On 28 Jan., 12:41, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > WM <mueck...@rz.fh-augsburg.de> writes: > > 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? I do > > *not* claim that there is a k such that all terminating decimals are > > shorter. I claim that every length n of digits is finite. That is a > > huge difference. > > We're not talking about finding a single length for "all terminating > decimals".
Just that is what you try when you look for a fixed k. 0.777... is terminating as long as it belongs to the set of all terminating decimals, i.e., does not surpass the domain of finite indices.
> I'm asking about a single number, 0.777..., which you > claim is terminating. > > Please show that there is a k and f as required by the definition of > terminating decimal representation so that
By *your* definition of terminating, this may be required. Not by the mathematical definition which relies upon the fact that every natural number is finite and that a non-terminating decimal cannot be defined by digits but only by the finite definition: "This decimal will never end".
My definition of terminating says only that the terminating decimal belongs to the set of all terminating decimals, which is as well defined as the set of all natural numbers. In order to show that a decimal is not terminating, you are obliged to find a digit, the index of which doe not belong to a FISON (1, 2, ..., n). Of course you will fail. But nevertheless try it.