Intereting Posts

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

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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.

If $\pmod m$ just means that you subtract or add multiples of $m$ until you end up in the range $0,1,2,…,m-1$ then the RHS is easily compared to the LHS by collecting multiples of $m$.

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