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