In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 28 Jan., 22:52, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > > > >> Well, you agreed to that definition. > > > > > Of course, but not as you now put it without restriction, but under > > > the proviso that every decimal is terminating. In that case for every > > > decimal the required k exists by definition. > > > > Er. So. > > > > You accepted the definition under the assumption that every decimal is > > terminating? > > I said that we are working in the terminating decimals.
YOU do not get to decide where Jesse, or anyone else of that matter, is constrained to work. He doe snot appear to want to work under the constraints you wish to impose on him, and you have no power to make him do so.
> I let open the > question whether there might be non-terminating decimals.
It is a question that has already been answered affirmatively everywhere outside Wolkenmuekenheim. > > > Well, not to put to fine a point on it, but you started > > by claiming that the set of terminating decimals is countable. > > Obviously true. > > > Now, > > if every real has a terminating decimal representation, then... > > I did not say that every real has a terminating representation.
But does every real accessible real have a decimal expansion, at least in theory?
Does every rational have a decimal expansion?
If yes to either of the above then there can be no objection to infinite decimal expansions. > > > > Look, you're just making my head hurt. I really don't have any clue > > what you think you're doing here, but it is the worst impression of > > mathematical reasoning I've ever seen, and that's saying something. > > As you see above you misunderstand. That what may look worse to you is > simply your wrong impression. I don't know whether you cannot > comprehend what I said.
WHat none of us can comprehend is why you say it, and keep on saying it, when it is so patently obvious that no one else believes it, and that no one will believe it from anyone as incompetent at logical argument as you have repeatedly proved your serf to be.
> > > AGH! You just rejected my definition and said that a decimal is > > terminating iff it is in the set of terminating decimals, and you > > define that set with reference to the definition of terminating > > decimal. > > > > Never mind. The point remains. The number 0.777... is not in T. > > Then show it! Show a digit that lies beyond all digits of terminating > decimals.
Since in 0.777... = sum_(n in |N) 7/10^n there is a digit for every natural n in |N ,, what need is there for any digit BEYOND every natural?
WM seems to be arguing that there is something in |N that is beyond every n in |N but still in |N. > > > The number 0.777... is the usual real number, namely > > > > 0.777... = sum_i=1^oo 7 * 10^-i. > > > > Now, is that a terminating decimal or not? > > Show a digit that lies beyond all digits of terminating decimals.
First you show that there needs to be some n in |N a member which lies beyond every other m in |N.
> Or do you claim that it is impossible to distinguish terminating and > non-terminating by digits?
Terminating digit sequences have a last digit, non-terminating don't.
> Or do you claim that it is impossible to > work in the domain of terminating decimals?
It is perhaps possible but inappropriate for the problem of 0.777...
In working with something like 0.777..., it is much better to work in a system in which 0.777... can actually occur that to work in a system where it cannot occur.
But that would make WM'sa foolishness obvious to everyone, so that WM insists on being an ass anyway.
> Why can't you support your claims?
Jesse makes no claims, but only asks questions about 0.777... which WM fails to answer. > > Regards, WM --