Prove uniqueness in division algorithm I have been asked to prove the following:

Let $n \in \Bbb{N}$. For every $m \in \Bbb{Z}$ there are unique $q,r
> \in \Bbb{Z}$ such that $m=qn+r$ and $0 \leq r \leq n-1$.

I am mostly comfortable with proving existence, but am less so with proving uniqueness. I guess I don't really understand how to prove uniqueness for really anything.
The "proof" we are given for uniqueness goes as follows:
Suppose that
$$m=qn+r= q'n+r'.$$
Thus $(q'-q)n+r'-r=0.$
Proof:
We see that
$$(q'-q)n=-(r'-r) = r - r'.$$
If $r=r'$, then $q=q'$, then we are done.
Otherwise, by multiplying by $-1$, we may assume that $(r-r') \in \Bbb{N}$. But since $r-r'$ is a natural number, $q'-q$ is also a natural number. But then $(q'-q)n \geq n$. We conclude that $(r-r') \geq n$. On the other hand $r \leq n-1$, so $r-r' \leq n-1$. This is absurd. We conclude that $q=q'$ and $r=r'$, i.e., that $q$ and $r$ are unique.
 A: If $r\ne r'$, it's not restrictive to assume $r>r'$ (otherwise exchange their role, or repeat the argument).
Then $0<r-r'<n$. From $r-r'=q'n-qn=(q'-q)n$, we obtain
$$
0<(q'-q)n<n
$$
and, canceling $n$,
$$
0<q'-q<1
$$
which is impossible.
A: I think your difficulty is just a point of logic.  When we want to prove uniqueness of a thing, it's a standard technique to assume there are two different such things and then show that they are not different after all. 
Suppose I needed to prove that there is a unique integer between 3 and 5.  I would assume that there were two such integers, $x$ and $y$ such that $3<x<5$ and $3<y<5$.  Then I would try to argue that $x=y$, and probably that argument would be "by contradiction."  That is, I would assume $x \neq y$ and try to get an absurdity.  
Since one of $x$ and $y$ is bigger than the other (because we're assuming they're not equal) and since it makes no difference which one is bigger, we might as well assume (WLOG, as they say) that $x>y$.  One way to say this is $x-y$ is a natural number (because natural numbers are positive.)
Now that things are set up nicely, I can get on with my (silly) proof:
Since $3<x<5,$ we have $3-y < x-y <5-y$.  And since $y>3$, $5-y <5-3 =2$,
so $5-y\leq 1.$  Combining these inequalities gives us $x-y < 5-y \leq 1$,
so $x-y$ is a natural number less than one, absurd.  I conclude that $x-y$.
A: Consider this.
$N=qn+r$ and $M =pn+s ;0\le r <n$ and $0\le s < n $.  We want to figure out if and when $N=M $.
Case 1: $q > p $
The $N -(q-p)n = pn+r $
$N-(q-p)n +(s-r)=pn+s = M $
So $N = M + (q-p)n +r -s$
$ \ge M+(q-p) -s $
$\ge M+1*n -s $
$=M+(n-s) \ge M +1 >M $.
So $N\ne M $
Likewise 
Case 2: if $p > q $ then $M>N $ be the exact same reason.
So $N\ne M $
Case 3: I'd $p=q $ but $r \ne s $
Then $N=qn+r\ne qn+s = pn+s =M $
So $N\ne M $.
So the only possible way for $N=M $ will be if both $q=p $ and $r=s $.
So if 
$m=qn+r=q'n+r';0\le r <n;0\le r'<n$ (and $m=m $ of course), the only way this is possible is if $q=q'$ and $r=r'$.  In other words if $m=qn+r $ then $q,r $ are unique.
A: Here is a different rendering of it. To show that there is only one way to write
$$
m=qn+r
$$
with $0\leq r\leq n-1$, assume it could also be written as
$$
m=q'n+r'
$$
where again $0\leq r'\leq n-1$. Our goal is to show that those two are in fact indentical, meaning $(q,r)=(q',r')$. Assume without loss of generality that $r\leq r'$. Then we have
$$
qn+r=q'n+r'\iff (q-q')n=r'-r
$$
Now since $r\leq r'$ we have $0\leq r'-r\leq n-1$. Finally, the only multiple of $n$ in this interval is $0$. Thus
$$
(q-q')n=r'-r=0
$$
showing that $q=q'$ and $r=r'$. We have proven that the two are identical.
A: We are given that
$$qn+r=q'n+r'$$ with $0\le r,r'\le n-1$.
Then by subtraction
$$|q-q'|n\le|r-r'|\le n-1.$$
The only possibility is $q-q'=0$.
A: Here is my simple proof which took me way too long to figure out:
Suppose that
$$n=mq+r=mq'+r' \ ,\  q,q',r,r'\in \Bbb{Z} \\ \ 0\leq r< m \ \  , \ 0\leq r'< m $$
WLOG, suppose $q'>q$ then $$q'=q+a \ \ , \ a\in \Bbb{Z^+}$$So $$n=mq'+r'=m(q+a)+r'\\ =mq+ma+r'  \geq mq+ma\\ \geq mq+m > mq+r=n$$
Aha! It's a contradiction, thus $q'=q$ and of course proving $r'=r$ is trivial!
Explanation of the inequality chains above:
The first inequality involving $r'$ is true since $r'\geq0$ , the second one involving $ma$ is justified by the fact that $a\in \Bbb{Z^+}$ i.e. $a\geq 1$ so $ma\geq m$ , finally $r<m$ so $mq+m>mq+r$
