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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?
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