I had to show that a' + b' = a + b mod n, when a mod n = a' and b mod n = b' I had to show the following:
For $\,\, n,a,b,a',b' \in N$;
if: $ a'\equiv a\mod{n}$ and $ b'\equiv b\mod{n} $
than $ a' + b' = \,(a+b)\mod{n}$
My try:
$\frac{a}{n} = k +a'$ , where $k \in N $ and $k$ is divisable by $n$.
$\frac{b}{n} = p + b'$ , where $p \in N$ and $p$ is divisable by $n$.
(NOTE: I think (please correct me if I am wrong) that the fact that $a',b'$ are both named  to be solutions to a modulor function, that both are not divisable by n. (or do I need to explicitly state this again))
$\frac{a+b}{n} = \, (k  + a')+(p+b')$
since $p$ and $k$ both are divisable by $n$; $(k + p)$ can be written as $n*h, h\in N$
$\Longrightarrow \frac{a+b}{n} = \, (n*h + (  a'+b')) $
$\Longrightarrow (a+b)\mod{n} = a' + b'$
It would be very helpful when you could say if the logical reasoning is correct, or where its flawed, and especially help me with formulations. So if the formulation of a certain part of how I tried to show what was asked for is bad or improveable, please give me as many advices as you are willing to share :)

$\dfrac{a}{n} = k +a'$  is wrong, the correct relation is
$$a=kn+a',0\le a'<n.$$
Then
$$b=jn+b',0\le b'<n$$
and by addition
$$a+b=(k+j)n+a'+b'$$
so that
$$a'+b'\equiv a+b\mod n.$$

Note that we don't necessarily have
$$a'+b'=(a+b)\bmod n$$ because it may by that $a'+b'\ge n.$
