4 ms·

> Imaginary numbers are not so scary if you just think of them as 2d vectors with a different multiplication operation added to it. Perhaps, but that view mak

by gspr 6y ago

> Imaginary numbers are not so scary if you just think of them as 2d vectors with a different multiplication operation added to it. Perhaps, but that view makes them seem arbitrary. It makes it seem as if you could just do the same thing in 3d, 4d, etc. and they'd all be equally impacful and meaningful.

leereeves 6y ago
You can (almost). You can do the same thing again to get quaternions, and again to get octonions, and so on. https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_constru... That's different from extending 1d to 2d, 3d, etc as vectors.
- gspr 6y ago
  I know. But these are not as fundamentally important as the complex numbers.
- AnimalMuppet 6y ago
  "And so on"? I thought octonions are where it stops? At least, anything past the octonions is not able to be a normed division algebra, if I understand correctly.
  - leereeves 6y ago
    It doesn't stop, but as you said, the next step (sedenions), and all steps after that have non-trivial zero divisors and aren't division algebras.
    - gspr 6y ago
      Exactly. This is what I was hinting at when critical at people who say complex numbers are "just 2d reals with fancy multiplication". It's true, of course, but it makes it seem as if the 2 in "2d" is unimportant/arbitrary.
bikenaga 6y ago
Representing complex numbers as 2-dim. real vectors is just one way of constructing complex numbers. It is convenient because you can draw 2-dim. vectors, and it makes complex numbers feel more tangible. Something like the original motivation (which you might find less "arbitrary") goes like this. We know how to solve some quadratic equations like x^2 = 9: We get x = 3 and x = -3 (though it was a nontrivial thing to realize those negative solutions are there). What about x^2 = 2? We can "see" the solution should be the length of the hypotenuse of a right triangle whose legs are both 1, so maybe encouraged by the geometry we don't feel too bad about having solutions (+/- sqrt(2)) which are, however, not rational numbers. (Notice that we keep "extending" our idea of "number". Well, if we didn't have feature creep no one would buy the new version.) Since we now have negative numbers, we ask about solutions to x^2 = -1. The first response is there are "no solutions", because there are no real numbers that square to -1. Rather than give up, we decide to "extend" the real numbers to provide solutions to this equation. The idea is actually pretty easy if you know some modular arithmetic. For instance, mod 6 arithmetic means that you take the integers and consider things that differ by multiples of 6 to be "the same". As a consequence, 17 = 5, 4 = -14, and (importantly) 6 = 0. Thinking in the same way, to get solutions to x^2 = -1, we want x^2 + 1 = 0. That's like "6 = 0", so "x^2 + 1" should be our "modulus", but what "number system" do we "mod out of" (instead of the integers, with the integers mod 6)? Since x^2 + 1 is a real polynomial, it looks like our "base number system" should be R[x], the (ring of) polynomials with real coefficients (stuff like -17 x^3 + 1.374 x^2 - 31/42 and so on). So we should take real polynomials, but agree that two polynomials differing by a multiple of x^2 + 1 are "the same". We conclude that the complex numbers "should be" R[x] mod <x^2 + 1> where the "<x^2 + 1>" means "all (polynomial) multiples of x^2 + 1". For instance, x^3 + 4 x + 1 = (x^3 + x) + (3 x + 1) = x(x^2 + 1) + (3 x + 1) = 3 x + 1 (mod x^2 + 1) (the multiple of x^2 + 1 is 0) (To agree that some things should be regarded as "the same" is to construct an equivalence relation, and the "bags" of "same things" are equivalence classes. I'm omitting the details from abstract algebra, but this construction is called "forming the quotient ring R[x]/<x^2 + 1>".) By this construction, "3 + 4 i" appears as 3 + 4 x (mod x^2 + 1) or (3 + 4 x) + <x^2 + 1>. In the construction of the complex numbers via 2-dim. real vectors, "3 + 4 i" appears as (3, 4). You can also construct the complex numbers as a subring of the 2 x 2 real matrices, so "3 + 4 i" appears as 3 [1 0] + 4 [0 1] [0 1] [-1 0] . Notice that [0 1]^2 = [-1 0] = - [1 0] (like i^2 = -1) [-1 0] [0 -1] [0 1] So those are three ways of constructing the complex numbers, and each has pros and cons. No one of them "is" the complex numbers. "The" complex numbers are really an isomorphism class of rings, and we're just giving different representatives from that class.

