[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).

[davrosthedalek]

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]

[olddustytrail]

Of course you can. What do you mean?

[Scarblac]

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

[davrosthedalek]

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)

[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)