Define the multiplication operation on the following equivalence class I am having trouble forming the proof for the following:
Let m>0. We can define operation * on the equivalence classes of m as follows:
[a]m*[b]m=[a*b]m  (The m's are subscripts) 

As an example, we are given the proof for the "+" operation:
[a]m+[b]m=[a+b]m
Proof: We have to show the fns we've claimed to define are well-defined. Suppose a1≡a2 (congruent mod m) and b1≡b2 (congruent mod m). We have to show a1+b1≡a2+b2 (congruent mod m).
By assumption m|a1−a2, and m|b1−b2. Therefore m|(a1−a2)+(b1−b2), which after reorganizing the right-hand side gives m|(a1+b1)−(a2+b2), i.e., a1+b1≡ma2+b2, as required. 

I understand what "congruent mod m" means, but I'm unsure what an equivalence class and equivalence relation are. Based on the given example, what I thought so far was I need to find steps that will lead to m|(a1*b1)-(a2*b2).
 A: What you're looking at here is the algebraic way of defining modular arithmetic.
You start with the integers, $\mathbb{Z}$.  And then, you define an equivalence relation: a way of declaring numbers 'equivalent' or 'not equivalent'.  An equivalence relation must satisfy a few properties: it has to be symmetric ($a\equiv b$ if and only if $b\equiv a$), it must be reflexive ($a\equiv a$ for all $a$), and it must be transitive ($a\equiv b$ and $b\equiv c$ implies $a\equiv c$). An equivalence relation breaks a set up into subsets of elements that are all mutually equivalent.
In this case, your equivalence relation is "differ by a multiple of $m$": $a\equiv b\pmod m$ means that $a-b=km$ for some $k\in\mathbb{Z}$. This breaks $\mathbb{Z}$ up into $m$ equivalence classes: $\{0,m,2m,\ldots\}$, $\{1,m+1,2m+1,\ldots\}$, and so on up through $\{m-1,m+m-1,2m+m-1,\ldots\}$.
We usually represent this equivalence classes by something like $[a]$, where $a$ is some specific member of the class.
Now, here's where it gets interesting. You can certainly try to define $[a]\cdot[b]=[a\cdot b]$.  But, what would happen if you had chosen a different representative for the equivalence classes? After all, $[a+m]=[a]$ and $[b+2m]=[b]$.  If we had chosen $a+m$ and $b+2m$ as our representatives, our 'rule' (we haven't yet proved it is a function) would give us $[(a+m)(b+2m)]$.  To declare this a function, it has to be the case that (among other things) $[(a+m)(b+2m)]=[ab]$.
For instance, if $m=5$, is it necessarily true that the equivalence class containing $2\cdot3$ is the same as the equivalence class containing $7\cdot13$?  In order for this operation to be well-defined, you must get the same equivalence class back no matter which representative you choose.
So, what you are asked to show is this: if $a_1\equiv a_2\pmod{m}$ and $b_1\equiv b_2\pmod{m}$, is it necessarily true that $a_1b_1\equiv a_2b_2\pmod{m}$?  
This boils down, of course, to exactly what you said you need to show: that $a_1b_1-a_2b_2$ is divisible by $m$.
As a first step, note that $a_1\equiv a_2\pmod{m}$ means there is a number $k$ so that $a_2=a_1+km$.  Similarly, there is $h$ so that $b_2=b_1+hm$.  Try plugging those into $a_1b_1-a_2b_2$, and see if you can find a way to show it is divisible by $m$.
A: Suppose , $x$ is in the class $a[m]$ and $y$ in the class $b[m]$.
Then, there are integers $u,v$ with $x=um+a$ and $y=vm+b$
Then, we have $x\cdot y=uvm^2+ubm+avm+ab\equiv ab\mod m$. 
Hence, $x\cdot y$ is in the class $(ab)[m]$
This shows that the multiplication $$a[m]\cdot b[m]=(ab)[m]$$ is well-defined.
A: Assume as before:
\begin{align}
a_1\equiv a_2 \pmod m\\
b_1\equiv b_2 \pmod m
\end{align}
We have to show:
$$a_1b_1\equiv a_2b_2 \pmod m$$
We have:
\begin{align}
m|(a_1-a_2)\\
m|(b_1-b_2)
\end{align}
and so:
$$m|(a_1-a_2)(b_1-b_2)$$
which gives:
$$m|(a_1b_1-a_1b_2-a_2b_1+a_2b_2)\tag{1}$$
Now $-a_1b_2+a_2b_2=-b_2(a_1-a_2)$, and hence is divisible by $m$.
Similarly for $-a_2b_1+a_2b_2$.
We can therefore conclude that:
$$m|(-a_1b_2+a_2b_2-a_2b_1+a_2b_2)$$
or:
$$m|(-a_1b_2-a_2b_1+2a_2b_2)$$
Subtract this from $(1)$ to give:
$$m|(a_1b_1-a_2b_2)$$
