4 ms·

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 notatio

by sclv 8y ago

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.

a_wild_dandan 8y ago
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 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 8y ago
  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.
  - deleted 8y ago
    [deleted]
  - mdpopescu 8y ago
    > 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 https://www.mathgoodies.com/articles/numbers for a bunch of uncommonly-defined numbers.
pron 8y ago
> 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 8y ago
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 8y ago
  > 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.

by sclv 8y ago

a_wild_dandan 8y ago

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 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 8y ago
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.
- deleted 8y ago
  [deleted]
- mdpopescu 8y ago
  > 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 https://www.mathgoodies.com/articles/numbers for a bunch of uncommonly-defined numbers.

pron 8y ago

> 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 8y ago

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 8y ago
> 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.