       Re: Truth in inequalities

• To: mathgroup at smc.vnet.net
• Subject: [mg29375] Re: Truth in inequalities
• From: "Orestis Vantzos" <atelesforos at hotmail.com>
• Date: Fri, 15 Jun 2001 02:23:44 -0400 (EDT)
• Organization: National Technical University of Athens, Greece
• References: <9g9mra\$fpt\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```I am afraid I have to agree with you...x might as well be Infinity in a
future evaluation of the x-Infinity expression, and so can not be readily
simplified.
A possible correction could be to use the following rule:
Infinity + n_?NumericQ -> Infinity
Orestis Vantzos
PS. The rule does not work for Infinity+Infinity since
NumericQ[Infinity]==False.
It does simplify to 2*Infinity though, which can be handled by a different
rule.

"Jack Goldberg" <jackgold at math.lsa.umich.edu> wrote in message
news:9g9mra\$fpt\$1 at smc.vnet.net...
> Hi group,
>
> Can someone explain the logic of the following:
>
> x < Infinity   returns x < Infinity
> while
> x - Infinity < 0  returns  True
>
> I should mention that I am aware of the fact that x - Infinity simplifies
> automatically to -Infinity which is then compared to 0 and found wanting.
> The issue I'm raising is why should a CAS that has  x-Infinity < 0 return
> True not also return True for the  x < Infinity?  One awkwardness of
> having this difference of behavior can be seen in the example,
>
> MyFunction[x_,y_]/;(x<y) :=  blah
>
> and
>
> MyFunction[x_,y_]/;(x-y<0)  := blah
>
> do not do the same thing when, say, y=Infinity.
>
> Just curious :-)
>
> Jack
>
>
>

```

• Prev by Date: Odd behavior of InterpolationFunction derivative
• Next by Date: Re: Truth in inequalities
• Previous by thread: Truth in inequalities
• Next by thread: Re: Truth in inequalities