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From: |-|ercules on 9 Jun 2010 17:17
"David Bernier" <david250(a)videotron.ca> wrote...
> |-|ercules wrote:
>> "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ...
>>> |-|ercules says...
>>>
>>>> Are you saying you find a new_sequence_of_digits using
>>>> diagonalization on
>>>> infinite lists?
>>>>
>>>> Like 260?
>>>>
>>>> How so when ALL (INFINITELY MANY) digits of ALL digit sequences are
>>>> computable?
>>>
>>> Yes, all finite sequences of digits are computable. So let's
>>> go ahead and assume that we have a list that includes every
>>> finite sequence of digits. For definiteness, let's use the following
>>> rule: Let real number n be the decimal representation of n, preceded
>>> by a decimal point. So, we have:
>>>
>>> L_0 = 0.0
>>> L_1 = 0.1
>>> L_2 = 0.2
>>> ...
>>> L_10 = 0.10
>>> L_11 = 0.11
>>> ...
>>> L_100 = 0.100
>>> ...
>>>
>>> The way I've enumerated them, some numbers appear more than once. For
>>> example, L_1 is equal (as a real number) to L_10. But that doesn't
>>> matter.
>>>
>>> With this list, every finite decimal expansion occurs somewhere on the
>>> list.
>>> But not every *infinite* decimal expansion occurs. So let's apply our
>>> diagonalization procedure: we get
>>>
>>> r = 0.5555555555...
>>>
>>> This number is not on the list. It's not equal to L_0, it's not equal
>>> to L_1, it's not equal to L_2, etc.
>>>
>>> The finite approximations are on the list, however. 0.5 is equal to
>>> L_5. 0.55 is equal to L_55. 0.555 is equal to L_555. etc. But the
>>> full number 0.555... (which happens to be the decimal representation
>>> of the fraction 5/9) is not anywhere on the list.
>>
>>
>> You split my 2 related questions and gave trivial solutions to each.
>>
>> Try again!
>>
>> ALL (INFINITELY MANY) digits of ALL digit sequences are computable!
>> Are you saying you can find a new_sequence_of_digits using
>> diagonalization on a computable reals list?
>> Like 260 in the first example?
>>
>>
>> ________________________________________
>>
>>
>> I can't find the post, but someone posted yesterday that there is a
>> DIFFERENT DIGIT at N=something etc. etc.
>>
>> And there can be numerous (infinite) different digits along the
>> expansion, for some real not on the computable reals list.
>>
>> If someone can find that post great.
>>
>> If not, it's utter BS!
>>
>> It's MEANINGLESS. Different digit to what?
>>
>> IF there are NUMEROUS different digits then there must be some FINITE
>> substring between them (inclusive)
>> that is not a computable digit sequence.
>
> To the digit d, one can associate
> anti(d) = d+3 (for d = 0, 1, 2, 3 or 4)
> = d-3 (for d = 5, 6, 7, 8 or 9). ($$$)
>
> anti(0) = 3
> anti(1) = 4
> anti(2) = 5
> anti(3) = 6
> anti(4) = 7
> anti(5) = 2 (***)
> anti(6) = 3
> anti(7) = 4
> anti(8) = 5
> anti(9) = 6
>
> Let's say we have a list where the 10th number is 55/123.
> The decimal notation for 55/123 is:
>
> 0.4471544715 4471544715 4471544715 44715................
>
> The tenth decimal after the point in the number above is '5'.
>
> anti(5) = 2 by (***) above or by ($$$).
>
> In the absence of the first nine numbers on the list,
> we can just put a question mark '?' for that position
> of ANTI, the anti-diagonal (or an anti-diagonal)
> for some list (whether finite or infinite).
>
> So as of now, we have:
> ANTI = 0.?????????2 ?????????? ...................
>
> ANTI can't be the same as 55/123 because two
> numbers that differ by 3 units in the tenth decimal
> position can't possibly be the same.
>
> Another thing is that blanks in terminating decimals
> on the list should be interpreted as zeros.
>
> David Bernier



Some time ago you said unless I come up with solid proof disputing Cantor's proof
nobody would accept it.

If you can help me find the post from yesterday that stated there is a different
digit at such and such, and possibly numerous other different digits.

THIS is provably false, and it should sway some of you, I'll go through all the posts
again I missed it last time I searched.

Herc

From: Daryl McCullough on 9 Jun 2010 18:21
|-|ercules says...
>
>"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ...

>>>ALL (INFINITELY MANY) digits of ALL digit sequences are computable!
>>
>> Okay, that's false, but you seem to believe it. Why?
>
>
>What about
>
>ALL digit sequences are computable to ALL finite lengths.

Once again, your statement is a muddled mess. What is true is
the following:

For every real number r, for every natural number n,
there is a computable real r' such that r' agrees with
r in the first n digits.

Do you understand what the above means? By saying
"ALL digit sequences are computable to ALL finite lengths"
do you mean something different, or do you mean the same
thing?

>I just disagree that specifying
>"it's different at digit N to the Nth real and it's different..."
>literally gives any new sequence of digits that are not computable.

Let's go through it more carefully then. Let L be our list of computable
reals, and let L_1, L_2, ... be the reals on the list. Let r be the
antidiagonal of L.

Do you agree that r is not equal to L_1?
Do you agree that r is not equal to L_2?
Do you agree that r is not equal to L_3?

We can prove the following general statement:

For all n, r is not equal to L_n.

From this it follows that:

r is not on the list L.

Since L is presumed to be the list of all computable reals, it
follows that r is not a computable real.

>Examples don't prove it for the entire domain.

That's why Cantor gave a proof. He didn't just give examples.

--
Daryl McCullough
Ithaca, NY

From: |-|ercules on 9 Jun 2010 18:28
"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote
>>ALL digit sequences are computable to ALL finite lengths.
>
> Once again, your statement is a muddled mess.

What's not to get?

here's a digit sequence:
31415926...
It's the expansion from arctan(1)*4/10.

Is this digit sequence computable to 5 digits?
Is it computable to 6 digits?
Is it computable to any finite number of digits?
Is it computable to ALL finite lengths?

Where are you getting muddled?

Herc

From: Daryl McCullough on 9 Jun 2010 18:50
|-|ercules says...
>
>"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote
>>>ALL digit sequences are computable to ALL finite lengths.
>>
>> Once again, your statement is a muddled mess.
>
>What's not to get?

Why can't you learn how to use quantifiers? Those make
what you are trying to say much more precise. You need
to take an elementary course in mathematics and logic.
Until then, you can't even ask a question without getting
muddled.

--
Daryl McCullough
Ithaca, NY

From: |-|ercules on 9 Jun 2010 19:16
"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote
> |-|ercules says...
>>
>>"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote
>>>>ALL digit sequences are computable to ALL finite lengths.
>>>
>>> Once again, your statement is a muddled mess.
>>
>>What's not to get?
>
> Why can't you learn how to use quantifiers? Those make
> what you are trying to say much more precise. You need
> to take an elementary course in mathematics and logic.
> Until then, you can't even ask a question without getting
> muddled.
>

NO dipsh1t. I have a degree in computer science and I am perfectly
fine using quantifiers.

Now let's go through your idiotic denial one question at a time.

here's a digit sequence:
31415926...
It's the expansion from arctan(1)*4/10.

Is this digit sequence computable to 5 digits?


How do you expect to see your error if you refuse to put your formula into words
so you can understand them?

Herc
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