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|-|erc · Guest

When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number. You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
....

What does that ... mean? It means Omega is computable
to infinite decimal places. OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger than oo.

FACT: All sequences of digits are computable to infinite length.

Herc
--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Mariano Suárez-Alvarez · Guest

On Jul 29, 10:16 pm, "|-|erc" <h...@r.c> wrote:
[quote]"fishfry" <BLOCKSPAMfish...@your-mailbox.com> wrote

In article <JNMjk.23889$IK1.3...@news-server.bigpond.net.au>,
"|-|erc" <h...@r.c> wrote:

When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number.

No, that>s just not true. There are uncountablyl many such sequences,
but only countably many computable numbers.

You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What do you mean here by Omega?

Isn>t the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise
something like that
[/quote]
Are you supposed to *know* what Omega is? That>s a
very mild requirement for your original post to have
any sense at all...

Now you are asking what the first bit is?

[quote][ ... yet another instance of the apparently endemic
sci.math disease causing people to `pass to the limit'
with no regards to meaningfulness, correctness or even
manners, snipped. ]
[/quote]
-- m

Robert J. Kolker · Guest

|-|erc wrote:
[quote]When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number. You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What does that ... mean? It means Omega is computable
to infinite decimal places. OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger than oo.

FACT: All sequences of digits are computable to infinite length.
[/quote]
Not true. The set of possible infinite sequences is a non-countable set.
The set of computable (i.e. Turing computable) infinite sequences is a
countable set.

Bob Kolker

fishfry · Guest

In article <JNMjk.23889$IK1.3755@news-server.bigpond.net.au>,
"|-|erc" <h@r.c> wrote:

[quote]When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number.
[/quote]
No, that>s just not true. There are uncountablyl many such sequences,
but only countably many computable numbers.

You might
[quote]argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...
[/quote]
What do you mean here by Omega?

Anyway, your reasoning is flawed. 10 is a finite natural number; and
10^100 is a finite natural number; and 10^googol is a finite natural
number; but the set of all finite natural numbers, N, is not a fiinite
natural number.

You have to be careful when passing to infinity.

[quote]
What does that ... mean? It means Omega is computable
to infinite decimal places.
[/quote]
You haven>t used any properties of computable numbers here. Whatever you
mean by Omega (do you mean lower-case omega, 'w' in ASCII, to represent
the order type of N?), you haven>t shown anything about its
computability.

[quote]OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger than oo.

FACT: All sequences of digits are computable to infinite length.

[/quote]
Clearly not.

|-|erc · Guest

"Robert J. Kolker" <bobkolker@comcast.net> wrote in message
[quote]|-|erc wrote:
When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number. You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What does that ... mean? It means Omega is computable
to infinite decimal places. OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger than oo.

FACT: All sequences of digits are computable to infinite length.

Not true. The set of possible infinite sequences is a non-countable set.
The set of computable (i.e. Turing computable) infinite sequences is a
countable set.

Bob Kolker
[/quote]
OK, the set of possible infinite sequences are computable to how many digits?

Herc

|-|erc · Guest

"fishfry" <BLOCKSPAMfishfry@your-mailbox.com> wrote
[quote]In article <JNMjk.23889$IK1.3755@news-server.bigpond.net.au>,
"|-|erc" <h@r.c> wrote:

When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number.

No, that>s just not true. There are uncountablyl many such sequences,
but only countably many computable numbers.

You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What do you mean here by Omega?
[/quote]
Isn>t the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise
something like that

[quote]
Anyway, your reasoning is flawed. 10 is a finite natural number; and
10^100 is a finite natural number; and 10^googol is a finite natural
number; but the set of all finite natural numbers, N, is not a fiinite
natural number.

You have to be careful when passing to infinity.
[/quote]
How many digits is Omega computable to?

[quote]

What does that ... mean? It means Omega is computable
to infinite decimal places.

You haven>t used any properties of computable numbers here. Whatever you
mean by Omega (do you mean lower-case omega, 'w' in ASCII, to represent
the order type of N?), you haven>t shown anything about its
computability.

OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger than oo.

FACT: All sequences of digits are computable to infinite length.

Clearly not.
[/quote]
OK, the set of possible infinite sequences are computable to how many digits?

Herc

Peter Webb · Guest

"|-|erc" <h@r.c> wrote in message
news:NePjk.23952$IK1.8403@news-server.bigpond.net.au...
[quote]"Robert J. Kolker" <bobkolker@comcast.net> wrote in message
|-|erc wrote:
When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number. You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What does that ... mean? It means Omega is computable
to infinite decimal places. OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger than oo.

FACT: All sequences of digits are computable to infinite length.

Not true. The set of possible infinite sequences is a non-countable set.
The set of computable (i.e. Turing computable) infinite sequences is a
countable set.

Bob Kolker

OK, the set of possible infinite sequences are computable to how many
digits?

Herc

[/quote]
Any finite number.

