Re: Vincent's 5 Oct rad(xx) proof

["Siegfried M. Rump" <rump@xxxxxxxxxxxxx>]

You might want to consult Section 3 in the appended paper,
which contains some thoughts on midpoint and radius.

Best wishes,
Siegfried

--
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Prof. Dr. Siegfried M. Rump
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and

Visiting Professor at Waseda University
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----- Original Message ----- 
From: "John Pryce" <j.d.pryce@xxxxxxxxxx>
To: "Vincent Lefevre" <vincent@xxxxxxxxxx>
Cc: "stds-1788" <stds-1788@xxxxxxxxxxxxxxxxx>
Sent: Friday, October 19, 2012 1:15 PM
Subject: Vincent's 5 Oct rad(xx) proof


> Vincent, P1788
>
> On 5 Oct 2012, at 13:26, Vincent Lefevre wrote:
> > On 2012-10-04 15:36:34 +0100, John Pryce wrote:
> > > Example 2: for an arbitrary type
> > >    r = rad(xx) shall be increasing under set inclusion.
> > >    That is xx \subseteq yy implies rad(xx) <= rad(yy).
> > > That seems less obvious, even with Vladik's specification for
> > > inf-sup interval types. Vladik: is it true there?
> >
> > This seems true for any interval type.
> ...
> >  * |m - (a+b)/2| ≤ |m' - (a+b)/2|.
> >  * m ≤ m'.
>
> ... I believe your proof is correct, however you left out one step of 
> argument, and I was puzzled for a bit ...
>
> > If m' = m, then r = b-m < b'-m' ≤ r'. So, let us assume that m' > m.
> Here you need to point out that this implies m' > (a+b)/2 else m' is 
> between m and (a+b)/2, which contradicts the first * relation above. Hence 
> that relation reduces to what you give in the next line:
>
> > Then (a+b)/2 - m ≤ m' - (a+b)/2, i.e. r = b-m ≤ m'-a ≤ r'.
> >
> > So, in every case, r ≤ r'.
>
> Nice stuff. Simple, but not trivial.
>
> You write "...seems true for any interval type". That's not the point, is 
> it? You've proved it for any *number format*. The type never appears; 
> indeed your proof relies on being able to choose an arbitrary interval. 
> Otherwise you might have [a,b] contained in [a',b'], both T-intervals, but 
> both [a,b'] and [a',b] not T-intervals so your argument using transitivity 
> would fail.
>
> John

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