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. I let open the question whether there might be non-terminating decimals.
> Well, not to put to fine a point on it, but you started > by claiming that the set of terminating decimals is countable.
> Now, > if every real has a terminating decimal representation, then...
I did not say that every real has a terminating representation. > > 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.
> 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.
> 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. Or do you claim that it is impossible to distinguish terminating and non-terminating by digits? Or do you claim that it is impossible to work in the domain of terminating decimals? Why can't you support your claims?