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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 � 0, a ϵ
Reals, b ϵ Reals, cϵ Reals, d ϵ Reals, c ϵ 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

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-- 
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>