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> If we rely upon the projection product of Euclidean vectors, we've already granted Pythagorean theorem in our assumptions. The Pythagorean theorem in particu

> If we rely upon the projection product of Euclidean vectors, we've already granted Pythagorean theorem in our assumptions. The Pythagorean theorem in particular (and any theorem in general) is always a little bit “circular”; the relation is inherent in any definition of perpendicular in a model of Euclidean space. In geometric algebra (where multiplication of vectors distributes over addition), the two statements a² + b² = (a + b)² ⇔ ab + ba = 0 are obviously equivalent, so taking either of them as a definition for “perpendicular” immediately proves the other.

mlyle 7y ago
Eh, I guess any proof is circular, but it is, IMO, "more" circular when the thing we're trying to prove was one of the preconditions of how we defined the system--- properties that we wanted to obtain a priori. It's like defining negative exponents based on the properties we want for commutation, identity, etc... and then afterwards proving that something raised to a negative exponent times something raised to a positive exponent is 1 and patting ourselves on the back.
- jacobolus 7y ago
  > thing we're trying to prove was one of the preconditions This entirely depends on what you consider to be a definition of the dot product. There are many possible ways this could be set up. (e.g. if you wanted you could develop this whole theory of vector algebra within the system of Euclid’s axioms. Or you could set it up based on explicit coordinates and concrete arithmetic of numbers with no geometrical basis per se. Or ...) The hard work leading up to this proof is showing that the algebraic definition a·b = 0 corresponds to the usual notion of perpendicularity in Euclidean space. After that, the algebraic proof of the Pythagorean identity is trivial.
rsj_hn 7y ago
You need more than the definition of perpendicular, you need the definition of "angle" in such a way as to ensure flatness of your space, otherwise a triangle might not correspond to three vectors that sum to zero. In this case, you cannot obtain the pythagorean identity from the algebraic identities. Remember a triangle is a set of 3 curves living in your space that meet back up. But angles between curves are measured as angles between the tangent vector to the curves and do not live in your space, they live in the tangent space. A triangle on the sphere, for example, has angles that don't sum to 180 degrees and the three tangent vectors do not sum to zero even though the three curves meet back at the same point. So what's crucial here is an assumption of flatness, which allows you to associate a tangent space to the underlying space in a way consistent with the underlying metric. This allows you to make the association between geodesics (distance minimizing curves in your geometry) and vectors in your tangent space so that you can pretend that the straight lines actually live in your space and are also distance minimizing. This is what you need for the pythagorean theorem. This is not something that you can get just from the distributive law, you need the distributive plus the property that a triangle has angles that sum to 180, or equivalently that the tangent vectors to your triangle can be embedded in your space and sum to zero.
- jacobolus 7y ago
  Notice I mentioned “Euclidean space”. That inherently involves flatness. Obviously there are several premises/axioms needed to set it up, and a variety of ways to do so. In Euclid, we have the famous parallel postulate which helps us establish flatness. > angles that don't sum to 180 degrees Note that Euclid’s Elements nowhere mentions angle measures. It only describes the concept of a right angle (and angles more or less than right). The Pythagorean theorem does not depend on angle measures. If you ask me angle measures are a quite poor/confusing tool to introduce in introductory Euclidean geometry courses, since they are a type of logarithm, and much more inherently complicated than the rest of a typical geometry course. > triangle is a set of 3 curves This is one possible definition of “triangle”. For Euclid a “trilateral figure” is contained by three straight lines, and “A straight line is a line which lies evenly with the points on itself.” (Which has been rather hard for readers to interpret throughout time.) * * * Tangentially, I’ve been working a lot with spherical triangles in my ongoing project. :-) https://observablehq.com/d/5e75dd8e56fe255f https://observablehq.com/d/5e75dd8e56fe255f
  - rsj_hn 7y ago
    How you are going to define euclidean space without the pythagorean theorem? That's basically the definition of euclidean. But the advantage of the pythagorean theorem is it allows you to measure how you deviate from flatness by comparing the difference of c^2 with a^2 + b^2.
    - jacobolus 7y ago
      That was my point upthread (any proof of the Pythagorean identity is somewhat circular, since it is inherent in the structure). The way Euclid does it is to set up various axioms which imply flat space without explicitly declaring the Pythagorean identity to be an axiom. But you could easily do it the other way around. Euclid’s axioms (and other alternatives proposed over the years) were chosen specifically to make the Pythagorean identity true. Cf. Feynman https://www.youtube.com/watch?v=hxKw4xEEFHQ https://www.youtube.com/watch?v=hxKw4xEEFHQ (proximately relevant bit starts at about 22 minutes) Also Lakatos: https://math.berkeley.edu/~kpmann/Lakatos.pdf https://math.berkeley.edu/~kpmann/Lakatos.pdf

