Product of Negatives (2010) • Hacker News Board

Re: Product of Negatives (2010)

#71

Post by dang » Sun, Mar 01, 2020, 9:17 PM UTC

thaumasiotes wrote: post #70

dang wrote: post #68

We'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?

I wish I could spell it all out, but unfortunately that would help spammers.

Here's one thing though: multiple accounts submitting, commenting, and promoting the same person's sites, articles, and (importantly) repos.

https://hn.algolia.com/?dateRange=all&page=0&prefix=false&qu...

Re: Product of Negatives (2010)

#72

Post by akalin » Sun, Mar 01, 2020, 10:46 PM UTC

thaumasiotes wrote: post #40

akalin wrote: post #12

Earlier 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.…

That's precisely the subtlety I'm talking about, in which additive inverses and real numbers less than 0 can both be referred to as 'negatives', and that the operation of taking additive inverses and real numbers less than 0 both use the symbol '-'.

It's pretty standard, though, that a 'negative number' is one that is less than 0, and a 'positive number' is one that is greater than 0, where a 'number' is an element of some subring of the reals.

Re: Product of Negatives (2010)

#73

Post by thaumasiotes » Sun, Mar 01, 2020, 11:01 PM UTC

akalin wrote: post #72

thaumasiotes wrote: post #40

Earlier 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.…

That's precisely the subtlety I'm talking about, in which additive inverses and real numbers less than 0 can both be referred to as 'negatives', and that the operation of taking additive inverses and real numbers less than 0 both use the symbol '-'. It's pretty standard, though, that a 'negative number' is one that is less than 0, and a 'positive number' is one that is greater than 0, where a 'number' is an element o…

I don't understand the point you're trying to make. You claimed that negative and positive are defined by absolute reference to zero. I claimed that they aren't. But you're presenting an example that assumes I'm right and you're wrong!

> 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.

If negative numbers were defined by reference to a comparison to zero, then the expression (-a) * (-b) would be meaningless nonsense in Z mod 5 -- as you point out yourself, Z mod 5 is not ordered in that way. But it isn't nonsense, and you're not saying it is -- instead, you assume it's obviously valid when you observe that the equality (-a)(-b) = ab holds.

Re: Product of Negatives (2010)

#74

Post by Koshkin » Sun, Mar 01, 2020, 11:11 PM UTC

thaumasiotes wrote: post #66

Koshkin wrote: post #62

Earlier quoted context omitted.

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.

That was precisely the original point - the distinction only makes sense in an ordered set; otherwise, calling -2 "negative" may be devoid of any meaning (e.g. when -2 = 3).

Re: Product of Negatives (2010)

#75

Post by akalin » Sun, Mar 01, 2020, 11:40 PM UTC

thaumasiotes wrote: post #73

akalin wrote: post #72

Earlier quoted context omitted.

That's precisely the subtlety I'm talking about, in which additive inverses and real numbers less than 0 can both be referred to as 'negatives', and that the operation of taking additive inverses and real numbers less than 0 both use the symbol '-'. It's pretty standard, though, that a 'negative number' is one that is less than 0, and a 'positive number' is one that is greater than 0, where a 'number' is an element o…

I don't understand the point you're trying to make. You claimed that negative and positive are defined by absolute reference to zero. I claimed that they aren't. But you're presenting an example that assumes I'm right and you're wrong! > 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…

I guess I'm not being too clear, so I'll try again. There are two concepts:

1) Positive and negative numbers (defined in terms of comparison to 0)

2) The negation of a number (i.e., the additive inverse)

They're related in that when both concepts are defined, a negative number is the negation of a positive number. However, the two concepts don't coincide. I'm sure you know this, but even over the reals '-x' is the negation of a number, but not necessarily a negative number.

(-a) * (-b) = a * b is an equation about #2, and it holds in any ring/field, even ones where #1 doesn't make sense, e.g. Z mod 5. If #1 makes sense, then this immediately implies that the product of two negative numbers is positive.

My original point was that the blog post is talking about real numbers, for which #1 and #2 are both defined. However, if it's generalized to arbitrary rings/fields, where only #2 is defined, then you can't really refer to the equation '(-a) * (-b) = a * b' as 'the product of two negative numbers is positive'.

Re: Product of Negatives (2010)

#76

Post by trevyn » Mon, Mar 02, 2020, 11:17 AM UTC

dang wrote: post #68

We'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…

Manipulate HN into talking about math? I don’t have the full picture, but the discussion this generated is better than many posts.

Re: Product of Negatives (2010)

#77

Post by [deleted] » Mon, Mar 02, 2020, 6:07 PM UTC

trevyn wrote: post #76

dang wrote: post #68

We'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…

Manipulate HN into talking about math? I don’t have the full picture, but the discussion this generated is better than many posts.

