Is $f:\mathbb{Z}_{30}\longrightarrow\mathbb{Z}_{30}$ defined by $f([a])=[7a]$ well defined? To tell the truth, I'm not even sure what this means.
The professor gave an example saying that $\mathbb{Z}_m=\{[0],[1],[2],\dots,[m-1]\}$, and I sort of understand that.. but I have no idea what this means.
Well, I have maybe some idea, but I don't know how to determine if it's well defined or not. It looks like the function defines something to do with mod 7, but don't ask me to elaborate beyond that.

Edit:
OK, so my current understanding of the problem is as such:
$\mathbb{Z}_{30}$ is a set of 30 equivalence classes such that 0~30, 1~31, 2~32, etc... It breaks up the integers into groups such that $x\equiv x+30\ (mod\ 30)$.
$\mathbb{Z}_{30}=\{[x]\mid a\in\mathbb{Z},\ x\sim a\iff x\equiv a\ (mod\ 30)\}$
$f([a])=[7a]$ is eluding me a bit though... Is seems like it's saying that when you give $f$ an equivalence class, it transforms it to the equivalence class $7$ times the one given.. so $f([2])=[14]$, $f([29])=[203]$... But what do I do with that? $203\ mod\ 30=23$, which is not the same equivalence class as $[29]$.
 A: For a function to be well-defined, you need it to be defined on each element in the domain and to give a unique element in the range as its result.  Your function will be on the equivalence classes $\pmod {30}$ and the result will be an equivalence class, perhaps not the same one.  The threat to being well-defined would be if it took two representatives of the same class to representatives of different classes.  To take your example, you would think $f([29])=[23]$.  But $29, 59, 89,$ and $329$ are all in $[29]$.  If you multiply them all by $7$ and take the result $\pmod {30}$ you will get $23$, so we haven't found a counterexample.
To prove it, let $[n]$ be one of our equivalence classes.  An arbitrary element of this class is $n+30k$ for $k$ some integer.  Then that element is sent to $7n+210k$, which is in the same class as $7n$, so it doesn't matter which representative you take, you get the same result.
A: If you consider $\mathbb Z_{30}$ as a ring, you want to prove that the ring structure is maintained.  For example, is $f(2\cdot 3 + 8)\equiv f(2) \cdot f(3) + f(8) \pmod {30}?$  Let's try this one: $f(2 \cdot 3 + 8)=f(14)=98 \equiv 8 \pmod {30}.$ But $f(2) \cdot f(3) + f(8)=14\cdot 21+56=350 \equiv 20 \pmod {30}$.  Failure.  If you consider just the group structure of $\mathbb Z_{30}$ you will succeed.
