Re: a^n*b^n != (a*b)^n
- To: mathgroup@smc.vnet.net
- Subject: [mg12378] Re: a^n*b^n != (a*b)^n
- From: "Clifford J. Nelson" <nelsoncj@gte.net>
- Date: Sun, 10 May 1998 02:05:00 -0400
- Organization: gte.net
- References: <6itjo9$1h0@smc.vnet.net>
On Thu, May 7, 1998 4:31 PM, Murray Eisenberg <mailto:murray@math.umass.edu> wrote: >Michael Milirud (mmichael@idirect.com) wrote: > >: complex and n is real. However: > >: 6 == Sqrt[36] == Sqrt[-4*-9] == Sqrt[-4]*Sqrt[-9] == 2i*3i == 6*i^2 == >: -6 > >: Hence 6 == -6 > >: ARGHHH!!!! > >: After quite some time, I found the problem to be in the step: > >: Sqrt[-4*-9] == Sqrt[-4]*Sqrt[-9] > >: which as Mathematica claims does NOT equal to each other!!! > >The root of the difficulty is that, once one enters the complex domain, >there is really no such thing as THE square root of a number (and >similarly for higher roots). You are now really talking, in the case >of square roots, about a set of two square roots. > >-- > Murray Eisenberg Internet: >murray@math.umass.edu > Mathematics & Statistics Dept. Voice: 413-545-2859 (W) > University of Massachusetts 413-549-1020 (H) > Amherst, MA 01003 Fax: 413-545-1801 > > Read the Mathematica documentation of PowerExpand. (a*b)^n -> a^n*b^n is considered dangerous. It has nothing to do with the fact that the relation: x^2 = y has two solutions, because Sqrt[x] is not a relation, it is a function. Old books written before the word function was defined with set theory speak of "multi-valued functions" but that is actually like "military intelligence". Functions only have one value, but, relations can have many values. (see Logic and Set Theory--- Schaum's Outline Series). Cliff Nelson