Our recent research has arrived at probable uniqueness conditions for the floor (integer part) function. Is it true that:
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfies the conditions:
(a) $f(0)=0$;
(b) The image of $g(x):=x-f(x)$ is $[0,1)$ (i.e., $g(\mathbb{R})=[0,1))$;
(c) For every $x_1,x_2,x_3,x_4\in \mathbb{R}$, $f(x_1)+g(x_2)=f(x_3)+g(x_4)$ implies $f(x_1)=f(x_3)$ and $g(x_2)=g(x_4)$,
if and only if $f$ is the floor function (i.e., $f=\lfloor\; \rfloor$)?
Achievements: It is obvious that the floor function $f=\lfloor\; \rfloor$ satisfies (a), (b), and (c).
Conversely, let $f$ be a real function satisfying the conditions. First, we have $f(\mathbb{R})\cap (-1,1)=\{0\}$. For if $f(x)\in (-1,1)$, then $f(x)+g(0)=f(0)+g(y)$ or $f(x)+g(z)=f(0)+g(0)$ for some $y,z\in \mathbb{R}$ (since $f(0)=g(0)=0$). Hence, (c) implies $f(x)=f(0)=0$.
Now, if $0\leq x<1$, then $f(x)=x-g(x)\in (-1,1)$ and so $f(x)=0$
Also, if $1\leq x<2$, then $g(x-1)=x-1$ and $$ f(x)+g(x)=x=g(x-1)+1 $$ On the other hand $f(1)=1-g(1)\in (0,1]$ thus $f(1)=1$ (since $f(\mathbb{R})\cap (0,1)=\emptyset$).
Therefore $f(x)+g(x)=x=g(x-1)+f(1)$ and so $f(x)=f(1)=1$.
Up to now, we have proved that $f=\lfloor\; \rfloor$ on $[0,2)$. The proof should be completed if and only if we can prove that $f(k)=k$, for all integers $k$ (similar to $0,1$). Because, if this is the case, then $f(x)+g(x)=f(\lfloor x \rfloor)+g(x-\lfloor x \rfloor)$.