# Prove that the equation x^2 - [x]x + [x] - {x} = 0 has only one solution

$\\x^2-[x]x+[x]-\left \{ x \right \}=0\\Since,x=[x]+\left \{ x \right \}\\Our\:equation\:become\\x^2-(x-\left \{ x \right \})x+x-\left \{ x \right \}-\left \{ x \right \}=0\\x(\left \{ x \right \}+1)-2\left \{ x \right \}=0\\0\leq \left \{ x \right \}<1$