How Can I prove (a+b) mod m = (a mod m) + (b mod m)) mod m? [closed]

modular arithmetic

How Can I prove (a+b) mod m = (a mod m) + (b mod m)) mod m ?

Let $a\mod m= a_1$ and $b\mod m = b_1$.

Then

$$\begin{cases}m\mid a-a_1\\m\mid b-b_1\end{cases}\implies m\mid a+b-(a_1+b_1)$$

$$\iff a+b\equiv a_1+b_1\pmod{m}\iff (a+b)\mod m=(a_1+b_1)\mod m\\\iff (a+b)\mod m =((a\mod m)+(b\mod m))\mod m\ \ \ \square$$

Using $\,\ c\equiv d\pmod m\color{#c00}\iff \bar c = \bar d,\ $ where $\,\ \bar n := (n\bmod m)\ $

${\rm mod}\ m\!:\,\ a\color{#c00}\equiv \bar a$
$\qquad\qquad\, b\color{#c00}\equiv \bar b$
$\ \, \Rightarrow\,\ a+b\equiv \bar a + \bar b\,\ $ by the $ $ Congruence Sum Rule

$\color{#c00}\Rightarrow\, (a+b)\ {\rm mod}\ m\, \equiv\, (\bar a + \bar b)\ {\rm mod}\ m,\,\ $ which is the titled claim.

Remark $ $ Generally, as above, to prove an identity about mod as an operator it is usually easiest to first convert it into the more flexible congruence form, prove it using congruences, then convert back to operator form.