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G R A V I T Y
All numbers are equal ! Theorem: All numbers are equal.
Proof: Choose arbitrary a and b, and let t = a + b. Then

a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b

So all numbers are the same, and math is pointless.
Oct 25, 2008 7:14 PM
Comments · 8
 (a - t/2)^2 = (b - t/2)^2 <=> a = b  OR a-t/2 = t/2-b. These are the decisions of the system of equations

a + b = t              (1)
(a-b)^2=t(a+b)   (2)

But it's not a proof that a=b allways, it's only one of the decisitions of this system, that all, nothing else.



November 10, 2008
Is it true that 1 = -1?
Yes, because 1^2 = 1 = (-1)^2.
That's the essence of the argument.

t=a+b. So a - t/2 = t/2 - b. And he "showed" by the above argument that a - t/2 = b - t/2, hence a = b.
Is this too serious? haha
November 9, 2008

** laugh

;)

November 7, 2008

ohhh.. hahaha.... im trying to deal with them !

If  it is...  x+y = (3x + 5).(y/3) ..., how many times can a person laughs in a day? 

=))))))

November 7, 2008
Yep, but you have to pay first !
October 26, 2008
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