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).
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
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.
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.
Very good visual way of explaining.Thank you for sharing
Thanks!
@@graphicmaths7677How is it known for sure Einstein came upmwotjbthis and not someone else? Is there written proof or something? Thanka for sharing
I don't get how you get from C=A+B to c^2=a^2+b^2.
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
Neither does he. (Binomial Expansion for the case n=2) And thus proof of Fermat's Last Theorem for n > 1.
Very nice!
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).
You are confusing capital C = A + B with lowercase c = a + b. Clearly someone else besides Pythagoras is confused
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
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.
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.
@@nguyenn7729 The hypotenuse of the right triangle is sqr(cc*) (i.e., 5)
I can not see the PROOF in this video.