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I have been looking for proofs for the pythagorean theorem that don't use area calculation but calculus, complex numbers or any other interesting ways to proof it.

I would love to see any interesting proof, Shay

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    Well, that depends on what you're willing to call the Pythagorean theorem, as well as what kind of assumptions you're willing to start with. One might argue that the machinery of certain parts of calculus and complex numbers _depends on_ the Pythagorean theorem, so that any such proof is circular. One might argue that there are various equivalent statements of the Pythagorean theorem, and that the proofs of their equivalence are not trivial.2011-07-22
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    Echoing Qiaochu: it really depends on what tools you have. For example, there is a "Pythagorean Theorem" that holds for any inner product space (in particular, for the real and complex planes): if $\langle \mathbf{x},\mathbf{y}\rangle = 0$ (if $\mathbf{x}$ and $\mathbf{y}$ are orthogonal), then $$\lVert \mathbf{x}\rVert^2 + \lVert \mathbf{y}\rVert^2 = \lVert \mathbf{x}+\mathbf{y}\rVert^2,$$using the definition $\lVert\mathbf{z}\rVert = \sqrt{\langle \mathbf{z},\mathbf{z}\rangle}$. How interesting that is, though, seems rather subjective...2011-07-22
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    Thank you for your comment. I have seen a "proof" of the pythagorean theorem that uses complex numbers but it is circular and therefore invalid. All I am looking for is a proof that for a right-angled triangle a^2+b^2=c^2, not using geometric areas calculation.2011-07-22
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    To second and amplify Qiaochu's statement, the very definition of a norm of a complex number effectively states the Pythagorean Theorem as an axiom. I suppose you could consider the multiplicativity of the norm - that is, the fact that it behaves as expected under scaling and rotations - as a 'proof' that it acts like a length, and thus as an indirect justification for the Pythagorean theorem, but that seems somewhat roundabout...2011-07-22
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    Arturo Magidin, I have just seen your proof. You said it yourself, it is not interesting because you have defined the distance using the result of the pythagorean theorem.2011-07-22
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    @shayfalador: Again: the problem is that if you don't say what you are *assuming* and what you are not assuming, then your question is meaningless. "Interesting" is subjective, so it is not something that anyone but you can answer (not a good question for this site). If you don't say **what** you are willing to assume and what you aren't, then you won't get good answers. Even writing "$a^2+b^2=c^2$" is meaningless without you saying exactly what you are assuming and what you aren't, or what the operations *mean* in that context.2011-07-22
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    I am willing to see any proof (especially if you find it interesting) that doesn't assume things such as the distance between $(a,b)$ to $(0,0)$ is $\sqrt(a^2+b^2)$. Thanks again!2011-07-22
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    @shayfalador: You aren't addressing the problem. For one, if you aren't assuming anything about distances, then just what does "$a^2+b^2=c^2$" even *mean*? You need to say what you **are** assuming, not just what you are not.2011-07-22
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    I can not be anymore specific than that. Any axiom, but those that define distance as I said, is ok.2011-07-22

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