-------------------------------------s-o-s------------------------------------
"John Savard" wrote in
> On Mon, 17 Jan 2005 10:51:55 +1000, "|-|erc" wrote, in part:
>
> >But have you shown there is no contradiction "for real Random and list Computables"?
>
> By showing that there is no contradiction between 1/3 and SimpleCount,
> as I put it, I do show that, for the same reason, if you have the list
> of computables, a random real number can be on it to an infinite number
> of digits, and yet not be on the list. Because showing there is no
> contradiction in the case that is easier to understand means that the
> general principle used to _claim_ a contradiction exists is false.
>
> In other words, I've shown you haven't shown there is a contradiction
> "for real Random and list Computables".
>
> John Savard
> http://home.ecn.ab.ca/~jsavard/index.html

Exactly, but you are the 1st to admit

"a random real number can be on it to an infinite number of digits"

HAHAHAHAHAHAHAHAHAHAHAHAHAHHAHA

HAHAHAHAHAHAHAHAHAHAHAHHAHAHAHAHA

50 people here said that proposition is false, since about September 2004.

wide crosspost merely to highlight the moment.

The question (5 months ago) was.

An infinite amount of people each flip coins infinite times each. Can you
come up with a new sequence of flips?

Herc

From:	Bill Smythe
Subject:	Re: How many digits is pi computable to?
Date:	Mon, 17 Jan 2005 10:49:48 -0600

"|-|erc" wrote:
> The question (5 months ago) was.
> An infinite amount of people each flip coins infinite times each.
Can you
> come up with a new sequence of flips?

If both instances of "infinite" above mean "countably infinite", then I'd
say yes. The total number of coin flips so far is countable times
countable, which is still countable. The number of possible countably
infinite coin flips is 2 to the countable, which is uncountable.

Bill Smythe

From:	Bill Smythe
Subject:	Re: How many digits is pi computable to?
Date:	Mon, 17 Jan 2005 10:59:10 -0600

"|-|erc" wrote:
> > The question (5 months ago) was.
> > An infinite amount of people each flip coins infinite times each.
> Can you
> > come up with a new sequence of flips?

Or, instead of using a countability argument (but still assuming the number
of sequences is countable and each sequence is countable), you can construct
a new sequence of coin flips as follows:

If the first flip by the first person was heads, define the first flip in
the new sequence to be tails, and vice versa.

If the second flip by the second person was heads, define the second flip in
the new sequence to be tails, and vice versa.

Et cetera.

That way, the new sequence will be different from the Nth previous sequence
at the Nth flip. Hence, the new sequence will be different from all the
previous sequences.

Bill Smythe

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 11:59:33 +1000

"Bill Smythe" wrote in ...
> "|-|erc" wrote:
> > > The question (5 months ago) was.
> > > An infinite amount of people each flip coins infinite times each.
> > Can you
> > > come up with a new sequence of flips?
>
> Or, instead of using a countability argument (but still assuming the number
> of sequences is countable and each sequence is countable), you can construct
> a new sequence of coin flips as follows:
>
> If the first flip by the first person was heads, define the first flip in
> the new sequence to be tails, and vice versa.
>
> If the second flip by the second person was heads, define the second flip in
> the new sequence to be tails, and vice versa.
>
> Et cetera.
>
> That way, the new sequence will be different from the Nth previous sequence
> at the Nth flip. Hence, the new sequence will be different from all the
> previous sequences.
>
> Bill Smythe

So this techniqe is correct?

Infinite people each flip coins infinite times.
Can you always find a different sequence of heads and tails?

Take one of the people, whatever his 1st flip was, reverse it! If he
flipped a head you select tail, if he flipped a tail, heads. That's
your first outcome, cross him off and select someone else, whatever was
their second flip, reverse it! Keep on going and you have an infinite
sequence that is different to everyone's sequence in atleast one flip.

X 0/10 WRONG F YOU'RE KIDDING! FAIL :(

its really quite simple, infinite people all doing the same thing you
are doing dispells any possibility of you being unique.

Surely rec.puzzlers can see the blind error sci.math is making with this
simple puzzle??

Herc

From:	Ed Murphy
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 05:24:17 GMT

On Tue, 18 Jan 2005 11:59:33 +1000, |-|erc wrote:

> Infinite people each flip coins infinite times. Can you always find a
> different sequence of heads and tails?

[snip diagonal argument]

> its really quite simple, infinite people all doing the same thing you are
> doing dispells any possibility of you being unique.

"Infinite" is insufficiently precise; some are larger than others.

P = number of people
C = number of coin flips per person
S = number of possible sequences of coin flips

C is countably infinite, but S is uncountably infinite.

If P is countably infinite:
* It can't cover all of S.
* Your comment is false.
* The diagonal argument works.

If P is uncountably infinite:
* It can cover all of S.
* Your comment can be true.
* The diagonal argument doesn't work.

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 15:32:40 +1000

"Ed Murphy" wrote in
> On Tue, 18 Jan 2005 11:59:33 +1000, |-|erc wrote:
>
> > Infinite people each flip coins infinite times. Can you always find a
> > different sequence of heads and tails?
>
> [snip diagonal argument]
>
> > its really quite simple, infinite people all doing the same thing you are
> > doing dispells any possibility of you being unique.
>
> "Infinite" is insufficiently precise; some are larger than others.
>
> P = number of people
> C = number of coin flips per person
> S = number of possible sequences of coin flips
>
> C is countably infinite, but S is uncountably infinite.
>
> If P is countably infinite: ./
> * It can't cover all of S.
> * Your comment is false.
> * The diagonal argument works.

Say you have an (countable) infinite set of people, and they only toss coins a finite number of times.

<>
P C
1 HTHT
2 HHTT
3 TTHH
4 TT
5 H
6 T
....

they are given the constraint to only toss 4 times maximum.

you can construct any sequence you want once I show you the list, but
first you have to tell me how long your sequence is going to be?

Herc

From:	Ed Murphy
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 06:42:18 GMT

On Tue, 18 Jan 2005 15:32:40 +1000, |-|erc wrote:

> Say you have an (countable) infinite set of people, and they only toss
> coins a finite number of times.
>
> <>
> P C
> 1 HTHT
> 2 HHTT
> 3 TTHH
> 4 TT
> 5 H
> 6 T
> ...
>
> they are given the constraint to only toss 4 times maximum.
>
> you can construct any sequence you want once I show you the list, but
> first you have to tell me how long your sequence is going to be?

Oh, you're not giving *me* the 4-tosses-maximum constraint? Fine,
then my sequence will be 5 tosses. My sequence is HHHHH. None of
your people got *that* sequence, did they?

