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MathGroup Archive 2004

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Re: FullSimplify with ForAll

  • To: mathgroup at smc.vnet.net
  • Subject: [mg48642] Re: [mg48624] FullSimplify with ForAll
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 9 Jun 2004 04:17:02 -0400 (EDT)
  • References: <200406080448.AAA28389@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 8 Jun 2004, at 13:48, Sharath wrote:

> Given the input (Mathematica 5.0)
>
> a = (d + 2)/d; b = (1/a)^(1/r^2);
> FullSimplify[ForAll[{r, d}, r>=1 && 0 < d && d < 1, 1 - b^(r^2) > 0]]
>
> the output is
> \!\(&#8704; \_\({r, d}, r &#8805; 1 && 0 < d && d < 1\)1 -
> \((\((d\/\(2 + \
> d\))\)\^\(1\/r\^2\))\)\^\(r\^2\) > 0\)
>
> which is the same expression with b expanded, that is,
>
> forall{r,d},r>=1 && 0<d && d<1 1-((d/d+2)^1/(r^2))^(r^2)>0
>
> Why is FullSimplify not cancelling 1/(r^2) and r^2? My interest is a
> more complex inequality but FullSimplify is just giving me the
> expression back. Then I tested with this simpler inequality and it
> still gives me the expression back.
>
> Could anyone know how I should give the input? I want to see if the
> expr is Ture or False.
>
> Thanks,
>
> Sharath
>
>


Because ForAll has an entirely different purpose. The right way to do 
this is:

a = (d + 2)/d; b = (1/a)^(1/r^2);



FullSimplify[1 - b^(r^2) > 0,r>=1 && 0 < d && d < 1]


True





Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/


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