Proof of $[(a \; \text{mod} \; n)+(b \; \text{mod} \; n)] \equiv (a+b)\; \text{mod}\; n$ I'm currently self-studying a course in cryptography, and i understand the importance of understanding modular arithmetic fully. I have proved many operations on modular arithmetic, but one i am stuck on is why:
$[(a \; \text{mod} \; n)+(b \; \text{mod} \; n)]=(a+b)\; \text{mod}\; n$, and the full proof of this.
I have had a few ideas on it, but not proved it fully.It may be obvious, but i am only $15$.
Thanks for any help.
 A: Let
$$a=q_an+r_a$$
$$b=q_bn+r_b$$
for quotients $q_a,q_b$ and remainders $0\le r_a,r_b<n$ of $a,b$ modulo $n$.
Then
$$\begin{align}
a+b&=(q_a+q_b)n+(r_a+r_b)\\
&=\left(q_a+q_b+\delta\right)n
+\left(r_a+r_b-\delta n\right)
\end{align}$$
for $$\delta=\left\lfloor\frac{r_a+r_b}{n}\right\rfloor$$ where
$\lfloor x\rfloor$ is the greatest integer (less than or equal to $x$) or
floor function
and $$\left(r_a+r_b-\delta n\right)=(a+b)\text{ mod }n.$$
Now equality only holds if $\delta=0$:
it is not true in general that $r_a+r_b=r_a+r_b-\delta n$.
For example, try $a=b=1,n=2$.
What does hold is congruence modulo $n$:
$$(a \; \text{mod} \; n)+(b \; \text{mod} \; n)
\equiv (a+b)\; \pmod n$$
which we have just proved.
In mathematics, we say
$$a\equiv b\pmod n \qquad\iff\qquad n | b-a$$
i.e. if their difference is divisible by $n$, but in some computer science contexts,
$$r_a=a\text{ mod }n$$
means that $r_a$ is the remainder on dividing $a$ by $n$ using the
division algorithm,
with either $0\le r_a<n$ or sometimes $-\lfloor\frac n2\rfloor\le r_a<\lfloor\frac n2\rfloor$ or even $-n<r_a<n$, which can be a source of confusion.
A: Hint:
It doesn't have to be proved: it's a definition, which is without ambiguity once you've proved this:

If $a\equiv a'\mod n$, and $b\equiv b'\mod n$, then $a+b\equiv a'+b'\mod n$

which results from observing that $a+b\bmod n=(a\bmod n+b\bmod n)\bmod n$.
