Re: a^n*b^n != (a*b)^n
- To: mathgroup@smc.vnet.net
- Subject: [mg12344] Re: a^n*b^n != (a*b)^n
- From: murray@math.umass.edu (Murray Eisenberg)
- Date: Thu, 7 May 1998 18:52:31 -0400
- Organization: University of Massachusetts, Amherst
- References: <6imi04$gvl@smc.vnet.net>
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