[gjulianm]

I think OP means that nothing breaks mathematically. It is not inconsistent and not false, so you can work with it. The only issue is to deal specially with the case of division by zero, which you have to do anyways. Code that assumes that (x/y) * y = x is wrong if you don't check for y = 0, independently of what you define x/0 to be.

[godelski]

You do realize that division is the inverse operation of multiplication, right? Like subtraction is the inverse of addition. By defining addition we define subtraction. By defining multiplication we define division. This is where the author fails. Division is multiplication of a fractional value. This is VERY important.

And just because it is mathematically a field does not mean it is particularly the right choice. A field is not strictly definitely by addition and multiplication, it is definitely by two operators where one is an abelian group (addition in our case) and the other forms and abelian group over the non identity term of the first (eg multiplication is an abelian group over non zero terms). The complex numbers create a field, which is really how we do addition and multiplication in 2D space. 3D space you can't form one, so you have a ring. Which is why quaternions are so important, because they form a field.

But the author is wrong because they think division is a different operation than multiplication. It's just a short hand.

[gowld]

In a field, division by zero is not the inverse operation of anything.

[gjulianm]

The usual definition of division is the multiplicative inverse, yes. But that does not mean that, as an intellectual curiosity (which is how I take this post) you can't define "division" as having other value. Sometimes that's useful, sometimes it's not. Most of the time it is fun to see how things break and don't break. For example, the infamous 1+2+3+... = -1/12 sum[1]: for most purposes it's a divergent sum but you can find ways to assign a finite value and make that useful.

1: https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B...

[godelski]

The OP says

> The field definition does not include division, nor do our definitions of addition or multiplication.

This is clearly a misunderstanding of not realizing that the definition of division is a symbolic shorthand for inverse multiplication.

Sure, you're right that we can define it another way. But I don't know what we would call an object with three operators. (Someone more knowledgeable in field theory please let me know, it's hard to Google) in the normal field we work with (standard addition and multiplication) we only have two unique operators. Everything else is a shorthand. In fact, even multiplication is a shorthand.

>For example, the infamous 1+2+3... = -1/12

Ramanujan Summation isn't really standard addition though. It is a trick. Because if you just did the addition like normal you would never end up at -1/12, you end up at infinite (aleph 0, a countable infinity). But the Ramanujan Summation is still useful. It just isn't "breaking math" in the way I think you think it is.

But I encourage you to try to break math. It's a lot of fun. But there's a lot to learn and unfortunately it's confusing. But that's the challenge :)

[gjulianm]

The fact that it's a shorthand means that it's not in the definition, just a convention. In fact, nowhere in my studies I saw anyone using 'division' when working explicitly with fields in algebra, it's always multiplicative inverse.

By the way, you cannot do the addition like normal on that series, you don't end up at anything. You can say it diverges and its limit is +infinity (not aleph0, aleph0 is used for cardinality of sets so I think no one would use it as the result of a divergent series). When I said it 'broke math' what I meant was that, in the same way that OP did, it is a way to assign values to things that usually don't have one. I know it does not actually break math.

[jle]

That's not necessarily the definition of division, there are many commonly used definitions in different contracts that are different.

[throwawaymath]

Bravo, that's a great explanation. I think you've done a better job here than my comments elsewhere in this thread. Nice tie in to dimensionality, the complexes and quaternions too.

There are legitimate reasons to advocate for defining division by 0 in the context of a programming language. But the attempt at mathematical rigor really distracts from those reasons. It feels like the author made a decision and reached for some odd mathematics that seems to support it but doesn't. Floating numbers don't even form a field. If we define division by 0 in a field it stops being a field and becomes a ring or some sort of weird finite field with a single element...

[rrampage]

> Division is multiplication of a fractional value

Thus, division by 0 is multiplying by (1/0). Does such a fraction exist? It can go along two potentially different paths depending on the limit we take.

Alternatively, there is information lost when you multiply something by 0: a x 0 = 0, b x 0 = 0. When you perform an inverse by dividing, will you get back a or b (or any number)? Thus, it is not invertible at least at 0.

Disclaimer: Not a mathematician

[godelski]

>Alternatively, there is information lost when you multiply something by 0: a x 0 = 0, b x 0 = 0. When you perform an inverse by dividing, will you get back a or b (or any number)? Thus, it is not invertible at least at 0.

>>A field is not strictly definitely by addition and multiplication, it is definitely by two operators where one is an abelian group (addition in our case) and the other forms and abelian group over the non identity term of the first (eg multiplication is an abelian group over non zero terms).

[pmiller2]

It is inconsistent. If 1/0 = 0, then 1 = 0*0.

[throwawaymath]

This should not be getting downvoted. The author of the article made an incorrect refutation of this point under the "Objections" heading. Division by 0 in fields is undefined precisely because it leads to contradictions like this. The full statement of a proof that 1/0 = 0 includes the temporary assumption that you have defined division by 0 such that it is no longer undefined.

You can't logically refute that proof by saying, "well no, you have yet to define division by 0 according to the field axioms, so you can't use that division as part of your proof." That's the point! The proof does not demonstrate that division by 0 results in 1, it demonstrates that you cannot define division by 0 while maintaining the algebraic structure of a field.

