
Re: Formal proof of the ambiguity of 0^0
Posted:
Nov 17, 2013 11:50 AM


On Sunday, November 17, 2013 10:19:33 AM UTC5, Julio Di Egidio wrote: > "Dan Christensen" <Dan_Christensen@sympatico.ca> wrote in message > > news:90552961bcc840d8856a6cbc94d8c52c@googlegroups.com... > > > On Saturday, November 16, 2013 12:41:35 PM UTC5, Bart Goddard wrote: > > <snip> > > >> (We'll leave aside the incorrect assumption that leaving > > >> 0^0 undefined is "common practice.") > > > > > > Every high school graduate knows that 0^0 is undefined, but not > > > necessarily why this is so. > > > > We were simply explained that 0^0 = 0/0, because x^0 = x/x when extending > > the notion of repeated multiplication to nonpositive exponents.
I can see the reasoning behind it, but I don't like the idea setting two undefined expressions equal to one another.
The latest elementary algebra textbook I have (Aufmann & Lockwood, 5th ed., 2011) simple states that 0^0.
More usual, I think, is to just point of the inconsistency in the patterns given by:
0^3 = 0 0^2 = 0 0^1 = 0 0^0 = ?
3^0 = 1 2^0 = 1 1^0 = 1 0^0 = ?
Dan Download my DC Proof 2.0 software at http://www.dcproof.com Visit my new math blog at http://www.dcproof.wordpress.com

