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.

> 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]

```

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