ID:111345
 
<!-- As a little side-project, I've been working on an algebra in a four-dimensional hyperspace. Formally, the algebra (which I've called f) is defined as
f = {(w,x,y,z) | w,x,y,z ∈ R}

where R is the set of real numbers. Thus, in this sense, it's topologically equivalent to R4 before the equivalence relation is applied. Furthermore, addition and subtraction of two vectors are defined the same way as in R4:

∀φ, ψ ∈ f : φ = (a,b,c,d), ψ = (w,x,y,z), then...
φ + ψ = (a,b,c,d) + (w,x,y,z) = (a+w, b+x, c+y, d+z)
φ - ψ = (a,b,c,d) - (w,x,y,z) = (a-w, b-x, c-y, d-z)

And thus, with this we can define the equivalence relation, as follows:

∀Σ, Π ∈ f, Σ = Π ⇔ (Σ ± Π) ∈ {Ψ = (m,n,o,p) | Ψ ∈ f ∧ m = n = o = p}.

To simplify things,

0* = {Ψ = (m,n,o,p) | Ψ ∈ f ∧ m = n = o = p}.
0 := any θ ∈ 0*
∀Θ ∈ f, Θ = 0 ⇔ Θ ∈ 0*

Thus, the equivalence relation can be rewritten as

∀Σ, Π ∈ f, Σ = Π ⇔ (Σ ± Π) = 0.

Additionally, it happens that f appears to be both a normed vector space and a metric vector space, though I have yet to prove either, and it seems non-trivial to prove:

∀φ ∈ f : φ = (w,x,y,z), ||φ|| = √((w² + x² + y² + z²) - (w+x+y+z)²/4).
∀φ, ψ ∈ f : φ = (w,x,y,z), ψ = (a,b,c,d), then d(φ, ψ) = ||φ - ψ|| = √(((w-a)² + (x-b)² + (y-c)² + (z-d)²) - ((w-a)+(x-b)+(y-c)+(z-d))²/4).

And two other operations can be defined, respectively called complementation and negation:

∀φ ∈ f : φ = (w,x,y,z)
-φ = -(w,x,y,z) = (-w,-x,-y,-z)
φ′ = (max(w,x,y,z) - w, max(w,x,y,z) - x, max(w,x,y,z) - y, max(w,x,y,z) - z)

It's also simple to demonstrate that negation and complementation are the same operation, which I will do so below.

Premise: ∀δ ∈ f : δ = (w,x,y,z), -δ = δ′

1. Proof that δ + -δ = 0:

δ + -δ = (w,x,y,z) + -(w,x,y,z)
= (w,x,y,z) + (-w,-x,-y,-z)
= (w + -w, x + -x, y + -y, z + -z)
= (w-w, x-x, y-y, z-z)
= (0,0,0,0)
= 0.

Therefore, δ + -δ = 0.

Proof that δ + δ′ = 0:

Assume M = max(w,x,y,z).
Then δ′ = (max(w,x,y,z) - w, max(w,x,y,z) - x, max(w,x,y,z) - y, max(w,x,y,z) - z)
= (M-w, M-x, M-y, M-z).

δ + δ′ = (w,x,y,z) + (M-w, M-x, M-y, M-z)
= ((M-w) + w, (M-x) + x, (M-y) + y, (M-z) + z)
= (M + (w-w), M + (x-x), M + (y-y), M + (z-z))
= (M + 0, M + 0, M + 0, M + 0)
= (M,M,M,M)
= 0.

Therefore, δ + δ′ = 0.

Proof that δ′ = -δ:

δ + -δ = 0 ∧ δ + δ′ = 0.

δ + δ′ = 0
δ + δ′ + -δ = -δ
(δ + -δ) + δ′ = -δ
0 + δ′ = -δ
δ′ = -δ

Therefore δ′ = -δ. ∎

And, lastly, is the vector product. Now I still have to try and work this out for sure, but I currently think that, assuming 0 is the nilpotent and that the vector product is distributive over addition, that this is the proper definition:

∀φ, ψ ∈ f : φ = (a,b,c,d), ψ = (w,x,y,z), then φ × ψ = (a,b,c,d) × (w,x,y,z) = (aw,az,ax,ay) + (by,bx,bz,bw) + (cz,cw,cy,cz) + (dx,dy,dw,dz) = (aw+by+cz+dx, az+bx+cw+dy, ax+bz+cy+dw, ay+bw+cx+dz).

This results in 0 being the nilpotent element, and the vector product being distributive over addition, but it doesn't have the normal product properties of commutativity and associativity. That is, A×B ≠ B×A and A×(B×C) ≠ (A×B)×C. I also believe that it has the interesting property of lacking, in a general case, a multiplicative identity element. That is, ∄β ∈ f : ∀φ ∈ f, φ×β = φ, but it does appear that ∀φ ∈ f ∃γ ∈ f : φ×γ = φ. This is probably one of the stranger properties of the algebra, and I don't know of another with it. --> NOPE.
You sir, are one big mind fuck. Well done though.
It's all Greek to me.

...That joke's not square. It's a hypercube!
That should actually be hypocube. :p
Oh, you tesseract like you know everything. ;)
You know Fizz, I've been wondering about this. There's no need to define the complement specifically. Any number M works there. You're really proving ((k,k,k,k)-(w,x,y,z))+(w,x,y,z)=(w,x,y,z)-(w,x,y,z)=0, which is the trivial, general case. So why define delta-tag specifically?
You know what, now that I think about it, this entire algebra is just four unions of three dimensional vector spaces: Sp{(1,0,0,0),(0,1,0,0),(0,0,1,0)} U Sp{(1,0,0,0),(0,1,0,0),(0,0,0,1)} U ..., since for every vector (w,x,y,z), where t=min{w,x,y,z}, the identity (w,x,y,z)-(t,t,t,t)=(w,x,y,z) holds. It's a partition of R4.