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.
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
What the fuck are you talking about? That's incorrect as a matter of simple fact.
Associativity is a property possessed by a single operation, whereas distribution is a property possessed by pairs of operations. Non-associative algebras aren't even generally ones that posses multiple operations, so how the hell do you think one implies the other?
Edit: actually, while we're on it, your first comment was nonsense too; you don't know what an identity is and you think that there's no notion of inverses without an identity. While that's generally the case there are exceptions like in Latin Squares, which describe the Cayley Tables of finite algebras for which every element can be operated with some other element to produce any one target element. In this way we can formulate a notion of "division" without using an identity.
Algebras have two operations by definition and the one thing they have in common is that the multiplication distributes over addition.
Yes, there is no notion of inverses without an identity, the definition of an inverse is in terms of an identity.
Stop posting.
Do you think a group isn't an algebra? What, by your definitions make an "Algebra" different from a "Ring"?
A group is not an algebra. A group consists of a single associative binary operation with an identity element and inverses for each element.
A ring is an abelian (commutative) group under addition, along with an additional associative binary operation (multiplication) that distributes over addition. The additive identity is called zero.
A field is a ring in which every nonzero element has a multiplicative inverse.
A vector space over a field consists of an abelian group (the vectors) together with scalar multiplication by elements of the field, satisfying distributivity and compatibility conditions.
A non-associative algebra is a vector space equipped with a bilinear multiplication operation that distributes over vector addition and is compatible with scalar multiplication.
An (associative) algebra is a non-associative algebra whose multiplication operation is associative.
You can read more about these definitions online and in textbooks - these are standard definitions. If you are using different definitions, then it would help your case to provide them so we can better understand your claims.
my brain hurts
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.
How did you correct for parallax amete-gramejons?
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.