```Date: Jan 27, 2013 5:27 PM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

WM <mueckenh@rz.fh-augsburg.de> writes:> On 27 Jan., 19:56, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:>> WM <mueck...@rz.fh-augsburg.de> writes:>> > On 27 Jan., 19:21, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:>> >> WM <mueck...@rz.fh-augsburg.de> writes:>> >> > On 27 Jan., 18:44, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:>>>> >> >> Anyway, you haven't proved that there is a function>>>> >> >>   f:{1,...,k} -> {0,...,9}>>>> >> >> as required by *your* definition of terminating decimal, so you have>> >> >> not shown that 0.777... is a terminating decimal.>>>> >> > You are wrong. Can't you understand? All natural numbers are finite.>> >> > Why the heck should I define a single k?>>>> >> Because, of course, you accepted the following definition:>>>> >>   Let x be a real number in [0,1].  We say that x has a terminating>> >>   decimal representation iff there is a natural number k and a>> >>   function f:{1,...,k} -> {0,...,9} such that>>>> >>    x = sum_i=1^k f(i) * 10^-i.>>>> > I did not fix k but only assumed that it is a natural number.>>>> Pardon me?>>>> Look, you said that a number has terminating decimal representation>> iff there is some k and f such that blah blah blah.  Thus, to show>> that a number *does* have a terminating decimal representation is>> equivalent to showing that there is, in fact, a k and f satisfying the>> above.>> 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 notsimply prove that there is such a k and f?>>>> What could be more obvious than this?>> Remember the Binary Tree to answer your doubts. Do you believe that> the set of all FISONs is definable in ZF? Is every FISON finite? Is> there a fixed k limiting all FISONs? No.As you said before, let's keep analogies out of this.  >> > You have no cause to claim the contrary, since there is no index>> > (natural number) infinitely many counts away from the decimal point.>>>> Are you asking me to prove that 0.777... does not have a terminating>> decimal representation?  I would be happy to do so, if needed.>> I am doubting that when working in FISONs you can find an index of a> diagonal number that requires to leave the set of FISONs (with respect> to the indexes of the digits).>>> it follows *FROM THE AGREED DEFINITION* that 0.777... has no>> terminating decimal representation.>> Show that in your 0.777..., and in particular in the anti-diagonal of> a list of terminating decimals, there is an index k that does not> belong to a FISON {1, 2, ..., n}. Utterly irrelevant to the matter at hand.  Evidently, you don'tunderstand the definition you agreed to.> As long as you refuse there is no reason to believe you, because> every index in a FISON belongs to a FISON (finite initial set or> sequence of natural numbers). Why do you try to deny that?Who ever denied that?My interest is waning here.  We have a definition, but you refuse toapply it (or, perhaps, are too stupid to understand it) in order toprove what you claim.Please, just show me that there is a function  f:{1,...,k} -> {0,...,9}such that  0.777... = sum_i=1^k f(k) * 10^-i, and I will concede that 0.777... has a terminating decimalrepresentation.  And if you cannot do that, then I will not concede the point butrather conclude you cannot show what you claim.-- "This sucks," said a Pennsylvania State University student [...] " Whycan't the college let me do what I want to do with my computer? Theschool computer security guys are being way more annoying than thespyware was." -- A student pines for his disabled spyware
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