# Linear Algebra - Associative property

1. Dec 11, 2011

### Karamata

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

2. Relevant equations

3. 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?

2. Dec 11, 2011

### JHamm

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

3. Dec 11, 2011

### 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 $$y+z\geq 0$$