What is the correct name for "complement of modulus" function, $f(n) =m - n\ \mathrm{mod}\ m$?

Question

What is the name of the function $f$, defined below?

$$f(n) = m - n\bmod m$$

or

$$n\bmod m + f(n) = m $$

If I were to make up a name, I would probably choose "complement modulus," and maybe $\mathrm{dom}$ or $\overline\bmod$ ("$\bmod$" with an overbar) for an abbreviation.

For example, the "$\bmod 4$" of the first eight natural numbers would be $(1, 2, 3, 0, 1, 2, 3, 0)$; while the "$\mathrm{dom}\ 4$" of the same would be $(3, 2, 1, 4, 3, 2, 1, 4)$.

@conditionalMethod "$\mathrm{round}$", as in "round to the nearest integer?" That's not what this is. — Ana Nimbus, Oct 23 '19 at 17:51
There are many rounds. When you round you remove the remainder modulo $1$. There are many versions of this (nearest integer, away from zero, ...) depending on how the 'remainder modulo 1' is defined. You can round to the nearest 10th, in which case you remove the remainder modulo $10$. etc. In your case, the only thing changing is the modulo. — conditionalMethod, Oct 23 '19 at 17:59
@conditionalMethod see edited version. Sorry for the typos in the original. — Ana Nimbus, Oct 23 '19 at 18:01
Looks like the additive inverse of $n$ modulo $m$, where the residue representatives are ${1,2,...,m}$, instead of the more common ${0,1,...,m-1}$ — J. W. Tanner, Oct 23 '19 at 18:02
@J.W.Tanner I think that is close. Elsewhere I found "modular additive inverse." I suppose that is equivalent to what I am asking, considering n mod n = 0 . — Ana Nimbus, Oct 23 '19 at 18:12
This is not a complement of a modulus; it's a complement relative to a modulus. — Michael Hardy, Oct 23 '19 at 18:23
@MichaelHardy would you elaborate on the terminology "of" vs "relative to?" — Ana Nimbus, Oct 23 '19 at 19:48
@conditionalMethod : This is not about making something close to something. — Michael Hardy, Oct 23 '19 at 20:27
@conditionalMethod : ok, I've read the original question. As I said, it's not about making something close to something. If the modulus is $6,$ then the operation is this: $$ \begin{array}{c} 0 \mapsto 0 \ 1 \mapsto 5 \ 2 \mapsto4 \ 3 \mapsto 3 \ 4\mapsto2 \ 5 \mapsto 1 \end{array} $$ How is that rounding? $\qquad$ — Michael Hardy, Oct 23 '19 at 20:36
@MichaelHardy Take your time thinking about it. You will get to it. If not keep fixing LaTeX around. — conditionalMethod, Oct 23 '19 at 20:42
Related question: Name for integer "quotient" rounded up (ceiling) instead of down (floor), and the "remainder" we'd subtract? — Silverfish, Mar 18 '23 at 21:18

score 3 · Accepted Answer · answered Oct 23 '19 at 18:16

3

What you are describing is simply the additive inverse of $n$ modulo $m$, where the residue representatives are $\{1,2,...,m\}$ rather than the more common $\{0,1,...,m-1\}$.

answered Oct 23 '19 at 18:16

J. W. Tanner

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J.G. · Answer 2 · 2019-10-23T18:00:58.280

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I don't know a name for it, but $f(n)=m\lfloor\frac{n}{m}\rfloor$. Your unlikely to get anything better than @conditionalmethod's suggestion, just as $x\mapsto2\exp\frac{x}{2}$ is unnamed.

Edit: now we've clarified $m-n+m\lfloor\frac{n}{m}\rfloor$ was meant, I don't think that has a specific name either, but I like what you suggested for it.

edited Oct 23 '19 at 18:00

answered Oct 23 '19 at 17:43

J.G.

118,053

I don't think so. Consider very large $n$ and $m = 2$. – Ana Nimbus Oct 23 '19 at 17:46
@AnaNimbus It looks based on your example that you really wanted $\color{blue}{m}-n\mod m$, i.e. $m-n+m\lfloor\frac{n}{m}\rfloor$. – J.G. Oct 23 '19 at 17:50
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yes. Thank you. edited question to reflect this. – Ana Nimbus Oct 23 '19 at 17:57

What is the correct name for "complement of modulus" function, $f(n) =m - n\ \mathrm{mod}\ m$?

2 Answers2

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