The Question
Solve for $(x,y,z)$
$$\left\{\begin{matrix} x+\left \lfloor y \right \rfloor + \left \{ z > \right \}=1,1\\ z+\left \lfloor x \right \rfloor + \left \{ y \right > \}=2,2\\ y+\left \lfloor x \right \rfloor + \left \{ z \right \}=3,3 \end{matrix}\right.$$
My Understanding
From the definition of floor function we know that $n = \left \lfloor n \right \rfloor + \left \{n \right \}$ and if you play with equation a bit $(1,1+,2,2=3,3)$ and do some cancelling it's easy to see that
$$x+z=0 \Rightarrow x = -z.$$
After that you set $z=-x$ and rewrite the equation. You'll see from $[2(1,1)=2,2]$ and a bit cancelling $\left \{ y \right \} - \left \{ x \right \}=2,2$. Also an important note that $\left \lfloor x \right \rfloor \neq \left \lfloor -x \right \rfloor , \left \{ x \right \} \neq \left \{ -x \right \}$ example: $\left \lfloor 5,35 \right \rfloor = 5$ but $\left \lfloor -5,35 \right \rfloor = -6$. I got this so far and need help.
These equations have no solution since the LH sides of equations (2) and (3) have the same fractional part, formed from the sum of the fractional parts of $y$ and $z$, and yet differ by $1.1$.
N.B. Looking at the comment of @cosmo5 I suspect that the last equation should be as in that comment. Then there is the solution $x=0.1,y=1.2,z=2$.