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Re: a^n*b^n != (a*b)^n
*To*: mathgroup at smc.vnet.net
*Subject*: [mg13761] Re: a^n*b^n != (a*b)^n
*From*: murray at math.umass.edu (Murray Eisenberg)
*Date*: Mon, 24 Aug 1998 05:07:12 -0400
*Organization*: University of Massachusetts, Amherst
*References*: <6imi04$gvl@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Michael Milirud (mmichael at 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 at 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
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