Dan Christensen writes: > On May 7, 10:54 am, Jussi Piitulainen wrote: > > Dan Christensen writes: > > > set theory are consistent), but wouldn't a system that defined 0^0=k > > > for ANY integer k also be consistent? > > > > > Presumably, x^(m+n) = x^m * x^n would still hold. Then 0^(m+0) = 0^m > > > * 0^0 = 0 * k. In this equation, ANY value would work for k. On what > > > basis can we say that k MUST be 1? > > > > The consequences of setting 0^0 to any number other than 0 or 1 are > > considered highly undesirable. Work it out: use the above rule of > > exponents to show that 0^0 = 3 implies 3 = 9. > > Thanks! We would have 0^0 = 0^(0+0) = 0^0 * 0^0. With 0^0 = 3, > that's 3 = 3*3.
> So, why not 0^0 = 0? Can this be ruled out using only natural number > arithmetic?
Not directly, as far as I can see, but almost nothing supports it while much requires x^0 = 1, either for all x or for all natural x.