alt Hacker NewsThis isn't unique, or even the least compute way to do this. For example, let f(x,y) = 1/(x-y). This too is universal. I think there's a theorem stating for any finite set of binary operators there is a single one replacing it.
write x#y for 1/(x-y).
x#0 = 1/(x-0) = 1/x, so you get reciprocals. Then (x#y)#0 = 1/((1/(x-y)) - 0) = x-y, so subtraction.
it's common problem to show in any (insert various algebraic structure here ) inverse and subtraction gives all 4 elementary ops.
I haven't checked this carefully, but this note seems to give a short proof (modulo knowing some other items...) https://dmg.tuwien.ac.at/goldstern/www/papers/notes/singlebi...
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I think this is the novel bit:
> This includes constants such as e, pi, and i; arithmetic operations including addition, subtraction, multiplication, division, and exponentiation as well as the usual transcendental and algebraic functions.
I don't think this can do any of the "standard" constants or what we generally consider to be closed-form expressions, though ! (E.g., no e, pi, exp, log, etc.)
> This too is universal
Could that be used to derive trigonometric functions with single distinct expressions?
You win the internet today!
yes, but are you currently experiencing both hypergraphia and chatbot AI induced psychosis while also thinking about this problem?
Good find.
It cites a paper from 1935: https://www.pnas.org/doi/10.1073/pnas.21.5.252
Here is a bit more: https://mathoverflow.net/questions/57465/can-we-unify-additi...