Linear Algebra - Associative property

[Karamata]

Homework Statement

G=\{x\in R|0\leq x<1\} and for some x,y\in G define x*y=\{x+y\}=x+y-\lfloor x+y \rfloor

Homework Equations

The Attempt at a Solution

I want to proof Associative property:
x*(y*z)=(x*y)*z \Leftrightarrow x*(y+z-\lfloor y+z \rfloor)=(x+y-\lfloor x+y \rfloor)*z
\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
\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
What now?

 

[JHamm]

The nested floor functions can be taken outside of their encapsulating floor functions since they are integers

 

[Karamata]

\lfloor x+y+z-\lfloor y+z \rfloor \rfloor=\lfloor x+y+z\rfloor-\lfloor y+z \rfloor
Right? That seems well.

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


Sorry for bad English.