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Re: ZFC and God
Posted:
Jan 27, 2013 1:21 PM
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WM <mueckenh@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.
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.
>> > Note, there is another meaning of infinite, namely "actually
>> > infinite". Those who adhere to that notion *in mathematics* should
>> > show that it differs from "potentially infinite" *in mathematics*,
>> > i.e., expressible by digits.
>>
>> Well, I don't understand why anyone would wish to show that.
>
> Perhaps in order to show that matheology is not complete nonsense?
>
>> But,
>> regardless, this is beside the point. I'm asking for a proof that
>> 0.777... is terminating according to the definition of terminating
>> that you agreed to.
>
> I did this in my last posting. Please look it up there. Well as I have
> it just at hand, here it is again:
> 0.7 is terminating.
> if 0.777...777 with n digits is terminating, then also 0.777...7777
> with n+1 digits is terminating. Therefore there is no upper limit for
> the number of digits in a terminating decimal. This is written as
> 0.777...
>
> This is the definition that I agreed to.
Er, no.
The definition that you agreed to is reproduced above. You have to
show that the definition above is actually satisfied, i.e., that there
is a natural number k and a function f satisfying the appropriate
conditions.
You've done no such thing.
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.
--
"Being who I am, I know that's a solution that will run in polynomial
time, but for the rest of you, it will take a while to figure that out
and know why [...But] it's the same principle that makes n! such a
rapidly growing number." James S. Harris solves Traveling Salesman
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