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 > > >