A visual proof that a^2 – b^2 = (a + b)(a – b)
futilitycloset.com
futilitycloset.com
A few years ago I re-drew a bunch of these in latex with my PhD advisor and another colleague [3]. We planned to print them as posters and hang them for a Pi day event that unfortunately never happened because the pandemic broke out.
[1] https://www.amazon.com/Proofs-without-Words-Exercises-Classr...
[2] https://en.m.wikipedia.org/wiki/Proof_without_words
[3] https://www.antonellaperucca.net/didactics/proof-without-wor...
I think it would be useful for people downloading this and then not remembering where they got it from. It would be cool to be able to give credit where its due :).
In this case, as someone else pointed out below, this proof has unjustified assumptions in it (at least it assumes that b < a).
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.
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.
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 ;)
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).
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.
https://www.dbai.tuwien.ac.at/proj/pf2html/proofs/pythagoras...
I find this much more "useful" since the Pythagorean theorem isn't immediately intuitive to me.
As for the proof in the original post, it seems really redundant to me. it follows from a (b+c) = ab + ac.
And while building intuition for this distributive property of multiplication is extremely essential when teaching maths, I feel that the intuition for why this is true is better built without leaning on geometry.
Even Pythagoras applies to any dimension, although admittedly it doesn't quite fit its usual statement in terms of triangles for higher dimensions: if a vector v has components (v₁, v₂, ...) then its length squared equals v₁² + v₂² + ...
Do you mean that x.y = x₁y₁ + x₂y₂ and x.x = |x|², so it follows directly from that? If you define the dot product to be the first of those identities then you need Pythagoras's theorem to prove the second, so your argument is circular.
(Or you can prove that x.y = |x| |y| cos θ but that's even further removed from the component-wise definition than Pythagoras's theorem. Or you can define the dot product that way, but then you still have to prove the component-wise formula from it.)
In an orthogonal basis this is trivial because cos(Pi/2) = 0 though…
Sorry, but you can't bypass proving Pythagoras's theorem by definition of the dot product or anything else.
(emphasis mine)
Well that's not wrong (because proving Pythagoras' theorem is pretty straightforward anyway) but at the same time the one trigonometric formula you need (cos(a-b) = cos(a)cos(b)+sin(a)sin(b)) “follows from a (b+c) = ab + ac” if you start from Euler's formula.
To be honest, your comments have been quite low effort. They amount to "yeah but that bit's pretty easy too" while leaving it to me to work out how (and whether) your points fit into a coherent proof. I do get why this stuff all feels so trivial: we usually skip over it in higher-level proofs. But the only reason we can is that we can use nice abstractions like the dot product with its equivalent definitions, and that's thanks to the foundation these lower-level theorems provide.
I does follow from the definition and common properties of the dot product, which was my original point. But you claimed it was circular because these properties derived from Pythagora's theorem, and so we've ended up showing it doesn't need to. And this later part was obviously much more involved than just “using the dot product”. But that's as if we had to prove that real numbers' multiplication is actually distributive over addition when saying “it follows from a (b+c) = ab + ac”, it's far from trivial if you want to go this far…
> To be honest, your comments have been quite low effort. They amount to "yeah but that bit's pretty easy too" while leaving it to me to work out how (and whether) your points fit into a coherent proof. I do get why this stuff all feels so trivial: we usually skip over it in higher-level proofs. But the only reason we can is that we can use nice abstractions like the dot product with its equivalent definitions, and that's thanks to the foundation these lower-level theorems provide.
I agree with you here, even on the low-effort part, I'm not particularly comfortable writing math on a keyboard and especially not on an English speaking forum because the notations are very different than the ones we use in France.
Then Pythagoras's theorem falls out of that - if we have two vectors u, v that are orthogonal, we can use the conditions in the definition of the dot product to prove that ||u + v||^2 = ||u||^2 + ||v||^2 (where ||u|| is the norm of u, defined as sqrt(<u,u>)).
(It's not a hard proof, because the definition of dot product says it's additive in the first slot, meaning <u+v, w> = <u,w> + <v, w>. So it's easy to prove things about sqrt(<u+v, u+v>) by splitting out the u's and v's. It's a bit hard to write this on HN because no mathjax though.)
