Date: Jan 27, 2013 9:49 AM Author: Jesse F. Hughes Subject: Re: ZFC and God WM <mueckenh@rz.fh-augsburg.de> writes:

> On 27 Jan., 13:10, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

>> WM <mueck...@rz.fh-augsburg.de> writes:

>> > On 26 Jan., 23:19, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

>>

>> >> > It is unclear why you apparently are unable to understand, that we are

>> >> > working in the set of terminating decimals. Therefore the diagonal

>> >> > cannot be actually infinite, although there is no last digit.

>>

>> >> Let me ask you a very simple question.

>>

>> >> Is 0.777.... a terminating decimal representation or a

>> >> non-terminating decimal representation?

>>

>> > That depends on the domain where you work in. We have started to work

>> > in the domain of terminating decimals. Since the diagonal consists

>> > only of (changed) digits of these decimals, it is obviously a

>> > terminating decimal.

>> > Now, to answer your question: You did not say where you take 0.777...

>> > from. And obviously that cannot be determined from the digits, as I

>> > jusr explained.

>>

>> When I write 0.777..., I mean the number

>>

>> sum_i=1^oo 7 * 10^-i

>>

>> That is, for each i in N, the i'th digit of 0.777... is defined and is

>> 7.

>

>

> And do you have problems to find this confirmed as possible in the

> complete set of terminating decimals? Any digit or index missing?

I've no idea what you mean when you ask whether I can "find this

confirmed as possible". But, for each i in N, the i'th digit of

0.777... is defined and equals 7. Is there anything more I need to

know in order to claim that it is a non-terminating decimal?

>>

>> Do you agree that there is only one number satisfying that

>> description? Or are there two numbers that satisfy that description

>> and one of the numbers is terminating and the other non-terminating?

>

> I agree that this is a finite definition. But I said that we are

> working in the set of terminating decimals and identify numbers by

> their digits, indices or nodes. Is that hard to understand?

I didn't ask whether it was a finite definition. I asked whether it

was terminating. And it is not sufficient to note that each t_i is a

terminating decimal to conclude that d defined by

d(j) = 7 if j > k or t_j(j) != 7

d(j) = 6 if j <= k and t_j(j) = 7.

is also terminating. This simply does not follow.

This is analogous to the fact that limits of sequences of rational

numbers may be irrational. We begin with a list of terminating

decimals, but it doesn't follow that the anti-diagonal is also

terminating. YOU HAVE TO PROVE THAT.

>>

>> Let's suppose there *are* two different numbers, corresponding to the

>> terminating 0.777... and the non-terminating 0.777... . Then

>>

>> term. 0.777... = sum_i=1^oo 7*10^-i

>>

>> and also

>>

>> non-term. 0.777... = sum_i=1^oo 7*10^-i,

>>

>> but then, of course, term. 0.777... = non-term. 0.777... ! Oops!

>>

>> Moreover, neither term. 0.777... nor non-term 0.777... satisfy the

>> definition of terminating decimal that you previously agreed to,

>> namely

>>

>> 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.

>>

>> The "terminating" 0.777... has no finite length.

>

> Please let me know when you will have succeded in finding a 7 that is

> not in the set of all terminating decimals.

I don't understand your question.

I reiterate: is there both a terminating and non-terminating decimal

0.777...? In what way do they differ?

If you insist that 0.777... is a terminating decimal, please prove

that there is a function f:{1,...,k} -> {0,...,9} such that

sum_i=1^k f(i)*10^-i = sum_i=1^oo 7*10^-1.

Otherwise, you have no argument that it is a terminating decimal,

because this is the definition of terminating decimal according to

*YOU*.

So, please give me a proof that there is, indeed, a function f with

finite domain such that the above holds.

--

"You can do 'math' until you drop dead, but if you're wrong, you're

still wrong, and the world will keep on turning regardless. Heaven

will not open up and God shout at you to stop, you'll just live in

error until one day you die." -- A James S. Harris self-reflection