On 28 Jan., 23:46, Virgil <vir...@ligriv.com> wrote: > In article > <a635f398-4127-40df-a437-23364ce53...@r14g2000yqe.googlegroups.com>, > > WM <mueck...@rz.fh-augsburg.de> wrote: > > On 27 Jan., 23:27, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > > > > > Of course. But why should we agree on a special k? Every natural > > > > number will do. So we only have to know that k is one of those natural > > > > numbers that belong to FISONs. As long as we work in FISONs we cannot > > > > have a non-terminating decimal. > > > > I honestly have no idea what you're trying to say here. Why not > > > simply prove that there is such a k and f? > > > Because every natural number is finite. Why fix one of them? > > You claim there must be such a natural, so you are obligated to prove it.
I claim that every natural belongs to a FISON. And I can prove it. Give me a natural, and I tell you one and, on request, several FISONs where it belongs to. Therefore, as long as you define digits by natural indexes, there is no chance to leave the domain of terminating digits.