Earlier quoted context omitted.
Complex numbers exist just as much as the Euclidean plane. Furthermore, they are also scalings and rotations of said plane.
The Euclidean plane also doesn't exist.
It does not hurt to assume that it does.
Product of Negatives (2010)
61–70 of 79 posts
Re: Product of Negatives (2010)
#62Earlier quoted context omitted.
There's a subtle point to keep in mind when generalizing to rings/fields. The concept of 'positive' and 'negative' are defined in terms of an order relation, e.g., 'positive' means >0 and 'negative' means For example, the integers mod n is a ring, so (-a) * (-b) = a * b holds, but it doesn't make sense to call a number mod n positive or negative, since -a mod n effectively means n - a mod n. (posted an earlier versio…
> The concept of 'positive' and 'negative' are defined in terms of an order relation, e.g., 'positive' means >0 and 'negative' means I thought the concept of "negative" was defined by reference to an operation. "Negative 5" is whatever value Q satisfies the equation 5 + Q = 0. That definition immediately tells you that the negative of a negative is a positive. Once we know 5 + Q = 0, we ask what the negative of Q is.…
In a finite numeric field, then, negatives are the same as positives. (For example, in Z mod 5, you get 2 + 3 = 0 and, sure, 2 + (-2) = 0, too.)
Re: Product of Negatives (2010)
#63Earlier quoted context omitted.
> Also, this post conflates the unary negation operator with negative numbers. -a is a standard way to represent additive inverse of an element in field. The point about "unary negation operator" seems irrelevant. In the real number field, additive inverse of a positive real number is indeed the negative of that number. The negative of that number is also obtained by the application of unary negation operator on the…
> In the real number field, additive inverse of a positive real number is indeed the negative of that number. The negative of that number is also obtained by the application of unary negation operator on the positive number. This is a fact that follows from the definition of +. But + needs to be defined before you can start making assumptions about what the additive inverse is. The set over which the field is defined…
> The set over which the field is defined (Z or R)
What? The set of all integers, Z, is not a field! R is. But Z isn't. Z is a ring, a commutative ring. I doubt you understand what a field is!
> already contains -3, -2 etc.
Yes, and those elements are literally the additive inverses of their positive counterparts. If you disagree with this, then the numbers -3, -2, etc. literally have no meaning.
> It then turns out that -3 is the additive inverse of 3.
Are you making this all up with your original research or do you have any proper literature written by a professional mathematician to back it up?
Re: Product of Negatives (2010)
#64Earlier quoted context omitted.
> In the real number field, additive inverse of a positive real number is indeed the negative of that number. The negative of that number is also obtained by the application of unary negation operator on the positive number. This is a fact that follows from the definition of +. But + needs to be defined before you can start making assumptions about what the additive inverse is. The set over which the field is defined…
> The set over which the field is defined (Z or R) What? The set of all integers, Z, is not a field! R is. But Z isn't. Z is a ring, a commutative ring. I doubt you understand what a field is! > already contains -3, -2 etc. Yes, and those elements are literally the additive inverses of their positive counterparts. If you disagree with this, then the numbers -3, -2, etc. literally have no meaning. > It then turns out…
To be fair, you can have a field that only has integers. For example, Z mod 5 is a field.
Re: Product of Negatives (2010)
#65Earlier quoted context omitted.
> The set over which the field is defined (Z or R) What? The set of all integers, Z, is not a field! R is. But Z isn't. Z is a ring, a commutative ring. I doubt you understand what a field is! > already contains -3, -2 etc. Yes, and those elements are literally the additive inverses of their positive counterparts. If you disagree with this, then the numbers -3, -2, etc. literally have no meaning. > It then turns out…
To be fair, you can have a field that only has integers. For example, Z mod 5 is a field.
I am aware. It is typically represented as Z_5. They are called prime fields.
I highly doubt wsxcde meant prime fields in their comment though. wsxcde seemed to be talking about the set of all integers and the set of all real numbers in their comment. Only the latter is a field (and a ring) whereas the former is only a ring.
And (-a)(-b) = ab holds in rings (and thus fields).
Re: Product of Negatives (2010)
#66Earlier quoted context omitted.
> The concept of 'positive' and 'negative' are defined in terms of an order relation, e.g., 'positive' means >0 and 'negative' means I thought the concept of "negative" was defined by reference to an operation. "Negative 5" is whatever value Q satisfies the equation 5 + Q = 0. That definition immediately tells you that the negative of a negative is a positive. Once we know 5 + Q = 0, we ask what the negative of Q is.…
In a finite numeric field, then, negatives are the same as positives. (For example, in Z mod 5, you get 2 + 3 = 0 and, sure, 2 + (-2) = 0, too.)
Yes?
In Z mod 5 using your notation, you have that 3 = -2. It doesn't make sense to distinguish two classes of "negative" and "positive" numbers in that case, but it still makes sense to talk about -2.
Re: Product of Negatives (2010)
#67If both 1 * -1 = -1 and -1 * -1 = -1, then -1 / -1 has two solutions.
Sure, but having a unique solution is not a required property of integer division.
Parallel to your statement: If both 1 * 0 = 0 and 2 * 0 = 0, then 0 / 0 has two solutions.
Re: Product of Negatives (2010)
#68We've banned the submitter, the site, and dozens of other accounts for using a ring of accounts to manipulate HN. Such abuse is not tolerated.
All: if you notice fishy things (as a user did in this case), please let us know at hn@ycombinator.com. We catch a lot of abuse between software and moderation, but unfortunately not all. Vigilant users make a huge difference, and protecting the integrity of HN is a community effort.
(Please don't post insinuations about abuse in the threads, though, since most suspicions don't end up leading to real evidence. Send them to hn@ycombinator.com. This is in the site guidelines: https://news.ycombinator.com/newsguidelines.html)
Re: Product of Negatives (2010)
#69Earlier quoted context omitted.
It shows that subtracting a minus one is equivalent to adding a plus one. The one logical leap that isn't explicitly spelled out is that subtracting X is the same as adding (-1)X. But I'm pretty sure that's the definition of integer multiplication.
I see how it’s intuition for addition/subtraction but that doesn’t tell us much about multiplication. You’re asserting that negative one times X is itself negative which is in fact what the article is attempting to prove in the first place so by explicitly supposing that, your analogy isn’t useful.
> You’re asserting that negative one times X is itself negative which is in fact what the article is attempting to prove in the first place
Absolutely not. The article explicitly postulates this:
> We also take for granted the fact that the product of a positive real number and a negative real number is a negative real number
You're right that that's the interesting part of the question, but as far as the article is concerned, it's just an uninteresting assumption.
Re: Product of Negatives (2010)
#70We've banned the submitter, the site, and dozens of other accounts for using a ring of accounts to manipulate HN. Such abuse is not tolerated. All: if you notice fishy things (as a user did in this case), please let us know at hn@ycombinator.com. We catch a lot of abuse between software and moderation, but unfortunately not all. Vigilant users make a huge difference, and protecting the integrity of HN is a community…
> All: if you notice fishy things like this, please let us know at hn@ycombinator.com
What would we have noticed, in this case?