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What do you mean by 'is defined by group operation'? Reals/integers with multiplication don't form a group. 'Multiplying 0 by itself 0 times' is what Wikipedia
by g_ 17y ago
What do you mean by 'is defined by group operation'? Reals/integers with multiplication don't form a group.
'Multiplying 0 by itself 0 times' is what Wikipedia calls 'empty product'. For me is is good intuition, but that's a matter of taste.
AFAIK the usual convention is:
* in contexts where the exponent is varying continuously and a is a real number, a^b is undefined for a=b=0
* in contexts where the exponent is varying discretely (as an integer/natural number), a^b = 1 for b=0 and any a
Haskell quite nicely distinguishes between these two:
(^) :: (Num a, Integral b) => a -> b -> a
(* * ) :: (Floating a) => a -> a -> a
but it still gives 0 * * 0 and 0^0 as 1. (There shouldn't be spaces between asterisks, HN treats them as italics)
- profquail 17y agoReals form a group under multiplication. Integers don't, but they do form a commutative ring (addition, negation, multiplication).
- jibiki 17y ago> Reals form a group under multiplication. I think his point was that 0 has no inverse. So in the literal sense, the reals are not a group under multiplication (although obviously, when somebody says "the reals are a group under multiplication", one usually interprets it as "R\{0} is a group under multiplication".) This is vaguely relevant to the issue at hand because we are doing 0 to the 0.