Re: Ok so I just proved 1 = -1. Someone help me find my error.
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Here's another "proof" in this vein that I've always found enjoyable:
x^2 = x + x + . . . x (x copies of x)
Differentiate both sides:
2x = 1 + 1 + 1 . . . = x
Divide by x:
2 = 1.
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Good one. The problem here is that
x^2 = x + x + . . . x (x copies of x)
is only true for integer x, and if this function is only defined for integer x, then the function is discreet and is not differentiable.
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