Perhaps the orbital mind control lasers are interfering with your
ability to say what you mean in a precise fashion.

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 19:12:59 +1000

"Ed Murphy" wrote in ...
> On Tue, 18 Jan 2005 15:32:40 +1000, |-|erc wrote:
>
> > Say you have an (countable) infinite set of people, and they only toss
> > coins a finite number of times.
> >
> > <>
> > P C
> > 1 HTHT
> > 2 HHTT
> > 3 TTHH
> > 4 TT
> > 5 H
> > 6 T
> > ...
> >
> > they are given the constraint to only toss 4 times maximum.
> >
> > you can construct any sequence you want once I show you the list, but
> > first you have to tell me how long your sequence is going to be?
>
> Oh, you're not giving *me* the 4-tosses-maximum constraint? Fine,
> then my sequence will be 5 tosses. My sequence is HHHHH. None of
> your people got *that* sequence, did they?
>
> Perhaps the orbital mind control lasers are interfering with your
> ability to say what you mean in a precise fashion.

that's fine, note that when competing against infinite other people you had to break their contraint.

Herc

From:	Ed Murphy
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 10:11:47 GMT

On Tue, 18 Jan 2005 19:12:59 +1000, |-|erc wrote:

> "Ed Murphy" wrote in ...
>> On Tue, 18 Jan 2005 15:32:40 +1000, |-|erc wrote:
>>
>> > Say you have an (countable) infinite set of people, and they only toss
>> > coins a finite number of times.
>> >
>> > <>
>> > P C
>> > 1 HTHT
>> > 2 HHTT
>> > 3 TTHH
>> > 4 TT
>> > 5 H
>> > 6 T
>> > ...
>> >
>> > they are given the constraint to only toss 4 times maximum.
>> >
>> > you can construct any sequence you want once I show you the list, but
>> > first you have to tell me how long your sequence is going to be?
>>
>> Oh, you're not giving *me* the 4-tosses-maximum constraint? Fine, then
>> my sequence will be 5 tosses. My sequence is HHHHH. None of your
>> people got *that* sequence, did they?
>>
>> Perhaps the orbital mind control lasers are interfering with your
>> ability to say what you mean in a precise fashion.
>
> that's fine, note that when competing against infinite other people you
> had to break their contraint.

Ah, now I see what you're attempting to stumble toward.

If the maximum length of a sequence of coin flips is finite, then the
number of possible sequences is also finite, and an infinite number of
people can choose them all.

However, if the maximum length of a sequence of coin flips is countably
infinite, then the number of possible sequences is *uncountably*
infinite. At this point, it is no longer adequate to refer simply to
an "infinite" number of people; we must specify either "countably
infinite" (in which case they cannot choose all possible sequences)
or "uncountably infinite" (in which case they can).

Here is a reiteration of the diagonal argument, which shows how to
construct a sequence missed by a countably infinite number of people.

If a set is countably infinite, then there is a function that maps
each element of that set to exactly one natural number. If the number
of people (other than me) is countably infinite, and the number of coin
flips per person is countably infinite, then the following functions
exist:

* A function f(P) that maps each person P (other than me) to exactly
one natural number.

* For every person P (other than me), a function g_P(C) that maps each
of their coin flips C to exactly one natural number.

* A function g_M(C) that maps each of my coin flips C to exactly one
natural number.

Let f'() be the inverse of f(), g_P'() be the inverse of g_P(), and
g_M'() be the inverse of g_M().

I construct my sequence as follows:

For every natural number N,
find the Nth person P = f'(N),
find their Nth coin flip g_P'(N),
and set my Nth coin flip g_M'(N) opposite.

If any person P (other than me) chose the same sequence as me, then
let N = f(P); but then their Nth coin is different from my Nth coin,
contradiction. Thus no person P (other than me) chose the same
sequence as me. QED.

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 20:43:03 +1000

"Ed Murphy" wrote in ...
> On Tue, 18 Jan 2005 19:12:59 +1000, |-|erc wrote:
>
> > "Ed Murphy" wrote in ...
> >> On Tue, 18 Jan 2005 15:32:40 +1000, |-|erc wrote:
> >>
> >> > Say you have an (countable) infinite set of people, and they only toss
> >> > coins a finite number of times.
> >> >
> >> > <>
> >> > P C
> >> > 1 HTHT
> >> > 2 HHTT
> >> > 3 TTHH
> >> > 4 TT
> >> > 5 H
> >> > 6 T
> >> > ...
> >> >
> >> > they are given the constraint to only toss 4 times maximum.
> >> >
> >> > you can construct any sequence you want once I show you the list, but
> >> > first you have to tell me how long your sequence is going to be?
> >>
> >> Oh, you're not giving *me* the 4-tosses-maximum constraint? Fine, then
> >> my sequence will be 5 tosses. My sequence is HHHHH. None of your
> >> people got *that* sequence, did they?
> >>
> >> Perhaps the orbital mind control lasers are interfering with your
> >> ability to say what you mean in a precise fashion.
> >
> > that's fine, note that when competing against infinite other people you
> > had to break their contraint.
>
> Ah, now I see what you're attempting to stumble toward.
>
> If the maximum length of a sequence of coin flips is finite, then the
> number of possible sequences is also finite, and an infinite number of
> people can choose them all.
>
> However, if the maximum length of a sequence of coin flips is countably
> infinite, then the number of possible sequences is *uncountably*
> infinite. At this point, it is no longer adequate to refer simply to
> an "infinite" number of people; we must specify either "countably
> infinite" (in which case they cannot choose all possible sequences)
> or "uncountably infinite" (in which case they can).
>
> Here is a reiteration of the diagonal argument, which shows how to
> construct a sequence missed by a countably infinite number of people.
>
> If a set is countably infinite, then there is a function that maps
> each element of that set to exactly one natural number. If the number
> of people (other than me) is countably infinite, and the number of coin
> flips per person is countably infinite, then the following functions
> exist:
>
> * A function f(P) that maps each person P (other than me) to exactly
> one natural number.
>
> * For every person P (other than me), a function g_P(C) that maps each
> of their coin flips C to exactly one natural number.
>
> * A function g_M(C) that maps each of my coin flips C to exactly one
> natural number.
>
> Let f'() be the inverse of f(), g_P'() be the inverse of g_P(), and
> g_M'() be the inverse of g_M().
>
> I construct my sequence as follows:
>
> For every natural number N,
> find the Nth person P = f'(N),
> find their Nth coin flip g_P'(N),
> and set my Nth coin flip g_M'(N) opposite.
>
> If any person P (other than me) chose the same sequence as me, then
> let N = f(P); but then their Nth coin is different from my Nth coin,
> contradiction. Thus no person P (other than me) chose the same
> sequence as me. QED.
>

How many flips of your 'new' coin sequence have other people done from flip 1 to flip X?

