Linear Algebra - Associative property

  • Thread starter Karamata
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  • #1
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Homework Statement


[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]

Homework Equations





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?
 

Answers and Replies

  • #2
390
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The nested floor functions can be taken outside of their encapsulating floor functions since they are integers
 
  • #3
60
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[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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