sqrt question

[nsg]

(04-07-2017 09:15 PM)Claudio L. Wrote:  Assuming clear and consistent conventions (like always choosing the positive root for sqrt(), even if complex),

You cannot do this because there is no such thing as "negative" for complex numbers.
"Greater than" and "less then" are only defined for reals (number line).

So, if you stay in real land, then sqrt(-9)*sqrt(-4) is undefined and if you expand into complex plane then it is +-6. Square root in complex plane IS a multivalued function, there is simply no way around it.

On a somewhat related note, i and -i are algebraically indistinguishable. C is extension field of R, extended by roots of irreducible polynomial x^2+1=0. It is second degree polynomial, has 2 roots, we pick one of them arbitrarily, call it "i" and continue. We might as well could have picked another one (it is equal to -i in our newly established convention) and have called THAT one "i" and would not notice any difference. This contrasts with 1 and -1, which are clearly algebraically distinct and whole bunch of equations will break if we replace 1 with -1 and -1 with 1.