Problem with Floor Function

Question

I have $f_k(x)=kx-\lfloor kx \rfloor$, where $k\in \mathbb N$ and $x\in\{0,1\}$ and $x\in \mathbb Q$. When I plug in some numbers it seems obvious that $$ f_k(x)+f_k(1-x)=1 $$ for example $k=28, x=\frac{21}{38}$ gives $$ 28\frac{21}{38} - \lfloor 28\frac{21}{38} \rfloor + 28\frac{17}{38} - \lfloor 28\frac{17}{38} \rfloor = \frac{294}{19} - 15 + \frac{238}{19} - 12 = \frac{532}{19} -27=1 $$ but I have trouble to prove this in general. I get $$ (x+(1-x))k-\lfloor kx\rfloor - \lfloor k(1-x)\rfloor=k -\lfloor kx\rfloor- \lfloor k(1-x)\rfloor $$ How to continue?

Answer 1

Continuing your calculation: \begin{align}k-\lfloor kx\rfloor-\lfloor k-kx\rfloor=k-\lfloor kx\rfloor-k-\lfloor-kx\rfloor=-\lfloor-kx\rfloor-\lfloor kx\rfloor=\lceil kx\rceil-\lfloor kx\rfloor\end{align} So it's $0$ if $kx\in\mathbb Z$, else it's $1$.

Answer 2

$f_k(x) = \{kx\}$. Then let $f_k(x) = a$. Note that $f_k(1-x) = \{k(1-x)\} = \{-kx\} = \{1-kx\}$ since $\{k\} = 0$ so the result follows.