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Topic: Formal proof of the ambiguity of 0^0
Replies: 6   Last Post: Nov 17, 2013 2:04 PM

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 Dan Christensen Posts: 6,760 Registered: 7/9/08
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 UTC-5, Julio Di Egidio wrote:
> "Dan Christensen" <Dan_Christensen@sympatico.ca> wrote in message
>
>

> > On Saturday, November 16, 2013 12:41:35 PM UTC-5, 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 non-positive 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