It's not hard to show all that, but by the time you've done it you'll have accidentally proved Pythagoras's theorem along the way. It's like you've gripped on a tube of toothpaste and said "see, there's nothing there" but really you just squeezed it all to the other end.
As a fun illustration of this, note that <x,y> := 2x₁y₁ + x₂y₂ satisfies the axioms but doesn't give you the normal distance measure (e.g. it's not rotationally invariant: |(1,0)| = 2 while |(0,1)| = 1) and has a different notion of "orthogonal".
If you want to prove Pythagoras's theorem on Euclidean space, aren't there about a thousand proofs from that? Including the semi-original from Euclid? I assume it was proved there. And yes of course, you have to start with a bunch of axioms and earlier proofs about Euclidean space, but that's always true isn't it?
> It's not hard to show all that, but by the time you've done it you'll have accidentally proved Pythagoras's theorem along the way. It's like you've gripped on a tube of toothpaste and said "see, there's nothing there" but really you just squeezed it all to the other end.
Funny metaphor. Yes, I don't think you can "simply" prove Pythagoras's theorem without a bunch of background assumptions, it's just that usually these are all assumptions we've already learned (explicitly or implicitly). And if you want to start without assumptions, like in the case of defining inner product space from scratch, then you are by definition starting abstractly and therefore left with the problem of showing this maps onto Euclidean space, somehow.
And I certainly don't mean to imply anything about the importance of abstract inner product spaces. In fact my masters thesis was about Hilbert spaces. And I find it pretty interesting that you can prove something like Pythagoras on the inner product form I mentioned at the end of my last comment.
> In fact my masters thesis was about Hilbert spaces. And I find it pretty interesting that you can prove something like Pythagoras on the inner product form I mentioned at the end of my last comment.
That's pretty cool, you're definitely more knowledgeable than I am, I'm just a math amateur :)
EDIT: Yes, I think it’s clear considering that the sum of the non-rectangular angles in a triangle is 90 degrees.
Someone who was thinking about a problem and drawing something would always with a drawing like that either intend the angle to be the same, or otherwise highlight the fact that one triangles is 8/3 and the other is 5/2 so that the slope is obviously not the same.
Good visual proofs simply use lines and figures to talk about actual algebra instead of symbols; but every outcome is still in a sense algebraic -- like the one linked and the popular about Pythagoras. Once you pull our your ruler and measure you are obviously lost. Every result should be algebraic, not visual, but it's fine to express the algebra in figures instead of letters.
(Edited due to formatting fail!)
That’s not to say the example is bad, or good, more to marvel at how differently people think.
Now take each of your groups and arrange its 5 tiles into a vertical line. Each line is now 5cm long and 1cm wide.
Arrange the lines side by side. You now have a rectangle whose height is 5cm and whose width is 20cm. You already know its area 100 cm^2, because you made it out of 100 tiles that were 1 cm^2 each. And now you can see that its area also corresponds to the multiple of its side lengths.
Like, if you have a graph showing power consumption over time, it’s great to be able to mentally recognize that, say, if the time units are hours and the power units are Watts, that the area under the graph will be counted in Watt hours; that a rectangle one hour wide by one Watt tall is one Watt hour, and so on.
Rather, area IS multiplication.
The unit of "square meter" quite literally means "meter multiplied by meter".
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.
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
(You can exchange a and b in, say a^2+b^2, because 2^2+3^2=3^2+2^2)
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.
That's not what anyone is saying.
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.
use the same visual proof but with a and b switched to get
-(b + a)(b - a) = (a + b)(a - b)
https://www.youtube.com/watch?v=DjI1NICfjOk (Fermat's sum of two squares)
https://www.youtube.com/watch?v=rr1fzjvqztY (Ptolemy's theorem)
https://www.youtube.com/watch?v=yk6wbvNPZW0 (Irrational numbers)
It shows many nice visualizations. Amongst others it shows my my favorite proof of the Pythagorean Theorem
https://www.matematicasvisuales.com/english/html/geometry/tr...
But it can also be viewed as translating the quadratic expression along the “X-axis” so that (at its new origin) it is left/right symmetric.
