An algebra problem by Wildan Bagus Wicaksono

$\large \sqrt { x+y } =\sqrt { x } +\sqrt { y } \\ x+y={ \left( \sqrt { x } +\sqrt { y } \right) }^{ 2 }\\ x+y=x+y+2\sqrt { xy } \\ 0=2\sqrt { xy } \\ 0=xy$ Thus, $x = 0$ or $y = 0$ or $x = y = 0$ .

The smallest value of $\left| \left\lfloor x+y \right\rfloor \right|$ will be obtained if $x = y = 0$ . So, $x + y = 0$ , $\\ \left| \left\lfloor x+y \right\rfloor \right| =0$