My Proof of Pythagoras' Theorem is NOT circular!

[MathTheBeautiful]

A few thoughts on the comments below:
First of all, props to @samueldeandrade8535 for his passion and desire to dig deeper.
Second, one of the comments said that I'm proponent of the coordinate-free approach. That's true, but only insofar as I'm a proponent of all approaches - the more the better, and the more different the better - the goal being to recognize the relative strengths and weaknesses of each approach. Case in point, I'm a huge proponent of the coordinate approach. In fact, in this course, my emphasis on the coordinate-free approach is mean to help illustrate the need for coordinates and to be able to introduce them properly.

[juliavixen]

Unrelated to the topic of this video, you need to turn off autofocus on your camera. Lock the focal distance to either the board or just in front of the board when you start recording. Narrow the aperture (set the f-stop to a higher number like f/4 or f/8, etc.) to increase the depth of field, if you have enough light.

[jacksonstenger]

I haven’t seen your other video, so you might have mentioned this. But this proves the law of cosines immediately, also it generalized easily to 3d involving the diagonal of a tetrahedron, and appears to give a 3d version of law of cosines

[ruairidhwilliams]

Your definition of the dot product does require the notion of a metric on your vector space. That metric in this case is pythagoras, however i believe in different vector spaces (like the space of cts functions f:[0, 1]->R] with f•g=int(fg) ) we can derive the same pythagorean property. So i guess the point is that being in a hilbert space gives you pyhtagoras, but then euclidean pythagoras relies on the notion of the length of a vector (which you require for your dot product) and in this case youre measuring your vector with pythagoras.

So in summary, unless im overlooking something, you have proved a very awesome property in a hilbert space, but to relate the norms to lengths in your euclidean vector space you need to define the lengths which are defined by pythagoras.

[BusyBeaver]

The proof isn't circular, it's just that vectors and dot products are more advanced tools that pack a lot of geometric information into a small expression. For example, if you want to add two vectors u and v geometrically, you have to construct a parallelogram, which is a lot more "physical work" than just writing u+v. Each line of this proof has an equivalent geometric interpretation, it's just that it takes more work to draw it:

we can interpret the dot product of vector u with itself (u*u) as the area of the square with its side length. If v is orthogonal to u, u*(u-v) can be interpreted as an area-preserving shear transformation.

when you do c*c = (a-b)*(a-b) = a*(a-b) - b*(a-b) (distributing just once), you are effectively shearing the smaller squares into parallelograms and squishing them together, like the animation here:

[hlee]

Your definition of dot product requires that the "length of the projection" being unique. That's the case only if your vector space is Euclidean.

[marceloviolato]

First, thank you for help me understand math. In the future, if you can use rational trigonometry instead the standard trigonometry, I would appreciate.

[theoremus]

Beautiful. It seems to me that much interest has been given to proving Pythagoras Theorem via trigonometry lately, as a result of the work of Johnson-Jackson challenging the claim of Elisha Loomis. I have made videos on this and on Jason Zimba's proof. We want to avoid circular reasoning.

[jacksonstenger]

This is a great proof. I see no flaws. Was expecting to see a sneaky equivocation between the component formula for the dot product and the cos theta formula for the dot product, but since it does not have this I believe it is valid

[samueldeandrade]

Man, someone has to be ... in a very special way to insist on such idea. Let's understand:

given a field K, a vector space V and a symmetric bilinear form

• : V×V → K

we can consider the quadratic form associated with •,

Q: V → K,
Q(u) := u•u

And for all that we can easily demonstrate, there is NOTHING new about this, the Pythagoras Theorem for • :

"If u, v in V are such that u•v=0, then

Q(u-v) = Q(u)+Q(v)"

IF K and • HAVE MORE PROPERTIES, not going to tell all details here, you can also define length of a vector and angle between two vectors by

|u| = √Q(u)

and

Angle(u, v) = (u•v)/(|u|*|v|)

This is all fine. Now ... the problem with this guy's idea is that it seems he thinks he is demonstrating the classical Pythagoras Theorem, instead of demonstrating it for the abstract dot product. HE IS NOT. To do that he would have to establish a "correspondence" between triangles and pairs of linear independent vectors, in a way that the usual length from classic Geometry corresponds to the abstract length AND the usual angle correspond to Arccos(Angle). THIS is what makes the proof INVALID. If you want to call it circular, you can, because to show such correspondence you will need Pythagoras Theorem.

So, for the Math of Euler, let's stop this insanity. It is extremely disappointing to see this whole situation. The guy thinks that using abstract functions he can prove THE theorem that is the reason for the definition of said abstract functions.

[KaiseruSoze]

Nicely done. Kudos for taking the time. I don't think I would have paid attention to the peanut gallery :)

[punditgi]

The only thing that's circular is the perfect shape of these videos! ❤🎉😊

[samueldeandrade]

I will make a last, serious, argument against this:

accepting this "proof" is basically the same as accepting to prove

lim_{x→0} sin(x)/x = 1

using L'Hôpital's Rule. That's it.

[samueldeandrade]

Oh my Euler ... how tiresome. Let's watch this ... questionable lecture.