4 ms·

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

by imh 8y ago

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 8y ago
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 https://news.ycombinator.com/item?id=17599087#17601806
- imh 8y ago
  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 8y ago
    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 8y ago
      >Since y = x/0, it follows that the product of y and 0 is equal to x, because division is the inverse of multiplication. Can you explain how this follows? I thought division was only the inverse of multiplication for all nonzero denominators, which would mean we can't use that definition for deduction in x/0. It might hinge on your next sentence: >By the field axioms, division does not exist if there is no multiplicative inverse with which to multiply. but I don't understand why that's necessarily true. I don't understand how the field axioms require division by x to require the existence of a multiplicative inverse of x when x is zero. Sorry to take a bunch of your Friday, but I'm very curious now. Explanation much appreciated. ------- edit: Come to think of it, couldn't I define x/y as cotton candy for all x,y in field F and still satisfy the field axioms? They just don't refer to division. Any connection between x/y and y's multiplicative inverse is just a nice convention. That convention states that x/y = x * mult_inv(y) when y != 0, but nothing else. That definition has nothing to do with the field axioms and changing it doesn't require that I change anything about multiplicative inverses. That means I don't touch the field axioms and my field is still a field.

by imh 8y ago

throwawaymath 8y ago

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

imh 8y ago
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 8y ago
  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 8y ago
    >Since y = x/0, it follows that the product of y and 0 is equal to x, because division is the inverse of multiplication. Can you explain how this follows? I thought division was only the inverse of multiplication for all nonzero denominators, which would mean we can't use that definition for deduction in x/0. It might hinge on your next sentence: >By the field axioms, division does not exist if there is no multiplicative inverse with which to multiply. but I don't understand why that's necessarily true. I don't understand how the field axioms require division by x to require the existence of a multiplicative inverse of x when x is zero. Sorry to take a bunch of your Friday, but I'm very curious now. Explanation much appreciated. ------- edit: Come to think of it, couldn't I define x/y as cotton candy for all x,y in field F and still satisfy the field axioms? They just don't refer to division. Any connection between x/y and y's multiplicative inverse is just a nice convention. That convention states that x/y = x * mult_inv(y) when y != 0, but nothing else. That definition has nothing to do with the field axioms and changing it doesn't require that I change anything about multiplicative inverses. That means I don't touch the field axioms and my field is still a field.