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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.


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


a and b have to be real numbers, whereas the identity works for any commutative ring.


And how exactly did you come to the conclusion that this is relevant here?


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.


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


> (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.


I'm not sure I quite understand what the visual proof looks like if they're negative


Check the quadrant which the result ends up in.


b<a Edit: and b,a element of R, but ok...


Really? It even holds true for either a=0 or b=0.


But not for b > a.


Just rename a to b and b to a.


Is that allowed? You will prove another equation. You cannot swap pi and e either.


You are not swapping the values, you are swapping the names.


I forgot the i for irony...


pi and e aren't names; they're values. Of course you can't swap values.


Why the downvote? That's a correct argument.


It is not. a and b are not symmetric in this equation, you can't just swap them.


You can swap them without loss of generality (WLOG).


No, this is not correct. WLOG means: I assume one of the possible cases, but the proof works the same way for other cases. But that's not true here. The proof, as shown, only works for a>b>0, it does not work (without extra work or explanation) for a<b. The proof for a<b is similar, but not the same. [And it certainly does not show it for a,b element of C]


WLOG just means the other cases follow from the one case. There is no implication about how hard it is to get to the other cases, although generally it is easy and you don't bother spelling it out exactly.


Of course you can. What do you mean?


The answer is going to be negative regardless of the names, so this geometric proof won't work.


3^2-2^2 =!= 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)


This is not what I meant. What is being proved is: a^2-b^2 - (a+b)(a-b) = 0. If you swap a and b you end up with a sign switch on the lhs which is inconsequential.


That is not what the proof proves. The proof proves the equivalence how it was originally stated, and assumes for that b<a.

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.


Did you think that I meant you can switch them on one side of the equation but not the other?

That's not what anyone is saying.


No, of course not.


But that's literally what you just did in your example.


I did not show the right side at all, so I am not sure how you can make that statement.

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.


(a^2 - b^2) = -(b^2 - a^2)

use the same visual proof but with a and b switched to get

-(b + a)(b - a) = (a + b)(a - b)




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