[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.

[throwawaymath]

> If something is explicitly undefined, there's nothing mathematically wrong with defining it, as long as the definition doesn't lead to inconsistency.

The definition does lead to inconsistency...you can't look at the field axioms, observe that 0 has no multiplicative inverse, then proceed to define a special, one-off division rule that doesn't involve multiplicative inverses for that one element. Either your division rule is pathological and breaks a fundamental field property or you've introduced a division rule which is just a syntactical sugar, not a real operation (in the latter case you've introduced confusing notation, not a new division function). Why do you think mathematicians explicitly state that the real field with the augmentation of positive and negative infinity (which allow division by 0) is not a field?

I don't understand why there is so much resistance to this idea in this thread, but the simple fact remains that if you define division by an additive identity (0) in any way, the field containing that unit ceases to be a field. This is because all elements cease to be unique. You can quickly prove that every element is equal to every other element, including (critically) the additive and multiplicative identity elements. Fields are defined by closure under the operations of addition and multiplication, and that closure requires uniqueness of their respective identities. Upend that and your entire field structure breaks down, because all you're left with is a field with a single element 0.

Stating that you've defined division by 0 using a one-off case that permits all other field identities to remain consistent is like saying you've turned the complex field into an ordered field using lexicographic ordering. You haven't, because i admits no ordering, much like 0 admits no multiplicative inverse.

Onlookers reading these comments probably think those of us harping on this point are anal pedants with a mathematical stick up our ass. But this thread is increasingly illustrating my central point, which is that the author shouldn't have tried to justify numerical operation definitions in a programming language using field axioms of all things.

[Dylan16807]

> Onlookers reading these comments probably think those of us harping on this point are anal pedants with a mathematical stick up our ass. But this thread is increasingly illustrating my central point, which is that the author shouldn't have tried to justify numerical operation definitions in a programming language using field axioms of all things.

You can't use your own stubbornness to justify itself.

I'm waiting for you to justify your claim that this extension to division breaks any of the field axioms. pron even made you a nice list of them.

Just name one equation/theorem that the new axiom would break.

I'm completely open to being convinced! But so far you've only given arguments about giving a multiplicative inverse to zero. Everyone agrees on that. It's the wrong argument.

[pron]

> you can't look at the field axioms, observe that 0 has no multiplicative inverse, then proceed to define a special, one-off division rule that doesn't involve multiplicative inverses for that one element

Why not? What mathematical axiom does this break?

> Either your division rule is pathological and breaks a fundamental field property

It doesn't, or you could show one here: https://news.ycombinator.com/item?id=17738558

> or you've introduced a division rule which is just a syntactical sugar, not a real operation

Is this math? We define symbols in formal math using axioms. I am not aware of a distinction between "syntactical sugar" and a "real operation."

> Why do you think mathematicians explicitly state that the real field with the augmentation of positive and negative infinity (which allow division by 0) is not a field?

That's simple: because unlike the addition of the axiom ∀x . x/0 = 0, adding positive and/or negative infinity does violate axioms 2, 7 and possibly 11 (depending on the precise axioms introducing the infinities).

> I don't understand why there is so much resistance to this idea in this thread, but the simple fact remains that if you define division by an additive identity (0) in any way, the field containing that unit ceases to be a field.

Because a field is defined by the axioms I've given here (https://news.ycombinator.com/item?id=17738558) and adding division by zero does not violate any of them. If you could show what's violated, as I have for the case of adding infinities and as done in actual math rather than handwaving about it, I assume there would be less resistance. I don't understand your resistance to showing which of the field axioms is violated.

> You can quickly prove that every element is equal to every other element, including (critically) the additive and multiplicative identity elements.

So it should be easy to prove using the theory I provided, which is the common formalization of fields. Don't handwave: write proofs, and to make sure that the proofs aren't based on some vagueness of definitions, write them (at least the results of each step) formally. The formalization is so simple and straightforward that this shouldn't be a problem.

> Stating that you've defined division by 0 using a one-off case that permits all other field identities to remain consistent is like saying you've turned the complex field into an ordered field using lexicographic ordering. You haven't, because i admits no ordering, much like 0 admits no multiplicative inverse.

Stop handwaving. Show which axioms are violated.

