I don't think the maths checks out
this post was submitted on 12 Mar 2025
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Maybe, maybe not. 0/0 can actually approach any value, so if the OP says it's perfect, well, I can't disagree.
Proof by comic
0/0 is undefined because everything would be a valid result. 0/0 is basically finding solutions for 0*x=0 and everything you put into x would satisfy the equation.
It should be 0/2 because out of the 2 things, they have 0.
Thank you! I've always wondered what the answer would be and why. 0 divided by a number is 0. A number divided by itself is 1. A number divided by 0 is undefined. I never knew which one took precedence and why
This doesn't clearly identify a problem IMO. Division by a number is defined as multiplication by the multiplicative inverse, and 0 has no multiplicative inverse because 0x = 1 has no solutions.
0/0 is undefined, not zero.
Indeterminate, to be precise.
Undefined is more precise. 0/0 being an "indeterminate form" refers to expressions of the form lim(x->c) f(x)/g(x) where lim(x->c)f(x) = lim(x->c)g(x) = 0.
After some quick research, I tend to agree with you.
MATH DOES NOT WORK THAT WAY.
GOODNIGHT
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"?
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
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.
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.
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If no work is beyond your means you can opt for the equally balanced shit work and shit life.
my favorite
Is that Asmongold?
You mean the nazi supporter?
This dude does have a job, it’s playing whatever game that is everyday
Westerners are late to the game. Chinese people are already doing it: https://en.wikipedia.org/wiki/Tang_ping
😎
thats my purpose in the future
im working now, to have no work and no life
is this about applying for jobs?
Is could be argued that 0/0=1
since x/x=1
Subbing in x=0 gives the above.
Complicating things is x/0=infinity
Subbing in x=0 gives a contradictory answer.
Anything divided by zero is undefined.
Yep, infinity is not a defined number.
1/0=?......how many times does 0 fit into 1?