# Boost :

Subject: Re: [boost] boost interval arithmetic
From: Thijs van den Berg (thijs_at_[hidden])
Date: 2015-01-15 08:14:10

On Thu, Jan 15, 2015 at 1:45 PM, ÐŸÐ°Ð²ÐµÐ» ÐšÑƒÐ´Ð°Ð½ <coodan_at_[hidden]> wrote:

>
>
>
> Thu, 15 Jan 2015 13:16:19 +0100 Ð¾Ñ‚ "Thijs (M.A.) van den Berg" <
> thijs_at_[hidden]>:
> >
> >> OK, but that is enough to conclude that [-inf, inf] IS( [-inf, -1] U
> [1, inf]) is not true and this operator returns not correct result, no
> matter is the way it corrupts own result documented or not.
> >>
> >
> >No, you are wrong here again, you claim so much but demonstrate nothing.
> You seem to lack the ability to reason precisely and mix up set theory with
> numerical interval computations.
> >
> >You should stick to pen and paper math, learn about logical inference
> >
> >(A and NOT(A))->proves the earth is flat
> >
> > ,lean about numerical representation theory (like floats) and don't
> touch a computer or post in forums until you do that!
> >
> >What in your opinion is the value of "d" in the following statement?
> >
> >
> >double d=1/3;
> >
> >
> >_______________________________________________
> >Unsubscribe & other changes:

>
>
>
> I would like you say me first that you really think that [-inf, inf] IS
> EMPTY, as I asked first.
>
> But, still, OK, double d=1/3 will contain a result of int operator/ as
> values you are dividing are int. So what? Next question you will ask will
> be a result of 'double d=1%3;' ???!
>
>
Since we are discussin boost interval, I'm going to use that definition of
interval and not some irrelevant set theoretical definition.

[-inf, inf] thus means: a computation using boost intervals as arguments
returned another boost interval, and thus the value must be somewhere in
that interval.

Don't you dare returning with a non-boost interval definition interval
argument! (something involving set theory, multi-interval arguments, or
Banachâ€“Tarski )