this post was submitted on 12 Mar 2025
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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.
Interesting. I think it isn't unital either otherwise Ω=0.
0=Ω+Ω=Ω+ΩΩ=Ω(1+Ω)=ΩΩ=Ω
Someone else had the same observation, but it is unital. Keep in mind that it isn't associative; you can't pull out the Omega like that.
The definition I'm aware of for non associative algebras has them distributive by default, so I believe the chain of equations is valid.
my brain hurts
That seems interesting. Do you have any material/link/blog on this?
No, I'm pretty shy about my work in-person and I don't like linking my online and IRL self. Do you have any recommendations for places to put my work?
Sadly, no. However, you could maybe do a personal blog, similar to how Terrence Tao does.
I really encourage you to try, it could help you find new stuff, check for mistakes, clarify ideas, and maybe even hear ideas from others.
I appreciate your encouragement; it's an extremely rare occurrence when I discuss my ideas with others. I'll think about what you've said and if I follow through I hope to remember to send you a message. I'm favouriting this comment so I can find it again.
That'll make me really glad :)
We can also exchange contact information via private message if you want to.
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.
No; 1 is the multiplicative identity.
1Ω=Ω, and for all x in C 1x=x. Thus, 1 fulfills the definition of an identity.
1 = Ω0 = Ω(Ω + Ω) = ΩΩ + ΩΩ = Ω + Ω = 0
so distributivity is out or else 1 = 0
Correct; multiplying by Ω doesn't distribute over addition.
Distributivity is a requirement for non associative algebras. So whatever structure is left is not one of those