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