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Linear Algebra - Associative property

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]G=\{x\in R|0\leq x<1\}[/tex] and for some [tex]x,y\in G[/tex] define [tex]x*y=\{x+y\}=x+y-\lfloor x+y \rfloor[/tex]

    2. Relevant equations



    3. The attempt at a solution
    I want to proof Associative property:
    [tex]x*(y*z)=(x*y)*z \Leftrightarrow x*(y+z-\lfloor y+z \rfloor)=(x+y-\lfloor x+y \rfloor)*z [/tex]
    [tex]\Leftrightarrow x+y+z-\lfloor y+z \rfloor-\lfloor {x+y+z-\lfloor y+z \rfloor}\rfloor=x+y+z-\lfloor x+y \rfloor-\lfloor x+y+z-\lfloor x+y \rfloor\rfloor[/tex]
    [tex] \Leftrightarrow\lfloor y+z \rfloor+\lfloor {x+y+z-\lfloor y+z \rfloor}\rfloor=\lfloor x+y \rfloor+\lfloor x+y+z-\lfloor x+y \rfloor\rfloor[/tex]
    What now?
     
  2. jcsd
  3. Dec 11, 2011 #2
    The nested floor functions can be taken outside of their encapsulating floor functions since they are integers
     
  4. Dec 11, 2011 #3
    [tex]\lfloor x+y+z-\lfloor y+z \rfloor \rfloor=\lfloor x+y+z\rfloor-\lfloor y+z \rfloor[/tex]
    Right? That seems well.

    Can we do that if [tex]x+y+z<x+y[/tex] (ok, that isn't true in this task), and if [tex]x+y+z\geq 0[/tex] and [tex]y+z\geq 0[/tex]


    Sorry for bad English.
     
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