mlyle 7y ago

Eh, I guess any proof is circular, but it is, IMO, "more" circular when the thing we're trying to prove was one of the preconditions of how we defined the system--- properties that we wanted to obtain a priori. It's like defining negative exponents based on the properties we want for commutation, identity, etc... and then afterwards proving that something raised to a negative exponent times something raised to a positive exponent is 1 and patting ourselves on the back.

jacobolus 7y ago
> thing we're trying to prove was one of the preconditions This entirely depends on what you consider to be a definition of the dot product. There are many possible ways this could be set up. (e.g. if you wanted you could develop this whole theory of vector algebra within the system of Euclid’s axioms. Or you could set it up based on explicit coordinates and concrete arithmetic of numbers with no geometrical basis per se. Or ...) The hard work leading up to this proof is showing that the algebraic definition a·b = 0 corresponds to the usual notion of perpendicularity in Euclidean space. After that, the algebraic proof of the Pythagorean identity is trivial.

rsj_hn 7y ago

You need more than the definition of perpendicular, you need the definition of "angle" in such a way as to ensure flatness of your space, otherwise a triangle might not correspond to three vectors that sum to zero. In this case, you cannot obtain the pythagorean identity from the algebraic identities. Remember a triangle is a set of 3 curves living in your space that meet back up. But angles between curves are measured as angles between the tangent vector to the curves and do not live in your space, they live in the tangent space. A triangle on the sphere, for example, has angles that don't sum to 180 degrees and the three tangent vectors do not sum to zero even though the three curves meet back at the same point. So what's crucial here is an assumption of flatness, which allows you to associate a tangent space to the underlying space in a way consistent with the underlying metric. This allows you to make the association between geodesics (distance minimizing curves in your geometry) and vectors in your tangent space so that you can pretend that the straight lines actually live in your space and are also distance minimizing. This is what you need for the pythagorean theorem. This is not something that you can get just from the distributive law, you need the distributive plus the property that a triangle has angles that sum to 180, or equivalently that the tangent vectors to your triangle can be embedded in your space and sum to zero.

jacobolus 7y ago
Notice I mentioned “Euclidean space”. That inherently involves flatness. Obviously there are several premises/axioms needed to set it up, and a variety of ways to do so. In Euclid, we have the famous parallel postulate which helps us establish flatness. > angles that don't sum to 180 degrees Note that Euclid’s Elements nowhere mentions angle measures. It only describes the concept of a right angle (and angles more or less than right). The Pythagorean theorem does not depend on angle measures. If you ask me angle measures are a quite poor/confusing tool to introduce in introductory Euclidean geometry courses, since they are a type of logarithm, and much more inherently complicated than the rest of a typical geometry course. > triangle is a set of 3 curves This is one possible definition of “triangle”. For Euclid a “trilateral figure” is contained by three straight lines, and “A straight line is a line which lies evenly with the points on itself.” (Which has been rather hard for readers to interpret throughout time.) * * * Tangentially, I’ve been working a lot with spherical triangles in my ongoing project. :-) https://observablehq.com/d/5e75dd8e56fe255f https://observablehq.com/d/5e75dd8e56fe255f
- rsj_hn 7y ago
  How you are going to define euclidean space without the pythagorean theorem? That's basically the definition of euclidean. But the advantage of the pythagorean theorem is it allows you to measure how you deviate from flatness by comparing the difference of c^2 with a^2 + b^2.
  - jacobolus 7y ago
    That was my point upthread (any proof of the Pythagorean identity is somewhat circular, since it is inherent in the structure). The way Euclid does it is to set up various axioms which imply flat space without explicitly declaring the Pythagorean identity to be an axiom. But you could easily do it the other way around. Euclid’s axioms (and other alternatives proposed over the years) were chosen specifically to make the Pythagorean identity true. Cf. Feynman https://www.youtube.com/watch?v=hxKw4xEEFHQ https://www.youtube.com/watch?v=hxKw4xEEFHQ (proximately relevant bit starts at about 22 minutes) Also Lakatos: https://math.berkeley.edu/~kpmann/Lakatos.pdf https://math.berkeley.edu/~kpmann/Lakatos.pdf