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I'm suspicious of vector algebraic proof of the the Pythagorean Theorem. Don't vector operation properties themselves follow from the Pythagorean Theorem (at l
by avodonosov 7y ago
I'm suspicious of vector algebraic proof of the the Pythagorean Theorem.
Don't vector operation properties themselves follow from the Pythagorean Theorem (at least in their application to space and geometric objects)? If so, using them to prove the theorem doesn't make sense.
I'm not sure right now whether such circularity exists, but one should be careful.
- mkl 7y agoIt seems circular to me. It relies on ||x||² = x·x, but the fact that ||x||² = x₁² + x₂² = x·x comes from the Pythagorean theorem.
- posterboy 7y agothere are hundreds of proofs for the pythagorean theorem, surely some of them using the axioms of geometry
- mkl 7y agoThis isn't just using the axioms of geometry, though, it's trying to prove the Pythagorean theorem using the Pythagorean theorem as an axiom. Hence circularity.
- posterboy 7y agono, not exactly. the euclidean norm can be defined to fulfill a couple basic properties or equations (like composibility, or schwarz' inequality, but I don't recall exactly), and it's pure coincidence, if you will, that the norm is equal to the root mean square. That's not circular reasoning. The name euclidean norm implies that the geometric angle (no pun intended) was the motivation, but what's really central is a question of epistemology. There's not much of a point to prescribe a certain approach over another, without a good argument.
- soVeryTired 7y agoI don't think it's quite circular, but in a sense you're right. if c = -a + b, then for any inner product you have <c,c> = <a,a> + <b,b> - 2<a,b> as a result of bi-linearity of the inner product. By definition if <a,b> = 0 then a and b are orthogonal and a version of pythagoras' theorem holds. But you're right in saying the choice of inner product <x,y> = x_1y_1 + ... + x_ny_n is motivated by pythagoras' theorem in Euclidean space.
- avodonosov 7y agoBy definition? In the abstract world of algebra we are free to choose any definition (different rules will give different algebras). But if we want our algebraic manipulations to prove the theorem about geometric objects we need to prove isomorphism between our algebra and the geometric objects and operations on them. I doubt distributivity and other properties of operations on geometric vectors can be proven without the the Pythagorean theorem.
- soVeryTired 7y ago> By definition? Yes, in a Hilbert space (i.e. an abstract vector space with an inner product), the definition of orthogonality is that the inner product of two nonzero vectors is zero. I'm not sure I really know what you mean by geometric vectors.
- avodonosov 7y agoWell... think harder, what can I say :)
- soVeryTired 7y agoFrom that comment I'm not sure you do either.
- avodonosov 7y agoI suspected trolling in your question about geometric vectors. Sides of a triangle and elements of your algebra are different domains. In order to translate results between them one needs to prove this makes sense. In the article the author only shows that inner product of c by itself equals to sum of inner product squires of a and b, if a and b are orthogonal. Who told you this has anything to do with lengths of triangle sides?