Herc

From:	Ed Murphy
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 05:37:23 GMT

On Tue, 18 Jan 2005 20:43:03 +1000, |-|erc wrote:

> "Ed Murphy" wrote in ...
>> On Tue, 18 Jan 2005 19:12:59 +1000, |-|erc wrote:
>>
>> > "Ed Murphy" wrote in ...
>> >> On Tue, 18 Jan 2005 15:32:40 +1000, |-|erc wrote:
>> >>
>> >> > Say you have an (countable) infinite set of people, and they only
>> >> > toss coins a finite number of times.
>> >> >
>> >> > <>
>> >> > P C
>> >> > 1 HTHT
>> >> > 2 HHTT
>> >> > 3 TTHH
>> >> > 4 TT
>> >> > 5 H
>> >> > 6 T
>> >> > ...
>> >> >
>> >> > they are given the constraint to only toss 4 times maximum.
>> >> >
>> >> > you can construct any sequence you want once I show you the list,
>> >> > but first you have to tell me how long your sequence is going to
>> >> > be?
>> >>
>> >> Oh, you're not giving *me* the 4-tosses-maximum constraint? Fine,
>> >> then my sequence will be 5 tosses. My sequence is HHHHH. None of
>> >> your people got *that* sequence, did they?
>> >>
>> >> Perhaps the orbital mind control lasers are interfering with your
>> >> ability to say what you mean in a precise fashion.
>> >
>> > that's fine, note that when competing against infinite other people
>> > you had to break their contraint.
>>
>> Ah, now I see what you're attempting to stumble toward.
>>
>> If the maximum length of a sequence of coin flips is finite, then the
>> number of possible sequences is also finite, and an infinite number of
>> people can choose them all.
>>
>> However, if the maximum length of a sequence of coin flips is countably
>> infinite, then the number of possible sequences is *uncountably*
>> infinite. At this point, it is no longer adequate to refer simply to an
>> "infinite" number of people; we must specify either "countably infinite"
>> (in which case they cannot choose all possible sequences) or
>> "uncountably infinite" (in which case they can).
>>
>> Here is a reiteration of the diagonal argument, which shows how to
>> construct a sequence missed by a countably infinite number of people.
>>
>> If a set is countably infinite, then there is a function that maps each
>> element of that set to exactly one natural number. If the number of
>> people (other than me) is countably infinite, and the number of coin
>> flips per person is countably infinite, then the following functions
>> exist:
>>
>> * A function f(P) that maps each person P (other than me) to exactly
>> one natural number.
>>
>> * For every person P (other than me), a function g_P(C) that maps each
>> of their coin flips C to exactly one natural number.
>>
>> * A function g_M(C) that maps each of my coin flips C to exactly one
>> natural number.
>>
>> Let f'() be the inverse of f(), g_P'() be the inverse of g_P(), and
>> g_M'() be the inverse of g_M().
>>
>> I construct my sequence as follows:
>>
>> For every natural number N,
>> find the Nth person P = f'(N),
>> find their Nth coin flip g_P'(N),
>> and set my Nth coin flip g_M'(N) opposite.
>>
>> If any person P (other than me) chose the same sequence as me, then let
>> N = f(P); but then their Nth coin is different from my Nth coin,
>> contradiction. Thus no person P (other than me) chose the same sequence
>> as me. QED.
>>
>>
> How many flips of your 'new' coin sequence have other people done from
> flip 1 to flip X?

I'm not sure what you're asking.

For any finite X, there may well be a person whose first X flips are the
same as my first X flips. *However*, for any person other than me, there
is some finite X such that *that person's* first X flips are *not* the
same as my first X flips.

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 15:59:09 +1000

"Ed Murphy" wrote in message

> >> >> > Say you have an (countable) infinite set of people, and they only
> >> >> > toss coins a finite number of times.
> >> >> >
> >> >> > <>
> >> >> > P C
> >> >> > 1 HTHT
> >> >> > 2 HHTT
> >> >> > 3 TTHH
> >> >> > 4 TT
> >> >> > 5 H
> >> >> > 6 T
> >> >> > ...
> >> >> >
> >> >> > they are given the constraint to only toss 4 times maximum.
> >> >> >
> >> >> > you can construct any sequence you want once I show you the list,
> >> >> > but first you have to tell me how long your sequence is going to
> >> >> > be?
> >> >>
> >> >> Oh, you're not giving *me* the 4-tosses-maximum constraint? Fine,
> >> >> then my sequence will be 5 tosses. My sequence is HHHHH. None of
> >> >> your people got *that* sequence, did they?
> >> >>
> >> >> Perhaps the orbital mind control lasers are interfering with your
> >> >> ability to say what you mean in a precise fashion.
> >> >
> >> > that's fine, note that when competing against infinite other people
> >> > you had to break their contraint.
> >>
> >> Ah, now I see what you're attempting to stumble toward.
> >>
> >> If the maximum length of a sequence of coin flips is finite, then the
> >> number of possible sequences is also finite, and an infinite number of
> >> people can choose them all.
> >>
> >> However, if the maximum length of a sequence of coin flips is countably
> >> infinite, then the number of possible sequences is *uncountably*
> >> infinite. At this point, it is no longer adequate to refer simply to an
> >> "infinite" number of people; we must specify either "countably infinite"
> >> (in which case they cannot choose all possible sequences) or
> >> "uncountably infinite" (in which case they can).
> >>
> >> Here is a reiteration of the diagonal argument, which shows how to
> >> construct a sequence missed by a countably infinite number of people.
> >>
> >> If a set is countably infinite, then there is a function that maps each
> >> element of that set to exactly one natural number. If the number of
> >> people (other than me) is countably infinite, and the number of coin
> >> flips per person is countably infinite, then the following functions
> >> exist:
> >>
> >> * A function f(P) that maps each person P (other than me) to exactly
> >> one natural number.
> >>
> >> * For every person P (other than me), a function g_P(C) that maps each
> >> of their coin flips C to exactly one natural number.
> >>
> >> * A function g_M(C) that maps each of my coin flips C to exactly one
> >> natural number.
> >>
> >> Let f'() be the inverse of f(), g_P'() be the inverse of g_P(), and
> >> g_M'() be the inverse of g_M().
> >>
> >> I construct my sequence as follows:
> >>
> >> For every natural number N,
> >> find the Nth person P = f'(N),
> >> find their Nth coin flip g_P'(N),
> >> and set my Nth coin flip g_M'(N) opposite.
> >>
> >> If any person P (other than me) chose the same sequence as me, then let
> >> N = f(P); but then their Nth coin is different from my Nth coin,
> >> contradiction. Thus no person P (other than me) chose the same sequence
> >> as me. QED.
> >>
> >>
> > How many flips of your 'new' coin sequence have other people done from
> > flip 1 to flip X?
>
> I'm not sure what you're asking.
>
> For any finite X, there may well be a person whose first X flips are the
> same as my first X flips. *However*, for any person other than me, there
> is some finite X such that *that person's* first X flips are *not* the
> same as my first X flips.
>

antidiag =
|<------ How Many flips ? ------->|

Infinite flippers
1
2
3
4
5
....