[svat]

Consider the following two functions, that we can define for any field F:

• Our first function f(x, y) is defined for all x in F, and all nonzero y in F. That is, the domain of this function f is (F × F\{0}). The definition of this function f is:

f(x, y) = x times the multiplicative inverse of y

(Note that in the definition of f(x,y) we could say "if y≠0", but whether we say it or not the meaning is the same, because the domain of f already requires that y≠0.)

• Our second function g(x, y) is defined for all x in F, and all y in F. That is, the domain of this function g is (F × F). To define this function g, we pick a constant C in F, say C=0 or C=1 (or any C∈F, e.g. C=17 if 17 is an element of our field). And having picked C, the definition of this function g is:

• g(x, y) = x times the multiplicative inverse of y, if y ≠ 0, and C otherwise [that is, g(x, 0) = C].

Note that this function is defined for all (x, y) in F×F, and when y≠0 it agrees with f, i.e. g(x,y) = f(x,y) when y≠0.

Would you agree that both of these are functions, on different domains?

Next, we have the notation x/y. To assign meaning to this notation (and to the word “division”), there are two conventions we could adopt:

• Convention 1: When we say "x/y", we will mean f(x,y) (as defined above) — that is, x * y^{-1}.

• Convention 2: When we say "x/y", we will mean g(x,y) (as defined above).

The point of the post is that we can well adopt Convention 2: with such a definition of "division", all the properties that were true of f (on the domain F×F\{0}) continue to be true of g, except that g is defined on a wider domain.

----------------------------------------------

Now, maybe Convention 2 offends you. Maybe you think there is something very sacred about Convention 1. In all your posts in this thread, you seem to be insisting that "division" or "/" necessarily have to mean Convention 1, and the provided justification for preferring Convention 1 seems circular to me — here are some of your relevant comments, with my comments in [square brackets]:

> Since there is no multiplicative inverse of 0, division by 0 is undefined behavior [This seems to be saying: Because of Convention 1, we cannot adopt Convention 2.]

> Since y = x/0, it follows that the product of y and 0 is equal to x, because division is the inverse of multiplication [Here, your reasoning is "because we adopt Convention 1"]

> in algebraic fields division by x is equivalent to multiplication by 1/x. This is precisely why you cannot have a field that admits division by 0: because 0 has no multiplicative inverse. [Again, you're stating that Convention 1 has to hold, by fiat.]

> If F is a field with elements x, y, then the quotient x/y is equivalent to the product x(1/y). If 0 has no multiplicative inverse, there is no division by 0. The two concepts are one and the same [This is just stating repeatedly that we have to adopt Convention 1.]

> A multiplicative inverse is a division. [This is merely insisting that Convention 1 has to be adopted, not Convention 2]

> divisor cannot exist unless it is a multiplicative inverse [Again, stating Convention 1.]

> you can't look at the field axioms, observe that 0 has no multiplicative inverse, then proceed to define a special, one-off division rule that doesn't involve multiplicative inverses for that one element. [Why not?] ...you've introduced a division rule which is just a syntactical sugar [yes the same is true of Convention 1; what's the problem?]

All these comments, which emphatically insist on Convention 1, seem to ignore the point of the article, which is that Convention 2 has no more mathematical problems than Convention 1, because the function g is no less a “valid” function than the function f.

In mathematics when we use words like “obvious” or insist that something is true because it just has to be true, that's usually a hint that we may need to step back and consider whether what we're saying is really mathematically justified. What we have here is a case of multiple valid definitions that we can adopt, and there's no mathematical reason for not adopting one over the other. (There's a non-mathematical reason, namely “it breaks convention”, but the entire point is that we can break this convention.)

[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.

[sclv]

I know what he's doing. The problem is when you make it a different function (even by just extending it) then you change its equational properties. So equational properties that held over the whole domain of the function no longer hold over the extended domain. This is repaired by modifying the equational properties. But the modified equational properties mean that you now have a different system than before. So the whole thing is just playing around with words.

[mdpopescu]

> So the whole thing is just playing around with words.

Er... that's what mathematics is. It's a word game - we build systems from arbitrary rules and then explore the results.

Look through https://www.mathgoodies.com/articles/numbers for a bunch of uncommonly-defined numbers.

[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.