complete this sequence

>>8577209
the imaginary number

the end of the sequence

whatever space quaternions make

>>8577217
But quaternions are an expansion of complex plane, i^2 = j^2 = k^2 = ijk = —1

The third axis will have nothing to do with that, right? I realize this is undefined maths, maybe something to do with division by zero

>a + bi + ct
where t^3 gives an expression involving i

>>8577252
The real number line is a field of real dimension 1, the complex plane is a field of real dimension 2, and the quaternions are a structure of real dimension 4.
This doubling can be continued forever, e.g. there are octonions.
There isn't any field much nicer than the copmlex numbers.
A very nice real 3 dimenional structure is SU(2), which is a group.

well it looks like it is tending towards the _____ solid/space

>>8577209
Barnett space

>>8577209
This is it -> >>8577217
Null point -> 0th dimension. all information is packed into a single point with no height, width, length, etc.
Real number line -> 1st dimension. a line composed of points.
Complex plane -> 2 dimensions. x is the real line and y is the complex line: we note X as 1,2,3,4,5 etc and Y as i,2i,3i,4i,5i etc. A point is written as a + bi
Quaternion -> 4 dimensions. As far as i know, to describe 3 dimensional space you would need 4 dimensional numbers (j and k). So a point is noted as a + bi + cj + dk

>>8577429
If x axis is an expression of reals, y axis is an expression of i's, why cant there be a z axis that expresses something else?

Why is it necessary to use 4 dimensions a la quaternions? Ive been told that it must be so but never told why

File: 3manifold-volume.jpg (79KB, 960x720px) Image search: [Google]

79KB, 960x720px

3-manifold, ainnit?!

>>8577446
Basically if you had a 3-dimensional "division algebra" (ie. a multiplication on a vector space of dimension 3 such that every nonzero element is invertible), then either it would be commutative, which is impossible because it would imply the existence of an irreducible real polynomial of degree 3 (primitive element theorem), or it would contain a copy of C, which is also impossible because it would then have a C-vector space structure, and would then have an even dimension over R

>>8578396
Suppose i have developed such an algebra that maintained commutativity. What should i do with it? Apply to reimann zeta and unveil it to collect 1million?

>>8577446
Simply put, in that case the group won't close.
Let's try your idea: consider a point a + i b + j c, where the basis is given by {1, i , j}, with i and j satisfying i^2 = -1 and j^2 = -1. For this to be closed, we require the product i*j to be in {1, i, j}, or some linear combination thereof. But it's clear this can't happen. For example, if we assume i*j = 1, then we need (i*j)j = j, but we also know (i*j)j = i*j^2 = -i, from which we find that i is not linearly independent from j, contrary to our initial assumptions. For that reason we have to expand the group by defining a fourth basis element k = i*j.

>>8578934
Those assumtions seem redundant. If both i and j are equivilentthere is no reason for a third axis, so of course it doesnt close

>>8578945
What's your definition of "equivalent"? That [math]x^2 = y^2?[/math] Because if so, then there are plenty of examples where [math](x^2 = y^2 \land x \neq y).[/math] For instance, take [math]x = \sqrt3[/math] and [math]y = -\sqrt3.[/math] Then we have that [math](x^2 = 3 = 3 = y^2) \land (x = \sqrt3 \neq -\sqrt3 = y).[/math]

Hilbert Space

>>8578953
I mean that both i^2 and j^2 resolving to -1 does not seem intuitive.

1st dimensionality => real numbers
2nd dimensionality => accounts for shortcomings of real numbers, is defined in terms of the thing lacking in previous dimension
3rd dimension=> should account for shortcomings of previous dimension and be defined in terms of i, right?

>>8578971
>shortcomings of previous dimension
What shortcomings? [math]\mathbb{C}[/math] is algebraically closed.

>>8577209
God plane

>>8578974
Not everything is always defined, for instance a/0

In the similar way sqroot —1 is undefined in the reals

The complex plane accomodates that shortcomong, perhaps anothrr dimension can account for a/0 and in doing so reveal intuition

>>8578971
What would you choose instead of j^2 = -1 then? Say you chose it to be -a for some other real number a. Then the same remarks as above hold, you've just changed the normalization of j. Instead, say you chose it to be -a*i. Again the same thing happens, since i^(1/2) is just a complex number (up to some subtleties which are not relevant here)

>>8579001
Interesting, can you tell me the name of those subtleties so i can check them out?

>>8577209
[math]\text{*}\mathbb C^3[/math]