4 ms·

Pardon my mathematical ignorance here, but I've always been curious as to whether negative numbers are the only alternate set that extends from the origin of ze

Pardon my mathematical ignorance here, but I've always been curious as to whether negative numbers are the only alternate set that extends from the origin of zero, or whether they are merely the only set that is diametrically opposed to their positive counterparts, and there are in fact an infinite number of these lines extending from the origin at zero, radiating from it in all directions, and what we think of as the positive and negative numbers are simply one pair of rays extending from this origin arbitrarily chosen as our base units.

ajkjk 3y ago
I often think that a lot of math would work out more easily if we _only_ used polar coordinates and regarded the negative numbers as a "separate number line" rather than a continuation of the positive numbers. (In particular, if you know about delta functions... a lot of weirdness around x=0 goes away if you write everything in terms of r \sgn (r) and take the derivatives of both terms. e.g. This gives "for free" the fact that the divergence of 1/r^2 is 4 pi delta(r).) I have heard of systems in which one sticks more lines out from 0 than just the positive and negative numbers. At some level that's what R^2 is, with four copies of the positive number line, but I don't see a strong reason why in principle you couldn't have an odd number of lines, which would correspond to... uh... R^1.5. But you have to define how these lines rotate into each other, and it is gonna be weird.
ithinkso 3y ago
If you want the equation x + 1 = 0 to have a solution, you need to invent negative numbers, now if you want the equation x^2 + 1 = 0 to have a solution, you need to invent complex numbers and 'i'. (Also, complex numbers, turns out, are enough for higher powers as well) The line and plane are just convenient representations of R and C but there is nothing inherently profound about them, in my opinion
- ajkjk 3y ago
  This is one take, and one that has become very popular, but it's not necessarily the only take. In particular it presupposes that your number-like indeterminates can be both (a) multiplied and (b) added to numbers (and, implicitly, divided). Naturally the solution has to be a division algebra. If instead you asked the question "for what values of O would O^2 (v) = -v", or even just O^4 (v) = v, then you would be more content having the answer live in a different space, of operators on vectors rather than vectors themselves, instead of in a field extension of the present space. Of course they are basically isomorphic but I think the alternate interpretations are useful to keep in mind so that we don't accidentally assume our way into a box of our own making. edit: I should add, by O^2 I mean O ∘ O, so there's no definition of "multiplication" on these necessarily, just composition.
pizza 3y ago
Check out the article "A Unified Mathematical Language for Physics and Engineering in the 21st Century" [0]. [0] http://geometry.mrao.cam.ac.uk/wp-content/uploads/2015/02/00RSocMillen.pdf http://geometry.mrao.cam.ac.uk/wp-content/uploads/2015/02/00...
- ajkjk 3y ago
  I.. don't think that has anything to do with their comment.
  - pizza 3y ago
    I thought they were asking about whether a coordinate system with basis elements e_1, e_2, ..., could be re-parameterized after rotation, and whether the re-parameterizations are infinite. The answer is simple via geometric algebra: yes. Given (* is geometric product, . is scalar product, ^ is wedge product) e_i * e_i = e_i . e_i + e_i ^ e_i = 1 + 0 = 1 e_i * e_j = e_i . e_j + e_i ^ e_j = 0 + -e_j ^ e_i = -(e_j * e_i) where the e_i are the basis elements along the infinite rays mentioned. To move to a rotated basis, make a rotor R = cos(theta/2) - e_i e_j sin(theta/2) ~R = cos(theta/2) + e_i e_j sin(theta/2) then you get the relationship in the new coordinate system: x' = R * x * ~R Since it's parameterized for any theta in [0..4pi], there's infinite of them, furthermore you get to pick which path you are taking to do the transformation along the way - either the 'negative' direction [0..2pi] or 'positive' direction [2pi..4pi]

ajkjk 3y ago

I often think that a lot of math would work out more easily if we _only_ used polar coordinates and regarded the negative numbers as a "separate number line" rather than a continuation of the positive numbers. (In particular, if you know about delta functions... a lot of weirdness around x=0 goes away if you write everything in terms of r \sgn (r) and take the derivatives of both terms. e.g. This gives "for free" the fact that the divergence of 1/r^2 is 4 pi delta(r).) I have heard of systems in which one sticks more lines out from 0 than just the positive and negative numbers. At some level that's what R^2 is, with four copies of the positive number line, but I don't see a strong reason why in principle you couldn't have an odd number of lines, which would correspond to... uh... R^1.5. But you have to define how these lines rotate into each other, and it is gonna be weird.

ithinkso 3y ago

If you want the equation x + 1 = 0 to have a solution, you need to invent negative numbers, now if you want the equation x^2 + 1 = 0 to have a solution, you need to invent complex numbers and 'i'. (Also, complex numbers, turns out, are enough for higher powers as well) The line and plane are just convenient representations of R and C but there is nothing inherently profound about them, in my opinion

ajkjk 3y ago
This is one take, and one that has become very popular, but it's not necessarily the only take. In particular it presupposes that your number-like indeterminates can be both (a) multiplied and (b) added to numbers (and, implicitly, divided). Naturally the solution has to be a division algebra. If instead you asked the question "for what values of O would O^2 (v) = -v", or even just O^4 (v) = v, then you would be more content having the answer live in a different space, of operators on vectors rather than vectors themselves, instead of in a field extension of the present space. Of course they are basically isomorphic but I think the alternate interpretations are useful to keep in mind so that we don't accidentally assume our way into a box of our own making. edit: I should add, by O^2 I mean O ∘ O, so there's no definition of "multiplication" on these necessarily, just composition.

pizza 3y ago

Check out the article "A Unified Mathematical Language for Physics and Engineering in the 21st Century" [0]. [0] http://geometry.mrao.cam.ac.uk/wp-content/uploads/2015/02/00RSocMillen.pdf http://geometry.mrao.cam.ac.uk/wp-content/uploads/2015/02/00...

ajkjk 3y ago
I.. don't think that has anything to do with their comment.
- pizza 3y ago
  I thought they were asking about whether a coordinate system with basis elements e_1, e_2, ..., could be re-parameterized after rotation, and whether the re-parameterizations are infinite. The answer is simple via geometric algebra: yes. Given (* is geometric product, . is scalar product, ^ is wedge product) e_i * e_i = e_i . e_i + e_i ^ e_i = 1 + 0 = 1 e_i * e_j = e_i . e_j + e_i ^ e_j = 0 + -e_j ^ e_i = -(e_j * e_i) where the e_i are the basis elements along the infinite rays mentioned. To move to a rotated basis, make a rotor R = cos(theta/2) - e_i e_j sin(theta/2) ~R = cos(theta/2) + e_i e_j sin(theta/2) then you get the relationship in the new coordinate system: x' = R * x * ~R Since it's parameterized for any theta in [0..4pi], there's infinite of them, furthermore you get to pick which path you are taking to do the transformation along the way - either the 'negative' direction [0..2pi] or 'positive' direction [2pi..4pi]