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 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 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 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^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 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