Re: modal intervals
Sorry, by "nonzero interval" I meant an interval a that does not
involve zero, that is neither 0 \in a, not a \in 0. I cannot understand
why it is so bothering that a*x = 1 is not solvable for intervals
involving zero.
Svetoslav
On 12 Nov 2008 at 15:27, Arnold Neumaier wrote:
> Svetoslav Markov wrote:
> >
> > a*x=1 is solvable for all nonzero Kaucher intervals a,
> > and the solution is alpha*dual (a), where alpha is a real
> > number:
> >
> > alpha = 1/(a^-a^+) in inf/sup presentation a=[a^-, a^+]
> > here dual [a^-, a^+] = [a^+, a^-], resp. dual (a';a") = (a'; -a")
>
> If I apply this to a=[-1,1], which is nonzero, I get
> x = alpha*dual(a)=(-1)*[1,-1] = [1,-1]
> which satisfies
> a*x = [1,-1]*[-1,1] = [0,0]
> So it is not a solution.
>
>
> Arnold Neumaier