Re: JSH: Operator ambiguity, Escultura

From: *** T. Winter (***.Winter_at_cwi.nl)
Date: 11/29/04 

Date: Mon, 29 Nov 2004 00:32:21 GMT

In article <qfcjq0lu2gdkir58dnvsfoo24hs4pv5dvi@4ax.com> ullrich@math.okstate.edu writes:
> On 27 Nov 2004 13:59:51 -0800, jstevh@msn.com (James Harris) wrote:
...
> >Naively then, you may believe that you can just say, take the positive
> >of the square root but as Escultura showed, that doesn't work:
> >
> >i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction.
> >
> >You see, the ambiguity in the square root operator still remains,
> >despite the convention.
...
> Finally, you seem to have missed the most important point
> in the "contradiction" above, which is that it's impossible
> to define sqrt(z) for complex z in such a way that
> sqrt(zw) = sqrt(z) sqrt(w).

He is missing one more point. That you can not algebraically distinguish
i and -i is also a result from Galois theory (you can not algebraically
distinguish the various roots of a primitive polynomial). 

-- 
*** t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~***/