Re: Bug in Reduce?

*To*: mathgroup at smc.vnet.net*Subject*: [mg60432] Re: [mg60406] Bug in Reduce?*From*: Chris Chiasson <chris.chiasson at gmail.com>*Date*: Fri, 16 Sep 2005 03:49:23 -0400 (EDT)*References*: <dfrhi4$g4l$1@smc.vnet.net> <dg8lfv$r8g$1@smc.vnet.net> <200509140926.FAA01590@smc.vnet.net> <200509150916.FAA15875@smc.vnet.net>*Reply-to*: chris.chiasson at gmail.com*Sender*: owner-wri-mathgroup at wolfram.com

On Mathematica 5.2 on Windows, this calculation returns: (a == 0 && b == 0) || (c == 0 && d == 0) On 9/15/05, Kennedy <jack at realmode.com> wrote: > The source of this apparent bug could be my misunderstanding of the middle, > "vars" parameter of Reduce, but it sure seems like the following output > indicates that c must be 0 for my two equations to be satisfied, when in > fact if a and b are both 0, c does not need to be 0. > > Regards, > Jack > > In[1]:= > Reduce[{a c - b d == 0, a d + b c == 0}, {a, b, c, d}, Reals] // > FullSimplify > > Out[1]= > c == 0 && (d == 0 || (a == 0 && b == 0)) > > (version 5.1 for Windows) > > -- Chris Chiasson http://chrischiasson.com/ 1 (810) 265-3161

**References**:**Re: Simplify and Noncommutativity***From:*Robert Schoefbeck <schoefbeck@hep.itp.tuwien.ac.at>

**Bug in Reduce?***From:*"Kennedy" <jack@realmode.com>