

0^0 = 1, 0, Undefined? - MikeMKH
http://comp-phil.blogspot.com/2009/06/00-1-0-undefined.html

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profquail
This article is totally wrong. 0^0 = 1 because it is defined by a group
operation, not by "multiplying 0 by itself 0 times". It has to do with
combinatorics or set theory (there are different ways to look at the problem):

Source 1:
[http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero...](http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power)

Source 2: <http://en.wikipedia.org/wiki/Empty_product>

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g_
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)

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profquail
Reals form a group under multiplication. Integers don't, but they do form a
commutative ring (addition, negation, multiplication).

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jibiki
> 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.