That is,
Q(x) = ax^2 + b x + c
With the right substitution, x’ = (x - B), the linear term vanishes. So when you re-write in terms of “x”, you get:
Q (x) = a (x - B)^2 + C
So the intuition is that the linear term in the original quadratic is the thing that shifts the “symmetry axis” of the quadratic.
I have found this helpful when “X” is a vector and you have a quadratic form. In this case, the coordinate shift centers the quadratic “bowl” about some point in R^n.
*
The chain rule for differentiation is another one with simple geometry but cumbersome notation. It’s like: we know[*] that
f(x) = g(h(j(k(x))))
must have a linear approximation about some point x0. The only possible thing it could be is the product of all the little local curve slopes of k, j, h, and g, at the “correct” point in each.
Thinking about little slopes also clarifies derivatives like
f(x) = g(x^3, x^2)
where g is an arbitrary function of two variables.
[*] Because we read it in baby Rudin, ofc
Geometric “proofs” like this are neat, but are no real substitute for the algebraic ones. I’d argue that in cases like the present one they also don’t provide any deeper insights. You’re just moving geometric shapes around instead of algebraic symbols. They might give you the feeling that the theorem isn’t as arbitrary as you thought, but it isn’t arbitrary in algebra either.
I’m putting “proof” in quotes here because there are many examples of incorrect geometric “proofs”, and there is generally no formal geometric way to verify their correctness.
There are formal models of synthetic (i.e. axiom-and-proof based) Euclidean geometry where proofs can in fact be verified. This is accomplished by rigorously defining the set of allowed "moves" in the proof and their semantics, much like one would define allowed steps in an algebraic computation.
https://mathoverflow.net/questions/168888/who-invented-diagr...
E.g.
>6 points: I'd say God invented it. – Fernando Muro Commented Jun 3, 2014 at 6:38
Approved answer: Arthur Cayley. (Inspired by Sophie?)
>I believe diagrammatic algebra started with 19th century invariant theorists
a=0 and b is non-zero
b>a (result is negative. If you swap the variables, you have to imagine the area is a negative area)
either variable is negative
1. What constitutes a proof?
2. In what context is a "proof" embodied?
3. Why is scribbling lines on a paper (that look like math to humans) 'more' of a proof than visual diagrams, if at all?
4. If you came across a proof that was persuasive to alien intelligences -- and led them to conclude true things were true and false things were false -- but, alas, you did not understand it, does that make it less of a proof?
An irrefutable demonstration of a conclusion, possibly via sequence of steps or combination of elements.
2. In what context is a "proof" embodied?
Do you mean what the range or domain of the proof are? Not sure on the "embodied". I think you mean the communication of the proof and the expected base knowledge to understand the proof.
3. Why is scribbling lines on a paper (that look like math to humans) 'more' of a proof than visual diagrams, if at all?
You seem to be focusing on the representation of the proof in a particular notation, rather than the actual logic of the proof.
The graphical demonstration leads to false conclusions. For example, if a=0, it implies that a^2 - b^2 is 0 (or it requires some unfamiliar graphical representation of negative areas)
4. If you came across a proof that was persuasive to alien intelligences -- and led them to conclude true things were true and false things were false -- but, alas, you did not understand it, does that make it less of a proof?
Again, the representation is not the proof, it is a means to record or communicate the proof.
If the representation implies that false things are true (e.g., if a==0), then it is not a proof.
- When I said "embodied" I roughly mean the ground rules that someone needs to know to check the proof. In the case of symbolic logic I mean the symbols and the transformation/rewrite rules. But I'm not sure yet how I would formalize the analogous concepts for visual proofs.
- Re: "You seem to be focusing on the representation of the proof in a particular notation, rather than the actual logic of the proof." ... yes, but maybe not necessarily. The "logic" of a visual proof is quite different than the "logic" of a symbolic proof.
