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MathGroup Archive 2004

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Re : fullsimplify problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52922] Re : [mg52908] fullsimplify problem
  • From: "Jaccard Florian" <Florian.Jaccard at he-arc.ch>
  • Date: Fri, 17 Dec 2004 05:18:31 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Why do you consider the output as incorrect?
I'd say it is elementary algebra.
If -a/b^2 == a/b^2 , then 2a/b^2==0 , so a==0 and b!=0   or   b==Infinity 
a/b==0  is equivalent! 
Don't you think so ?

regards
 
F.Jaccard


-----Message d'origine-----
De : symbio [mailto:symbio at has.com] 
Envoyé : jeudi, 16. décembre 2004 09:41
À : mathgroup at smc.vnet.net
Objet : [mg52908] fullsimplify problem

Given ( - a / b^2) = =  ( + a / b^2), it can be written as (-2 a / b^2), but 
when I use FullSimplify on that equation, I get the incorrect result ( a / 
b ) = = 0.  How is this possible?

In[2]:=
-a/b^2 == a/b^2
% // FullSimplify
Out[2]=
\!\(\(-\(a\/b\^2\)\) == a\/b\^2\)
Out[3]=
\!\(a\/b == 0\)

In[5]:=
-2a/b^2 == 0 // FullSimplify
Out[5]=
\!\(a\/b == 0\)


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