Its not a hard question, remember John Savards comment
" a random real number will be on it to an infinite number of digits"

Herc

From:	Ed Murphy
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 08:48:28 GMT

On Wed, 19 Jan 2005 15:59:09 +1000, |-|erc wrote:

> "Ed Murphy" wrote in message
>
>> >> >> > Say you have an (countable) infinite set of people, and they
>> >> >> > only toss coins a finite number of times.
>> >> >> >
>> >> >> > <>
>> >> >> > P C
>> >> >> > 1 HTHT
>> >> >> > 2 HHTT
>> >> >> > 3 TTHH
>> >> >> > 4 TT
>> >> >> > 5 H
>> >> >> > 6 T
>> >> >> > ...
>> >> >> >
>> >> >> > they are given the constraint to only toss 4 times maximum.
>> >> >> >
>> >> >> > you can construct any sequence you want once I show you the
>> >> >> > list, but first you have to tell me how long your sequence is
>> >> >> > going to be?
>> >> >>
>> >> >> Oh, you're not giving *me* the 4-tosses-maximum constraint? Fine,
>> >> >> then my sequence will be 5 tosses. My sequence is HHHHH. None of
>> >> >> your people got *that* sequence, did they?
>> >> >>
>> >> >> Perhaps the orbital mind control lasers are interfering with your
>> >> >> ability to say what you mean in a precise fashion.
>> >> >
>> >> > that's fine, note that when competing against infinite other people
>> >> > you had to break their contraint.
>> >>
>> >> Ah, now I see what you're attempting to stumble toward.
>> >>
>> >> If the maximum length of a sequence of coin flips is finite, then the
>> >> number of possible sequences is also finite, and an infinite number
>> >> of people can choose them all.
>> >>
>> >> However, if the maximum length of a sequence of coin flips is
>> >> countably infinite, then the number of possible sequences is
>> >> *uncountably* infinite. At this point, it is no longer adequate to
>> >> refer simply to an "infinite" number of people; we must specify
>> >> either "countably infinite" (in which case they cannot choose all
>> >> possible sequences) or "uncountably infinite" (in which case they
>> >> can).
>> >>
>> >> Here is a reiteration of the diagonal argument, which shows how to
>> >> construct a sequence missed by a countably infinite number of people.
>> >>
>> >> If a set is countably infinite, then there is a function that maps
>> >> each element of that set to exactly one natural number. If the
>> >> number of people (other than me) is countably infinite, and the
>> >> number of coin flips per person is countably infinite, then the
>> >> following functions exist:
>> >>
>> >> * A function f(P) that maps each person P (other than me) to exactly
>> >> one natural number.
>> >>
>> >> * For every person P (other than me), a function g_P(C) that maps
>> >> each
>> >> of their coin flips C to exactly one natural number.
>> >>
>> >> * A function g_M(C) that maps each of my coin flips C to exactly one
>> >> natural number.
>> >>
>> >> Let f'() be the inverse of f(), g_P'() be the inverse of g_P(), and
>> >> g_M'() be the inverse of g_M().
>> >>
>> >> I construct my sequence as follows:
>> >>
>> >> For every natural number N,
>> >> find the Nth person P = f'(N),
>> >> find their Nth coin flip g_P'(N),
>> >> and set my Nth coin flip g_M'(N) opposite.
>> >>
>> >> If any person P (other than me) chose the same sequence as me, then
>> >> let N = f(P); but then their Nth coin is different from my Nth coin,
>> >> contradiction. Thus no person P (other than me) chose the same
>> >> sequence as me. QED.
>> >>
>> >>
>> > How many flips of your 'new' coin sequence have other people done from
>> > flip 1 to flip X?
>>
>> I'm not sure what you're asking.
>>
>> For any finite X, there may well be a person whose first X flips are the
>> same as my first X flips. *However*, for any person other than me,
>> there is some finite X such that *that person's* first X flips are *not*
>> the same as my first X flips.
>>
>>
> antidiag =
> |<------ How Many flips ? ------->|

I have been assuming "a countably infinite number". I'm not sure
whether it's meaningful for this number to be uncountably infinite.

> Infinite flippers
> 1
> 2
> 3
> 4
> 5
> ...

I have addressed both a countably-infinite and uncountably-infinite
number of flippers.

> Its not a hard question, remember John Savards comment
> " a random real number will be on it to an infinite number of digits"

I assume you are referring to Message-ID:
<41eb3f42.795692@news.ecn.ab.ca><354kurF4dasldU1@individual.net>

Problem is, John is also misusing "infinite"; he should have said
"arbitrarily large".

As I stated previously: For any arbitrarily large but finite N,
the first N digits of D (where D is the number constructed by the
diagonal argument) are matched. But D has an infinite number of
digits, and Infinity Is Weird, and you will bloody well be Wrong,
Wrong, Absolutely Brimming Over With Wrongability until you buckle
down and accept that and deal with it.

You have not succeeded in overturning the diagonal argument unless
you identify an element of the set that has *no* digits mismatched
with D. However, if the size of the set is countably infinite,
then the diagonal argument shows that every element of the set has
a digit that mismatches; and if the size of the set is uncountably
infinite, then the diagonal argument no longer places a claim on it.

To summarize:

P = number of people
C = number of coin flips per person

If P is countably infinite and C is countably infinite, then the
diagonal argument works.

If P is countably infinite and C is uncountably infinite, then the
diagonal argument still works. (Consider a countably infinite
subset of each person's sequence of coin flips. This case now
reduces to the previous case.)

If P is uncountably infinite, then the diagonal argument doesn't
work. (It relies on a bijection from the set of people to the
natural numbers; if P is uncountably infinite, then by definition
there isn't one.)

