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]= \!\(a\/b == 0\)