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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 tha

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.
    - quietbritishjim 2y ago
      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.
      - littlestymaar 2y ago
        > 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.
edanm 2y ago
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.)
- quietbritishjim 2y ago
  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".
  - edanm 2y ago
    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.

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.
  - quietbritishjim 2y ago
    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.
    - littlestymaar 2y ago
      > 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.

edanm 2y ago

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.)

quietbritishjim 2y ago
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".
- edanm 2y ago
  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.