We're on the same page, but if I were to take it one level deeper, I would add:
a. While I know what you mean by "irrefutable", I wouldn't use that word, because it sounds too much like "untestable". The whole idea of a proof is that each step can be verified. With a big enough lookup table, a proof can be checked in linear time (right?). If a step does not "obviously" follow then the step is not well-explained (in the "trivial to verify" sense).
b. Your choice of "demonstration" is key here. An essential aspect of a proof is that it is easier to check than generate. Simply "read off" each line of the proof and check against some known set of facts and transformation rules.
c. It is useful to distinguish between a verified proof (which is subject to correction!) and just a proof (which is a form, a way of communicating how something can be verified). See this Stack Overflow page: "Widely accepted mathematical results that were later shown to be wrong?" [1]
[1]: https://mathoverflow.net/questions/35468/widely-accepted-mat...
- irrefutable is "cannot be denied or disproven" - it means absolutely proven. Very different from untestable (which means you can't verify or prove it).
This is precisely the distinction we've been calling out from the original - it's a demonstration that works for some cases. The diagrams fail as a proof because they can be refuted by the negative/0 cases where they don't work.
- testing or verifying with a set of data is also very different from a proof. This "checking" or "demonstrating" provides some assurance of correctness or utility for the test domain.
- demonstrating and checking against known facts is not sufficient for a proof - "I've tested my division function for millions of positive and negative integers and real numbers! I even verify by multiplying the quotient by the divisor and I've proven it is correct!" did you happen to test a divisor of 0? (dividing by 0 can also invalidate proof attempts that do not exclude 0 as a divisor)
It needs to be proven to hold for all cases, not just a sampling of cases (though it is valid to define the range of a proof - the example could have stated, "this is a proof of the equation for positive values of a and b, and were b < a" maybe that could constitute a visual proof for that domain?)
And why is that a problem? Just reverse the equation before and after the proof.
That's not the point. To me, it's useful to already know that an algebraic proof of that equation exists, but to see it work out visually. I don't need to see it worked out visually for every single possible value for this to be helpful for understanding.
It also nicely illustrates how algebra and geometry are linked. And that multiplication is geometrically taking you from 1 dimension to 2.
They aren’t used later on because, for more complicated expressions, manipulating the algebraic formulas is easier than the geometric method, or completely breaks down (e.g. when the powers aren’t integers or are variables)
For example, I think anybody who can visualize what (a+2b)³ or (a+b)⁴ looks like geometrically can also, and easier, do the expansion algebraically.
No, or at least they were not in the high-school math books I was assigned. There were no explicit referrals in any way to visual representations of the algebra involved.
a ( b+c) = ab + ac.
I agree it's important that students intuitively feel that
a ( b+c) = ab + ac
is true.
But once we are at that, why would we even bother to teach or "proof" all variations of that expression?
It doesn't mean it is not possible to have rigorous visual proofs, the one in the article is, but doing so can be deceptively hard. It is also common for visual proofs to be less complete, for instance, here, it only covers the cases where a>0, b>0 and a>b, there are no such limitations when doing so with algebra. I guess you can tweak the visual proof to account for these cases, but it will become far less elegant.
So I understand why teachers avoid visual proofs in math classes, they actually want you to stay away from them. They have entertainment value, that's why you see them so much in pop-science, and I think it is important, and also some historical value, as they were much more present in ancient times, but to actually learn maths, not so much. That is unless you are studying visual proofs specifically, but I think it is an advanced topic you would tackle well after you can master simple equations like a^2 – b^2 = (a + b)(a – b).
Intuitive but formally wrong is still good for intuition.
> No, “formally wrong” means that it fails formal verification.
The word "formal" has meaning in mathematics independent of formal verification. The latter builds upon the former. Agree?
> In the context of hardware and software systems, formal verification is the act of proving or disproving the correctness of a system with respect to a certain formal specification or property, using formal methods of mathematics. - Wikipedia https://en.wikipedia.org/wiki/Formal_verification
Here is how I'm using the terms:
- "Formal" in mathematics refers to rigorous logical reasoning with precise definitions and deductive steps. This is the older, more general meaning.
- "Formal verification" emerged later as a specific term in computer science referring to automated/mechanical verification of system properties. This is now the standard meaning in software/hardware contexts.
So, putting these terms into use... If a human verifies a formal proof "by hand", I think it is fair to say that does not comprise "formal verification". On the other hand, if an automated system verifies the proof, then I would say "formal verification" has happened. Perhaps you will agree this is how many, if not most, experts use the terms.
Are we mostly playing language games -- or is there a key insight you think I don't understand?