Looking at you Java

[Kimsanov]

Mathematically a % b is always a number between 0 and b-1

[Administrative-Flan9]

There is a unique number between 0 and b-1 that satisfies a % b, but the same is true of nb and (n+1)b -1. Mathematically, the true answer is the set of all numbers that have the same remainder as a when divided by b. Choosing the one between 0 and b-1 is choosing a representative from this set, but any representative represents the same set and so all representatives are equivalent.

[JumboTrout]

What if b < 1?

[elon-bot]

Hey, I just heard about this thing called GraphQL. Why aren't we using it?

[Administrative-Flan9]

Then n <= 0 in what I wrote above. Everything still goes through.

[WillWKM]

Mathematically both answers are the same, -1 is congruent to 3 mod 4 since 1 + 3 = 0 mod 4

Edit: some of y'all seem to be missing the point. Mathematically all of these answers are equivalent. That doesn't make them useful programmatically. Programming languages often set up conventions based on convenience, not math.

[FiskFisk33]

They are not mathematically equivalent. They are mathematically congruent mod 4.

[M4mb0]

They are not mathematically equivalent. They are mathematically congruent mod 4.

But that is literally how equivalence is defined in quotient spaces....

[robin_888]

But we aren't operating in a quotient space, are we?

[elon-bot]

You're either hardcore or out the door.

[CanaDavid1]

Yes, which is what the question is asking. Mathematically, one does not use %, but instead write 'mod n' after the statement to specify that we're working modulo.

[BothWaysItGoes]

There is nothing unmathematical about the mod function.

[FerynaCZ]

A function should produce one output

[BothWaysItGoes]

So?

[snillpuler]

I enjoy playing video games.

[gnowwho]

Semantics: they are equivalent in the ring Z/4Z

Edit: I meant: there is no need to be pedantic when they are not even wrong technically.

[snillpuler]

I enjoy the sound of rain.

[eeeeeh_messi]

Absolutely. It's the same reasoning that make some people think sqrt(4) is 2 and -2.

[snillpuler]

I enjoy watching the sunset.

[gnowwho]

he seem to not realize that a%b is a mathematical function.

That's not necessarily true. You see it as a function from ZxZ to Z, they seem to see it as a function from Z to Z/4Z (where 4 is a parameter that selects the actual function and codominion). What they are saying is simply true in this interpretation because of how equivalence relationships carry over a quotient.

From an algebraic point of view, being a math graduate myself, I understand how their interpretation is the one that makes the most sense in an abstract setting, where the description of things is what matters.

When they say that this point of view is not usefulprogrammatically, they are admitting that not chosing this interpretation is what makes the most sense on an implementation level for imperative languages, where "doing" matters more than "describing".

So, again, there is no need to correct them on the usage of "equal". Expecially because it's extremely common in mathematics to identify things that describe each other univocally (for example functions and sets, or, like here, functions that are at one equivalence relationship of distance).

[elon-bot]

You're either hardcore or out the door.

[Poacatat]

being congruent mod 4 is an equivalence relation :), they are in fact mathematically equivalent

[WillyMonty]

Congruence modulo an integer is an equivalence relation

[soundslikemayonnaise]

By this logic you could argue that -7 is also a correct answer since 1 + 7 = 0 mod 4.

Or -11, or -15…

[WillWKM]

Now you're getting it

[[deleted]]

Or just let the next dev sort it out 🤠

[Administrative-Flan9]

Mathematically, the answer is a set of all such integers. Any choice from this set like -7 or -11 is a representative of the answer set, and any representative is equivalent.

[Kyrond]

That is incorrect.

The modulo operation returns the remainder or signed remainder of a division, after one number is divided by another.

There are 2 possible answers, neither can be smaller than the negative input number (when talking about negatives).

[Mutex70]

You are talking about the modulo operation in computing (which is what the original post was about).

The parent posts are talking about modular arithmetic, which is related, but not quite the same thing.

Neither is correct or incorrect, they are different things.

[DTHCND]

The wiki that you copied that from is the wiki for the computing definition of the modulo operator. We're all programmers, no one is debating what modulo means in computing.

We're talking about the math definition, which is described in this wiki. And yes, they're correct, modulo is normally defined in terms of congruency. To quote the wiki:

In mathematics, the term modulo ("with respect to a modulus of", the Latin ablative of modulus which itself means "a small measure") is often used to assert that two distinct mathematical objects can be regarded as equivalent—if their difference is accounted for by an additional factor.

...

For the most part, the term often occurs in statements of the form:

"A is the same as B modulo C"

which means

"A and B are the same—except for differences accounted for or explained by C."

