
Re: ZFC and God
Posted:
Jan 27, 2013 1:56 PM


WM <mueckenh@rz.fhaugsburg.de> writes:
> On 27 Jan., 19:21, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> WM <mueck...@rz.fhaugsburg.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.
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 terminating decimal 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 Cantorlist, as long as we work in the > domain of terminating decimal representations. > > If you insist in a nonterminating one, please show it!
0.777... is a nonterminating decimal representation. It is a sum of the 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 to 0.777... . 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^i
or
sum_i=1^k f(k) * 10^i >= sum_i=1^k1 7 * 10^i + 8 * 10^k.
In other words, either 0.f(1) f(2) f(3) ... f(k) <= 0.777...7 ^^^^^^^ length k
or
0.f(1) f(2) f(3) ... f(k) >= 0.777...8 ^^^^^^^ length k
Do you agree?
It is then a simple matter to show that
0.777... > 0.777...7 ^^^^^^^ length k
and a little harder to show that
0.777... < 0.777...8 ^^^^^^^ length k
The 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...8
Since 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 no terminating decimal representation.
 "Looking at their behavior I see them endangering not only their own futures, but that of their families, and now, considering my latest result, the future of people all over the world."  James S. Harris, on the shortsightedness of his mathematical critics

