Re: 0^0

"Ulrich Sondermann" <usondermann@xxxxxxxxxxxxx> writes:

> Compilers, etc use definitions given to them by their authors,
> depending on which camp they are in, you will get a result. If they
> symbolically try to solve equations first a^0, a=0, may indeed end
> up 1, just by definition. However, when using numbers in variables,
> you will get a NaN. The danger is symbolic division by zero, which
> remains undefined.

Complete and utter nonsense. No division is involved here whatsoever.

With your kind of argumentation, 0*5 would be undefined in the
arithmetic of positive integers, since it is (3+(-3))*5, and -3 is not
a positive integer.

x^0 is a product containing exactly 0 factors of x, so the value of x
is completely irrelevant to the result. Just like 0*x is a sum
containing exactly 0 summands of x, making the value of x completely
irrelevant to the result.

Read the sci.math FAQ at
<URL:http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/>.

--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
.

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