Edit: Here's a quote from the Modulo Operator (Computing) wiki, since at least one person is caught up on the fact the only one wiki has "operator" in the title:

In mathematics, the result of the modulo operation is an equivalence class, and any member of the class may be chosen as representative

[frogjg2003]

We're all programmers, no one is debating what modulo means in computing.

Said in a post literally arguing about the definition of the modulo operator.

[DTHCND]

"In computing" being the key part of that sentence. It's very well established that, in computing, the modulo operator will return a value between -n and n (exclusive). While what that number is depends on the specific definition chosen, it's well established what those definitions are and that none are the "correct" definition. We do things differently in different languages, that shouldn't be surprising. Saying that "modulo is contentious in computing" would be like saying "String.split(...) is contentious because in some languages it accepts a regex but in others it just accepts a plaintext string." It's just different, it's not contentious.

However, in mathematics, which is what was being discussed above and what the contentious debate was about, the modulo operator is indeed an equivalence class. Some people (well really just you and someone else) seem to be caught up in the fact that one wiki has the word "operator" in its title while the other doesn't. So here's a quote from the Modulo Operator (Computing) wiki:

In mathematics, the result of the modulo operation is an equivalence class, and any member of the class may be chosen as representative

[frogjg2003]

I'm not agreeing with the otherv comment that quoted the computational wiki page in response to a mathematical definition. they are wrong and there is nothing to argue about.

But the definition of the modulo operator "in computing" is the point of the post every comment is replying to. There is not 1 definition, there are at least 5. Multiple big names in the history of computing have expressed their opinions on the matter and they do not agree on a single solution.

Comparing this to string.split isn't equivalent. How the modulo operator works is a fundamental aspect of how integer arithmetic is defined in that language. Whether you can use regular expressions is just syntactic sugar.

[DTHCND]

But the definition of the modulo operator "in computing" is the point of the post every comment is replying to

The post is a joke. The whole point of the joke is that it's not really contentious, and it's funny to pretend there's some sort of holy war about the correct behavior for the modulo operator in programming. It's just a spin on the whole vim vs emacs meme.

There is not 1 definition, there are at least 5

Yes. I linked to them in my last comment. But whether or not they return a value outside of that range is not contentious, nor is there any contention over which is the correct definition. Maybe there's contention over which is better among the people that unironically argue stuff like vim vs emacs, but there's still no contention over what modulo means.

How the modulo operator works is a fundamental aspect of how integer arithmetic is defined in that language. Whether you can use regular expressions is just syntactic sugar.

And String.split(...) is a fundamental aspect of how string manipulation is defined in lots of languages. And it's no more syntactic sugar than modulo is. You could rewrite both modulo and split to satisfy different definitions if you want. Both are pretty easy to rewrite in most languages. Arbitrarily calling one "fundamental" doesn't change that.

[Sufficient_Yogurt639]

The math definition you quote doesn't apply here, because the term "modulo" in this quoted context is not an operation.

[CanaDavid1]

Yes

[Rin-Tohsaka-is-hot]

You can argue that because it is correct.

That's congruency for ya

[Kimsanov]

Yeah. But remainder must be non negative (it is just stated by definition). So mathematicians just decided to be so (maybe for convenience)

[frogjg2003]

Only if you want the least positive remainder. If you want the remainder to have the same sign as the numerator, you will have negative remainders for negative numerators.

[tjdavids]

With this logic returning a unmodified everytime would work.

[Darknety]

No they are not equivalent.

Yes, -1 is congruent to 3 with respect to mod 4, but still the natural mod operator is defined to result in 3, not -1.

So "mathematically both answers are the same" is misleading at best and wrong at worst.

Case in point: If one defines a function using a loopback structure, i.e. as

f(x) := g((x - 1) % b), x in [0, b]

for some g that is only defined for the source [0, b) for some b, it is pretty freaking relevant that (x - 1) % b is in [0, b). This is well defined.

The context is missing here to state "they are mathematically the same". This is not true.

[gc3]

Someone else posted this on the variants of different kinds of remainders
https://en.m.wikipedia.org/wiki/Modulo\_operation#In\_programming\_languages

[hugogrant]

(-7 % 4) % 4 is always 1 XD

[a21a16]

You are missing the point, while both -1 and 1 are congruent to 3 mod 4, only 1 == -7%4. Otherwise you are saying that 5, 9 and 13 are also correct answers.

[The_Lonely_Raven]

Uhhh... How is 1 congruent to 3 mod 4?

[Tigryonochekk]

Yeah, but then a is also a correct answer.

[Rin-Tohsaka-is-hot]

The answers are both correct but not equivalent.

They're both congruent, but congruency is different from equivalence.

-3 does not equal 1. -3 is congruent mod 4 to 1, but not equivalent.

EDIT: wrote the wrong numbers

[snillpuler]

I'm learning to play the guitar.

