In this case, as someone else pointed out below, this proof has unjustified assumptions in it (at least it assumes that b < a).
There are ways to reformulate the above so that "infinity" isn't involved but this is the clearest way to think of it. It isn't much different than the Kronecker delta function delta(t), which is 1 at t=0 and 0 elsewhere. We have lim t->0 delta(t) ≠ delta(lim t->0 t).
But you can fairly assume that wlog.
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"
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.
.. 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.
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.
a^2 - b^2 = a^2 - (-c)^2 = a^2 - c^2 = (a + c)(a - c) = (a + b)(a - b)
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.
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.
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).
It is not necessarily bad but explains why every step isn't always proven. To avoid tedium and long papers.
A visual proof is supposed to appeal to our visual intuition - I don't know about you, but negative areas are not something that is visually intuitive to me.
Kids learn subtraction before they learn negative numbers - once you learn negative numbers, you know that addition and subtraction are almost interchangeable, but this is not necessarily intuitive to begin with.
But, intuition is subjective, so you may need to adjust the terminology to fit the visualization.
In addition to that, for all I know there could be some pitfalls involved with negative areas which I'm not aware of. Even if there aren't any pitfalls, this isn't immediately obvious to someone who isn't familiar with the concept of negative area.
If I'm willing (or forced) to think in such abstract terms, I would much prefer an algebraic proof to this visual proof.
I think that build up to tensor fields should be in every school program. If you can’t think of a field, you’re mathematically disabled and too many basic ideas about real world are inaccessible to you. This limits the ability to vote on a set of topics and participate in non-local decisions that involve systemic understanding. Same for formal logic and statistics.
Once familiarized with that, you can easily start thinking of nonlinearly signed areas, complex areas and areas simultaneously positive and negative by an attribute.
https://youtu.be/60z_hpEAtD8?si=HHs_9m0IJ43nfI3S (~50m video)
TLDW: Yes, the concept is there, makes much more sense than a cross product (which is just an oriented area) and generalizes really nicely.
Alternatively, read: https://en.m.wikipedia.org/wiki/Bivector
Maybe if you viewed an animation where `a` starts larger than `b`, and steps till it's smaller.
Then you could see where the negativity happens. And seeing is nice.
You know position is the integral of velocity right? So say you walk in a straight line from your starting point, then you keep slowing down until you come to a stop and walk backwards past your starting point.
If you were to graph your velocity vs time at some point it would dip below the t axis because your velocity would be negative. Ok cool. If you integrate from the point you came to a stop and started walking backwards you’re calculating the area above the negative velocity curve(between it and the time axis). You’ll find it is a negative area. You know it has to be negative because you walked backwards past your starting point so it gets so negative that it cancels out all the positive area from when you were walking forwards.
You can do all sorts of factorizations the same way and handle negative areas by drawing them in a different color.
The only "but what if..." would be if a=b, which has no geometric proof, but also doesn't need one because "zero = zero times anything" is (by definition) true for fields.
Diagram doesn't show proof for "shorter2-longer2". I believe showing negative area would spark more controversy (imagine that negative area is painted orange, visual area would still be positive).
Shorter2-longer2 gives you same absolute value with reverse signed, so it feels symmetric to me (can't remember what the formal definition is), i.e:
1. a2 - b2 = ( a-b)(a+b) # * -1
2. -(-a2 + b2) = -(-a+b)(a+b)
3. -(b2 - a2) = -( b-a)(b+a) # / -1
4. b2 - a2 = ( b-a)(b+a)
1. (a² - b²) = -(b² - a²), because of even powers
2. (a - b) = -(b - a)
3. (a² - b²) = (a - b)(a + b)
4. -(b² - a²) = -(b - a)(b + a)
3a. (a² - b²) = (a - b)(a + b), a > b
4a. -(b² - a²) = -(b - a)(b + a), b > a
4b. (b² - a²) = (b - a)(b + a), a > b, by inversion
4c. (a² - b²) = (a - b)(a + b), b > a
3a. (a² - b²) = (a - b)(a + b), a > b
4c. (a² - b²) = (a - b)(a + b), b > a
The visual proof is the neat part that people literally can't think of unless you show it to them, after which things might suddenly click for them. The algebraic proof is boring AF and doesn't make for a good maths hook ;)
This is not a visual proof but a nice visualization, like a written explanation that is not a proof.
For an actual novel proof, nobody would imagine that they could eyeball it for a few minutes and conclude it was complete, correct, and consistent - maybe with the exception of professional mathematicians examining for simple proofs. You might eyeball it and follow its logic and not see any immediate flaws, but that's different.