And now let's set aside all these side issues that may or may not
pertain, and revisit your original question. "How many digits is
pi computable to?" Or, to avoid digressions into practical physics,
and also to explicitly address the difference between countable and
uncountable infinities: "Does the complete decimal expansion of pi
consist of a countably or uncountably infinite number of digits?"

I don't know. Discuss.

From:	Kent Paul Dolan
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 11:49:44 +0000 (UTC)

"Ed Murphy" wrote:

> "Does the complete decimal expansion of pi consist
> of a countably or uncountably infinite number of
> digits?"

But "the complete" of an infinite sequence, is, I
would think, a "meaningless noise", at least until
you shift from cardinals to ordinals.

Infinity (of the cardinal kind at least) is about
"becoming", not "being", and the trap of "right
after infinite steps, my next step is", is an
"infinitely" beguiling trap [of which Russ Easterly
seems (to me) to be the poster child perennial
victim, and Herc merely a newly-arrived one].

So, "right after I list the digits of pi, I count
them" is one step too far / "one toke over the
line, sweet Mary". You can't get there from here.

Moreover, since the sequence of digits of pi can be
produced constructively in an operation

(dozens of such operations are known (some of
them, or at least approximations with the same
net characteristic of converging in countable
steps, have been known for millennia))

with countable steps, it is hard to conceive of that
operation somehow having net uncountable output,
unless you also have some handwaving bijection
between the integers and whatever the/a "smallest"
uncountable set might be

(still, I believe, an open (and/or perhaps by
now proved to be undecidable, or to be decidable
either way and produce consistent theories for
each choice) problem),

a bijection known to all but the Hercs of the
universe to be long ago proved impossible.

> I don't know. Discuss.

You _did_ ask, remember; all circular arguments
swallow their own tails, unfortunately.

Perhaps, as in Heinlein's _Glory Road_, we can now
convince Herc to swallow his, with a similar happy
outcome/outgo.

xanthian.

--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 22:07:58 +1000

u snipped the guy you're talking to
serendipity for him

--
Have you now or have you ever been a member of the antidisestablishmentarianism party?
"Kent Paul Dolan" wrote in message
> "Ed Murphy" wrote:
>
> > "Does the complete decimal expansion of pi consist
> > of a countably or uncountably infinite number of
> > digits?"
>
> But "the complete" of an infinite sequence, is, I
> would think, a "meaningless noise", at least until
> you shift from cardinals to ordinals.
>
> Infinity (of the cardinal kind at least) is about
> "becoming", not "being", and the trap of "right
> after infinite steps, my next step is", is an
> "infinitely" beguiling trap [of which Russ Easterly
> seems (to me) to be the poster child perennial
> victim, and Herc merely a newly-arrived one].
>
> So, "right after I list the digits of pi, I count
> them" is one step too far / "one toke over the
> line, sweet Mary". You can't get there from here.
>
> Moreover, since the sequence of digits of pi can be
> produced constructively in an operation
>
> (dozens of such operations are known (some of
> them, or at least approximations with the same
> net characteristic of converging in countable
> steps, have been known for millennia))
>
> with countable steps, it is hard to conceive of that
> operation somehow having net uncountable output,
> unless you also have some handwaving bijection
> between the integers and whatever the/a "smallest"
> uncountable set might be
>
> (still, I believe, an open (and/or perhaps by
> now proved to be undecidable, or to be decidable
> either way and produce consistent theories for
> each choice) problem),
>
> a bijection known to all but the Hercs of the
> universe to be long ago proved impossible.
>
> > I don't know. Discuss.
>
> You _did_ ask, remember; all circular arguments
> swallow their own tails, unfortunately.
>
> Perhaps, as in Heinlein's _Glory Road_, we can now
> convince Herc to swallow his, with a similar happy
> outcome/outgo.
>
> xanthian.
>
>
>
> --
> Posted via Mailgate.ORG Server - http://www.Mailgate.ORG

From:	Bill Smythe
Subject:	Infinite number of infinite coin flips
Date:	Wed, 19 Jan 2005 10:46:48 -0600

"Ed Murphy" wrote:
> To summarize:
>
> P = number of people
> C = number of coin flips per person
>
> If P is countably infinite and C is countably infinite, then the
> diagonal argument works.
>
> If P is countably infinite and C is uncountably infinite, then the
> diagonal argument still works. (Consider a countably infinite
> subset of each person's sequence of coin flips. This case now
> reduces to the previous case.)
>
> If P is uncountably infinite, then the diagonal argument doesn't
> work. (It relies on a bijection from the set of people to the
> natural numbers; if P is uncountably infinite, then by definition
> there isn't one.)

What if P and C are both uncountably infinite (say, to the same degree of
aleph-whatever)? Is there a variation of the diagonal argument which will
work in this case?

> And now let's set aside all these side issues that may or may not
> pertain, and revisit your original question. "How many digits is
> pi computable to?" ....

You're right, we should have changed the thread title a long time ago -- as
I have now done.

> .... Or, to avoid digressions into practical physics,
> and also to explicitly address the difference between countable and
> uncountable infinities: "Does the complete decimal expansion of pi
> consist of a countably or uncountably infinite number of digits?"

Huh?? The first digit after the decimal point is number 1. The next is
number 2. Etc. There's your bijection.

Bill Smythe

From:	Matthew Russotto
Subject:	Re: Infinite number of infinite coin flips
Date:	Wed, 19 Jan 2005 13:21:17 -0600

In article ,
Bill Smythe wrote:
>
>What if P and C are both uncountably infinite (say, to the same degree of
>aleph-whatever)? Is there a variation of the diagonal argument which will
>work in this case?

If P times C is of a cardinality equal to or higher than that of the
continuum, the proposition that one can generate a sequence that they
do not is false. If it's less than that of the continuum, the
proposition is true, no diagonal argument needed. If a diagonal
argument worked, it would prove the existence of cardinals between
aleph-0 and c, and there's no such proof, so I would guess that no
diagonal argument would work in this case either.

From:	Bill Smythe
Subject:	Re: Infinite number of infinite coin flips
Date:	Fri, 21 Jan 2005 09:05:35 -0600

I wrote:
> What if P and C are both uncountably infinite (say, to the same degree of
> aleph-whatever)? Is there a variation of the diagonal argument which will
> work in this case?

I think I have answered my own question. If P and C are the same degree of
infinity (i.e. there is a bijection between P and C), then the diagonal
argument still works, almost without modification.

For each element of P, simply look at the corresponding element in the
corresponding version of C, and reverse it. The set of all those will be a
new version of C which differs from each old one in at least one spot, and
there will be an obvious bijection between P and this new C.

