[seqfan] Re: Without any discussion

franktaw at netscape.net franktaw at netscape.net
Tue Sep 1 07:28:12 CEST 2009


Actually, there is a discrete context where 0^0 = 0.

In a commutative ring, define b to be the pseudo-inverse of a if a^2*b 
= a and a*b^2 = b.  (In the non-commutative case, the makes b a right 
pseudo-inverse of a, and a likewise a left pseudo-inverse of b.  As 
with the ordinary inverse, a number might have multiple left inverses 
or multiple right inverses, but it is has both, they are equal.)  
Obviously, if a has an inverse, that is also its pseudo-inverse.

One would then naturally define a^(-k) to be b^k, and a^0 should then 
be a^1 * a^(-1) = a*b.

Now since 0 is always its own pseudo-inverse, we have 0^0 = 0.

-----
As for A061862, there should be a version where 0^0 = 1 is allowed, and 
one where it is not.  Neil has chosen A061862 to be the one where it is 
not allowed; so submit the other one as a new sequence.

Franklin T. Adams-Watters




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