Quotient group with functions and relations question 
For the set $\mathbb Z/5 \mathbb Z $ (the quotient group  of $\mathbb Z$ with the relation R that is defined by $xRy$ if $5|y-x$)
We'll define the following operations (both are $\cdot, +$ functions $(\mathbb Z/5 \mathbb Z) \times (\mathbb Z/5 \mathbb Z) \to \mathbb Z/5 \mathbb Z$):
$$[x]+[y]=[x+y] \\ [x]\cdot[y]=[x\cdot y]$$
  
  
*
  
*Show that these operations are well defined, i.e they aren't dependant on representetives for the qoutient set. 
  
*Prove that for all $a,b,c \in \Bbb Z$: $$[a]\cdot ([b]+[c])=[a]\cdot [b]+[a]\cdot[c]$$
  
*Calculate the product of the element $[6789]$ in itself $12345$ times ($[6789]\cdot...\cdot[6789]$).
  



*

*Well for showing the operations are well defined I don't suppose it's like T. Bongers do it here, I don't really know how to work with these functions...

*This is really similar to showing associativity but I guess we need to use the relation here somehow, I'm not really sure what to do here.

*I assume we need to use both multiplication and addition functions above to solve it and finding a $(y+x)\cdot 5=6789$ I guess $6785/5+4$ can work ? But then multiplying it $12345$ times, maybe take $(6785/5+4)=x$ and $12345=y$  ?
Please share your thoughts on how to solve this.
Thanks.
 A: For the first question, you have to prove that $[x]+[y]=[x']+[ y]$, for all $x,x'$ with $[x]=[x']$. Same for $[x]\cdot[y]=[x'][y]$.
For the second question, i think that is just part of the definition of a ring. You could also just bruteforce this part by checking it for alle equivalency classes $[a]$, $[b]$ and $[c]$
The third part: You know that $[6789]=[4]$, so you just have to evaluate $[4]\cdots [4]$. You can calculate the order of $[4]$ in the multiplicative group $\mathbb Z/5\mathbb Z$. It is $2$. (Because $[4]\cdot[4]=[16]=[1]$ and $[1]$ is the identity. So $[4]^{12345}=[4]^1=[4]$.
EDIT
The group $\mathbb Z/5\mathbb Z$ consists of the following set of equivalency classes: $\{[0],[1],[2],[3],[4]\}$. (Two elements of $\mathbb Z$ $a$ and $b$ are equivalent when $a-b\in5\mathbb Z$. You can see that elements $a$, $b$ with $a\equiv b\mod 5$ are in the same equivalency class.)
To show that addition is 'well defined', you have to show that $[a']+[b]=[a]+[b]=[a+b]=[a'+b]$ for all $[a]=[a']$, i.e. $a\equiv a'\mod 5$. This is not hard to see, because when $[a]=[a']$, it is clear that $[a]+[b]=[a']+[b]$. Also, $a'+b\equiv a+b\mod 5$, so $[a'+b]=[a+b]$.
