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From: Daryl McCullough on 9 Jun 2010 12:54
|-|ercules says...
>
>"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote...

>>>1 Start with a list containing all sequences.
>>>2 Find a NEW sequence
>>>3 CONTRADICTION
>>
>> You assume that there exists a list containing
>> all sequences, and then you find out that assumption
>> leads to a contradiction. So the assumption is false.
>
>not necessarily. I demonstrated that all sequences occur and your
>'new sequence' is merely self reference and negation.

There is no self-reference in the definition of the anti-diagonal
function. The anti-diagonal function is a two-place function

f(L,n)

which returns a digit for each list L and for each natural number
n. It's a simple function:

First define a transformation on digits c(d) as follows:
c(5) = 4. If d is not equal to 5, then c(d) = 5.

Now, we define f(L,n) as follows:

f(L,n) = c(L(n,n))

where L(n,n) = the nth digit of the nth real in the list L.

Now, the antidiagonal real is defined by:

antiDiag(L) = that real r such that the integer part of r is 0,
and forall n, the nth digit of r is equal to f(L,n).

This is a function that given any list of reals L, returns
another real, antiDiag(L), which is guaranteed to not be on
the list L.

There is nothing self-referential about this definition.

>Are you saying you find a new_sequence_of_digits using
>diagonalization on infinite lists?

Yes, I am absolutely saying that. Let's try it out with
a simple example.

Let's let L be the list containing the falling reals:

L_0 = 0.000...
L_1 = 0.300...
L_2 = 0.3300...
L_3 = 0.3330...
etc.

Then the antidiagonal will be
0.555...

This is clearly not on list L.

You can try yourself with different choices for the list L, as
long as it is possible to determine what the first real is,
what the second real is, etc.

--
Daryl McCullough
Ithaca, NY

From: Daryl McCullough on 9 Jun 2010 13:03
|-|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.

--
Daryl McCullough
Ithaca, NY

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: |-|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

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