This article from wikipedia defines a Fano variety as a complete variety whose anticanonical bundle is ample. It also states that:
A Fano surface is also called a del Pezzo surface. Every del Pezzo surface is isomorphic to either P1 × P1 or to the projective plane blown up in at most 8 points, which must be in general position. As a result, they are all rational.
After looking it up, I've found that some authors define a del Pezzo surface as either $\Bbb{P}^1\times\Bbb{P}^1$ or the blowup of $\Bbb{P}^2$ at $8$ points at most, while others define it as a projective surface with ample anticanonical divisor (i.e. Fano, as in wikipedia).
This appears to be an equivalence, and I'm trying to understand it.
I was able to prove that $-K_{\Bbb{P}^1\times\Bbb{P}^1}$ is ample, and if $S$ is the blowup of $r$ points ($1\leq r \leq 8$) I can show that $(-K_S)^2=9-r>0$. This last result is expected from an ample divisor, but I still can't prove that $-K_S$ is ample.
Conversely, I still can't see how $-K_S$ being ample implies it must be some blowup of $r\leq 8$ points. How do I do that?
Any help will be appreciated, thank you!