[ujikoluk]

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]

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

[xigoi]

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

[martin293]

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

[davrosthedalek]

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]

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

[phoe-krk]

> (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]

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

[phoe-krk]

Check the quadrant which the result ends up in.

[davrosthedalek]

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

[notorandit]

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

[Scarblac]

But not for b > a.

[wcrossbow]

Just rename a to b and b to a.

[nuancebydefault]

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

[Ma8ee]

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

[nuancebydefault]

I forgot the i for irony...

[kelnos]

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

[kleiba]

Why the downvote? That's a correct argument.

[supernewton]

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

[henry2023]

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

[olddustytrail]

Of course you can. What do you mean?

[Ma8ee]

(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)