That seems to show that there exist a and b such that the equality holds. But not that it holds for all a and b.
You can chart a and b on a 2D coordinate system, where they're allowed to be negative. Even positivity is not strictly required here.
How is it not obvious to the dullest of the dull that this visual proof is not supposed to work for goddamn commutative rings lmao
It's probably not even supposed to work for negative reals, 0 or the case b>a. It's supposed to demonstrate the central idea of the visual proof. Also yes, by choosing suitable ways to interpret the lengths shown in the diagrams it's absolutely possible to extend the proof to all reals but I'm not convinced it's meant to be interpreted like that.
But bringing commutative rings into this... man you're funny
(You can exchange a and b in, say a^2+b^2, because 2^2+3^2=3^2+2^2)
Your rewriting is of course true for all a,b and might be used in an algebraic proof. But this transformation is not at all shown in the geometric proof.
That's not what anyone is saying.
The point is that a+b is symmetric in a <-> b and a-b is anti-symmetric. Both left and right side are anti-symmetric.
use the same visual proof but with a and b switched to get
-(b + a)(b - a) = (a + b)(a - b)