Re: TI-83 BUG?????
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Re: TI-83 BUG?????
On Tue, 10 Sep 1996, jfhill wrote:
>
>
> On Tue, 10 Sep 1996, Dave Wollenberg wrote:
>
> > On Tue, 10 Sep 1996, P. Kolbus wrote:
> >
> > > No, according to all my textbooks (that mention it), 0^0=1.
> >
> > Anything to the 0 power is 1. I've never been able to figure out why, but
> > it is.
> >
> Well, according to the division property of exponents, for any non-zero
> real number r ( (a^m) / (a^n) = a^(, where a is real and m,n are
> integers.
> This can be illustrated in the specific by (2^5) / (2^3) = (2*2*2*2*2)
> /(2*2*2) = (2*2*2)(2*2) / (2*2*2) = (2*2)[ where the (2*2*2)'s factor
> (cancel) out]= 2^2 or 2^(5-3). Replace the 2's with a's and you have the
> general case.
> So, if m=n then (a^m)/(a^n)=(a^m)/(a^m)=a^(m-m)=a^0. But
> (a^m)/(a^m)
> represents a number divided by itself, then (a^m)/(a^m) must also =1.
> If (a^m)/(a^m)=a^0 and (a^m)/(a^m)=1 then a^0=1.
> I'd always thought that 0^0 was undefined, in general.
>
> John
>
>
That second line should have read "for any non-zero real number, a, where
m and n are integers, (a^m)/(a^n)=a^(m-n)"
Something weird happened between the typing and the sending.
jfh
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