Re: motioin43 amended
Dmitry & all
On 24 Apr 2013, at 06:32, Dmitry Nadezhin wrote:
> Both original Jürgen's original definition and simplified definition
> clearly determine that
(*)
> divPair([1,1],Entire)=([-inf,0],[0,+inf])
>
> Do you say that the result should be (Entire,Empty) ?
I don't think Vincent is saying that at all. He is asking for clarity.
I think the definition should be based on the "powerset" version of the solution set. In the example (*) above, if one returns Entire,Empty one is needlessly discarding useful information.
I propose this amendment to the document. (Could it be renamed divpair, I can't see the point of the 2 in div2pair; a version number in the text would also help.)
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1. The "a \bullet b ..." equation on line 3 be numbered (1) so the current (1) becomes (2).
2. Replace "If the solution set consists of only 1 interval, the second output will be the empty set." by something like
"The set defined by (1) or (2) may be (a) the union of two disjoint nonempty intervals C1, C2 or (b) a nonempty interval C or (c) empty. If (a), order them so that C1<C2, and return (C1bar, C2bar) where bar means closure (or hull). If (b), return (Cbar, Empty). If (c), return (Empty, Empty)."
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The C1<C2 removes another possible ambiguity in the operation.
John
> ----- Исходное сообщение -----
> От: vincent@xxxxxxxxxx
> Кому: stds-1788@xxxxxxxxxxxxxxxxx
> Отправленные: Среда, 24 Апрель 2013 г 6:52:29 GMT +04:00 Абу-Даби, Маскат
> Тема: Re: motioin43 amended
>
> On 2013-04-23 18:48:07 -0700, Dmitry Nadezhin wrote:
>> divPair([a],[b]) can be simplified to three cases:
>> (3) ([a],[b])->([-oo,sup(a)/sup(b)] , [sup(a)/inf(b),+oo]) if ([a] < 0 and 0 \subset [b])
>> (4) ([a],[b])->([-oo,inf(a)/inf(b)] , [inf(a)/sup(b),+oo]) if ([a] > 0 and 0 \subset [b])
>> (8) ([a],[b])->(div([a],[b]),\empty) otherwise
>
> Yes: either one gets 2 non-empty intervals, or one gets the interval
> returned by the conventional div and Empty.
>
> But there are cases that are not clearly determined, for instance
> divPair([1,1],Entire). The powerset result is Entire \ {0},
> i.e. [-inf,0) U (0,+inf]. But with closed intervals, this gives
> [-inf,0] U [0,+inf] as mentioned above. However this is equivalent
> to Entire. So, should the result be ([-inf,0],[0,+inf]) or
> (Entire,Empty)? With the former choice, the closed intervals
> are not disjoint. The motion must be clear on which context
> "disjoint" is used.