Why not? Either b < a, b = a or b > a. If b = a both sides are zero and the result is trivial. If b > a just swap the names of a and b and multiply both sides by -1.

With this additional justification I accept it, yes. But (imo) now you are doing algebra. And then you might as well prove it just by distributing the RHS of the original expression.

This additional justification is so trivial it's just left out for brevity. In general a lot of statements are accepted even if not each and every detail is spelled out.

Let me fully spell out the proof that generality is not lost

Assume that b > a.

1. Swap the names a and b.

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

2. Multiply both sides by minus one

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

3. Absorb the minus into the second factor

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

This is the same as the original equality we wanted to prove.

Now compare that to the purely algebraic proof for the whole theorem.

1. Distribute over the first parenthesis

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

2. Distribute again

(a+b)(b-a) = a^2-ab + ba-b^2

3. Cancel equal terms

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

The proof that generality is not lost is of similar length to a full proof of the theorem!

So if you think the former can be skipped, then you must accept a proof of the whole theorem that simply reads "Follows from trivial algebraic manipulation"

They may be equally long, but length when written is IMO not a good measure of complexity.

The 2nd I would need pen and paper for to keep my head straight doing the distributive law. The 1st I can do in my head as:

"If b is larger than a, the left side will be negative instead of positive, but also "b-a" gets a negative sign, and these cancel each other, so it is the same"

But I agree with you anyway ...this is indeed "doing algebra".

I agree that the figure would have been better if it said that in that figure it displays the case where a is larger than b. And we should probably call it visualization of an algebraic proof, instead of visual proof.

The entire proof is trivial.

Sorry, but no. First, "just switch a and b" is clearly wrong, and we had multiple people propose that (Without the adjustment of sign). So not that trivial, I would argue. At least not more trivial than the underlying equality one wants to prove in the first place.

> just swap the names

Then you've just skipped the case when a^2 - b^2 is negative. The diagram does not prove that case and swapping the names still doesn't prove it.

> Then you've just skipped the case when a^2 - b^2 is negative.

Not really. If b > a, then swap them to conclude that b^2 - a^2 = (b + a)(b - a), which is what the visual proof demonstrates.

Your conclusion is equivalent to saying that a^2 - b^2 = (a + b)(a - b).

if you just extend the metaphor in the diagram, and imagine a negative length to just refer to direction, sure it does :)

personally, I love visual proofs because they can communicate an idea efficiently, sure they have their pitfalls, but its less about the actual mechanism of the proof and more about the core idea that lets me appreciate how its working- and visual proofs add a pseudo-physical intuition that helps me appreciate it.

Trying the proof with a < b, with the b square from the bottom-right as in the diagram, I get a region to the top and left, and moving a piece (differently to the diagram) I get (a + b)(b - a) as a positive area for that region, and then flip the sign because it's negative.

What is a is positive and b is negative? How do you extend the visual proof?

If b is negative, define c = -b, and use the linked result to show that

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

You can do this if you extend your concepts of length and area to signed quantities. You have to be clear to explain how the setup of the drawing works, but instead of cutting away a square of side b out of the corner of the square of side a, you might end up appending a negatively-signed square of side −b.

I am not convinced since in this world you can have the following rects:

1 x 1

1 x -1

-1 x -1

Is just the middle one a cut out?

I reckon you need the axiom from the field to do generalize this to negative numbers.

https://en.m.wikipedia.org/wiki/Field_(mathematics)

There’s no need for this extension: https://news.ycombinator.com/item?id=42425681

Well the whole point here is that we're doing geometry. If you just want to do symbol twiddling you don't need the picture at all.

The purely geometric proof works just fine without the case analysis if you use the oriented area.

You do it 3 times, a=b, a<b, a>b.

.. now that I think about it, the "visual proof" only 'proves' the statement for a specific 'a' and 'b'. Probably there is a proof, that can handle all 'a' and 'b' pairs at once.

Turn your monitor sideways?