by gspr 6y ago

leereeves 6y ago

You can (almost). You can do the same thing again to get quaternions, and again to get octonions, and so on. https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_constru... That's different from extending 1d to 2d, 3d, etc as vectors.

gspr 6y ago
I know. But these are not as fundamentally important as the complex numbers.
AnimalMuppet 6y ago
"And so on"? I thought octonions are where it stops? At least, anything past the octonions is not able to be a normed division algebra, if I understand correctly.
- leereeves 6y ago
  It doesn't stop, but as you said, the next step (sedenions), and all steps after that have non-trivial zero divisors and aren't division algebras.
  - gspr 6y ago
    Exactly. This is what I was hinting at when critical at people who say complex numbers are "just 2d reals with fancy multiplication". It's true, of course, but it makes it seem as if the 2 in "2d" is unimportant/arbitrary.

bikenaga 6y ago

Representing complex numbers as 2-dim. real vectors is just one way of constructing complex numbers. It is convenient because you can draw 2-dim. vectors, and it makes complex numbers feel more tangible. Something like the original motivation (which you might find less "arbitrary") goes like this. We know how to solve some quadratic equations like x^2 = 9: We get x = 3 and x = -3 (though it was a nontrivial thing to realize those negative solutions are there). What about x^2 = 2? We can "see" the solution should be the length of the hypotenuse of a right triangle whose legs are both 1, so maybe encouraged by the geometry we don't feel too bad about having solutions (+/- sqrt(2)) which are, however, not rational numbers. (Notice that we keep "extending" our idea of "number". Well, if we didn't have feature creep no one would buy the new version.) Since we now have negative numbers, we ask about solutions to x^2 = -1. The first response is there are "no solutions", because there are no real numbers that square to -1. Rather than give up, we decide to "extend" the real numbers to provide solutions to this equation. The idea is actually pretty easy if you know some modular arithmetic. For instance, mod 6 arithmetic means that you take the integers and consider things that differ by multiples of 6 to be "the same". As a consequence, 17 = 5, 4 = -14, and (importantly) 6 = 0. Thinking in the same way, to get solutions to x^2 = -1, we want x^2 + 1 = 0. That's like "6 = 0", so "x^2 + 1" should be our "modulus", but what "number system" do we "mod out of" (instead of the integers, with the integers mod 6)? Since x^2 + 1 is a real polynomial, it looks like our "base number system" should be R[x], the (ring of) polynomials with real coefficients (stuff like -17 x^3 + 1.374 x^2 - 31/42 and so on). So we should take real polynomials, but agree that two polynomials differing by a multiple of x^2 + 1 are "the same". We conclude that the complex numbers "should be" R[x] mod <x^2 + 1> where the "<x^2 + 1>" means "all (polynomial) multiples of x^2 + 1". For instance, x^3 + 4 x + 1 = (x^3 + x) + (3 x + 1) = x(x^2 + 1) + (3 x + 1) = 3 x + 1 (mod x^2 + 1) (the multiple of x^2 + 1 is 0) (To agree that some things should be regarded as "the same" is to construct an equivalence relation, and the "bags" of "same things" are equivalence classes. I'm omitting the details from abstract algebra, but this construction is called "forming the quotient ring R[x]/<x^2 + 1>".) By this construction, "3 + 4 i" appears as 3 + 4 x (mod x^2 + 1) or (3 + 4 x) + <x^2 + 1>. In the construction of the complex numbers via 2-dim. real vectors, "3 + 4 i" appears as (3, 4). You can also construct the complex numbers as a subring of the 2 x 2 real matrices, so "3 + 4 i" appears as 3 [1 0] + 4 [0 1] [0 1] [-1 0] . Notice that [0 1]^2 = [-1 0] = - [1 0] (like i^2 = -1) [-1 0] [0 -1] [0 1] So those are three ways of constructing the complex numbers, and each has pros and cons. No one of them "is" the complex numbers. "The" complex numbers are really an isomorphism class of rings, and we're just giving different representatives from that class.