Peter Webb · Guest

"|-|erc" <h@r.c> wrote in message
news:UhPjk.23954$IK1.14233@news-server.bigpond.net.au...
[quote]"fishfry" <BLOCKSPAMfishfry@your-mailbox.com> wrote
In article <JNMjk.23889$IK1.3755@news-server.bigpond.net.au>,
"|-|erc" <h@r.c> wrote:

When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number.

No, that>s just not true. There are uncountablyl many such sequences,
but only countably many computable numbers.

You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What do you mean here by Omega?

Isn>t the 1st digit of Omega 1 if the 1st Turing machine halts, 0
otherwise
something like that

Anyway, your reasoning is flawed. 10 is a finite natural number; and
10^100 is a finite natural number; and 10^googol is a finite natural
number; but the set of all finite natural numbers, N, is not a fiinite
natural number.

You have to be careful when passing to infinity.

How many digits is Omega computable to?

[/quote]
That depends upon exactly how Omega is defined, ie the TM rules.

You might be able to work out the first n digits (with n finite) if you can
prove that for all n digits you can use some argument to prove that all of
these definitely halt or don>t halt. However, no such proof is possible in
general; at some point in the expansion you will just have to say that you
don>t know whether the TM terminates, and at that point you don>t know what
the rest of Omega looks like.

[quote]

What does that ... mean? It means Omega is computable
to infinite decimal places.

You haven>t used any properties of computable numbers here. Whatever you
mean by Omega (do you mean lower-case omega, 'w' in ASCII, to represent
the order type of N?), you haven>t shown anything about its
computability.

OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger than oo.

FACT: All sequences of digits are computable to infinite length.

Clearly not.

OK, the set of possible infinite sequences are computable to how many
digits?

Herc

[/quote]
As I said before, any finite number.

|-|erc · Guest

"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote >
[quote]"|-|erc" <h@r.c> wrote in message
news:NePjk.23952$IK1.8403@news-server.bigpond.net.au...
"Robert J. Kolker" <bobkolker@comcast.net> wrote in message
|-|erc wrote:
When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number. You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What does that ... mean? It means Omega is computable
to infinite decimal places. OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger than oo.

FACT: All sequences of digits are computable to infinite length.

Not true. The set of possible infinite sequences is a non-countable set.
The set of computable (i.e. Turing computable) infinite sequences is a
countable set.

Bob Kolker

OK, the set of possible infinite sequences are computable to how many
digits?

Herc

Any finite number.
[/quote]
None, some, or all?

Herc

|-|erc · Guest

"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote
[quote]OK, the set of possible infinite sequences are computable to how many
digits?

Herc

Any finite number.
[/quote]
Bill Goates has so much money in the bank.
If you asked whether he had any finite number of money in the bank
the answer was always yes, how much money total does he have?

Infinite.

Herc

|-|erc · Guest

"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote >
[quote]"|-|erc" <h@r.c> wrote in message
news:UhPjk.23954$IK1.14233@news-server.bigpond.net.au...
"fishfry" <BLOCKSPAMfishfry@your-mailbox.com> wrote
In article <JNMjk.23889$IK1.3755@news-server.bigpond.net.au>,
"|-|erc" <h@r.c> wrote:

When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number.

No, that>s just not true. There are uncountablyl many such sequences,
but only countably many computable numbers.

You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What do you mean here by Omega?

Isn>t the 1st digit of Omega 1 if the 1st Turing machine halts, 0
otherwise
something like that

Anyway, your reasoning is flawed. 10 is a finite natural number; and
10^100 is a finite natural number; and 10^googol is a finite natural
number; but the set of all finite natural numbers, N, is not a fiinite
natural number.

You have to be careful when passing to infinity.

How many digits is Omega computable to?

That depends upon exactly how Omega is defined, ie the TM rules.

You might be able to work out the first n digits (with n finite) if you can
prove that for all n digits you can use some argument to prove that all of
these definitely halt or don>t halt. However, no such proof is possible in
general; at some point in the expansion you will just have to say that you
don>t know whether the TM terminates, and at that point you don>t know what
the rest of Omega looks like.

[/quote]
In a binary system of numbers, roughly 1/2 of all computable reals (0<=cr<1)
begin with 1, and the other 1/2 begin with 0.

Say Omega starts with 0.1

1/2 of CR also begin with 0.1.

Say Omega starts with 0.10

1/4 of CR also begin with 0.10

Omega then becomes 0.101

1/8 of CR also begin with 0.101

No matter how many digits of Omega you consider, a percentage of computable
reals will compute that string of digits.

As the number of reals in the list of computable reals approaches infinity,
the number of digits of Omega that are computed approaches infinity.

Considering the entire infinite set of computable reals, it contains the sequence
of digits contained in Omega (to infinite length)

Herc

Mariano Suárez-Alvarez · Guest

On Jul 30, 5:33 am, "|-|erc" <h...@r.c> wrote:
[quote]"Virgil" <Vir...@gmale.com> wrote in...

