Prove that the given set is a finite set. $A = \{(a,b) \in N^2 | (2+a)(2-b) \ge 2(a-b)\}$. Prove that A is a finite set.
I'm having problem with proving A is a finite set. So far,
Let $x, y \in A$. Then,
\begin{equation*}
\begin{aligned}
    (2+x)(2-y) &\ge 2(x-y)\\
    4-2y+2x-xy &\ge 2x-2y\\
    4-xy &\ge 0\\
    xy &\le 4
\end{aligned}
\end{equation*}
This is what I have, but I'm not sure that showing $xy \le 4$ is enough to show that we have finite set. The reason why I thought this was enough because x and y are natural numbers. I searched other proofs and saw someone used contradiction to prove this is a finite set by assuming A is a infinite set. Do you guys think I need contradiction? or is it enough to prove A is a finite set?
 A: Of course it's fine to show directly it's a finite set if that is what you're trying to prove! Your proof is fine. If we write it out very explicitly: you have shown shown for an arbitrary element $(x,y) \in A$, we have $xy \leq 4$. Since we know that $x$ and $y$ are natural numbers, this then implies that ($x=1 \text{ and } y=4$) or ($x=2 \text{ and } y=2$) or ($x=4 \text{ and } y=1)$ etc. There will only be a finite number of terms here (you can write them all out explicility). Hence you've shown whenever $(x,y)$ is any element of $A$, it must be either $(1,4)$ or $(2,2)$ or $(4,1)$ or one of these other finitely many solutions.
I.e. you have proven that $A\subset \{(1,1),(1,2),(1,3),(1,4),(2,2),(4,1),(3,1),(2,1)\}$ (in fact we have exact equality by the contrapositive statement that not being one of these pairs means you are not in $A$). So obviously $A$ is a finite set!
Also, I can't see a way to prove this by contradiction that is more straightforward than what you've done. What you've done proves that the number of solutions for the equation, and hence the elements of $A$, is a finite number. You could say that if $A$ were infinite, then $A$ would have to contain a solution that is not one of the possible ones. But this is entirely pointless, as the result is already proved.
