Can we have modulo a negative number?

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Main Question or Discussion Point

Why can't we have modulo negative number? I have never seen this.

PeroK
Homework Helper
Gold Member
Why can't we have modulo negative number? I have never seen this.
You would simply have to define what you mean by it.

In terms of factorisation and remainders, everything can be seen with a positive divisor, so I wouldn't expect any new insights.

The same goes for allowing negative numbers to be prime.

fresh_42
Mentor
Why can't we have modulo negative number? I have never seen this.
Because a negative number $-n=(-1)\cdot n$ differs from its positive counterpart only by a unit, $\varepsilon=-1$. So the ideals $n\mathbb{Z}$ and $\varepsilon n\mathbb{Z}$ are the same. This implies that the factor rings $\mathbb{Z}[x]/n\mathbb{Z} = \mathbb{Z}_n = \mathbb{Z}[x]/\varepsilon n\mathbb{Z}$ are equal, too. And as it makes no difference, it is a matter of convenience to choose the positive version. Otherwise the calculations were full of unnecessary signs.

FactChecker
Gold Member
I don't see any conceptual difference between modulo of a positive versus negative integer. I am not sure if all computer languages would implement it. Perl does:

Integers 0..9 mod -3:
0 mod -3 = 0
1 mod -3 = -2
2 mod -3 = -1
3 mod -3 = 0
4 mod -3 = -2
5 mod -3 = -1
6 mod -3 = 0
7 mod -3 = -2
8 mod -3 = -1
9 mod -3 = 0

fresh_42
Mentor
I don't see any conceptual difference between modulo of a positive versus negative integer. I am not sure if all computer languages would implement it. Perl does:

Integers 0..9 mod -3:
0 mod -3 = 0
1 mod -3 = -2
2 mod -3 = -1
3 mod -3 = 0
4 mod -3 = -2
5 mod -3 = -1
6 mod -3 = 0
7 mod -3 = -2
8 mod -3 = -1
9 mod -3 = 0
I had been impressed, if it would have calculated the results in the standard representation, i.e. $[0],[1],[2]$. This way its more an incidental result.

The essential point is really, that "up to units" is often dropped in the definitions, be it as sloppiness or because it is self-evident. As @PeroK mentioned
The same goes for allowing negative numbers to be prime.
the situation with primes is the same: "up to units", because $-3$ is as prime as $3$ is. And that $1$ isn't a prime, is only because the definition starts with "A number is prime, if it is no unit and ..." It makes absolute sense to exclude units. E.g. nobody ever asked, whether the fundamental theorem of arithmetic is wrong, since we can always add units as many as we want to: $6=2\cdot 3= 1\cdot 2\cdot 3= (-1)\cdot (-1)\cdot 2 \cdot 3 = 1\cdot 1\cdot 3\cdot 2\cdot 1 =\ldots$ Units simply don't change the game, so why bother them?

FactChecker
Gold Member
Modular arithmetic sets up equivalences between negative numbers (the additive inverse) and positive numbers. -3 mod 3 = 0; -2 mod 3 = 1; -1 mod 3 = 2; etc. Modulo 3 and modulo -3 works out to be consistant:

-4 mod 3 = 2
-3 mod 3 = 0
-2 mod 3 = 1
-1 mod 3 = 2
0 mod 3 = 0
1 mod 3 = 1
2 mod 3 = 2
3 mod 3 = 0
4 mod 3 = 1

-4 mod -3 = -1
-3 mod -3 = 0
-2 mod -3 = -2
-1 mod -3 = -1
0 mod -3 = 0
1 mod -3 = -2
2 mod -3 = -1
3 mod -3 = 0
4 mod -3 = -2

Last edited:
fresh_42
Mentor
Sure. But this isn't evidence for a smart programming of Perl as you implicitly suggested, but simply a logic outcome of ordinary arithmetic. An automated conversion into standard results would have been smart. The above is more or less an unintended result of a correct division.

