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#math
<Syzygy-> MatiaB: I didn't understand that last comment.
<Hansfire> ? If you went around Earth at orbital velocity--one circuit in 90 minutes--the centrifugal force would just balance gravity. Our rotation speed (one circuit in about 24 hours) is nowhere near that.
<MatiaB> Syzygy-, oki, What is would like is to proove that if s is a permutation of f, then is fs := f o s o f^(-1) is a permutation of Y
<Syzygy-> Yes.
<Syzygy-> MatiaB: Now, PLEASE, state precisely what your working definition of a permutation is.
<Syzygy-> Without knowing that it'll be quite difficult to suggest a proof method.
<Hansfire> my problem is to find the most suitable frame of reference grid or reference point
<MatiaB> Syzygy-, hehe :D That would be a bejective map sigma : X -> X, a map from X into X
<Syzygy-> Right. Thank you.
<Hansfire> is there a channel where they do physics as well?
<Crito> #physics
<Syzygy-> MatiaB: Do you see how to prove that if f:X->Y is bijective and g:Y->Z is bijective, then g o f:X->Z is bijective.
<Hansfire> thanks
<MatiaB> Syzygy-, Nope, i think it there i go wrong. I dont know that the g: Y -> Z is for
<Syzygy-> Suppose you have three sets, X,Y,Z. And two functions, f:X->Y and g:Y-Z.
<Syzygy-> Then the composition of f and g is bijective.
<Syzygy-> (i.e. the two functions are supposed to be bijective)
<MatiaB> if i know f and g is bijective
<Syzygy-> If you know f and g are bijective, you can prove that their composition is bijective.
<MatiaB> okay, and that is what i actually would like to proove?
<Syzygy-> Yes.
<Syzygy-> Since if you have that fact, the rest is trivial.
<Syzygy-> The rest, after that, is just seeing that f o s o f^-1 goes between the right sets.
<toed> what about the converse? if f o g is bijective must f and g be?
<Syzygy-> toed: No.
<MatiaB> So f: X -> Y, would be f o s and from there it would be g: Y -> Z would be s o f^(-1), do i get that right? :)
<Syzygy-> Let f:{0,1} -> Integers be the identity, and g:Integers -> {0,1} be (mod 2). Then the composition is bijective, but each of them isn't.
<Syzygy-> MatiaB: Not quite.
<MatiaB> Syzygy-, damn :D
<Syzygy-> If composition preserves bijectivity, then any (finite) chain of bijections compose to a bijection, and not only pairs.
<toed> ah righto
<Syzygy-> And f, f^-1 and s are all bijections.
<MatiaB> Syzygy-, hmm, okay :D, Ill take some paper and see if i can get it working. Thanks for the help so far :)
<Syzygy-> kk
<pjd> hello
<pjd> is anyone familiar with the idea of representing rational numbers as single lists of prime exponents?
<pjd> such that [a, b, c, ...] corresponds to 2^a * 3^b * 5^c * ...
<pjd> where the exponents are possibly-negative integers
<Jafet> It's called the "prime-power factorization" of a number.
<Jafet> When the number is integral, of course. When it's rational... don't know a common name for it, though.
<Jafet> Got a question?
<pjd> well, i was thinking about representing rational numbers
<TRWBW> pjd: haven't seen that used, but it looks simple enough.
<TRWBW> pjd: do you see how to represent integers this way?
<pjd> and this representation has the useful property of being intrinsically normalized
<pjd> and making it easy to find numbers related by prime factors
<pjd> so i'm interested to know if they've been studied anywhere
<TRWBW> pjd: sure. it's just the reduced form of a rational, prime factor top and bottom, use positive exponenets for the top and negative ones for the bottom.
<pjd> TRWBW: exactly
<Jafet> Addition and subtraction would be an arse in that kind of representation, though.
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