<!-- 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.