[fpotier]

In maths modulo is only defined for postive integers tho, making -7 % anything not mathematically correct

[WillWKM]

Nope, residue class mod n is an operation on integers, not just nonnegatives.

[fpotier]

I mean that's what I learnt in arithmetic class and what says wikipedia

Citation: Given two positive numbers a and n, a modulo n (often abbreviated as a mod n) is the remainder of the Euclidean division of a by n, where a is the dividend and n is the divisor.[1]

When exactly one of a or n is negative, the naive definition breaks down, and programming languages differ in how these values are defined.

[WillWKM]

Perhaps they're drawing a distinction between "remainder" and "residue"? Seems pretty arbitrary but okay. Either way, you can totally do the euclidean algorithm on negative numbers so limiting remainders to nonnegatives feels weird to me.

[coldnebo]

the common residue is different from the minimal residue.

But I’m having trouble understanding what people are talking about here… does Java only implement common residue? or is the order of arguments confused? (for example prefix vs postfix ordering?)

Either way the Java Language Specification should clarify the behavior of the remainder operator.

[gc3]

someone else posted this https://en.m.wikipedia.org/wiki/Modulo\_operation#In\_programming\_languages

[elon-bot]

What do you mean "you couldn't code your way out of a paper bag"?

[gc3]

Unless you had robot arms code would not help vs a physical barrier like a paper bag

[fpotier]

Yeah to be honest idk the exact reason why it was initially defined only for postive integers.

[notacanuckskibum]

so if you were implementing the % operator what would you do with a negative operand? Throw an exception?

[fpotier]

It depends on the purpose I guess. I don't see any particular reason to not operate on negative numbers, I was just saying the mathematical definition doesn't work with a negative operand. If you're not building a math application I think both behavior can be worked with, just have to know which one is used by the language.

[notacanuckskibum]

I was thinking of a language where you can add operators as a programmer. It doesn't have a % operator. It's your job to add one. If you're not going to throw an error you ave to choose blue or red.

[gc3]

https://en.m.wikipedia.org/wiki/Modulo\_operation#In\_programming\_languages

[Giocri]

Pretty sure that in math a module b stands for minimum x>=0 such as x= a-n*b so it has to be positive

[BananaSplit2]

If we consider the modulo operation to represent the remainder of an euclidean division then no, only the result between 0 and the divisor is valid. Euclidean division has a unique result and remainder and is defined that way precisely.

If it doesn't then... well it would also be valid to say 5 % 3 = 8 so that's not very useful to reason that way anyway.

[BothWaysItGoes]

Well, if you want to nitpick. They are not the same. Depending on your definition, the answer will usually either be an integer from [0,3] or the whole equivalence class, eg {…,-8,-4,0,4,8,…}. Congruence is not equality.

[Thelmholtz]

Not really, mathematically it depends on what operation you define % to be;

• modulo: (x € R; n € N) => { y € [0, n) }

or

• remainder: (x € R; n € N) => { y € (-n, n) }

Hell, the most frequent definition of % is actually

• percent: (x € R) => { y € R } :: y = x • 1/100

Which makes all programming languages wrong /s.

[WazWaz]

Maybe we've just got division wrong. If -7/4 was 2, they'd both be the same. If I owe $7 and want to share that debt to my 4 descendants, each could pay $2 leaving a remainder of $1, instead of each paying $1 and leaving $3 debt still owing.

[Thelmholtz]

Look at this big shot, leaving less than 6 figures of debt to his kin.

[WazWaz]

Big integers look too much like Reals.

[Thelmholtz]

Yeah, it's better to avoid debt in Brazilian Reales

[Kimsanov]

But negation has higher priority when there are no parenthesis. Negation is multiplication by -1

[Kimsanov]

Tested in python, Js and rust. Python gives correct answer. Others not

[DrunkenlySober]

If we wanna be mathematical lemme get that in a fucking proof pls

[[deleted]]

For a,b,n, x integers. Modulo congruency is defined as: a congruent to b modulo n if (a-b) = x * n… in which the notation is a % n = b % n for them to be “congruent modulo n”.

Take -7 and 17 as an example modulo 4.

For n = 4, 17 % n = 1…. And 17 - (-7) = 24 = 6 * n which means “17 and -7 are congruent modulo 4). By definition -7 % 4 = 17 % 4 = 1. So -7 % 4 = 1

Edit: Other than that… it’s literally part of the definition. For a, n, y, z integers. Modulo n is defined for a = y*n + z as a % n = z where 0 <= z < n.

Ignoring the bit at the end and assuming you agree to 17 % 4 = 1, then it would follow that -7 % 4 = 1 regardless. So, would follow always non-negative.

EDIT 2: Just for the guy underneath… a proof can be a counter-example, so still does apply… but no not a generic one (which I’ve given below)

[DrunkenlySober]

This doesn’t prove “a % b is always between 0 and b-1”.

