Prove $+$ and $\times$ are well-defined on quotient rings I am asked to prove that the $+$ and $\times$ operations are well-defined on quotient rings. Having little experience with similar questions I looked up the proofs online and found that most of them feature a step I don't entirely understand. 
Normally, they would begin with something like this: 
Let $r+I=r'+I$ and $s+I=s'+I$, then $r-r',s-s' \in I$.
I have run through several examples in my head and realise that this is correct, but I don't really understand where the general conclusion comes from, especially since $r-r'$ or $s-s'$ isn't necessarily $0$. I'm probably missing something extremely obvious, but being new to quotient rings I just don't see why $r-r' \in I$, and a similar argument applies to the $\times$ proof.
 A: Let's do an example.  Consider the ring $\mathbb{Z}[x]$ (polynomials in $x$ with integer coefficients) and the ideal $I=(x)$ (polynomials with $0$ constant term).
Let's take $2+(x)$ and  $3 + (x)$ in the quotient ring $\mathbb{Z}[x]/(x)$.  Now, their product is defined as $$(2+(x))(3+(x)) = 6 + (x).$$  But notice, $2+(x) = 2+x + (x)$, so we could've also multiplied as $$(2+x+(x))(3+(x)) = 6 + 3x + (x).$$  Now, notice that $6+3x+(x)$ and $6+(x)$ are equal, so BOTH multiplications give the SAME answer.  This is what well-defined means.

To check that the multiplication is well-defined in general, you have to show that picking different representatives of an ideal $r+I$ gives the same multiplication/addition.
A: The key is that, for an ideal $I$ and element $r$ we simply have
$$r+I=I\ \iff\ r\in I$$
Now apply this to $r=r_2-r_1$. 
A: Suppose I have a ring $R$, it has $0, +, -, \times$ as required.
Let's additionally endow $R$ with an equivalence relation, $\simeq$ that is transitive, reflexive, and symmetric as required, but a priori unrelated to the structure of $R$ .
$(R, \simeq)$ is now a ring with an equivalence relation. Let's say that $(\simeq)$ respects the ring structure of $R$ if all of the following hold for all $a, b, c$ in $R$ .
$$ a \simeq b \implies (-a) \simeq (-b) \\ 
   a \simeq b \implies c + a \simeq c + b \\
   a \simeq b \implies a\times c \simeq b \times c \\
   a \simeq b \implies c \times a \simeq c \times b $$
In other words, once you have two ring elements $a, b$ that are equivalent, if you can construct a probe out of constant ring elements and ring functions that send $a$ and $b$ to nonequivalent ring elements, then $\simeq$ does not respect the ring structure of $R$.
Suppose $\simeq$ respects the ring structure of $R$ .
$(R, \simeq)$ is not our quotient ring, but we can construct a new ring $R' = R/\simeq$ whose elements are the equivalence classes that $\simeq$ partitions $R$ into. We adorn each equivalence class a designated canonical element, denoted $x_c$ where $x$ is in $R'$ . 
Now we need to define $0, +, -, \times$ in $R'$
$0$ is whatever class in $R'$ contains $0$.
$x + y = z \stackrel{\text{def}}{\iff} x_c + y_c \in z $
$-x = z \stackrel{\text{def}}{\iff} -(x_c) \in z $
$x \times y = z \stackrel{\text{def}}{\iff} x_c \times y_c \in z $ .
Because $\simeq$ respects the ring structure of $R$, it does not matter which elements of the equivalence classes we picked as canonical.
Ring Ideals give us an equivalence relation that can be used to construct a $\simeq$ that respects the ring structure.
In particular, let $I$ be a ring ideal and let $\simeq_I$ be the equivalence relation associated with the ideal $I$.
$$ x \simeq_I y \stackrel{\text{def}}{\iff} x - y \in I $$
A ring ideal is a subset of a ring $R$ satisfying two properties:


*

*Strong multiplicative closure. $\forall r \in R \mathop. r \times I \subset I$  and $\forall r \in R \mathop. I \times r \subset I $ . In words, $I$ is closed under multiplication by any ring element $R$. (I have never heard anyone else call it strong multiplicative closure, but it's a decent mnemonic). 

*Additive closure $\forall w \in I \mathop. w + I \subset I $. $I$ is closed under addition. Equivalently, $(I,+)$ is an Abelian group.


So, if we check that $(R, \simeq_I)$ respects the ring structure of $R$, we know that $R/I$ is a bona fide ring.
Checking negation:
$$ a \simeq b \\
   a - b \in I \\
   -(a - b) \in I \\
   -a - (-b) \in I \\
   -a \simeq -b $$
Checking addition
$$ a \simeq b \\
a - b \in I \\
a + (c - c) - b \in I \\
(a + c) - (c + b) \in I \\
(a + c) - (b + c) \in I \\
a + c \simeq b + c $$
Checking left multiplication
$$ a \simeq b \\
a - b \in I \\
c \times (a - b) \in I \\
c \times a - c \times b \in I \\
c \times a \simeq c \times b $$
Checking right multiplication
$$ a \simeq b \\
a - b \in I \\
(a - b) \times c \in I \\
a \times c - b \times c \in I \\
a \times c \simeq b \times c $$
A: Hint:
$r+I=r'+I$ means $r-r'\in I$. Similarly, $s-s'\in I$.
Now you have to show that, if $r-r'$ and $s-s'\in I$, then


*

*$(r+s)-(r'+s')\in I$. Observe that $(r+s)-(r'+s')=(r-r')+(s-s')$.

*$rs-r's'\in I$. Rewrite it as
$\;rs-r's'=(r-r')s+r's-r's'=\dots$
