Re: motioin43 amended
Vincent,
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) ?
-Dima
----- 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.
--
Vincent Lefèvre <vincent@xxxxxxxxxx> - Web: <http://www.vinc17.net/>
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