5 ms·
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.
by ujikoluk 2y ago
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.
- wrsh07 2y agoWhich constraints on a,b (besides positive) does this proof require?
- davrosthedalek 2y agob<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.
- xigoi 2y agoa and b have to be real numbers, whereas the identity works for any commutative ring.
- martin293 2y agoAnd how exactly did you come to the conclusion that this is relevant here?
- davrosthedalek 2y agoBy 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 agoOk 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
- notorandit 2y agoReally? It even holds true for either a=0 or b=0.
- Scarblac 2y agoBut not for b > a.
- deleted 2y ago[deleted]
- wcrossbow 2y agoJust rename a to b and b to a.
- kleiba 2y agoWhy the downvote? That's a correct argument.
- supernewton 2y agoIt is not. a and b are not symmetric in this equation, you can't just swap them.
- olddustytrail 2y agoOf course you can. What do you mean?
- Scarblac 2y agoThe answer is going to be negative regardless of the names, so this geometric proof won't work.
- davrosthedalek 2y ago3^2-2^2 =!= 2^2-3^2. (You can exchange a and b in, say a^2+b^2, because 2^2+3^2=3^2+2^2)
- deleted 2y ago[deleted]