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Topic:
2017 Nobel prize in Math o:^>___? goes to Franz of Germany;; For reversemath
Replies:
2
Last Post:
Oct 4, 2017 4:44 AM




2017 Nobel prize in Math o:^>___? goes to Franz of Germany;; For reversemath
Posted:
Oct 4, 2017 2:49 AM


John Gabriel:: here is an exciting day in Germany where Franz has just won the Nobel Prize in Math, and here is Alouatta, here to explain what Franz has achieved with ReverseMath.
Alouatta:: yes, John, Reverse Math is where you compose a proof in which everything, every statement is wrong, except for one sentence in the proof that is possibly true, all the rest of the sentences are crap.
John;; is there a need for such a mathematics? I mean, isn't it difficult enough to get someone to have a full page without errors?
Alouatta:: that is the boldness of Reverse Math, which is extremely difficult to do, to compose all lies, except for one sentence. Most math professors can only get 5 true sentences, whereas Franz has now eclipsed all fake sentences except one. See in Franz's classical report below.
On Tuesday, October 3, 2017 at 11:14:58 AM UTC5, Me (Franz of Germany) wrote: > > You obviously don't know the general equation for an ellipse. > > Hint: > > (1/ab)y^2 + (4/h^2)(x  h/2)^2 = 1 > > is the equation for an ellipse. > > If a = b = r and h = 2r we get the equation of a circle: > > (1/r^2)y^2 + (1/r^2)(x  r)^2 = 1 > > => (x  r)^2 + y^r = r^2 > > Are you really too dumb to understand these simple things, Archie? > > Here's the complete proof again: > > > > Cone/Cylinder (side view): > > > > > > /  \ (with b <= a) > > > /b  \ > > > /+´ <= x = h > > > / ´ \ > > > / ´  \ > > > / ´  \ > > > x = 0 => ´+\ > > > / a  \ > > > > > > (cone: b < a, cylinder: b = a = r) > > > > > > r(x) = a  ((ab)/h)x > > > d(x) = a  ((a+b)/h)x > > > > > > y(x)^2 = r(x)^2  d(x)^2 = ab  ab(2x/h  1)^2 = ab(1  4(x  h/2)^2/h^2 > > > > > > => (1/ab)y(x)^2 + (4/h^2)(x  h/2)^2 = 1 ...equation of an ellipse > > qed. > > Now lets just look at some "properties" of this ellipse: > > > > Some considerations: > > > > > > => y(h/2 + x')^2 = ab  ab(2(h/2 + x')/h  1)^2 = ab  ab(2x'/h)^2 > > > > > > => y(h/2 + x') = sqrt(ab) * (sqrt(1  (2x'/h)^2) ...symmetric relative to h/2 (hence Ec = cF) > > > > > > => y(h/2) = sqrt(ab) (= Gc = cH)



