How Wide Do All The Possible Computed Digit Sequences Go? [Math]

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From: George Greene on 24 Jun 2010 19:47

On Jun 24, 4:01 pm, Graham Cooper <grahamcoop...(a)gmail.com> wrote:
> On Jun 25, 1:13 am, George Greene <gree...(a)email.unc.edu> wrote:
>
> > On Jun 24, 2:16 am, Graham Cooper <grahamcoop...(a)gmail.com> wrote:
>
> > > It's the digit width of a SET of numbers.
>
> > IT *IS*NOT*.
>
> IT IS TOO!
>
> It's my definition it's my theorem.

YOU HAVE NOT GIVEN any definition!
HOW DO YOU define the width OF A SET of three things
when one of them has width 3, another has width 4, and another
has width 5?

You are of course going to say that in the infinite case, you define
it is infinity, but that IS OBVIOUSLY TOO wide since (if it is
an infinite list of finite things) EVERYthing on the list IS NOT AS
WIDE
as that! Things that apply to all those SMALLER FINITE widths are
NOT necessarily going to apply to the INFINITE width that you are
assigning to the whole list.
But you say they do because you're JUST THAT STUPID.

> All you have to do is put in your own words the corrollery
> of this short proof.

This IS NOT a proof, DUMBASS.
This is a bunch of gibberish.
You are talking about the width of a set of things
WHOSE ELEMENTS have widths, but YOU WON'T SAY
how that implies that THE SET ITSELF has any width.
Sets-of-reals and reals ARE DIFFERENT KINDS of things!
"Width" MEANS SOMETHING DIFFERENT for one than for
the other, and something different AGAIN when the set is infinite!
Unfortunately, YOU ARE NOT COMPETENT TO STATE ANY relevant
definitions here!

> Assumption. There exists a real number with a finite sequence
> of digits that is not computable. Contradiction.
>
> Therefore...... <take it away George!>

Therefore NOTHING.
THAT'S A CONTRADICTION.
EVERYthing follows TRIVIALLY from a contradiction, but the most
important thing that follows MEANINGFULLY is that ONE OF YOUR
PREMISES IS FALSE.
Every finite sequence is computable.
Every finite subsequence of every real is computable.
That is why you DO NOT NEED ANY "computable reals"
on your list! YOU CAN use JUST the finite sequences!
THAT list is computable and has the further advantage that if
you used it, YOU MIGHT ACTUALLY KNOW WHAT YOU WERE TALKING
ABOUT!

But even that list is too long.
There is a list OF ONE computable real that contains all the possible
finite digit-sequences. It's the real that looks like this:
..
012345678900010203040506070809190111213141516171819202122232425262728293031323334353637383940414243.....etc.

YOU DON'T EVEN *NEED* A LIST, PERIOD, let alone an infinite list!
ONE REAL
will do!

From: George Greene on 24 Jun 2010 19:49

On Jun 24, 4:14 pm, Graham Cooper <grahamcoop...(a)gmail.com> wrote:
> Here we go. George will come up with the most muddled
> quantified filled meaningless interpretaion with copious
> references to FINITE sequences.

DIPSHIT: YOUR version uses finite sequences!
Your point about "the list of all computable reals" is that
it matches every FINITE prefix of every real!

IF ALL you are trying to do IS MATCH FINITE PREFIXES
then OBVIOUSLY YOU DON'T NEED ANY INfinitely wide
"computable reals"!

> And still deny it clearly hc3

hc 3 is not even grammatical, so nothing can prove it.
I can't help it if you are too illiterate to tell the difference
between
infinite, infinity, and infinitely many.

From: George Greene on 24 Jun 2010 19:49

On Jun 24, 4:14 pm, Graham Cooper <grahamcoop...(a)gmail.com> wrote:
> Here we go. George will come up with the most muddled
> quantified filled meaningless interpretaion with copious
> references to FINITE sequences.

No, I won't. YOU have to do that.
You have made AN ATTEMPT in English but it fails with numerous syntax
errors.

