I define a binary operation $\oplus$ on $\mathbb{Z} _n$ as follows: $$[a] \oplus [b] = [a + b],$$ for all $[a], [b] \in \mathbb{Z} _n$. Then $(\mathbb{Z} _n, \oplus )$ is a finite cyclic group with order n.
This proposition in my notes says that the proof involves showing that the value of $[a] \oplus [b]$ is independent of the choice of equivalence class representatives.
Could someone please clarify to me why this is the case?
Kindest regards.
On
<<Could someone please clarify to me why this is the case?>>
Because you are trying to define an operation between classes ("$[a] \oplus [b]$") by using (arbitrary) representatives of them and the group operation ($a$, $b$ and "$+$" in the writing "$ = [a + b]$"). So, you want the result to be the same (i.e. the class $[a + b]$) if you pick any other representatives of the two classes-term (say $a'\in [a]$ and $b'\in [b]$).
This is the fee to pay to the impossibility to handle with the classes if not by means of their representatives.
Independence of choice of a representative means(as $\oplus$ is commutative):
$$\textrm{if} \; [a_1] = [a_2] \in \mathbb{Z} _n \; \textrm{ then, } \; [a_1] \oplus [b] = [a_2] \oplus [b] \; \; \; \forall \; [b] \in \mathbb{Z} _n$$
Example: See $\mathbb{Z} _5$, then $[1]=[6]$. If you calculate $$[1] \oplus [2] = [1+2] = [3]$$ and $$[6] \oplus [2] = [6+2]= [8]$$ you see that $[3] = [8]$, as we are in $\mathbb{Z} _5$. This means the result is the same, no matter which representative of the class $[1]= \{ ... ,-4, 1,6,11,... \} $ you chose.
Of course this was only an example, but this is the same way a proof will work.
You prove this, because you want your binary operation to be "well-defined".