this post was submitted on 01 Dec 2024
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[โ€“] [email protected] 43 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

Edit 0: Okay, so the Hamburger is the Integers Mod 2.

Edit 1: I can't be certain, but my guess is that ๐ŸŒญ is the nth power set of the reals. However, I'm unfamiliar with a topic that naturally contains both Z mod N and a power set of the reals, so I suspect my guess is wrong. Furthermore, I don't know what n could be referring to, other than an arbitrary integer.

Edit 2: The next line has a notation that I'm unfamiliar with, but my guess is that it has to do with Cartesian algebra. I don't know Cartesian algebra, and I'm not even confident I'm remembering the name correctly. I may look into this more later.

Edit 3: Okay, so the final line suggests the line referenced in Edit 2 describes a structure that's comparable to a polynomial ring on the integers Mod 2. Polynomial rings are relatively simple objects that (although most people don't realize it) people often start studying in middle school. Essentially, we're looking at something that behaves like algebra of objects like "2x+1" and "3x+2y+xy+0". I think the main question we should be asking is how many variables this ring has. Given the cardinality of an arbitrary power set of the reals, my assumption is that it's somehow similar to a polynomial ring with some uncountable number of variables. I cannot be certain however.

[โ€“] [email protected] 10 points 2 weeks ago (1 children)

It's (co)homology, not Cartesian algebra. There's also a typo in the meme. I have a fixed version and solution somewhere.

[โ€“] [email protected] 7 points 2 weeks ago

If you could post them here, I'd appreciate it. I find the problem weird and interesting.

[โ€“] [email protected] 24 points 2 weeks ago (2 children)

Leaving a comment to remember to check this post again, in case someone drops the answer to this.

[โ€“] [email protected] 14 points 2 weeks ago* (last edited 2 weeks ago)

Unless you know algebraic topology it's kind of hopeless (but that's the joke). If you're curious, it's the first example on the page on cohomology rings.
Edit: So, you could say H^โ—^(๐ŸŒญ;๐Ÿ”) = ๐Ÿ”[๐Ÿฅš] / (๐Ÿฅš^n+1^) where |๐Ÿฅš| = 1.

[โ€“] [email protected] 12 points 2 weeks ago (1 children)
[โ€“] Siegfried 5 points 2 weeks ago (2 children)
[โ€“] [email protected] 8 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

No, it's the integers Mod 2 (Notation "Z/2Z" where Z is the Integers) which is the only group of order 2 and the smallest non-trivial field.

[โ€“] Siegfried 4 points 2 weeks ago

Oh shit, you are right, i read it as ๐Ÿฅช being an interger. Shit goes deeper and faster.

[โ€“] HonoraryMancunian 2 points 2 weeks ago

๐Ÿ• = โ€“1/12

[โ€“] [email protected] 10 points 2 weeks ago

It's ๐ŸŸ, obviously.

[โ€“] EyIchFragDochNur 9 points 2 weeks ago

I don't even understand the question but my answer is mealtime=t-0sec

[โ€“] [email protected] 7 points 2 weeks ago

You got me and I love it.

[โ€“] [email protected] 4 points 2 weeks ago

It's obvious. A sealion in a hat!

(I hope at least one person gets the reference)

[โ€“] [email protected] 4 points 2 weeks ago* (last edited 2 weeks ago)

Hโ€ข(๐Ÿ‡, ๐Ÿ”) ~= Hโ€ข(๐Ÿ”,๐Ÿฅช)[๐Ÿ‡]