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Topic: 0^0=1
Replies: 145   Last Post: Jun 5, 2012 1:10 PM

 Messages: [ Previous | Next ]
 Dave Dodson Posts: 690 Registered: 12/13/04
Re: 0^0=1
Posted: Apr 22, 2012 2:22 AM

On Saturday, April 21, 2012 2:03:23 AM UTC-5, Peter Webb wrote:
> "Dave" <dave_and_darla@juno.com> wrote in message
> news:26122529.1302.1334983483051.JavaMail.geo-discussion-forums@ynlp3...

> > On Friday, April 20, 2012 11:25:42 PM UTC-5, Peter Webb wrote:
> >> "DonH" <donlhumphries@bigpond.com> wrote in message
> >> news:zKpkr.6017\$%E2.342@viwinnwfe01.internal.bigpond.com...

> >> > "Richard Tobin" <richard@cogsci.ed.ac.uk> wrote in message
> >> > news:jmsvh1\$nml\$1@matchbox.inf.ed.ac.uk...

> >> >> In article
> >> >> Tonico <Tonicopm@yahoo.com> wrote:
> >> >>

> >> >>>Or, since we're dealing with real or complex numbers, consider lim x^x
> >> >>>= lim e^{x log(x)} , when x --> 0 ...this is a good reason to define
> >> >>>0^0 = 1 .

> >> >>
> >> >> But equally one could consider the limit of 0^x.
> >> >>
> >> >> For integers the set mapping argument seems compelling, but when
> >> >> dealing with the reals it seems better to leave it undefined or define
> >> >> it explicitly if needed.
> >> >>
> >> >> -- Richard

> >> >
> >> > # Yes, 0^0 may be an arbitrary device to accommodate a computer, as
> >> > Barron's Maths Study Dictionary excludes zero(^0) (pg.62) when defining
> >> > the "zero exponent".
> >> > Which still leaves, eg. 5^0 = 1.
> >> > If I have five apples and multiply them by themselves zero times,
> >> > do I
> >> > end up with one apple?
> >> > Also, does 5^0 = 5 * 1/5?
> >> > Has x^0 = 1 ever been proved? Or, is it merely a convenient dogma?

> >>
> >> x^a * x^b = x^(a+b)
> >> x^a * x^0 = x^(a+0) = x^a
> >> x^0 = 1

> >
> > Your last statement assumes that x^a is defined and non-zero. But if x = 0
> > and a > 0, then x^a = 0, and if x = 0 and a < 0, then x^a is undefined.
> > Thus, your "proof" breaks down in the case under discussion: 0^0.
> >
> > Dave

>
> I was responding to the last line of the previous post, "Has x^0 = 1 ever
> been proved?". I am quite aware that there are restrictions on x and a. You
> didn't even manage to list them all correctly; if x<0 and a is irrational
> then x^a is undefined. (If you want to nit-pick, get it correct).
>
> The OP clearly knows very little maths, and I just wanted to show him why
> x^0 is defined as being equal to 1 in a simple way.

Whether x^a is complex or undefined when x < 0 and a is irrational depends on whether you say that sqrt(-1) is undefined or i, doesn't it?

Dave

Date Subject Author
4/20/12 Don H
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4/20/12 Richard Tobin
4/20/12 Don H
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