Conversation

QC (in Berlin 7/29-??? w/ COVID 😷)
@QiaochuYuan
like you *could* define 0 + 0 = 1 if you really wanted to, that is in some sense a choice, but nobody would do that because then addition would fail to be associative. that's the same sense in which this isn't a choice, to a category theorist
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Arbutus Tree
@aphercotropist
Depends on context. 0^0 is undefined as an operation on reals because there are limits with that indeterminate form with arbitrary values. 0^0 = 1 for natural numbers for the reasons given
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QC (in Berlin 7/29-??? w/ COVID 😷)
@QiaochuYuan
i'll concede that this is a defensible convention on the reals but the convention i prefer is that 0^0 = 1 always and that various functions simply can't be extended continuously in various ways, e.g. x^0 and 0^x are not continuous on [0, ∞)
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