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All Numbers Are Equal
j2 x( q' c$ A/ g( UTheorem: All numbers are equal. Proof: Choose arbitrary a and b, and let t = a + b. Then ) o" [! j K( Z) ^
& y& E% n2 Y, ^/ x( Z h( Ya + b = t) b/ J/ Q2 t# ^6 R7 n
(a + b)(a - b) = t(a - b)
8 |9 K# G' ~7 G1 k$ f8 |a^2 - b^2 = ta - tb( N; a& T; n7 |
a^2 - ta = b^2 - tb
, w# O, L S3 M# G! q* t, ^a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/41 j, [! M" B/ I1 ^$ C* i
(a - t/2)^2 = (b - t/2)^2
# E: F* K- h' R9 L1 h+ i5 L* @4 V& La - t/2 = b - t/2
2 f7 L2 J9 M, r' D p5 Ca = b # Q) l( k. H. M& L% r8 X
5 e& W# m, m7 ~/ N: W) Z& C
So all numbers are the same, and math is pointless. |
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