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Comment by konschubert | Hacker News Reader

Here is a visual proof for the Pythagorean theorem:

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.

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One time I was really sure that splitting a line into three equal segments let you draw lines from a point and split a 60 degree angle into three 20 degree angles, but this wasn't actually true.

I think that the visual proof is not complete. You have to demonstrate also that the quadrilateral in the right figure is a square.

Shouldn’t that be clear from considering angle properties?

EDIT: Yes, I think it’s clear considering that the sum of the non-rectangular angles in a triangle is 90 degrees.

I don't feel like it's more redundant that Pythagorean theorem though, as we can say that the later directly follows from the definition of dot product…

I guess my intuition for vector algebra is much weaker...

So you prove something in the 2d space via a 3d space intermezzo? Not very intuitive to me. Distribution on the other hand, can be explained by counting a handful of the same objects.

The dot product exists in any dimension

Indeed I was even taught it in 2d before 3d (and higher).

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 + ...

How does Pythagoras's theorem follow from the definition of the dot product?

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.)

> 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…

It may seem trivial, but to use that to prove the component-wise formula for general vectors you're assuming distributivity of the dot product over addition of one of its arguments. But if you're starting with the x.y = |x| |y| cos θ definition, how do you prove that (without first going via the component wise definition that you're still in the process of proving)? You end up needing trigonometric angle formulae that are at least as hard to prove as Pythagoras's theorem.

Sorry, but you can't bypass proving Pythagoras's theorem by definition of the dot product or anything else.

> You end up needing trigonometric angle formulae that are at least as hard to prove as Pythagoras's theorem.

(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.

It's so straightforward that you've used the exponential function on the complex plane to prove it?! You started by claiming that Pythagoras's theorem follows "directly" from the definition of the dot product. Can you admit that we're now quite far away from that?

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.

> It's so straightforward that you've used the exponential function on the complex plane to prove it?! You started by claiming that Pythagoras's theorem follows "directly" from the definition of the dot product. Can you admit that we're now quite far away from that?

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.

Normally you define the dot product (or inner product more generally) with a few conditions that it has to satisfy, and then define two vectors to be orthogonal if their dot product equals 0: <u,v> = 0.

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.)

If you mean you'd start with the axiomatic definition of an inner product space (bilinear, symmetric, positive definite) then I agree that proving that formula is trivial. The problem with a really abstract approach like this is that you haven't actually proved Pythagoras's theorem in Euclidian space. How do you know that the usual definition of dot product satisfies these axioms (or indeed anything does)? How do you show that the "distance" you've defined from the inner product is what we'd expect to be distance (e.g. that it's rotation invariant) or that what you've defined as "orthogonal" from your inner product is related to angle in Euclidean space?

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".

Yeah, I was talking about an inner product space.

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.

I agree that there's are lots of proofs of Pythagoras's theorem on Euclidean space. The comment I replied to said that it follows "directly" from the definition of the dot product. That's all I was disagreeing with. They had missed that they were using some or other property of the dot product that was actually proved from Pythagoras in the first place, or some other non-trivial fact about Euclidean geometry.

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.

Ah yes, you're right of course, and I haven't really thought about it in this way before - that either you're based on "real world" geometry, in which case things are a bit harder to prove but make sense, or you're more abstract, in which case you can define things to be easier to prove e.g. Pythagoras, but the complexity is in the mapping between your definitions and the "real world".

> 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 :)

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