,c,d \in R,\overline{a+b}=\overline{a}+\overline{b}=\overline{c}+\overline{d}=\overline{c+d}$ and $\overline{a+b}\not= \overline{c+d}$ which is absurd. You can think of it this way: You want to be able to define the sum and the product of two equivalence classes so that it is independent of the representatives you chose.
And the fact that $I$ is an ideal gives this equivalence property those property (compatibility with $+$ and $\times$).
