Date: Oct 17, 2013 4:15 PM
Author: Bart Goddard
Subject: Re: Formal proof of the ambiguity of 0^0
Dan Christensen <Dan_Christensen@sympatico.ca> wrote in
>> 3^2 = (0^0)^2 = 0^(2*0) = 0^0 = 3
> Good point! That's why I stipulate that a non-zero base for the Power
> of a Power Rule (Theorem 5) which you use in your 2nd step.
Which is why you have nothing but contradictions here. You're
asserting that 0^0 can be defined to be anything, and the
exponent rules still work. They don't. However, if 0^0=0
or 0^0 =1, they do. Everyone on this group gets this point
but you. You can't extend a definition without extending it.