[deleted]

Re: Product of Negatives (2010)

#78

Post by thaumasiotes » Mon, Mar 02, 2020, 7:04 PM UTC

crdrost wrote: post #47

stan_rogers wrote: post #6

Earlier quoted context omitted.

Complex numbers, the way they're used in most cases, is a tuple notation. They're a handy way of keeping your chocolate separate from your peanut butter, so to speak, as that little "times i " makes it difficult to accidentally get things mixed up. And that's the way I always explained it to my students: there are imaginary numbers in the original sense of fake roots that will go away if you ignore them long enough,…

You're describing some sort of type mismatch between two concepts I think, and I really don't understand it: and I feel like as someone who occasionally teaches these things I really would like to. Could you elaborate? For my part, I do like to think of adjoining numbers onto an existing system, but that immediately becomes matrices. So you decide to adjoin an ε such that ε² = 0. Your numbers are now vectors (a, b) a…

I'd like to understand what you're doing, but I'm missing something.

1. Adjoining ε such that ε² = 0.

If the numbers have the form a + bε, isn't the action of ε to map (a, b) to (0, a)? Did you mean to say that ε = [0,1; 0,0]?

2. Adjoining φ = (1 + √5)/2 = [0,1; 1,1].

I'm fascinated by the idea of determining that a 2x2 matrix is equal to a real number.

I see that if you take successive powers of the real number φ, and express them in the form aφ + b, the coefficients a and b will take on values from the Fibonacci sequence. So far so good.

I don't follow the claim that the matrix [0,1; 1,1] actually represents φ. This would imply that the formula F_n = [1, 0] · φ^(n+1) · [1, 0] means that F_N equals the real number 1 (= 1 + 0φ), times φ^{n+1}, times 1 again. But this isn't true. It certainly is true that [1,0][0,1; 1,1]^{n+1}[1,0] is equal to F_n, but I don't follow the interpretation as adjoined numeric values (as opposed to as coefficients of φ).

3. Equivalence of representations [a,b] and aI + bM, where M is a 2x2 matrix representing any adjoined number.

This looks to me like the claim that if [a,b]M = [c,d], then (aI + bM)M = cI + dM. I tried to work this out algebraically and I'm pretty sure it isn't true in general. I'm open to being told that I'm wrong about this. Have I interpreted the idea correctly? What are the conditions under which the equivalence holds?

Re: Product of Negatives (2010)

#79

Post by crdrost » Tue, Mar 03, 2020, 9:11 PM UTC

thaumasiotes wrote: post #78

crdrost wrote: post #47

Earlier quoted context omitted.

You're describing some sort of type mismatch between two concepts I think, and I really don't understand it: and I feel like as someone who occasionally teaches these things I really would like to. Could you elaborate? For my part, I do like to think of adjoining numbers onto an existing system, but that immediately becomes matrices. So you decide to adjoin an ε such that ε² = 0. Your numbers are now vectors (a, b) a…

I'd like to understand what you're doing, but I'm missing something. 1. Adjoining ε such that ε² = 0. If the numbers have the form a + bε, isn't the action of ε to map (a, b) to (0, a)? Did you mean to say that ε = [0,1; 0,0]? 2. Adjoining φ = (1 + √5)/2 = [0,1; 1,1]. I'm fascinated by the idea of determining that a 2x2 matrix is equal to a real number. I see that if you take successive powers of the real number φ, a…

(1) This gets a little into difficult notation but e.g. in Mathematica-style notation,

    {{0,0},{1,0}} {{a},{b}} = {{0}, {a}}.

I think that you are preferring to left multiply your matrices by your points so that your points remain horizontal, whereas I am just used to my vectors being column vectors that I write as points sometimes? So that is why we are getting transposes of each other's notation.

(2) So the case for φ is very similar to the case for ε: you want to start with the constitutive relation, in this case φ² = φ + 1, to build the matrix with first column {{0},{1}} (multiplying 1 by φ gives φ) and the second column {{1},{1}} (the above constitutive relation).

The isomorphism is then that if X represents this adjoined unit then the point {{a}, {b}} becomes a I + b X one way [or say if you have a cubic constitutive relation then {{a},{b},{c}} becomes a I + b X + c X² similarly] and M becomes M {{1},{0}} [or say M {{1},{0},{0}} etc].

I would definitely agree that this is probably a much narrower statement than your “if [a,b]M = [c,d], then (aI + bM)M = cI + dM” as the structure of X is very tight. It is always [e_2 e_3 ... e_n c] where e_i is a unit column vector with a 1 in the i’th place and 0s everywhere else, and c is the vector embodying the constitutive relation.

But if you want a condition, the isomorphism condition is probably the best place to go.

So the claim is that φ^n = F_{n-1} + F_n φ.