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fisherjeff
on June 25, 2016
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A Quirky Function
> And it equals its reciprocal, because 0^0 = 0^-0 = 1/0^0
Doesn't this already assume that 0^0 evaluates to 1 and not 0?
qf303rjr3
on June 25, 2016
[–]
No - you have the following identity because 0 = -0
0^0 = 0^(-0)
and by the definition of negative exponents,
0^(-0) = 1/(0^0)
So whatever number 0^0 is, it is equal to its own reciprocal.
fisherjeff
on June 25, 2016
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Ah, ok. I didn't have an issue with the first half, I just misunderstood: I thought he was stating that 0^0 == 1/0^0, not deriving it.
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Doesn't this already assume that 0^0 evaluates to 1 and not 0?