Modulo of a Sum/Summation

Question

Let me define $a~mod~N \triangleq a/N$. It is well known that,

$(a+b)/N = (a/N + b/N)/N$.

Can some one clarify how to translate this to simplify, $(a + b + c)/N$?

From (Link 1) and (Link 2), it is said that,

$(a + b + c)/N = (a/N + b/N + c/N)/N$ but this is not clear to me. From the case of $a + b$, of I set $b = b +c$, I should get,

$$\begin{align*} \left( {a + b + c} \right)/N & = \left( {a/N + (b + c)/N} \right)/N \hfill \\ & = \left( {a/N + \left( {b/N + c/N} \right)/N} \right)/N. \hfill \\ \end{align*}$$

This does not explain the linked results.

Can this also be used for functions $k_p(x)$,

$$\left( {\sum\limits_{p = 1}^P {{k_p}\left( x \right)} } \right)/N.$$

Your help is appreciated.

Using fraction notation for mod will cause much confusion. Better to stick to standard notation. These and related results are trivial if you first prove them as congruences, then do a final mod to get the remainder. — 2017-01-03

user id:403752 156 · Accepted Answer

Your result is actually true as well.

Their result uses that the "modulo" operation squares to itself, i.e. (a/N) = (a/N)/N for all a.

You should think of modulo in a different way.

You have your normal ring of numbers (say $\mathbb{Z}$ or $\mathbb{R}$) and then you have the modulo operation $\varepsilon_{n}: \mathbb{Z} \to \mathbb{Z}_{n}= \{[0]_{n}, [1]_{n}, [2]_{n}, ..., [n-1]_{n}\}$. $\mathbb{Z}_{n}$ is a new ring with only n elements. Each element is an equivalence class $[a]_{n}$ of elements $a \in \mathbb{Z}$ which have the same reminder dividing by n. So $[a]_{n}$ and $[a+n]_{n}$ are actually the same element in the ring $\mathbb{Z}_{n} = \{[0]_{n}, [1]_{n}, [2]_{n}, ..., [n-1]_{n}\}$. What your "/N" is is essentially applying $\varepsilon$ to get the equivalence class modulo n and then applying the inclusion $i: \mathbb{Z}_{n} \to \mathbb{Z}$ by choosing a representative, i.e. the smallest non-negative representative. (Note that this inclusion depends of the chose of a repr.)

Upshot: You can take as many modulo as you want as long as you take one at the end. While being in the modulo ring adding and subtracting n to the repr. a doesn't change the object $[a]_{n}$

Thanks for your help. But I still don't understand something: $(a+b)/N = (a/N + b/N)/N$. I can not simplify this further to: $(a/N + b/N)/N = a/N + b/N)$. Right? — 2017-01-03
If a/N for you is a number in $\mathbb{Z}$ (or $\mathbb{R}$) you can't. For example N=5, a=9, b=3. The normal modulo-operation you'll find in informatics would give $(a/N + b/N)/N = (4 + 3)/N = 7/N = 2 \ne 7 = 4 + 3 = (a/N + b/N)$. However, the correct module calc. in $\mathbb{Z}_{N}$ would be: $[a+b]_{N} = [12]_{5} = [2]_{5} = [7]_{5} = [4]_{5} + [3]_{5} = [9]_{5} + [3]_{5}$, which is a much clearer way of dealing with it. (note that these are classes and the number inside is just any representative of this class) — 2017-01-03

Modulo of a Sum/Summation

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