Complete this sequence

the null point, the real number line, the complex plane, ...

the imaginary number

the end of the sequence

whatever space quaternions make

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

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

Barnett space

This is it -> 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

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

3-manifold, ainnit?!

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

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?

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.

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

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

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?

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

God plane

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

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)

Ive been messing about with this

>ar + bi + ct

Where a, b, c are reals

r^2 = 1 = —t
i^2 = -1 = —r
t^2 = i^3 = —i

Multiplying ar + bi + ct by (r) constitutes a rotation about i axis
By (i) rotation about t axis
By (t) rotation about r axis

r x i = i
i x t = t
t x r = r

r = 1
i = sqrt(—1)
t = sqrt(—1 x sqrt(—1))

I havent worked it all out yet. pls no steal

To elaborate, i gave the real component a meta—signifier in the sane way the imaginary components have one. For the reals, it resolves to 1, the real identity.

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

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