All Numbers Are Equal / j5 P" Y+ a( f& gTheorem: All numbers are equal. Proof: Choose arbitrary a and b, and let t = a + b. Then ; x5 \$ K' N" \9 u3 e/ J- u) ~. {* o3 K. S$ }
a + b = t * K6 V. t3 P6 k; J# ^, M8 H9 V(a + b)(a - b) = t(a - b) 5 s. y8 q: o2 _, s0 H; \a^2 - b^2 = ta - tb 8 m$ x4 M5 o ?9 a8 h1 \1 d5 ?0 Wa^2 - ta = b^2 - tb / h4 h% C% U2 r1 I# `a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4) Y4 A: J9 p( o) N+ ]+ B) i
(a - t/2)^2 = (b - t/2)^2 3 Y! D2 X0 g$ K$ `a - t/2 = b - t/2 # i6 k6 p7 ~6 sa = b ' K& `: \4 z/ K: }* ^/ ?6 G2 |9 C0 ~4 u7 |- B. b$ }0 k0 {' j
So all numbers are the same, and math is pointless.