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 :)

$a\equiv a'$ and $b\equiv b'$ mod $n$ means that $a$ and $a'$ give the same remainder when divided by $n$, or equivalently, their difference $a-a'$ is a multiple of $n$. The same applies to $b$ and $b'$. So you have
$$
a-a'=kn;\qquad b-b'=rn
$$
for some integers $r$ and $k$. If you add the two equations above and arrange terms you get
$$
a+b-(a'+b')=(r+k)n
$$
and therefore, by the same token, $a+b\equiv a'+b'$ mod $n$.
