Ki Song
Posts:
221
Registered:
9/19/09
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Re: 0^0=1
Posted:
May 8, 2012 9:38 AM
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On Monday, May 7, 2012 6:05:38 PM UTC-4, Ross wrote:
> On May 7, 4:06 pm, PotatoSauce <kiwisqu...@gmail.com> wrote:
> > On Monday, May 7, 2012 4:51:38 AM UTC-4, Ross wrote:
> >
> > > I'm not a mathematician, and I can't read all 380 posts in this
> > > thread.
> >
> > > But, my understanding of the reason that x^0 = 1 is a consequence of
> > > negative exponents. E.g.
> >
> > Not quite, it's the other way around. The definition that
> >
> > x^-n = 1/x^n
> >
> > comes from
> >
> > 1) x^0 = 1.
> > 2) x^(n+m) = x^n x^m
> >
> >
>
> Thanks.
>
> I looked for proofs that x^0 = 1, and found two extremely informal
> proofs for people 'to explain to their children' how x^0 = 1.
> http://www.homeschoolmath.net/teaching/zero-exponent-proof.php
>
> Neither of the derivations on that page work when x = 0.
>
> The first is based on:
>
> x^n * x^0 = x^(n+0) = x^n
>
> Where x <> 0, then x^0 must be equal to 1. But, if x = 0, then it
> makes no difference what the value of x^0 is because it's multiplied
> by zero and equated to zero.
>
> The second is noting that:
>
> x^4 = x * x * x * x
> x^3 = x * x * x
> x^2 = x * x
> x^1 = x
> x^0 = ?
How about
x^4 = 1*1 * x * x * x * x
x^3 = 1*1 * x * x * x
x^2 = 1*1 * x * x
x^1 = 1*1 * x
x^0 = 1*1
I put two "1"s instead of just 1, in case you feel like multiplication can only be done when there are at least two numbers.
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