Einstein's proof of Pythagoras theorem

Einstein's proof of Pythagoras' theorem
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This video present's Einstein's simple and elegant proof of Pythagoras' theorem. Instead of constructing a square on each side of the right-angled triangle, this proof constructs triangles. The triangles are all similar to the main triangle, and it can be shown quite easily that the two smaller triangles have the same area as the main triangle, which is sufficient to prove Pythagoras.
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Pythagoras' theorem: • Pythagoras theorem
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This video was created using generativepy, a free and open source Python library for creating mathematical art and animations, available from github.com/mar...

[@saenzmath]

Very good visual way of explaining.Thank you for sharing

[@graphicmaths7677]

Thanks!

[@leif1075]

​@@graphicmaths7677How is it known for sure Einstein came upmwotjbthis and not someone else? Is there written proof or something? Thanka for sharing

[@wasgehtsiedasan8660]

I don't get how you get from C=A+B to c^2=a^2+b^2.

[@bilaalahmed8923]

he said C= c^2, A=a^2 and B=b^2 in the first place, assigning them to respective places results a^2+b^2=c^2

[@BuleriaChk]

Neither does he. (Binomial Expansion for the case n=2) And thus proof of Fermat's Last Theorem for n > 1.

[@jakobthomsen1595]

Very nice!

[@BuleriaChk]

c=a+b c^2 = (a+b)^2 = [a^2 + b^2] + [2ab] (binomial expansion) c^2 a^2+b^2 The "proof" in the video is only valid in the imagination. (Pythagoras was also confused).

[@nguyenn7729]

You are confusing capital C = A + B with lowercase c = a + b. Clearly someone else besides Pythagoras is confused

[@BuleriaChk]

What does capital C have to do with it? C = A + B C^2 = (A+B)^2 = [A^2 + B^2] + [2AB] C^2 [A^2 + B^2] C = A + iB C* = A - iB CC* = A^2 + B^2 only in the imagination. Note that A = 4. B=5 CC* = 25 49 49 = 7 ^2 = [49] = [25] + [24] = [CC*]+ 24

[@nguyenn7729]

The Pythagorean theorem, a^2 +b^2 = c^2 refer to side length lower case a,b,c of a right triangle. Capital A,B,C in this context refer to the areas of squares with side lengths a,b,c respectively. The areas of such squares are clearly a^2, b^2, and c^2 respectively. Hence it suffices to prove capital A + B = C.

[@nguyenn7729]

And you dont even understand imaginary numbers. If c = a + ib is a complex number, what you have described is the norm of c not c^2. c^2 = c*c in the normal understanding of powers.

[@BuleriaChk]

@@nguyenn7729 The hypotenuse of the right triangle is sqr(cc*) (i.e., 5)

[@BtMeszinet]

I can not see the PROOF in this video.