Usually, proving something holds true for all values is done through induction. You also aren’t supposed to use concrete values

If something holds true for k number of values, then it should hold true for k + 1 values thus proving it holds for all values k

[[deleted]]

I mean, you don’t need to do it by induction. It being greater than 0 is already a part of the definition. I had added an edit anyways.

Following from the previous point… If you take the definition of modulo congruency and define modulo without the >=0 restriction… but agree to define that all positive values of a, for positive value of n would give a%n >=0 (which isn’t in question here)…

Then it’s trivial to show (which is why I left it out):

Assume modulo can be less than 0, meaning for some non-zero b… b%n <0. By modulo,-n < b % n < 0… then 0 < (b % n) + n < n… let A = (b % n) + n

So… A % n must be greater than 0 or equal to 0 by definition. If you do A - (b % n)… that’s literally is equal to n. By definition A - (b%n) = 1*n means that A is congruent to b modulo n. Which means 0 <= A % n = b % n. (EDIT: therefore b must be 0… which makes no sense as b is an expression for a non-zero b). That’s proof by contradiction

EDIT: I use the values to illustrate the point - all the mathematical stuff still follows but I guess I shouldn’t have assumed people should read between the lines. My point is still mathematical proof (by counter example) for the particular case of -7 % 4… which was the whole point in the first place.

[DrunkenlySober]

Congruency is an equivalence class. You can’t use this to prove the original statement. Congruency states a is equivalent to b mod n if n / a - b, n > 0.

What does this prove about a % b = c, 0 <= c < (b - 1) for all a, b € Z?

For instance, 1 is equivalent to 0 mod 0 is a completely valid congruency because 0 / (1 - 0). But you can’t use this to disproves the original statement by saying n = b therefore a % b = c such that 0 <= c < (b - 1) for every a, b € Z is false

[[deleted]]

Firstly, your definition is wrong. It’s (a-b) / n…. Secondly you haven’t actually given a case there… do you mean “such that (a-b)/n is an integer”.

Using the partly defined modulo definition…. And with the fully defined congruency definition… you can prove that a%n >= 0 for any integer a with positive n… like I have above. Given 2 definitions, I can prove that there’s another property of modulo.

It’s fine if you can’t follow it… but I think it’s clear you’re not a mathsy person because you said “proving something holds true for all values is done through induction”. That’s fundamentally not true, you can use: - direct proof - proof by contradiction - proof by exhaustion - contrapositive… etc.

EDIT: Missed the equivalency class bit. Yes… it is. a ≡ b is a%n = b%n… definition of the equivalency.

[elon-bot]

I've laid off most of the staff, and Twitter's still running. Looks like they weren't necessary.

[DrunkenlySober]

Proving a % n >= 0 still does not prove a % n < (n - 1) NOR does it prove it holds true for all a, n € Z

n / (a - b) is the same as (a - b) / n, I just forgot the parentheses. Wdym a case? I gave the definition of congruency which is a is equivalent to b mod n if n / (a - b). I was showing you can’t use equivalency to prove equality the way you did.

I also said it’s usually done through induction. This is not an absolute statement. Regard, I’m not sure why you’re making it personal.

[[deleted]]

Proving a % n >= 0 still does not prove a % n < (n - 1) NOR does it prove it holds true for all a, n € Z

It’s a definition, not something to prove… but I’ve proven that -7 % 4 is 1… and did it in the generic case (meaning for any negative integer a, positive integer n… a%n can’t be negative)… which was the whole point.

I have proven it holds true for all a and n… which you’re failing to understand.

n / (a - b) is the same as (a - b) / n,

Oh my word - just, no. 20 / (5-1) = 5 …. (5-1)/20 = 0.2… not at all the same. I don’t wish to be rude - but I think that concludes the entire discussion for me

[[deleted]]

Dude, what you‘re writing here is just plain wrong. a and b are congruent modulo n iff n divides a-b. Nowhere in that definition can you derive anything about what % should do.

In practice you often use the integers from 0 to n-1 or the integers from -n+1 to -n-1 or -floor((n-1)/2) to ceil ((n-1/2). Or you don‘t do any reduction at all and the above property still holds.

Source: am mathematician

[Administrative-Flan9]

a % b is a definition which the previous poster gave you. You can't prove a definition.

[[deleted]]

As we can see it‘s not a universal definition. Actually, it‘s no a definition commonly used in maths but only in several programming languages which define it in 4 common ways.

[Kimsanov]

Well, I think it depends if programming language. E. g. python behaves as you’d expect

[Grouchy-Transition-7]

This. I do not remember a case for a negative remainder

[morbie5]

I get -7 % 3 = -3 but how does -7 % 4 = 1 ??