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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.

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  • $\begingroup$ 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. $\endgroup$ – Bill Dubuque Jan 3 '17 at 17:52
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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 $[a]_{n}$

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  • $\begingroup$ 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? $\endgroup$ – R. Johnson Jan 3 '17 at 16:28
  • $\begingroup$ 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) $\endgroup$ – Snow bunting Jan 3 '17 at 20:10
  • $\begingroup$ Thanks for the clarification. Very helpful indeed. $\endgroup$ – R. Johnson Jan 4 '17 at 1:35

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