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

*To*: mathgroup at smc.vnet.net*Subject*: [mg66632] Re: Interval[{a,b}]-Interval[{a,b}] = 0?*From*: "David W.Cantrell" <DWCantrell at sigmaxi.org>*Date*: Thu, 25 May 2006 02:57:47 -0400 (EDT)*References*: <e510r2$8eg$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"Richard Fateman" <fateman at cs.berkeley.edu> wrote: > Is this a bug or a feature? > Notice that Interval[{-1,1}]-Interval[{-1,1}] is Interval[{-2,2}]. Johan has already given you an excellent answer. Here's just a slightly different way of looking at the matter. I hope that each step seems intuitively obvious: -Interval[{-1,1}] is the same as Interval[{-1,1}]. (In other words, since Interval[{-1,1}] is symmetric about the origin, negating all of its elements, we get the same interval.) a - b is the same as a + (-b), even in interval arithmetic. So then we have Interval[{-1,1}] - Interval[{-1,1}] = Interval[{-1,1}] + (-Interval[{-1,1}]) = Interval[{-1,1}] + Interval[{-1,1}] = 2 Interval[{-1,1}] = Interval[{-2,2}]. --------------------- For better or worse, most operations in interval arithmetic do not have inverses. (Example: a/b = c does not imply that a = b*c.) -Interval[{-1,1}] is the _negative_ of Interval[{-1,1}], yes. But it's not an additive inverse of Interval[{-1,1}]. In fact, there can be no such additive inverse. Regards, David