If the author wants to talk about defining division by 0 in wheels or something esoteric they're more than welcome to. But among fields, and among the real numbers, it's not possible. This whole exercise of trying to refute what has been commonly accepted for over a century is frankly ridiculous for trying to justify undefined behavior in a programming language.

[imh]

But the point is you keep the structure of a field for everything that already has it, right? No structure is taken away. All statements valid on any subset of the reals are still valid without modification under this new definition. The algebraic structure is still a field obeying the same axioms. There are just some new valid statements too.

[throwawaymath]

No that's not correct, and this is why I think the author's entire point is pretty inane. If you're going to start off your argument with the full formalism of field axioms and consequent theorems, you need to be prepared to split hairs about whether or not your definitions constitute a field.

Mathematics is thoroughly pedantic about definitions for a reason. If those formalisms don't matter because what you've done is "close enough", then skip the song and dance about field definitions and stop trying to use it to justify the behavior of an undefined operation in a programming language. Just say you're defining 1/0 to be equal to whatever you want because the world doesn't break down. It actually detracts the author's point to so confidently (and incorrectly) refute something that is robustly proved in the first few weeks of an undergraduate analysis course. Why is this even in a blog post about a programming language?!

This is essentially the same point as the extended real (or complex) number systems. The sets of all real and complex numbers (respectively) form fields under the axioms of addition and multiplication. But you can define division by 0 and division by infinity in a way that works with familiar arithmetic (I explained how to do this in another comment barely two weeks ago [1]). But the key point here is that in doing this you sacrifice the uniqueness of real numbers.

The author tries to refute this observation by claiming the proof uses an undefined division operation, but that's a red herring. The real assertion is that you cannot define division as an inverse operation from multiplication to be inclusive of division by the unique unit (i.e. 0, in the real field) unless you are willing to state that every number is equal to every other number in the entire field. And you can do that, but it trivially follows that you no longer have a field without a nonzero element.

So really the actual proof is that division by 0 is undefined for any field with at least one nonzero element.

__________________________________

1. https://news.ycombinator.com/item?id=17599087#17601806

[imh]

Is there a proof or something elsewhere you can link to? To be honest I can't really tell the point you're trying to make.

[throwawaymath]

I can give you a simple proof by contradiction.

1. Let F be a field containing an element x =/= 0.

2. Suppose we have defined division by zero in F such that, for all x in F, there exists an element y = x/0 (i.e. F adheres to the field axiom of multiplicative closure). Note that at this point it does not matter how we have defined division by 0, we will just generously continue and assume you've done it in a way that maintains the other field axioms.

3. Since y = x/0, it follows that the product of y and 0 is equal to x, because division is the inverse of multiplication. By the field axioms, division does not exist if there is no multiplicative inverse with which to multiply.

4. But by the field axioms this implies that x = 0, which contradicts our initial assumption. Likewise, since we can repeat this procedure with any element x in F, this demonstrates that there exists no nonzero element x in F, and in fact F = {0}.

The failure in the article's refutation is that this proof is designed to permit you to assume you have suitably defined division by zero, then proceed to demonstrate without any loss of generality that you could not possibly have unless 1) F is not a field, or 2) F contains only 0. The fundamental algebraic property you sacrifice by defining division by zero is uniqueness, and uniqueness is a hard requirement in fields with nonzero elements.

[imh]

This is explicitly talked about in OP about 2/3 of the way down the page. "If 1/0 = 0, then 1 = 0 * 0" is untrue. Your argument is "1/0 = 0, so multiply both sides by zero: 1/0 * 0 = 0 * 0, and then take 1/0 * 0 = 1 * 0/0." That last step isn't the case for reasons covered in the article we're ostensibly discussing.

[sclv]

Right. So the multiplicative inverse property _breaks_! He just points out that it breaks, and thus you need to use a more complicated property instead. That doesn't mean that the property doesn't break.

[gmfawcett]

He's not saying that it breaks; quite the opposite, he repeatedly states that 0⁻ doesn't exist. He's just defining division in a specific way.

The MI property states that every element except 0 has a multiplicative inverse. He's defining division via two cases: If b≠0, then a/b = a*b⁻ (multiplicative inverse). If b=0, then a/b=0. This definition does not imply that 0⁻ exists, so there's no violation of MI.

[sclv]

The problem this and the other replies miss is that the standard definition of division is multiplication by the inverse. The entire argument rests on a notational slight of hand. The property that held before -- that _when defined_ division has the inverse property -- no longer holds. Thus many equational identities that otherwise would hold do not hold.

[pron]

> The problem this and the other replies miss is that the standard definition of division is multiplication by the inverse.

Try to state this definition formally. The statement: ∀ x,y . x/y = xy⁻¹ is not a theorem of fields or a definition of division. However, ∀ x, y . y ≠ 0 ⇒ x/y = xy⁻¹ is, but is completely unaffected by defining division at 0. Those who think they see a problem rely on informal and imprecise definitions. Could you formally state a theorem that is affected? That would help you get around issues that are merely artifacts of imprecision.

