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Re: 0^0 = 1?

  • To: mathgroup at
  • Subject: [mg95644] Re: 0^0 = 1?
  • From: Dave Seaman <dseaman at>
  • Date: Fri, 23 Jan 2009 05:09:45 -0500 (EST)
  • Organization: Information Technology at Purdue
  • References: <gl7211$c8r$> <gl9mua$ajr$>

On Thu, 22 Jan 2009 11:56:58 +0000 (UTC), dh wrote:

> Hi,

> 0^0 means the limit if both base and exponent go to zero. 

No, that is not how 0^0 is defined.  Does 2+2 mean the limit as both
summands go to 2?  The value may happen to be the same in that case, but
that is not how 2+2 is defined.

The value of x^y for cardinal numbers x and y is the cardinality of the
set of mappings from y into x.  In the case where x and y are the empty
set, there is exactly one such mapping.  Hence, 0^0 = 1.

It's a theorem of ZF (as stated in Suppes, _Axiomatic_Set_Theory_) that
m^0 = 1 for every cardinal number m.

Another way is to notice that 0^0 represents an empty product, whose
value is the identity element in the monoid of the integers (or the

	In[1]:= Product[0,{k,0}]

	Out[1]= 1

One might also consider the series expansion for Exp[0], which reduces to

	1 = 0^0/0! + (lots of terms that all reduce to zero).

Having x^y be discontinuous at (0,0) does not "cause problems" any more
than having the Sign function be discontinuous at 0 causes problems.
Anyone who works with limits should be aware that you can't just blindly
assume continuity when evaluating limits.  You have to consider the
actual definition of the limit.

Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.

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