On Apr 25, 4:06 pm, Jussi Piitulainen <jpiit...@ling.helsinki.fi> wrote: > Dan Christensen writes: > > I have always seen exponentiation defined recursively for integer or > > natural number exponents n as: > > > x^1 = x > > x^(n+1)= x^n * x. > > > For integer n, and non-zero x, we have x^n=x^(n+1)/x. Thus x^0 = 1 > > for non-zero x and x^0 is left undefined for x=0. > > ... > > > > Division by zero is quite different. Zero cannot have a > > > multiplicative inverse, call it w, because then we would have both > > > 0*w = 0 (zero does that to any number) and 0*w = 1 (the > > > multiplicative inverse does that to the number), which is a > > > contradiction. I'm taking for granted that (a = b and a = c) > > > implies a = c, and not 0 = 1. > > > Good point. x/y is is undefined for y=0. As a direct result, x^0 is > > undefined for x=0 (see above). > > I make no such point.
Didn't you make the point x/y is undefined for y=0? If not, I apologize.
> The failure to define x^0 by a rule that is not > valid when x=0 is entirely yours, and only you are impressed by it. >
As I recall, many sources leave 0^0 undefined.
> > > > (x+y)^n = x^n if y=0 and not x = n = 0 > > > > = y^n if x=0 and not y = n = 0 > > > > = 0 if x+y = 0 and n > 0 > > > > = sum for k = 0, ..., n of C(n,k) x^k y^(n - k) if x =/= 0 and > > > > y =/= 0 and not x+y = n = 0 > > > > > (Have I covered all the cases?) > > ... > > > > My source says the theorem is "too important to be arbitrarily > > > restricted", which is what happens when one leaves 0^0 undefined. > > > Too big to fail? I have argued here that BT does not require 0^0 to > > be defined. > > No, "too important to be arbitrarily restricted".
Do you think it is impossible to prove the binomial theorem with 0^0 being undefined?