i found this high-quality math shitpost screencap on my laptop and i have no idea who wrote it, but it should be shared with the world before i delete it from my hard drive

@metapianycist this proof is incomplete! from this data we cannot conclude that there are fucks with *irrational* coefficients, therefore we cannot confirm that the mapping from fucks to the complex numbers must be surjective, and cannot conclude that the mapping is isomorphic! Fucks may merely be isomorphic with the complex numbers with *rational* coefficients, and have an *injective* relationship with the complex numbers!

@snep i came here to say this and i cannot fucking believe i was beaten to it

@snep @metapianycist fucks are isomorphic to Q(i), and are therefore denumerable, rather than having the cardinality of the continuum

@metapianycist @Canageek so we’ve got complex fucks, but limited to the rationals. We can extend this into the irrationals with “some crazy fuck.”

@DialMforMara @metapianycist @Canageek or equivalently with "what the flying fuck", which implies that there are continuous paths in fuckspace

@serindelaunay @DialMforMara @metapianycist @Canageek I'm curious how functions fit in all this. Pretty sure they come in imperial and metric, but not certain.

@RecursiveRabbit @serindelaunay @metapianycist @Canageek The fuckton is inherently imperial, because you have to say "metric fuckton" to disambiguate

@lanodan @metapianycist We now have to ask if there an ISO standard for fucks
@aearil @metapianycist I wish all ISO standards would be Open and available easily (like RFCs).

@metapianycist Does this mean that there's algebraic fucks? Fuck which we can represent by polynomials, where the fucks are the roots of those polynomials?

I don't have a good joke for this, I just like algebraic numbers.

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