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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