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#math
<diseaser> do you just enter stuff in the main window?
<asphyxia> Now, the theorem is proven, and for (2) it says: A transposition (x_1,x_2) is of the form (ax) if either x_1 or x_2 is equal to a. If both x_1 and x_2 is different from, it follows that (x_1 x_2) = (a x_1)(a x_2)(a x_1)
<cyclicFifths> diseaser: pretty much
<cyclicFifths> diseaser: there is a command line version of mathematica
<asphyxia> I dont see how
<Jafet> Who owns mbot around here, anyway?
<Dacicus> Cale, I think.
<diseaser> cyclicFifths: how do you run that?
<cyclicFifths> Jafet: good question... i was thinking the same the other day
<diseaser> err is it a cmd line thing?
<diseaser> like a switch or something?
<Jafet> "there is a command line version of mathematica..."
<diseaser> "like a switch or something?"
<cyclicFifths> diseaser: check docs.. i used to know, but i don't use mathematica very often anymore
<diseaser> ok
<diseaser> is it like an interpreter?
<diseaser> that'd be pretty cool
<asphyxia> Is anyone able to explain to me how (x_1 x_2) = (a x_1) (a x_2) (a x_1)
<cyclicFifths> asphyxia: i'm not sure what you mean... these are cycles?
<asphyxia> yeah
<Jafet> I don't think Mathematica supports more than a little rudimentary computing, though.
<asphyxia> cyclicFifths: I stated it above
<asphyxia> Ok, I have a 'main theorem' in my book, which states: 'Let X be a finite set. Then (1) Any permutation o of X can be constructed as a product of transpositions (2) If a in X is a given element, then there can be chosen transpositions on the form (ax) for x in X\{a} for the construction'
<asphyxia> Now, the theorem is proven, and for (2) it says: A transposition (x_1,x_2) is of the form (ax) if either x_1 or x_2 is equal to a. If both x_1 and x_2 is different from, it follows that (x_1 x_2) = (a x_1)(a x_2)(a x_1)
<Jafet> Might want to try http://www.uoregon.edu/~noeckel/Mathematica.html for *nix and Mac OS
<diseaser> ahh its the kernel, I see :)
<diseaser> thanks for the info Jafet
<lokieee> hey guys
<asphyxia> hi lokieee
<cyclicFifths> asphyxia: i'm not sure how to explain this really
<cyclicFifths> asphyxia: Let S = {1,2,3}, then A(S), the group of all permutations of S, is the set of all bijections of S into S
<cyclicFifths> so there are 6 elements in A(S), these are the unique mappings
<asphyxia> ok
<asphyxia> yeah, I see that
<lokieee> i had a simple question... I have, Cycle time of X = 1/2 Ghz = 500ps.. well i looked up p, and it stands for peco, which is E-12....Ghz = E9. So i took 1/2E9 =5.E-10. Well if its 500ps...wouldn't I need to move that decimal 2 places to the left? and get .005E-12? or .005ps..i thought moving it to the right would make it E-8
<cyclicFifths> I would suggest making some examples of these mappings
<cyclicFifths> and try to break one of the mappings down to 2-cycles
<cyclicFifths> cycles of length two that is
<cyclicFifths> not sure if that helps really
<asphyxia> like:
<asphyxia> 1 2 3
<asphyxia> 1 3 2
<cyclicFifths> = (2 3)
<asphyxia> Yeah, I see that.
<cyclicFifths> hmm, i was hoping i could wrap back to your question...
<asphyxia> Oh, so the theorem states that I construct transpositions of transpositions
<asphyxia> ehm, maybe not. I just think I have to read it again
<wolfbone> ...a...x1...x2... -> ...x1...a...x2... -> ...x1...x2...a... -> ...a...x2...x1...
<asphyxia> wolfbone: If that is to my example, I dont get it. The way I read it, is that a -> x1 and x1 -> x2 and x2 -> a.
<asphyxia> But here you state (after the first ->) that x_1 -> a and a -> x_2.
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