3 ms·

Which constraints on a,b (besides positive) does this proof require?

• davrosthedalek 2y ago
  b<a Edit: and b,a element of R, but ok...
• phoe-krk 2y ago
  > (besides positive) 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.
  • wrsh07 2y ago
    I'm not sure I quite understand what the visual proof looks like if they're negative
    • phoe-krk 2y ago
      Check the quadrant which the result ends up in.
• xigoi 2y ago
  a and b have to be real numbers, whereas the identity works for any commutative ring.
  • martin293 2y ago
    And how exactly did you come to the conclusion that this is relevant here?
    • davrosthedalek 2y ago
      By the fact that the geometric proof in the link wants to proof the formula, but only does so for a small subset of all a,b for which the formula is correct. This makes it a partial proof, at best.
      • martin293 2y ago
        Ok nvm I can't resist wasting my time and typing stuff on the internet again, probably gonna regret it later. 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