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Here is a visual proof for the Pythagorean theorem: https://www.dbai.tuwien.ac.at/proj/pf2html/proofs/pythagoras/pythagoras/pythagoras3.gif https://www.dbai.tu

Here is a visual proof for the Pythagorean theorem: https://www.dbai.tuwien.ac.at/proj/pf2html/proofs/pythagoras/pythagoras/pythagoras3.gif 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.

littlestymaar 2y ago
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…
- konschubert 2y ago
  I guess my intuition for vector algebra is much weaker...
- nuancebydefault 2y ago
  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.
  - r0uv3n 2y ago
    The dot product exists in any dimension
    - quietbritishjim 2y ago
      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 + ...
- quietbritishjim 2y ago
  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.)
  - littlestymaar 2y ago
    > 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…
    - quietbritishjim 2y ago
      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.
      - littlestymaar 2y ago
        > 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.
        2y ago
alberto_ol 2y ago
I think that the visual proof is not complete. You have to demonstrate also that the quadrilateral in the right figure is a square.
- konschubert 2y ago
  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.
axus 2y ago
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.

littlestymaar 2y ago

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…

konschubert 2y ago
I guess my intuition for vector algebra is much weaker...
nuancebydefault 2y ago
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.
- r0uv3n 2y ago
  The dot product exists in any dimension
  - quietbritishjim 2y ago
    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 + ...
quietbritishjim 2y ago
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.)
- littlestymaar 2y ago
  > 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…
  - quietbritishjim 2y ago
    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.
    - littlestymaar 2y ago
      > 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.
      - 2y ago

alberto_ol 2y ago

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

konschubert 2y ago
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.

axus 2y ago

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.