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

[Vincent Lefevre <vincent@xxxxxxxxxx>]

On 2012-04-11 11:43:46 -0500, Nate Hayes wrote:
> Vincent Lefevre wrote:
> >So, basically, you get something that is equivalent to an unbounded
> >interval, but with a more complex definition!
> 
> Exactly.

I prefer a simple theory than a complex one.

> The reasons for doing it this way are:
>    -- unbounded intervals are then completely unnecessary at any level
>    -- the absence of unbounded intervals at Level 1 means A = B
> when A + X = B + X
>    -- valid ranges for things like 1/[0,1]=[1,+OVR] can still be computed
> at Level 2 (which is where for a computational standard like P1788 it
> matters in the real world).

Similarly, real numbers are unnecessary, because one can use
sequence of rationals. The absence of irrational numbers means
that one can write each rational under the form p/q. And one
can still compute everything at Level 2 with rational numbers.
Now I wonder why one uses real numbers in practice. :)

> >>>>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.
> >
> >Do you mean that you have an operation at Level 2 with no
> >corresponding operation at Level 1? This doesn't make sense.
> Sure it does.
> It all depends on the underlying axioms and definitions.

A math problem is always expressed at Level 1, not at Level 2.

> >I've given an example in another mail.
> Where?

The range of atan(1/x) over [0,1]. Could you give a complete proof
of the result with your axioms and definitions?

> >>All your points seem really academic to me.
> >
> >No, really, I need to compute ranges of functions (that can be defined
> >by arbitrary expressions), and I need a correct answer.
> So do I. I can compute them at Level 2 with 1/[0,1]=[1,+OVR].
> I've never written a computer program that operates at Level 1. That's
> nonsense.

for you only.

> >>The arithmetic can be closed at Level 2, anyhow, with overflown
> >>intervals.
> >
> >My point is that it must be closed at Level 1 too.
> Why? I haven't seen you give any compelling reason for this.

Because it makes everything easier. Do you have a proof of the FTIA
with your model?

> At Level 1, I see it is a choice betwee closure and cancellation.
> IMO and experience, cancellation is the much more important property.

Silly. You're going nowhere with such claims: similarly I can say
"I've never used cancellation, so IMHO, closure is the much more
important property."

2. Cancellation is still *valid* with the P1788 model when you do
not use unbounded intervals (which is your case). So, I don't see
why you're complaining.

BTW, cancellation is invalid at Level 2. That makes a good reason
not to consider it.

> I'm still waiting for you or anyone else to show why they are *necessary*.

See my example (again). And the range of tan over [0,10].
I want to use intervals, not useless complex things such as
familly of intervals.

-- 
Vincent Lefèvre <vincent@xxxxxxxxxx> - Web: <http://www.vinc17.net/>
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Work: CR INRIA - computer arithmetic / AriC project (LIP, ENS-Lyon)