Can the Pythagorean theorem be proved using imaginary numbers? The proof must avoid circular reasoning, of course.
I asked essentially the same question at MSE, but did not receive a definitive answer, so I thought I would ask here.
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5$\begingroup$ Actually I think it's not at all clear how to give any (modern) proof of the Pythagorean theorem that doesn't feel at least a little bit circular in spirit. The issue boils down to the question: where does the Euclidean distance formula come from, if it doesn't come from the Pythagorean theorem? I spent awhile thinking about this off and on and I still don't feel totally satisfied, but I think the most honest answer is that it must and can only come from our phenomenological experience of physical space (e.g. the fact that we can freely rotate our bodies). $\endgroup$– Qiaochu YuanCommented yesterday
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$\begingroup$ @QiaochuYuan What's your view of proving the Pythagorean theorem "synthetically" from Tarski's axioms? This makes no reference to $\mathbb{R}\times\mathbb{R}$ as a model for the axioms and so the Euclidean distance formula doesn't enter the picture. Right? Now, you can ask about where Tarski's axioms come from, and I agree that physical intuition plays a role, but at least the apparent circularity associated with the Pythagorean theorem itself has been eliminated. $\endgroup$– Timothy ChowCommented 7 hours ago
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Yes. Write a complex number $a+bi$ in polar form $ce^{it}$. Then $$a^2+b^2=(a+bi)(a-bi)=(ce^{it})(ce^{-it})=c^2.$$ This is the first theorem I prove in my complex analysis class (after defining complex multiplication via the polar form and checking that it agrees with the more traditional definition via $i^2=-1$ and distributivity).
Added. See the comments for more details what this proof is really based on, and why it is not circular.
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5$\begingroup$ According to my understanding, this proof has met with objections, here and here. I believe the gist of the objections is that, in this proof, distance is implicitly being defined using the Pythagroean theorem, thus the reasoning is circular. I'm not sure what to make of the objections. $\endgroup$– DanCommented yesterday
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4$\begingroup$ @Dan No. In my class, I base complex arithmetic on Euclidean geometry where distance is given (it is not a defined quantity but part of the model). It is a genuine proof just as any proof in Euclidean geometry. The difference is that here complex arithmetic is based on Euclidean geometry, and then the Pythagorean theorem becomes a one-liner as in my post. I leave you to fill in the details (or come to my class). $\endgroup$ Commented yesterday
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3$\begingroup$ @GHfromMO Do you have class notes on a website, so that we can see that approach fleshed out? $\endgroup$ Commented yesterday
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3$\begingroup$ @GHfromMO The context that you have defined complex arithmetic through Euclidean geometry is absolutely essential, and it should not be left in the comments. Please include it in your answer body, because one could interpret your answer as being just like many other identical circular answers that I’ve seen. At any rate, I might still have lingering questions e.g. it seems as if your primitive notion of Euclidean distance is real-valued, and I’d have to think a little more to find others $\endgroup$– FShrikeCommented yesterday
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1$\begingroup$ @FShrike One more thing. The linear algebra version of the above proof would go as follows. Consider $\mathbb{R}^2$ with the standard orthonormal basis. Rotation about the origin by angle $t$ has matrix $\begin{pmatrix}\cos t&-\sin t\\\sin t&\cos t\end{pmatrix}$, while rotation about the origin by angle $-t$ has matrix $\begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix}$. So the product of these two matrices is the identity matrix. Performing the multiplication of these two matrices, we infer that $\cos^2 t+\sin^2 t=1$, which is the Pythagorean theorem. This is not a circular proof! $\endgroup$ Commented yesterday