Timeline for Can the Pythagorean theorem be proved using imaginary numbers?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| 17 hours ago | comment | added | Emil Jeřábek | All right, thank you for the explanation. | |
| 18 hours ago | comment | added | GH from MO | @EmilJeřábek As we defined multiplication in polar coordinates, it follows that the map $z\mapsto wz$ is a rotation (by the angle of $w$) followed by a dilation (by the length of $w$). Now consider a triangle with side vectors $z_1,z_2,z_1+z_2$. Applying the above mentioned rotation and dilation to this triangle, we get a triangle with side vectors $wz_1,wz_2,w(z_1+z_2)$. Hence $wz_1+wz_2=w(z_1+z_2)$, which is the law of distributivity. Combining this with $i^2=-1$ (which follows from the definition of multiplication), we get that $1=(\cos t+i\sin t)(\cos t-i\sin t)=\cos^2 t+\sin^2 t$. | |
| 18 hours ago | comment | added | Emil Jeřábek | Sorry, I missed that multiplication is defined in polar coordinates. So this means the real proof happens in establishing the identity $a^2+b^2=(a+ib)(a-ib)$, i.e., distributivity. I have no idea what to make of "when you rotate and dilate a triangle, you get a triangle". | |
| 19 hours ago | comment | added | GH from MO | @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! | |
| 19 hours ago | comment | added | GH from MO | @FShrike It follows from Euclidean geometry that every nonzero complex number has a unique polar decomposition $re^{it}$. Of course $r$ is the length of the number (distance from the origin), while $t$ is the angle of the number (directed angle measured from the positive real axis). Here $e^{it}$ abbreviates the unit vector with angle $t$, which in Cartesian coordinates equals $(\cos t,\sin t)$; this is the definition of $\cos t$ and $\sin t$. At any rate, $re^{it}$ abbreviates $r(\cos t+i\sin t)$, that is, $r\cos t+ir\sin t$. See the "Added" section in my post. I stop here for lack of time. | |
| 19 hours ago | history | edited | GH from MO | CC BY-SA 4.0 |
added 112 characters in body
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| 19 hours ago | comment | added | FShrike | @EmilJeřábek It seems they have defined things so that $e^{it}e^{-it}$ is definitionally $e^{i(t+(-t))}=e^0=1$. Though, to connect the number ‘$c$’ as in polar form with the Euclidean distance seems to have the same content as $|e^{it}|=1$, so I too would like to see this fully fleshed out in a noncircular way. It could be horribly tedious though, as we ought to decide what foundations we are using to even phrase the Pythagorean theorem, and check if these are compatible with $\Bbb C$. In comments under Dan’s post on MSE I describe why I think we shouldn’t even try to prove this in $\Bbb C$ | |
| 19 hours ago | comment | added | GH from MO | @EmilJeřábek You did not pay attention to what I wrote. $e^{-it}e^{it}$ is, by the definition of complex multiplication I gave, equal to $e^{i0}=1$. The law of distributivity is deduced from Euclidean geometry: when you rotate and dilate a triangle, you get a triangle. Hence $e^{it}e^{-it}=(\cos t+i\sin t)(\cos t-i\sin t)$ also equals $\cos^2 t+\sin^2 t$. This is a proof of the Pythagorean theorem! | |
| 19 hours ago | comment | added | FShrike | @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 | |
| 20 hours ago | comment | added | GH from MO | @FedericoPoloni No, unfortunately. At any rate, I take for granted the representations $a+bi$ and $re^{it}$ of complex numbers, where $e^{it}$ abbreviates $\cos t+i\sin t$. These follow from Euclidean geometry. Then I define the sum of complex numbers as $(a+bi)+(a'+b'i):=(a+a')+(b+b')i$, and the product of complex numbers as $(re^{it})(r'e^{it'}):=(rr')e^{i(t+t')}$. So addition comes from translation, while multiplication comes from rotation and dilation. One can check with Euclidean geometry that we obtain a field this way. As $i=e^{i\pi/2}$, we have $i^2=e^{i\pi}=-1$, etc. Nice and easy. | |
| 21 hours ago | comment | added | Federico Poloni | @GHfromMO Do you have class notes on a website, so that we can see that approach fleshed out? | |
| 23 hours ago | comment | added | GH from MO | @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). | |
| 23 hours ago | comment | added | Dan | 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. | |
| 23 hours ago | history | answered | GH from MO | CC BY-SA 4.0 |