In article <5JTjk.24056$IK1.18...@news-server.bigpond.net.au>,
"|-|erc" <h...@r.c> wrote:

"Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote
"|-|erc" <h...@r.c> wrote in message
news:UhPjk.23954$IK1.14233@news-server.bigpond.net.au...
"fishfry" <BLOCKSPAMfish...@your-mailbox.com> wrote
In article <JNMjk.23889$IK1.3...@news-server.bigpond.net.au>,
"|-|erc" <h...@r.c> wrote:

When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number.

No, that>s just not true. There are uncountablyl many such sequences,
but only countably many computable numbers.

You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What do you mean here by Omega?

Isn>t the 1st digit of Omega 1 if the 1st Turing machine halts, 0
otherwise
something like that

Anyway, your reasoning is flawed. 10 is a finite natural number; and
10^100 is a finite natural number; and 10^googol is a finite natural
number; but the set of all finite natural numbers, N, is not a fiinite
natural number.

You have to be careful when passing to infinity.

How many digits is Omega computable to?

That depends upon exactly how Omega is defined, ie the TM rules.

You might be able to work out the first n digits (with n finite) if you can
prove that for all n digits you can use some argument to prove that all of
these definitely halt or don>t halt. However, no such proof is possible in
general; at some point in the expansion you will just have to say that you
don>t know whether the TM terminates, and at that point you don>t know what
the rest of Omega looks like.

In a binary system of numbers, roughly 1/2 of all computable reals (0<=cr<1)
begin with 1, and the other 1/2 begin with 0

Say Omega starts with 0.1

What requires "Omega" to have any binary representation at all?

its a real number. if the 1st Turing machine halts, Omega begins 0.1,
if the 1st Turing Machine runs forever, Omega begins 0.0. Similarly
for the second digit and so on, giving it a binary representation.

Is |-| uck claiming that omega is a computable real that falls between 0
and 1?

no. I>m claiming all the digits of Omega appear in sequence in the set of
computable reals.

Unless he is, the rest of his post is garbage.

True or False?
As the number of reals in the list of computable reals approaches infinity,
the number of digits of Omega that are computed approaches infinity.
[/quote]
There is another option: meaningless.

-- m

Mariano Suárez-Alvarez · Guest

On Jul 30, 5:36 am, "|-|erc" <h...@r.c> wrote:
[quote]"Mariano Suárez-Alvarez" <mariano.suarezalva...@gmail.com> wrote
On Jul 29, 10:16 pm, "|-|erc" <h...@r.c> wrote:

"fishfry" <BLOCKSPAMfish...@your-mailbox.com> wrote

In article <JNMjk.23889$IK1.3...@news-server.bigpond.net.au>,
"|-|erc" <h...@r.c> wrote:

When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number.

No, that>s just not true. There are uncountablyl many such sequences,
but only countably many computable numbers.

You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What do you mean here by Omega?

Isn>t the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise
something like that

-Are you supposed to *know* what Omega is? That>s a
-very mild requirement for your original post to have
-any sense at all...

-Now you are asking what the first bit is?

No we don>t need to know any values of Omega. We know by redundancy
that we have computed Omega to n digits because every possible sequence
to n digits has been computed.
[/quote]
So...

Say that I pick a number between 1 and 10, but
I do not tell you which. Then you make a list of all
thenumbers between 1 and 10. Do you think that after
making that list you know which number I picked?

-- m

kunzmilan · Guest

On Jul 30, 12:25 am, "|-|erc" <h...@r.c> wrote:
[quote]When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number. You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What does that ... mean? It means Omega is computable
to infinite decimal places. OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger than oo.

FACT: All sequences of digits are computable to infinite length.

Herc
--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
[/quote]
I tried to prove something similar.
But, when we define the rational matrix R, r(i,j) = j/i,
we see that there are as many rational numbers lesser than
1, as rational numbers greater than 1. Since in the first row of R,
all natural numbers n are enumerated, rational numbers greater than 1
overflow capacity of natural numbers to count, regardless to the
infinite length of the string of natural numbers. This infinity
increases the size of the rational matrix R, too.
kunzmilan

Dave L. Renfro · Guest

|-|erc wrote:

[quote]When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

Every possible sequence of digits to infinite length
is represented as a computable number. You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...

What does that ... mean? It means Omega is computable
to infinite decimal places. OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that>s good enough to
refute the existence of sets that are uncountably larger
than oo.

FACT: All sequences of digits are computable to infinite
length.
[/quote]
I believe the issue is the existence of a uniform
computability procedure for each n-digit truncation
of the number. Sure, there is such a procedure
for the first 100 digits, such a procedure for
the first 1000 digits, such a procedure for the
first 10000 digits, and so on, but these procedures
have to be "essentially the same" for all such
n-digit truncations if we>re going to use them
to verify the computability of the number. For
computable numbers this can be arranged, but for
non-computable numbers this can>t be arranged.
Kind of like uniform continuity vs. continuity,
or any situation in which a statement of the form
(for all x)(there exists y) is true, but not the
corresponding statement where the quantifiers
are switched, (there exists y)(for all x).

Anyway, I don>t know much about formal computability,
so maybe someone who does can comment on whether
this general idea has any bearing on the actual
issue at hand.

Dave L. Renfro


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