But let's entertain you, and state that what we really mean by the informal and vague statement, "division is the inverse of multiplication," could be stated formally as:

    ∀ x ∈ dom(1/t). x(1/x) = 1

You are absolutely correct that this equational theorem is broken by extending the domain of division. However, there is absolutely no way to say that the formalization of this theorem isn't actually

    ∀ x ≠ 0 . x(1/x) = 1

because the two are equivalent. You cannot then claim that something is necessarily broken if you choose to pick a formalization that is indeed broken, while an equivalent formalization exists, that is not broken (not to mention that the formalization that is broken requires a strictly richer language). All that means is that your formalization in this case is brittle, not that laws are broken.

[gmfawcett]

I have to disagree -- this isn't sleight of hand. The standard definition isn't being violated here, because standard division isn't a total function. The denominator's domain in Hillel's function is a proper superset of the standard domain: when restricted to the standard domain, the two functions are precisely equivalent. Therefore, every standard identity still holds under Hillel.

The hole that he is filling here isn't one that he bored into the standard definition, but a hole that the standard definition already admitted. If something is explicitly undefined, there's nothing mathematically wrong with defining it, as long as the definition doesn't lead to inconsistency.

[a_wild_dandan]

Look at it this way...

Standard definition of division function, d:

d(x, y) = x * y⁻, for all x and y EXCEPT 0

Author's modified, piecewise (https://en.wikipedia.org/wiki/Piecewise) definition:

d(x, y) = x * y⁻, for all x and y EXCEPT 0

d(x, y) = 0, for y = 0

He's just adding 0 to the domain of d(x, y) to extend the definition, and deliberately not using xy⁻ for that particular element of the domain. No inverse needed.

[improv32]

You missed the part where he defines division as a/c = a * inverse(c) for all c != 0

[imh]

Multiplicative inverse is already broken. x / 0 * 0 is either 0 or ⊥ (aka exception). It wasn't true to begin with. I happen to like the exception/partial function version, so I'm playing devil's advocate, but it really was already broken when we're talking about zero.

[ASalazarMX]

Wait, if 1 / 0 = infinity, then infinity * 0 = 1

This seems just as bizarre, since zero times anything shouldn't become 1, no matter how big or how many times you do it.

[umanwizard]

According to the standard definition of division, 1/0 does not equal infinity. It doesn't equal anything. It is undefined.

[hexane360]

Yes, that's correct. The fact that 1/0 shouldn't be defined as infinity isn't an argument that it should be defined as 0.

[jacobolus]

There are many contexts where defining either 1/0 = -1/0 = ∞ or 1/0 = +∞ and –1/0 = –∞ is better than the alternatives, especially when working in an approximate number system like floating point.

In geometric modeling kinds of applications, I would say that these definitions are typically desirable, with 1/0 = undefined only better in unusual cases. As a simple example, it is typically much more useful for the “tangent” of a right angle to be defined as ∞ than left undefined.

But anyhow, there are no “facts” involved here. Only different choices of mathematical models, which can be more or less convenient depending on context / application.

[gmfawcett]

Yes, exactly, I have no idea why you're being downvoted.

[milesvp]

Look up the dirac delta function. It's a spike thats infinitely tall and infinitely narrow, with an area of 1. It's established now as a very useful tool in EE. But many people fought it tooth and nail because of this logic.

[lmm]

> This seems just as bizarre, since zero times anything shouldn't become 1, no matter how big or how many times you do it.

It isn't bizarre, because there's an equal and opposite argument that anything times infinity is infinity, no matter how small the thing you multiply by.

If you actually do infinity * 0 you get NaN since there's no way to determine (without more information) whether the result should be 0, infinity, or anything in between.

[FlipperBucket]

1/0 = undef (cast to 0)

1 != 0*0

I don't see the inconsistency. Abstract math vs practical application.

[pmiller2]

You’re doing something different than real number arithmetic. Saying 1/0 = x, and then treating x like a real number is inconsistent. But just saying “we are going to augment the real numbers with an element x that is not a real number, and then define some properties of x and prove things about it” is not.

[FlipperBucket]

> You’re doing something different than real number arithmetic

Computer languages execute on rules that are not utilizing real number arithmetic. I didn't want to mention it, but there's these things called floats...

Edit: Pony took out the "normal" version of division by zero and suggest to write a wrapper to check beforehand.

[pmiller2]

Where did I say anything about computers? I’m referring to real numbers.

[FlipperBucket]

The topic is a computer language. I'm referring to utility.

[tzs]

It's not inconsistent. See this article [1] for an explanation. It's the first thing covered in the "Objections" section.

[1] https://www.hillelwayne.com/post/divide-by-zero/

[Confusion]

So according to you, 0*0 = 0, thus 0/0 = 0?

[mreome]

If division by zero is not undefined, yes. But how does this disagree with the assertion that it's mathematically inconsistent (as opposed to programmatically inconsistent)? If any case exists that leads to a logically false conclusion, then it's not logically consistent.

[pmiller2]

Nope. If 0/0 = 0, then x * 0/0 = x * 0, so 1/0 = 1, implying 1 = 0.

[yati]

You didn't read the article, did you?