Re: Interval[{a,b}]-Interval[{a,b}] = 0?

*To*: mathgroup at smc.vnet.net*Subject*: [mg66663] Re: Interval[{a,b}]-Interval[{a,b}] = 0?*From*: "David W.Cantrell" <DWCantrell at sigmaxi.org>*Date*: Fri, 26 May 2006 04:17:30 -0400 (EDT)*References*: <e510r2$8eg$1@smc.vnet.net> <e514do$96v$1@smc.vnet.net> <e53l4f$3r0$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"Richard Fateman" <fateman at cs.berkeley.edu> wrote: > I'm sorry I was not clear in my original note, since several people Including me. We thought you were _wanting_ Interval[{a,b}]-Interval[{a,b}] to be 0. I had no idea, until a moment ago, that Mathematica _gives_ Interval[{a,b}] - Interval[{a,b}] = 0. That's horrid. > seem to think I was objecting to [-1,1] - [-1,1] simplifying to [-2,2]. > This is universally agreed to be correct in the "reliable computing" > community, for the reason given by Johan, and I certainly > agree with it. > > What is likely to be a bug, in my view, is the treatment of > Interval[{a,b}]-Interval[{a,b}] , and I was just wondering if > anyone could defend simplifying it to 0 as a feature. If it's a feature, it's a terrible one. OTOH, it is, in a very perverse way, "consistent" with In[4]:= Interval[{a, b}]/Interval[{a, b}] Out[4]= 1 > The closest I got was a suggestion that "this is not the right tool". ??? > One correct answer might be to leave it alone. Not helpful. > or maybe after ascertaining that a,b are real and a<=b , > the interval [-2*a, 2*b]. No. It should be the interval [a-b, b-a]. Surely that should be easy to implement. David Cantrell > The answer 0 not only is wrong, but apparently leads to bugs > in Limit computations, where Interval notation is used /abused. > > "Johan Grönqvist" <johan.gronqvist at gmail.com> wrote in message > news:e514do$96v$1 at smc.vnet.net... > > .... snipped... explanation of [-1,1]-[-1,1]=[-2,2]