MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: fullsimplify problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52943] Re: fullsimplify problem
  • From: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>
  • Date: Fri, 17 Dec 2004 05:19:29 -0500 (EST)
  • References: <cprjjd$r57$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

This looks correct to me. Your equation says that something equals minus 
itself, so the something must be zero.

You could also try evaluating ( - a / b^2) == ( + a / b^2)//Reduce to obtain 
the result a\[Equal]0&&b\[NotEqual]0.

Steve Luttrell

"symbio" <symbio at has.com> wrote in message news:cprjjd$r57$1 at smc.vnet.net...
> 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]=
> \!\(a\/b == 0\)
> 



  • Prev by Date: Re : fullsimplify problem
  • Next by Date: Re: Re: Mathematica 5.0 on Mandrake 10.1?
  • Previous by thread: Re: fullsimplify problem
  • Next by thread: help mo on a recursive function