Re: Decorations and Motion 22
> From: "Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>
> To: "John Pryce" <j.d.pryce@xxxxxxxxxxxx>,
> "Arnold Neumaier" <Arnold.Neumaier@xxxxxxxxxxxx>
> Cc: "1788" <stds-1788@xxxxxxxxxxxxxxxxx>
> Subject: Re: Decorations and Motion 22
> Date: Sun, 31 Oct 2010 13:49:14 -0500
>
> John Pryce wrote:
>
> > What is the result of (bare decoration) op (bare interval)? E.g.
> > ok + [1,2]
>
> By section 2.3, the result has to be a bare decoration: ok.
>
> Nate Hayes
Nate,
You are not addressing the crux of John's question
here. You only answer in the case of add which is
everywhere defined.
And your answer seems to imply that a bare decoration
qualifies as a new type of idempotent interval along
the lines of empty & NaI.
>
> > Then I suppose "ok" promotes to "an arbitrary ok decorated interval"
> > (xx,ok). Whence
> > ok + [1,2] = "possibly everywhere defined"
> > because that's the worst that can happen with (xx,ok) + ([1,2],2); and
> > [1,2] / ok = "nowhere defined"
> > because xx might be [0,0].
John also asks what is to be done when such an
interval passes through an operator that is NOT
everywhere defined such as divide. He could just
as easily have asked what is the value of sqrt(ok).
I ask: What is the value of sqrt(<bareDecoration>)?
Dan