Re: fullsimplify problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg52938] Re: fullsimplify problem*From*: Robert <robert_dot_paynter_ at _eng_.ox.ac.uk>*Date*: Fri, 17 Dec 2004 05:19:09 -0500 (EST)*Organization*: Oxford University, England*References*: <cprjjd$r57$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

symbio wrote: > 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]= > \!\(a\/b == 0\) > I think what you have found is the "trivial" solution to an eigenvalue problem. R