Bill Smythe

From:	Ed Murphy
Subject:	Re: Infinite number of infinite coin flips
Date:	Wed, 19 Jan 2005 20:39:40 GMT

On Wed, 19 Jan 2005 10:46:48 -0600, Bill Smythe wrote:

>> .... Or, to avoid digressions into practical physics,
>> and also to explicitly address the difference between countable and
>> uncountable infinities: "Does the complete decimal expansion of pi
>> consist of a countably or uncountably infinite number of digits?"
>
> Huh?? The first digit after the decimal point is number 1. The next is
> number 2. Etc. There's your bijection.

My intuition says that this is correct, but I'm rusty enough at this
stuff that I don't quite trust my intuition on the subject. Especially
when the least mis-statement on my part is liable to result in Herc
jumping on it and saying "see? see? I was right all along!".

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 21:34:47 +1000

"Ed Murphy" wrote in message
> On Wed, 19 Jan 2005 15:59:09 +1000, |-|erc wrote:
>
> > "Ed Murphy" wrote in message
> >
> >> >> >> > Say you have an (countable) infinite set of people, and they
> >> >> >> > only toss coins a finite number of times.
> >> >> >> >
> >> >> >> > <>
> >> >> >> > P C
> >> >> >> > 1 HTHT
> >> >> >> > 2 HHTT
> >> >> >> > 3 TTHH
> >> >> >> > 4 TT
> >> >> >> > 5 H
> >> >> >> > 6 T
> >> >> >> > ...
> >> >> >> >
> >> >> >> > they are given the constraint to only toss 4 times maximum.
> >> >> >> >
> >> >> >> > you can construct any sequence you want once I show you the
> >> >> >> > list, but first you have to tell me how long your sequence is
> >> >> >> > going to be?
> >> >> >>
> >> >> >> Oh, you're not giving *me* the 4-tosses-maximum constraint? Fine,
> >> >> >> then my sequence will be 5 tosses. My sequence is HHHHH. None of
> >> >> >> your people got *that* sequence, did they?
> >> >> >>
> >> >> >> Perhaps the orbital mind control lasers are interfering with your
> >> >> >> ability to say what you mean in a precise fashion.
> >> >> >
> >> >> > that's fine, note that when competing against infinite other people
> >> >> > you had to break their contraint.
> >> >>
> >> >> Ah, now I see what you're attempting to stumble toward.
> >> >>
> >> >> If the maximum length of a sequence of coin flips is finite, then the
> >> >> number of possible sequences is also finite, and an infinite number
> >> >> of people can choose them all.
> >> >>
> >> >> However, if the maximum length of a sequence of coin flips is
> >> >> countably infinite, then the number of possible sequences is
> >> >> *uncountably* infinite. At this point, it is no longer adequate to
> >> >> refer simply to an "infinite" number of people; we must specify
> >> >> either "countably infinite" (in which case they cannot choose all
> >> >> possible sequences) or "uncountably infinite" (in which case they
> >> >> can).
> >> >>
> >> >> Here is a reiteration of the diagonal argument, which shows how to
> >> >> construct a sequence missed by a countably infinite number of people.
> >> >>
> >> >> If a set is countably infinite, then there is a function that maps
> >> >> each element of that set to exactly one natural number. If the
> >> >> number of people (other than me) is countably infinite, and the
> >> >> number of coin flips per person is countably infinite, then the
> >> >> following functions exist:
> >> >>
> >> >> * A function f(P) that maps each person P (other than me) to exactly
> >> >> one natural number.
> >> >>
> >> >> * For every person P (other than me), a function g_P(C) that maps
> >> >> each
> >> >> of their coin flips C to exactly one natural number.
> >> >>
> >> >> * A function g_M(C) that maps each of my coin flips C to exactly one
> >> >> natural number.
> >> >>
> >> >> Let f'() be the inverse of f(), g_P'() be the inverse of g_P(), and
> >> >> g_M'() be the inverse of g_M().
> >> >>
> >> >> I construct my sequence as follows:
> >> >>
> >> >> For every natural number N,
> >> >> find the Nth person P = f'(N),
> >> >> find their Nth coin flip g_P'(N),
> >> >> and set my Nth coin flip g_M'(N) opposite.
> >> >>
> >> >> If any person P (other than me) chose the same sequence as me, then
> >> >> let N = f(P); but then their Nth coin is different from my Nth coin,
> >> >> contradiction. Thus no person P (other than me) chose the same
> >> >> sequence as me. QED.
> >> >>
> >> >>
> >> > How many flips of your 'new' coin sequence have other people done from
> >> > flip 1 to flip X?
> >>
> >> I'm not sure what you're asking.
> >>
> >> For any finite X, there may well be a person whose first X flips are the
> >> same as my first X flips. *However*, for any person other than me,
> >> there is some finite X such that *that person's* first X flips are *not*
> >> the same as my first X flips.
> >>
> >>
> > antidiag =
> > |<------ How Many flips ? ------->|
>
> I have been assuming "a countably infinite number". I'm not sure
> whether it's meaningful for this number to be uncountably infinite.
>
> > Infinite flippers
> > 1
> > 2
> > 3
> > 4
> > 5
> > ...
>
> I have addressed both a countably-infinite and uncountably-infinite
> number of flippers.
>
> > Its not a hard question, remember John Savards comment
> > " a random real number will be on it to an infinite number of digits"
>
> I assume you are referring to Message-ID:
> <41eb3f42.795692@news.ecn.ab.ca><354kurF4dasldU1@individual.net>
>
> Problem is, John is also misusing "infinite"; he should have said
> "arbitrarily large".
>
> As I stated previously: For any arbitrarily large but finite N,
> the first N digits of D (where D is the number constructed by the
> diagonal argument) are matched. But D has an infinite number of
> digits, and Infinity Is Weird, and you will bloody well be Wrong,
> Wrong, Absolutely Brimming Over With Wrongability until you buckle
> down and accept that and deal with it.
>
> You have not succeeded in overturning the diagonal argument unless
> you identify an element of the set that has *no* digits mismatched
> with D. However, if the size of the set is countably infinite,
> then the diagonal argument shows that every element of the set has
> a digit that mismatches; and if the size of the set is uncountably
> infinite, then the diagonal argument no longer places a claim on it.
>
> To summarize:
>
> P = number of people
> C = number of coin flips per person
>
> If P is countably infinite and C is countably infinite, then the
> diagonal argument works.
>
> If P is countably infinite and C is uncountably infinite, then the
> diagonal argument still works. (Consider a countably infinite
> subset of each person's sequence of coin flips. This case now
> reduces to the previous case.)
>
> If P is uncountably infinite, then the diagonal argument doesn't
> work. (It relies on a bijection from the set of people to the
> natural numbers; if P is uncountably infinite, then by definition
> there isn't one.)
>
> And now let's set aside all these side issues that may or may not
> pertain, and revisit your original question. "How many digits is
> pi computable to?" Or, to avoid digressions into practical physics,
> and also to explicitly address the difference between countable and
> uncountable infinities: "Does the complete decimal expansion of pi
> consist of a countably or uncountably infinite number of digits?"
>
> I don't know. Discuss.
>

A lot of points so I'll only focus on one.

