Prove this affirmation: A necessary condition because, being $x≥0$ and $y≥0$, it follows that $x + y = 0$ is that $x = 0$ and $y = 0$. Prove this affirmation and said if it is a sufficient condition.

We assume that $x$ is strictly $> 0$. Then, as if $x + y = 0$, we have $y = -x <0$. But we have stayed that $y≥0$ !!!!! Therefore, $x = y = 0$ (if you do not see the half that I jumped, it is exactly the same as the symmetric argument).