Re: How can I evaluate this statement?

*To*: mathgroup at smc.vnet.net*Subject*: [mg89970] Re: How can I evaluate this statement?*From*: magma <maderri2 at gmail.com>*Date*: Wed, 25 Jun 2008 06:29:16 -0400 (EDT)*References*: <g3q7n2$akf$1@smc.vnet.net> <g3r4ml$q3c$1@smc.vnet.net>

On Jun 24, 5:43 pm, Jean-Marc Gulliet <jeanmarc.gull... at gmail.com> wrote: > pircdefense wrote: > > I'm trying to show for all real numbers whether the following statement is true or false: > > > If a > b, then a^2 > b^2. I've tried various syntax and configurations in Mathematica, but I can't get it to report true or false. Can anyone offer a solution? > > Functions such as ForAll, Exists, Resolve, Reduce, or FindInstance are > your friends here. For instance, > > In[1]:= Resolve[ForAll[{a, b}, a > b, a^2 > b^2]] > > Out[1]= False > > In[2]:= Resolve[ForAll[{a, b}, Inequality[a, Greater, b, GreaterEqual, 0], > a^2 > b^2]] > > Out[2]= True > > In[3]:= Reduce[a > b && a^2 > b^2] > > Out[3]= (b <= 0 && a > -b) || (b > 0 && a > b) > > In[4]:= FindInstance[a > b && a^2 > b^2, {a, b}] > > Out[4]= {{a -> 2, b -> -1}} > > Regards, > -- Jean-Marc Uhm....something is not clear to me. If we consider the given statement: expr = ForAll[{a, b}, a > b, a^2 > b^2] we know it is false: FullSimplify[expr] so its negation !expr is true FullSimplify[! expr] We can then ask Mathematica for instances FindInstance[! expr, {a, b}] and surprisingly we get a->0 and b->0, which do not respect the condition a>b. On the other end, Reduce[a > b && a^2 <= b^2] gives b < 0 && b < a <= -b which - correctly - excludes b->0 and a->0. So Reduce is correct, FindInstance seems not. Can somebody explain this?

**Follow-Ups**:**Re: Re: How can I evaluate this statement?***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>