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Virgil · Guest

In article <5JTjk.24056$IK1.18640@news-server.bigpond.net.au>,
"|-|erc" <h@r.c> wrote:

[quote]"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote
"|-|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.

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
[/quote]
What requires "Omega" to have any binary representation at all?
Is |-| uck claiming that omega is a computable real that falls between 0
and 1?

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

|-|erc · Guest

"Virgil" <Virgil@gmale.com> wrote in...
[quote]In article <5JTjk.24056$IK1.18640@news-server.bigpond.net.au>,
"|-|erc" <h@r.c> wrote:

"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote
"|-|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.

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?
[/quote]
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.

[quote]Is |-| uck claiming that omega is a computable real that falls between 0
and 1?
[/quote]
no. I>m claiming all the digits of Omega appear in sequence in the set of
computable reals.

[quote]
Unless he is, the rest of his post is garbage.
[/quote]
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.

Herc

|-|erc · Guest

"Mariano Suárez-Alvarez" <mariano.suarezalvarez@gmail.com> wrote
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?

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.

Herc

|-|erc · Guest

"Mariano Suárez-Alvarez" <mariano.suarezalvarez@gmail.com> wrote
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?

Now you get it! What I said was, in this context, is that I can write down your number.

Do you agree that the 1st 10 digits of Omega appear in sequence in the right position,
somewhere in the list of computable reals?

Herc

|-|erc · Guest

[Herc]
[quote]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]
[m]
-There is another option: meaningless.

So you>re saying no digits of Omega are ever computed, by any computer ever?

Can anyone else evaluate whether my proposition above is true or false?

Herc

Peter Webb · Guest

"|-|erc" <h@r.c> wrote in message
news:_PSjk.24045$IK1.16606@news-server.bigpond.net.au...
[quote]"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote
"|-|erc" <h@r.c> wrote in message
news:NePjk.23952$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.

None, some, or all?

Herc

[/quote]
Any = for all n e N. Which somewhat equates to "all", except in common
English "all" is slightly ambiguous in that it can mean either "any" or "the
set of".

But the deal is, you give me any finite number and I can work it out to that
precision.

Peter Webb · Guest

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

********************
Nicely argued.

Mariano Suárez-Alvarez · Guest

On Jul 30, 7:02 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
[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

********************
Nicely argued.
[/quote]
And he agreed!

-- m

hagman · Guest

On 30 Jul., 00:25, "|-|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]
Whoa! This looks a bit like the WM tree in disguise!

hagman

Dave L. Renfro · Guest

|-|erc wrote:

[quote]Do you agree that the 1st 10 digits of Omega appear
in sequence in the right position, somewhere in the
list of computable reals?
[/quote]
Virgil wrote:

[quote]I do not even agree that Omega has any digits appearing any place.
[/quote]
I believe he>s talking about Chaitin>s omega.

http://www.google.com/search?q=Chaitin>s-omega

Dave L. Renfro

Virgil · Guest

In article <tHVjk.24122$IK1.18122@news-server.bigpond.net.au>,
"|-|erc" <h@r.c> wrote:

[quote]"Virgil" <Virgil@gmale.com> wrote in...

Say Omega starts with 0.1

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

its a real number.
[/quote]
By what definition of "real number"?

The most commonly accepted definitions of "real number" involve either
Cauchy sequences or Dedekind cuts of rationals, neither of which allow
omega as a real number, so what is your definitions of "real number"
that includes omega?

[quote]

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.
[/quote]
You first have to establish that omega has digits.
[quote]

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

True or False?
[/quote]
No, garbage

[quote]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]
Not even false.

Virgil · Guest

In article <O%Vjk.24131$IK1.21296@news-server.bigpond.net.au>,
"|-|erc" <h@r.c> wrote:

[quote]Do you agree that the 1st 10 digits of Omega appear in sequence in
the right position, somewhere in the list of computable reals?
[/quote]
I do not even agree that Omega has any digits appearing any place.

Virgil · Guest

In article <K7Wjk.24133$IK1.7828@news-server.bigpond.net.au>,
"|-|erc" <h@r.c> wrote:

[quote][Herc]
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.

[m]
-There is another option: meaningless.

So you>re saying no digits of Omega are ever computed, by any computer ever?
[/quote]
One would be well advised to avoid use of any computer on which any such
digits have been computed.
[quote]
Can anyone else evaluate whether my proposition above is true or false?
[/quote]
I see questions, but no proposition in this posting.

Ralf Bader · Guest

hagman wrote:

[quote]On 30 Jul., 00:25, "|-|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. Â 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? Â It means Omega is computable
to infinite decimal places. Â OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. Â But the set of CR does
contain this sequence of digits. Â ALL sequences of digits
to oo are computable. Â 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
--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Whoa! This looks a bit like the WM tree in disguise!
[/quote]
He has been asked what he means by omega, but didn>t give a sensible answer.
Maybe it is something like what is called omega here:
http://plus.maths.org/issue37/features/omega/index.html
But who knows.

Ralf

Virgil · Guest

In article
<72ad002f-da47-4eda-8781-7ef8b13776da@f63g2000hsf.googlegroups.com>,
"Dave L. Renfro" <renfr1dl@cmich.edu> wrote:

[quote]|-|erc wrote:

Do you agree that the 1st 10 digits of Omega appear
in sequence in the right position, somewhere in the
list of computable reals?

Virgil wrote:

I do not even agree that Omega has any digits appearing any place.

I believe he>s talking about Chaitin>s omega.

http://www.google.com/search?q=Chaitin>s-omega

Dave L. Renfro
[/quote]
If he refuses to say which one, the he is hardly in a position to object
to my counter based on an omega rather more relevant to Cantor.


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