5 ms·

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 denom

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.
- pron 8y ago
  > 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 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 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.
- Dylan16807 8y ago

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.

pron 8y ago
> 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 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 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.
Dylan16807 8y ago