Comments: 
I like it! One idea that a young reader (I'm thinking about my 5th grade brother) might be unfamiliar with is that knowing two of the angles of a triangle tells you the third angle. Maybe showing a transformation that flips and then shrinks one of the roofs into another would explain it?
I don't know if I have time to make another figure, but I could definitely throw a "Since knowing the two angles tells you the third angle..." in there. Thanks for the feedback!
A pretty dull dinner party indeed!
I hadn't seen this proof before; I rather like it, although I think I share your skepticism about "intuitive" geometry proofs. As a reasonably mature mathematician, I still like knowing a lot of them as a form of memory compression. I know to really prove it, I'd do something else.
The reason I like this proof is it makes some sense if you don't even draw the canonical squaresonsidesoftriangles, and just focus on the right triangle and how it can be subdivided into two similar triangles by dropping that perpendicular to the hypotenuse. For you have to learn at some point that fundamental fact about ndimensionalness that when you scale up an ndimensional thingy by a factor of k, its volume goes up as k^n. And so just to see that you can find widgets "proportional to" the two legs of the right triangle that in (twodimensional) volume add up to the hypotenuse one means that you are really talking about squared things adding up to a third squared thing.
(And anyhow I actually believe that it's the central limit theorem that's the reason we live in a space where the pythagorean theorem (approximately, locally) holds in the first place. No matter whether you start with a square lattice or a hex lattice or whatever, the purely combinatorial numberofways you can get from point A to point B goes like e^(x^2) * e^(y^2) * e^(z^2) = e^((x^2 + y^2 + z^2)) and so similar physics obtains for the same sumofsquares, because QFT is all about weighted counts of paths mumble mumble rant...)
For this proof, you needn't know that the k^n thing is true in general  you just need to know that the area of a square is defined as the length of its side, squared. But there's no reason why they have to be squares  they could be any old similar polygons that have a side of the triangle as one of their sides. Actually, I don't think they even have to be polygons, but then you have to worry about defining similarity for arbitrary shapes, I guess.
I don't know if it really makes sense to say that this or that proof of a particular fact is the real reason that the fact is true. However, I do know that if I put e^(x^2) * e^(y^2) * e^(z^2) = e^((x^2 + y^2 + z^2)) on my homework, Doug would fire me. Or whatever it is that they do to grad students.
I don't know if it really makes sense to say that this or that proof of a particular fact is the real reason that the fact is true. However, I do know that if I put e^(x^2) * e^(y^2) * e^(z^2) = e^((x^2 + y^2 + z^2)) on my homework, Doug would fire me. Or whatever it is that they do to grad students.
Hee hee yes.
I mean, I use phrases like "the real reason" very fuzzily and intuitively without any claim that they really mean anything. But I think there is a deep connection between the squarings that show up in the e^x^2 of the central limit theorem and the squarings that show up in the Euclidean norm (of any dimension, of course, not just 2!). It's not the reason that the Pythagorean Theorem is true in Euclidean space; proofs such as the one you gave are explanations of that fact. But I suspect it has something to with why our universe is locally Euclidean in the first place.
 From: cos 20101012 02:24 pm (UTC)
 (Link)

That seems convincing, though I don't know how much of that is based on knowing things already. I mean, is the idea of "similar triangles" and proportional sizes "intuitive"? To whom?
That's a good question. It's irritating to me that we talk about intuitiveness a lot in this course without ever defining what it means. For example, today the claim was made that because human bodies are vertically symmetric, humans "relate better" to drawings that are vertically symmetric  which is why it's "easier" to bisect an angle that's pointed straight up or down than it is to bisect some other angle. That kind of made my blood boil. My body is mostly bilaterally symmetric, sure, but what if it weren't? What if, for instance, I was missing one leg? Would I then be less able to "relate" to a vertically symmetric diagram? Moreover, isn't part of the point of cognitive science to examine things that are "intuitively" "obvious" to a human and figure out why they are so? Sometimes the best way to do that is to try to take off our humancolored glasses. (Of course, we humans can't help looking at things from human points of view. But it's sure interesting to try not to, and this course seems to take the opposite approach.)
 From: pmb 20101013 04:28 pm (UTC)
You made me do my own!  (Link)

Nice! My favorite proof is the one I use in Calc class as a "fun fact". Just rearrange the four equal triangles: But I actually like yours quite a bit as well  I have seen that drawing of the triangle with the squares on its sides, but nobody ever was able (until now) to explain to me exactly how and why that managed to prove the theorem. Thank you!
Have you seen the ones that use tiling?
 From: pmb 20101013 04:47 pm (UTC)
Re: You made me do my own!  (Link)

Neat!  