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From: |-|ercules on 9 Jun 2010 13:15
"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.

Herc

From: Daryl McCullough on 9 Jun 2010 15:15
|-|ercules says...
>
>"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.

As I said, your questions are completely incoherent nonsense. I'm
trying to give coherent responses, but you have to meet my half-way,
and not continue to spout gobbledy-goop.

>ALL (INFINITELY MANY) digits of ALL digit sequences are computable!

Okay, that's false, but you seem to believe it. Why?

>Are you saying you can find a new_sequence_of_digits using
>diagonalization on a computable reals list?

I already answered. Absolutely I believe that. I gave you
an example.

>It's MEANINGLESS. Different digit to what?

You've already been told, many, many times. If L is
a list of reals, then antidiagonal(L) is a real that
whose first digit is different from that of the first
real in list L, whose second digit is different from
that of the second real in list L, whose third digit
is different from the third real in L, etc.

The antidiagonal is different from every single real
on the list.

>IF there are NUMEROUS different digits then there must be some FINITE >substring
>between them (inclusive) that is not a computable digit sequence.

Okay, that's false. But you seem to believe it. Why?

--
Daryl McCullough
Ithaca, NY

From: George Greene on 9 Jun 2010 15:54
On Jun 9, 1:52 am, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> So you think the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS

Of course it does.

> and this does not contradict that ALL sequences of digits are on the computable
> list of reals

OF COURSE it WOULD contradict that, if that were true.

> up to all (an infinite amount of) digit positions?

You keep SAYING this. That doesn't mean it MEANS anything.
It IS NOT THE CASE that "all" sequences of digits are on the
computable list of reals,
DUMBASS.
In particular, THE ANTI-DIAGONAL OF THAT LIST is NOT ON that list.


All FINITE sequences of digits are on that list, yes.
As soon as you say "up to" ANY "digit position", you ARE STOPPING at
that digit position
which means you are talking about a FINITE string.
If all I wanted was a list that contained all FINITE sequences, I
wouldn't even NEED anything as fancy as
"all computable numbers". The following list:
0
..1
..2
..3
..4
..5
..6
..7
..8
..9
..01
..11
..21
..31
..41
..51
..61
..71
..81
..91
..02
..12
..32
etc.
WOULD DO JUST FINE.
That list DOES NOT contain PLENTY of computable numbers like pi, e, or
sqrt(2),
but it DOES contain EVERY FINITE initial segment of those numbers, up
to ANY
(again, FINITE) place you require. So there are infinitely many
FINITE places
up to which all these numbers get approximated. But that does NOT
mean that
ANY of these numbers OCCURS on this list! EVERY element on this list
is finite
(unless you think of it has having infinitely many 0s on the end), so
NO infinite
decimal is on it. Hell, lots of RATIONAL numbers like 1/3, 1/6, and
1/7 ARE NOT ON this list!
But for ANY AND EVERY *FINITE* place (in other words infiniteLY MANY
DIFFERENT finite places),
the list approximates any real up to that place (and beyond).
From: |-|ercules on 9 Jun 2010 17:13
"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ...
> |-|ercules says...
>>
>>"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.
>
> As I said, your questions are completely incoherent nonsense. I'm
> trying to give coherent responses, but you have to meet my half-way,
> and not continue to spout gobbledy-goop.
>
>>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.

If you agree, you just took your first steps into a larger world.

You should contemplate for a moment that I do follow your diagonal argument.

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.

Examples don't prove it for the entire domain.

Herc

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

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