FactChecker
Gold Member
Sure. But this isn't evidence for a smart programming of Perl as you implicitly suggested, but simply a logic outcome of ordinary arithmetic. An automated conversion into standard results would have been smart. The above is more or less an unintended result of a correct division.
there is no reason for them to apologize for a "happy coincidence" that turns out to be mathematically correct.

fresh_42
Mentor
there is no reason for them to apologize for a "happy coincidence" that turns out to be mathematically correct.
Yes, but that's not the point. The point is, that similar to what led to the OP's question, the programmers completely ignored the role of units. In this sense, it is rather off-topic here than a support for the equation $n\mathbb{Z}=-n\mathbb{Z}$. The example brought you no millimeter closer to the answer of the question. It only shows that you can do calculate with negative numbers, not why they are usually ignored in books.

FactChecker
Gold Member
IMO, in any algebra, -x is the additive inverse of x. x + (-x) = 0. I see no issue with units.

lavinia
Gold Member
Why can't we have modulo negative number? I have never seen this.
You can do arithmetic using negative bases for instance arithmetic base -2.

fresh_42
Mentor
IMO, in any algebra, -x is the additive inverse of x. x + (-x) = 0. I see no issue with units.
Because you confuse addition with multiplication. We are talking about multiplication here, since the module is $z \,\,\cdot\,\, \mathbb{Z}\,$.
No ring, no question.

PeroK
Homework Helper
Gold Member
Given that it's the 21st century, perhaps mathematicians should be thinking about equal rights for negative numbers! Why should $-5$, for example, be denied the right to be called a prime number, just because it was born on the wrong side of the number line?

FactChecker
Gold Member
I guess I don't see the problem. It seems natural and uncomplicated to extend the equivalence classes in the non-negative integers defined by mod 3 ({0,3,6,9,...}, {1,4,7,10,...},{2,5,8,11,...}) to include the negative integers ({...,-9,-6,-3,0,3,6,9,...}, {...,-8,-5,-2,1,4,7,10,...},{...,-7,-4,-1,2,5,8,11,...}). Does this cause a problem when multiplication is introduced? Off-hand, I don't see it.

fresh_42
Mentor
The female version of many professions is obtained by a suffix "in" in German, e.g. a female carpenter would be a carpenterin. This lead to thousandths of texts which are written with constructions like "... we are looking for a carpenter/in with at least five years of experience ..."

I see me writing: Let $p/-p$ be a prime. ...

However, in rings which have more than these two units, things will become quite disturbing: Let $p/\varepsilon\cdot p$ be a prime, where $\varepsilon$ is a unit, ...

fresh_42
Mentor
I guess I don't see the problem. It seems natural and uncomplicated to extend the equivalence classes in the non-negative integers defined by mod 3 ({0,3,6,9,...}, {1,4,7,10,...},{2,5,8,11,...}) to include the negative integers ({...,-9,-6,-3,0,3,6,9,...}, {...,-8,-5,-2,1,4,7,10,...},{...,-7,-4,-1,2,5,8,11,...}). Does this cause a problem when multiplication is introduced? Off-hand, I don't see it.
No, it does not. It simply doesn't contribute anything to the answer of the question: Why are negative numbers ignored? It only shows, that the question has a justification, but nothing as to why this is the case.

I had the same problem as the OP when I saw -1(MOD 10) on the OEIS site.
Centred 10-gonal numbers
A062786
no idea why the server cannot now find the site

FactChecker
Gold Member
No, it does not. It simply doesn't contribute anything to the answer of the question: Why are negative numbers ignored? It only shows, that the question has a justification, but nothing as to why this is the case.
The original OP question was "Why can't we have modulo negative numbers?". The answer is that we can. It is a natural extension of the concept to include negative integers. Then the OP asks why he has never seen it. The answer to that is that there is not much benefit.

mathwonk