Does anyone else agree John Savards statement is faulty?

[John]
if you have the list
of computables, a random real number can be on it to an infinite number
of digits, and yet not be on the list

[Ed]
> Problem is, John is also misusing "infinite"; he should have said
> "arbitrarily large".
>
> As I stated previously: For any arbitrarily large but finite N,
> the first N digits of D (where D is the number constructed by the
> diagonal argument) are matched

Herc

From:	Will Twentyman
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 15:21:32 -0500

|-|erc wrote:
> "Ed Murphy" wrote in message
>
>>On Wed, 19 Jan 2005 15:59:09 +1000, |-|erc wrote:
>>
>>
>>>"Ed Murphy" wrote in message
>>>
>>>
>>>>>>>>>Say you have an (countable) infinite set of people, and they
>>>>>>>>>only toss coins a finite number of times.
>>>>>>>>>
>>>>>>>>><>
>>>>>>>>>P C
>>>>>>>>>1 HTHT
>>>>>>>>>2 HHTT
>>>>>>>>>3 TTHH
>>>>>>>>>4 TT
>>>>>>>>>5 H
>>>>>>>>>6 T
>>>>>>>>>...
>>>>>>>>>
>>>>>>>>>they are given the constraint to only toss 4 times maximum.
>>>>>>>>>
>>>>>>>>>you can construct any sequence you want once I show you the
>>>>>>>>>list, but first you have to tell me how long your sequence is
>>>>>>>>>going to be?
>>>>>>>>
>>>>>>>>Oh, you're not giving *me* the 4-tosses-maximum constraint? Fine,
>>>>>>>>then my sequence will be 5 tosses. My sequence is HHHHH. None of
>>>>>>>>your people got *that* sequence, did they?
>>>>>>>>
>>>>>>>>Perhaps the orbital mind control lasers are interfering with your
>>>>>>>>ability to say what you mean in a precise fashion.
>>>>>>>
>>>>>>>that's fine, note that when competing against infinite other people
>>>>>>>you had to break their contraint.
>>>>>>
>>>>>>Ah, now I see what you're attempting to stumble toward.
>>>>>>
>>>>>>If the maximum length of a sequence of coin flips is finite, then the
>>>>>>number of possible sequences is also finite, and an infinite number
>>>>>>of people can choose them all.
>>>>>>
>>>>>>However, if the maximum length of a sequence of coin flips is
>>>>>>countably infinite, then the number of possible sequences is
>>>>>>*uncountably* infinite. At this point, it is no longer adequate to
>>>>>>refer simply to an "infinite" number of people; we must specify
>>>>>>either "countably infinite" (in which case they cannot choose all
>>>>>>possible sequences) or "uncountably infinite" (in which case they
>>>>>>can).
>>>>>>
>>>>>>Here is a reiteration of the diagonal argument, which shows how to
>>>>>>construct a sequence missed by a countably infinite number of people.
>>>>>>
>>>>>>If a set is countably infinite, then there is a function that maps
>>>>>>each element of that set to exactly one natural number. If the
>>>>>>number of people (other than me) is countably infinite, and the
>>>>>>number of coin flips per person is countably infinite, then the
>>>>>>following functions exist:
>>>>>>
>>>>>>* A function f(P) that maps each person P (other than me) to exactly
>>>>>> one natural number.
>>>>>>
>>>>>>* For every person P (other than me), a function g_P(C) that maps
>>>>>>each
>>>>>> of their coin flips C to exactly one natural number.
>>>>>>
>>>>>>* A function g_M(C) that maps each of my coin flips C to exactly one
>>>>>> natural number.
>>>>>>
>>>>>>Let f'() be the inverse of f(), g_P'() be the inverse of g_P(), and
>>>>>>g_M'() be the inverse of g_M().
>>>>>>
>>>>>>I construct my sequence as follows:
>>>>>>
>>>>>>For every natural number N,
>>>>>>find the Nth person P = f'(N),
>>>>>>find their Nth coin flip g_P'(N),
>>>>>>and set my Nth coin flip g_M'(N) opposite.
>>>>>>
>>>>>>If any person P (other than me) chose the same sequence as me, then
>>>>>>let N = f(P); but then their Nth coin is different from my Nth coin,
>>>>>>contradiction. Thus no person P (other than me) chose the same
>>>>>>sequence as me. QED.
>>>>>>
>>>>>>
>>>>>
>>>>>How many flips of your 'new' coin sequence have other people done from
>>>>>flip 1 to flip X?
>>>>
>>>>I'm not sure what you're asking.
>>>>
>>>>For any finite X, there may well be a person whose first X flips are the
>>>>same as my first X flips. *However*, for any person other than me,
>>>>there is some finite X such that *that person's* first X flips are *not*
>>>>the same as my first X flips.
>>>>
>>>>
>>>
>>>antidiag =
>>> |<------ How Many flips ? ------->|
>>
>>I have been assuming "a countably infinite number". I'm not sure
>>whether it's meaningful for this number to be uncountably infinite.
>>
>>
>>>Infinite flippers
>>>1
>>>2
>>>3
>>>4
>>>5
>>>...
>>
>>I have addressed both a countably-infinite and uncountably-infinite
>>number of flippers.
>>
>>
>>>Its not a hard question, remember John Savards comment
>>> " a random real number will be on it to an infinite number of digits"
>>
>>I assume you are referring to Message-ID:
>><41eb3f42.795692@news.ecn.ab.ca><354kurF4dasldU1@individual.net>
>>
>>Problem is, John is also misusing "infinite"; he should have said
>>"arbitrarily large".
>>
>>As I stated previously: For any arbitrarily large but finite N,
>>the first N digits of D (where D is the number constructed by the
>>diagonal argument) are matched. But D has an infinite number of
>>digits, and Infinity Is Weird, and you will bloody well be Wrong,
>>Wrong, Absolutely Brimming Over With Wrongability until you buckle
>>down and accept that and deal with it.
>>
>>You have not succeeded in overturning the diagonal argument unless
>>you identify an element of the set that has *no* digits mismatched
>>with D. However, if the size of the set is countably infinite,
>>then the diagonal argument shows that every element of the set has
>>a digit that mismatches; and if the size of the set is uncountably
>>infinite, then the diagonal argument no longer places a claim on it.
>>
>>To summarize:
>>
>>P = number of people
>>C = number of coin flips per person
>>
>>If P is countably infinite and C is countably infinite, then the
>>diagonal argument works.
>>
>>If P is countably infinite and C is uncountably infinite, then the
>>diagonal argument still works. (Consider a countably infinite
>>subset of each person's sequence of coin flips. This case now
>>reduces to the previous case.)
>>
>>If P is uncountably infinite, then the diagonal argument doesn't
>>work. (It relies on a bijection from the set of people to the
>>natural numbers; if P is uncountably infinite, then by definition
>>there isn't one.)
>>
>>And now let's set aside all these side issues that may or may not
>>pertain, and revisit your original question. "How many digits is
>>pi computable to?" Or, to avoid digressions into practical physics,
>>and also to explicitly address the difference between countable and
>>uncountable infinities: "Does the complete decimal expansion of pi
>>consist of a countably or uncountably infinite number of digits?"
>>
>>I don't know. Discuss.
>>
>
>
> A lot of points so I'll only focus on one.
>
> Does anyone else agree John Savards statement is faulty?
>
> [John]
> if you have the list
> of computables, a random real number can be on it to an infinite number
> of digits, and yet not be on the list