From: Graham Cooper on 24 Jun 2010 21:12

On Jun 25, 9:49 am, George Greene <gree...(a)email.unc.edu> wrote:
> On Jun 24, 4:14 pm, Graham Cooper <grahamcoop...(a)gmail.com> wrote:
>
> > Here we go. George will come up with the most muddled
> > quantified filled meaningless interpretaion with copious
> > references to FINITE sequences.
>
> No, I won't. YOU have to do that.
> You have made AN ATTEMPT in English but it fails with numerous syntax
> errors.

You can't conclude ANYTHING from this?

Assumption. There exists a real with a finite
sequence that is not computable.
Contradiction.

Jesus George. All you have to do is describe
the negation of the assumption

that is not ANYTHING

You are a gutless evasive git!

Herc

From: Sylvia Else on 24 Jun 2010 22:55

On 25/06/2010 6:10 AM, Graham Cooper wrote:
> On Jun 24, 11:35 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>> On 24/06/2010 5:09 PM, Graham Cooper wrote:
>>
>>
>>
>>
>>
>>> On Jun 24, 5:00 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>> On 24/06/2010 1:12 PM, Tim Little wrote:
>>
>>>>> On 2010-06-23, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>>>> On 23/06/2010 5:00 PM, Tim Little wrote:
>>>>>>> It can't be conceded, as it is simply false. The predicate "list L
>>>>>>> contains x" means exactly that there exists n in N such that L_n = x.
>>>>>>> The list does not contain pi since there is no such n. It really is
>>>>>>> that simple.
>>
>>>>>> Well, OK. Though I can't see how it makes any difference to Herc's
>>>>>> argument, since I could never see what role it played anyway.
>>
>>>>> It is the very core of his argument: if pi were "contained in" the
>>>>> infinite list
>>
>>>>> 3
>>>>> 3.1
>>>>> 3.14
>>>>> 3.141
>>>>> ...
>>
>>>>> then it would also be "contained in" any other list sharing the "all
>>>>> prefixes" property, such as
>>
>>>>> 3.0000000
>>>>> 3.1000000
>>>>> 3.1428571. . .
>>>>> 3.1414141
>>>>> ...
>>
>>>>> It would even be "contained in" any list having that as a sublist, e.g.
>>
>>>>> 0.0000000
>>>>> 3.0000000
>>>>> 3.1000000
>>>>> 1.0000000
>>>>> 3.1428571. . .
>>>>> 2.0934953
>>>>> 0.5829345
>>>>> 3.1414141
>>>>> ...
>>
>>>>> The list of all computable reals is exactly such a list. Most
>>>>> importantly, no property specific to pi is used here. Every real
>>>>> would be "contained in" that list if we were to use Herc's broken idea
>>>>> of "contained in".
>>
>>>> Ok, but his next step - all finite prefixes implies all infinite
>>>> sequences - is false. So all it means is that his 'proof' contains two
>>>> invalid steps rather than just one.
>>
>>>> Sylvia.
>>
>>> I've told you atleast 3 times specifically that is NOT an implication
>>> of hc3. I put it in a implication formula -> are you amnesiac?
>>
>> I never said it was an implication *of* hc3. I say it's an implication
>> *in* hc3.
>>
>>
>>
>>> As long as all permutations oo wide are in the set, (segmented,
>>> appended to hitlers number, inverted, imputed, with any other
>>> numbers you can think of) I don't care!
>>
>> You need to prove that they're all in the set, or you have nothing.
>>
>>> But one of these days
>>> one of is going to confirm oo digits of every sequence are ALL THERE
>>
>> Sylvia.
>>
>>
>>
>>
> That is no excuse you are still wrong. Something that is implied
> is implied. You corrected the correction in place of admitting
> you have been told
>
> all finite prefixes -> all oo long sequences
>
> is N O T hc3
>
> you've been told 5 times now plus I explained it several
> more times in this thread and you still carry on.
>
> Learn to read!
>
> Herc

I think it's best if we kill this subthread, and instead try to deal
with the entire issue in the subthread where I've questioned your step 1.

Sylvia.