```Date: Jan 27, 2013 1:56 PM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

WM <mueckenh@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 representationiff there is some k and f such that blah blah blah.  Thus, to showthat a number *does* have a terminating decimal representation isequivalent to showing that there is, in fact, a k and f satisfying theabove.What could be more obvious than this?>>>> Thus, if you claim that 0.777... has a terminating representation,>> then you must show that there is a natural number k and a function f>> as above such that>>>>   0.777... = sum_i=1^k f(i) * 10^-i.>>>> Else, you have no cause to claim that 0.777... has a terminating>> decimal representation.>>> 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 terminatingdecimal representation?  I would be happy to do so, if needed.>>>> > This is the definition that I agreed to.>>>>> Frankly, I'm a bit stunned that you're arguing that 7/9 has a>> terminating decimal representation, but as long as you're claiming so,>> then you need to stick to the definition we've agreed on.>> I am not claiming that 7/9 ot 1/3 or sqrt(2) have decimal> representations at all.> Just the contrary. But I am claiming that all decimal representations> that exist in the domain of terminating decimals are terminating, in> particular the diagonal of a Cantor-list, as long as we work in the> domain of terminating decimal representations.>> If you insist in a non-terminating one, please show it!0.777... is a non-terminating decimal representation.  It is a sum ofthe form   sum_i=1^oo 7 * 10^-i,and this is *not* the same as a finite sum  sum_i=1^k f(k)*10^-i.Moreover, there is no terminating decimal representation equal to0.777... .  Let k and f:{1,...,k} -> {0,...,9} be given.  Now, either  sum_i=1^k f(k) * 10^-i <= sum_i=1^k 7 * 10^-ior   sum_i=1^k f(k) * 10^-i >= sum_i=1^k-1 7 * 10^-i + 8 * 10^-k.In other words, either     0.f(1) f(2) f(3) ... f(k) <= 0.777...7                                        ^^^^^^^                                length kor     0.f(1) f(2) f(3) ... f(k) >= 0.777...8                                 ^^^^^^^                                 length kDo you agree?It is then a simple matter to show that  0.777... > 0.777...7                       ^^^^^^^               length kand a little harder to show that   0.777... < 0.777...8               ^^^^^^^               length kThe latter claim is truly trivial if you allow me that   0.777... < 1.After all,   0.777... = sum_i=1^k 7*10^-i + 10^-k * 0.777...           < sum_i=1^k 7*10^-i + 10^-k * 1           = 0.777...8Since k and f were arbitrary, and we showed  0.777... != sum_i=1^k f(k) * 10^-i,it follows *FROM THE AGREED DEFINITION* that 0.777... has noterminating decimal representation.-- "Looking at their behavior I see them endangering not only their ownfutures, but that of their families, and now, considering my latestresult, the future of people all over the world." -- James S. Harris,                 on the shortsightedness of his mathematical critics
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