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.

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 $a+b\bmod n=(a\bmod n+b\bmod n)\bmod n$.
