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Re: Motion 31 draft text V04.4, extra notes

Vincent Lefevre wrote:

On 2012-04-11 08:46:42 -0500, Nate Hayes wrote:

Vincent Lefevre wrote:
>On 2012-04-05 15:35:18 -0500, Nate Hayes wrote:
>>Vincent Lefevere wrote:
>>>I agree. But there will be no "overflown" intervals at Level 2.
>>>AFAIK, the Level 2 choice for the midpoint on unbounded intervals
>>>has been done for practical reasons, not because of some notion
>>>of "overflown" intervals (if this is what you meant).
>>
>>Nope. That's not what I meant.
>>
>>What I meant is that at Level 1 there is no definition of midpoint for
>>unbounded intervals. So why include unbounded intervals in the Level 1
>>model?
>
>because the midpoint is not the only function. For instance:
>
> 1 / [0,1] = [1,+oo]
>
>at Level 1 and Level 2 (+ some decorations).

We don't actually perform arithmetic in a computer at Level 1, and at
Level 2 with overflow one has
   1 / [0,1] = [1,+OVR]


This is wrong. At Level 1, 1 / [0,1] is defined as the smallest
(closed) interval that contains { 1 / x | x in [0,1] and x <> 0 },
and this is [1,+oo]. At Level 2, the interval must contain [1,+oo]
entirely. If [1,+OVR] means some interval [1,K] with K finite (but
we don't which one), the result is incorrect, because for any K,
[1,K] does not contain [1,+oo].


In my recent e-mails I've said that the interval [1,+OVR] should be
considered as a family of intervals: the number of elements in the family is
inifinite, but each element is closed and bounded.





>>Especially when a similar treatment at Level 2 of "overflown"
>>intervals can provide the same practical benefits? IMO the Level 1
>>model is then cleaner and simpler.
>
>I disagree: without unbounded intervals, you wouldn't have a closed
>arithmetic

It seems academic to me: we don't perform arithmetic in a computer at
Level 1, (and the overflown arithmetic is closed at Level 2, btw).


But from the Level 2 results, one can deduce Level 1 results. This
is what matters.


When restricted to bounded intervals, the Level 1 arithmetic is closed
and
cancellative for addition, subtraction, multiplication and division with
0
not in the denominator.


This is not the definition of a closed arithmetic. If you can get
an interval as a result, say [0,1], you mustn't remove it from the
possible inputs of an operation.


At Level 2 its not.




This is the oldest interval arithmetic of Ramon Moore, etc. and has
withstood the test of time already. IMO this is better than the
current P1788 model with unbounded intervals.


P1788 needs a closed arithmetic.


The practice of interval arithmetic has thrived for over 50 years based on
Moore's Level 1 model. Where is the evidence to support your claim to the
contrary?

BTW, where is your answer to my original question about a concrete example?
All your points seem really academic to me.

The arithmetic can be closed at Level 2, anyhow, with overflown intervals.

For me, it is more important that the Level 1 arithmetic is cancellative,
and the current P1788 model with unbounded intervals does not have this
property.

Unbounded intervals are unnecessary.

Nate