Flippanarchy

292 readers

346 users here now

Flippant Anarchism. A lighter take on social criticism with the aim of agitation.

Post humorous takes on capitalism and the states which prop it up. Memes, shitposting, screenshots of humorous good takes, discussions making fun of some reactionary online, it all works.

This community is anarchist-flavored. Reactionary takes won't be tolerated.

Don't take yourselves too seriously. Serious posts go to [email protected]

Rules

If you post images with text, endeavour to provide the alt-text
If the image is a crosspost from an OP, Provide the source.
Absolutely no right-wing jokes. This includes "Anarcho"-Capitalist concepts.
Absolutely no redfash jokes. This includes anything that props up the capitalist ruling classes pretending to be communists.
No bigotry whatsoever. See instance rules.
This is an anarchist comm. You don't have to be an anarchist to post, but you should at least understand what anarchism actually is. We're not here to educate you.

founded 6 months ago

MODERATORS

[email protected]

499

Civilization (lemmy.dbzer0.com)

submitted 3 weeks ago by [email protected] to c/[email protected]

100 comments fedilink hide all child comments

you are viewing a single comment's thread
view the rest of the comments

[–] [email protected] 4 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

No. The standard field (that is, a ring where both operations are abelian groups) on the complex numbers doesn't have a multiplicative inverse of 0; rings can't have a multiplicative inverse for the additive identity. You can create an algebra with a ring as a sub-algebra with such, but it will no longer be a ring. My preferred method is to impose such an algebra on the one-point compactification of the Complex Numbers, where the single added point is denoted as "Ω".

I started this project when I was 12, and when I could show that the results were self-consistent this was what I had settled on:

let z be a complex number that is not otherwise specified by the following equations. Note: the complex numbers contain the Real numbers, and so the following equations apply to the them as well.

0Ω=Ω0=1

z+Ω=Ω+z=zΩ=Ωz=Ω=ΩΩ

Ω-Ω=0. Ω-Ω=Ω+(-Ω)=Ω+(-1Ω)=Ω+Ω=0

The algebra described above is not associative. That is to say, (AB)C does not always equal A(BC).

[–] [email protected] 1 points 2 weeks ago

Addendum: despite what my earlier statement implied, there is exactly 1 ring for which the additive identity has a multiplicative inverse: the trivial ring, which has only 1 element (that I will label as 0).

The operations are a such: 0+0=0=0*0. Note: this is also the only ring for which the additive and multiplicative identities are the same element.

I was originally going to mention it, but I didn't want to make my comment more complicated than in needed to be. Then I realized that the way I phrased it was technically inaccurate, and so this addendum exists.