Es sei ${\displaystyle {}x-\left\lfloor x\right\rfloor =y-\left\lfloor y\right\rfloor }$. Da ${\displaystyle {}\left\lfloor x\right\rfloor ,\left\lfloor y\right\rfloor }$ ganze Zahlen sind, ist ${\displaystyle {}n=\left\lfloor y\right\rfloor -\left\lfloor x\right\rfloor }$ ganzzahlig. Damit gilt

${\displaystyle {}{\begin{aligned}y&=\left\lfloor y\right\rfloor +(y-\left\lfloor y\right\rfloor )\\&=\left\lfloor y\right\rfloor +(x-\left\lfloor x\right\rfloor )\\&=x+\left\lfloor y\right\rfloor -\left\lfloor x\right\rfloor \\&=x+n.\end{aligned}}}$

Sei nun ${\displaystyle {}y=x+n}$ mit ${\displaystyle {}n\in \mathbb {Z} }$. Aus der definierenden Beziehung

${\displaystyle {}\left\lfloor x\right\rfloor \leq x<\left\lfloor x\right\rfloor +1\,}$

folgt

${\displaystyle {}\left\lfloor x\right\rfloor +n\leq x+n<\left\lfloor x\right\rfloor +n+1\,,}$

daher muss

${\displaystyle {}\left\lfloor x+n\right\rfloor =\left\lfloor x\right\rfloor +n\,}$

sein. Somit ist

${\displaystyle {}{\begin{aligned}y-\left\lfloor y\right\rfloor &=x+n-\left\lfloor x+n\right\rfloor \\&=x+n-(\left\lfloor x\right\rfloor +n)\\&=x-\left\lfloor x\right\rfloor .\end{aligned}}}$