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
1) x^0 = 1. 2) x^(n+m) = x^n x^m
> > if 2^-2 equals 1/2^2 and x^a * x^b = x^(a+b), then 2^2 * 2^-2 = 2^(2 + > -2) = 2^0. > > We can then expand the example to get 2^0 = 2^2 * 1/(2^2) = 1. > > However, if x = 0, then the derivation of a^0 = 1 no longer works: > > 0^2 * 0^-2 = 0*0 / (0*0) = 0/0 which is undefined.
0 to any negative exponent is not defined.
> > Hence x^0 = 1 is based on a derivation that doesn't work for x=0, and > hence to quote the consequence of that derivation to argue the value > of 0^0 is plain wrong. > > I presume that I'm "missing something" and someone will be along to > spank me soon.