Re: Bug in Reduce?

*To*: mathgroup at smc.vnet.net*Subject*: [mg60483] Re: [mg60406] Bug in Reduce?*From*: Pratik Desai <pdesai1 at umbc.edu>*Date*: Sat, 17 Sep 2005 02:32:22 -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>*Sender*: owner-wri-mathgroup at wolfram.com

Kennedy 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) > > > I get a more richer solution when I used assumptions (c notequal 0) and simplify Clear[a, b, c, d] expr1 = {a*c - b*d == 0, a*d + b*c == 0} $Assumptions = {c â? 0, a Ïµ Reals, b Ïµ Reals, cÏµ Reals, d Ïµ Reals, c Ïµ Reals} s1 = Reduce[expr1, {a, b, c, d}] // Simplify >>(a == 0 && (b == 0 || (c == 0 && d == 0))) || ((b*c)/a + d == 0 && a != 0 && (b == (-I)*a || b == I*a)) || (c == 0 && d == 0 && a^2 + b^2 != 0) Hope this helps If there are some ineligible characters those are meant to be element Esc Shift+1=Esc -- Pratik Desai Graduate Student UMBC Department of Mechanical Engineering Phone: 410 455 8134

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

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