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All Numbers Are Equal ' V) P1 l4 N# Z, u; m9 o& |
Theorem: All numbers are equal. Proof: Choose arbitrary a and b, and let t = a + b. Then # I. s' \1 D8 \' `, r! S8 |
, U$ \/ V- T+ ^: {2 qa + b = t# ~( t, U$ p0 c6 t7 H4 C: a& z
(a + b)(a - b) = t(a - b)
- X/ h( w8 u& z6 Ea^2 - b^2 = ta - tb6 @3 ?8 i- _2 ?$ T9 @5 ~
a^2 - ta = b^2 - tb, j! L2 J; g! w( t. t
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
# A: `, D9 ^4 L. A) {' n(a - t/2)^2 = (b - t/2)^2
|6 o6 P2 L# `$ oa - t/2 = b - t/2
1 A O8 h0 U; u0 H9 ua = b : H a1 L$ \1 b
M' F) L' Q6 u% I/ W( d2 v
So all numbers are the same, and math is pointless. |
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