Re: Re: Question about "Reduce"

*To*: mathgroup at smc.vnet.net*Subject*: [mg61936] Re: [mg61921] Re: Question about "Reduce"*From*: Chris Chiasson <chris.chiasson at gmail.com>*Date*: Sat, 5 Nov 2005 01:52:19 -0500 (EST)*References*: <dkco86$qcp$1@smc.vnet.net> <200511041011.FAA14948@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Sometimes I think I notice (in the output of Reduce) conditions that should evaluate to true given the assumptions I supplied. So, I am not suprized by your solution. On 11/4/05, Stanley Rabinowitz <stan.rabinowitz at comcast.net> wrote: > I changed my "Reduce" command to > > Reduce[ForAll[{u, v}, 0 <= u <= Sqrt[3] && 0 <= v <= Sqrt[3] && 0 <= 1 > - u v<= (u + v)Sqrt[3], ineq]] // FullSimplify > > and this time I got the correct result. > So I am thinking perhaps this is a bug in Mathematica. > Since u>=0 and v>=0, the conditions > 0 <= (1 - u v)/(u + v)<= Sqrt[3] and 0 <= 1 - u v<= (u + v)Sqrt[3] > should be equivalent. However, the latter condition refers only to > polynomials, > so may be easier for Mathematica. > > -- http://chrischiasson.com/contact/chris_chiasson

**References**:**Re: Question about "Reduce"***From:*"Stanley Rabinowitz" <stan.rabinowitz@comcast.net>