Okay, so I had a personal project for a long time that addressed the potential for an algebra that allowed for the multipicitive inverse of the additive identity.
In the context of the resulting non-associative algebra, 0/0=1, rather than 0.
For anyone wondering, the foundation goes as such: Ω0=1, Ωx=ΩΩ=Ω, x+Ω=Ω, Ω-Ω=Ω+Ω=0.
A fun consequence of this is the exponential function exp(x)=Σ((x^n)/n!) diverges at exp(Ω). Specifically you can reduce it to Σ(Ω), which when you try to evaluate it, you find that it evaluates to either 0 or Ω. This is particularly fitting, because e^x has a divergent limit at infinity. Specially, it approaches infinity when going towards the positive end and it approaches 0 when approaching the negative.
There’s more cool things you can do with that, but I’ll leave it there for now.
There’s not much coherent algebraic structure left with these “definitions.” If Ωx=ΩΩ=Ω then there is no multiplicative identity, hence no such thing as a multiplicative inverse.
Okay, so I had a personal project for a long time that addressed the potential for an algebra that allowed for the multipicitive inverse of the additive identity.
In the context of the resulting non-associative algebra, 0/0=1, rather than 0.
For anyone wondering, the foundation goes as such: Ω0=1, Ωx=ΩΩ=Ω, x+Ω=Ω, Ω-Ω=Ω+Ω=0.
A fun consequence of this is the exponential function exp(x)=Σ((x^n)/n!) diverges at exp(Ω). Specifically you can reduce it to Σ(Ω), which when you try to evaluate it, you find that it evaluates to either 0 or Ω. This is particularly fitting, because e^x has a divergent limit at infinity. Specially, it approaches infinity when going towards the positive end and it approaches 0 when approaching the negative.
There’s more cool things you can do with that, but I’ll leave it there for now.
There’s not much coherent algebraic structure left with these “definitions.” If Ωx=ΩΩ=Ω then there is no multiplicative identity, hence no such thing as a multiplicative inverse.
my brain hurts
How did you correct for parallax amete-gramejons?
Interesting. I think it isn’t unital either otherwise Ω=0.
0=Ω+Ω=Ω+ΩΩ=Ω(1+Ω)=ΩΩ=Ω
That seems interesting. Do you have any material/link/blog on this?