It's quite obviously faulty by simply reading it. Replacing "infinite"
with "arbitrary finite" would make it correct. At best it is imprecise,
at worst it is self-contradictory.

--
Will Twentyman
email: wtwentyman at copper dot net

From:	Michael Mendelsohn
Subject:	Re: How many digits is pi computable to?
Date:	Mon, 17 Jan 2005 18:53:16 +0100

Bill Smythe schrieb:
> "|-|erc" wrote:
> > > The question (5 months ago) was.
> > > An infinite amount of people each flip coins infinite times each.
> > Can you
> > > come up with a new sequence of flips?
>
> Or, instead of using a countability argument (but still assuming the number
> of sequences is countable and each sequence is countable), you can construct
> a new sequence of coin flips as follows:
>
> If the first flip by the first person was heads, define the first flip in
> the new sequence to be tails, and vice versa.
>
> If the second flip by the second person was heads, define the second flip in
> the new sequence to be tails, and vice versa.
>
> Et cetera.
>
> That way, the new sequence will be different from the Nth previous sequence
> at the Nth flip. Hence, the new sequence will be different from all the
> previous sequences.

IIRC Herc rejects any kind of diagonalization argument.
(Maybe that has changed, though.)

Cheers
Michael
--
Still an attentive ear he lent Her speech hath caused this pain
But could not fathom what she meant Easier I count it to explain
She was not deep, nor eloquent. The jargon of the howling main
-- from Lewis Carroll: The Three Usenet Trolls

From:	Bill Smythe
Subject:	Re: How many digits is pi computable to?
Date:	Mon, 17 Jan 2005 13:39:57 -0600

"Michael Mendelsohn" wrote:
> IIRC Herc rejects any kind of diagonalization argument.
> (Maybe that has changed, though.)

How can he reject a valid argument?

Of course, that argument depends on the number of coin-flip sequences being
countable. (And, by using the word "sequence", I guess I'm implying that
each set of coin flips is individually countable.)

If the number of coin-flip sequences is uncountable, then of course my proof
breaks down. For example, consider the set of ALL (countable) coin-flip
sequences. Obviously, it is not possible to create a sequence not in the
set. In fact, that's one way of PROVING that the set of all countable
coin-flip sequences is uncountable.

Bill Smythe

From:	Michael Mendelsohn
Subject:	Re: How many digits is pi computable to?
Date:	Mon, 17 Jan 2005 23:26:17 +0100

Bill Smythe schrieb:
> "Michael Mendelsohn" wrote:
> > IIRC Herc rejects any kind of diagonalization argument.
> > (Maybe that has changed, though.)
>
> How can he reject a valid argument?

I have no idea.
I haven't really gotten to understand the way he (?) thinks.
I merely recalled an exchange from around last April, and I thought I'd
warn you because you might not want to waste your time.

I see that Barb Knox, another poster from sci.logic (I am not, but
apparently Barb and Herc are) whom I recall taking part in that
exchange, has already provided you with a pointer.

Cheers
Michael
--
Still an attentive ear he lent Her speech hath caused this pain
But could not fathom what she meant Easier I count it to explain
She was not deep, nor eloquent. The jargon of the howling main
-- from Lewis Carroll: The Three Usenet Trolls

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 11:59:37 +1000

"Michael Mendelsohn" wrote
> Bill Smythe schrieb:
> > "Michael Mendelsohn" wrote:
> > > IIRC Herc rejects any kind of diagonalization argument.
> > > (Maybe that has changed, though.)
> >
> > How can he reject a valid argument?
>
> I have no idea.
> I haven't really gotten to understand the way he (?) thinks.

How do YOU think MM?

Not all infinite expansions are unique.

Say you have an infinite set of people, and they only toss coins a finite number of times.

<>
P C
1 HTHT
2 HHTT
3 TTHH
4 TT
5 H
6 T
....

they are given the constraint to only toss 4 times maximum.

you can construct any sequence you want once I show you the list, but
first you have to tell me how long your sequence is going to be?

Herc

From:	Barb Knox
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 10:56:08 +1300

In article <5eGdnbxpt4DqiXHcRVn-vw@rcn.net>,
"Bill Smythe" wrote:

>"Michael Mendelsohn" wrote:
>> IIRC Herc rejects any kind of diagonalization argument.
>> (Maybe that has changed, though.)

Oddly, he rejects ALL-BUT-ONE diagonalization argument. He DOES accept the
one that shows the unsolvability of the halting problem. And he doesn't
notice the logical difficulty with doing that.

>How can he reject a valid argument?

Search Google Groups on 'group:sci.logic author:|-|erc (god OR adam)' and
you'll have your answer...

[snip]

--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 11:59:35 +1000

"Barb Knox" wrote in messag
> In article <5eGdnbxpt4DqiXHcRVn-vw@rcn.net>,
> "Bill Smythe" wrote:
>
> >"Michael Mendelsohn" wrote:
> >> IIRC Herc rejects any kind of diagonalization argument.
> >> (Maybe that has changed, though.)
>
> Oddly, he rejects ALL-BUT-ONE diagonalization argument. He DOES accept the
> one that shows the unsolvability of the halting problem. And he doesn't
> notice the logical difficulty with doing that.

Diagonalisation DISPROVES a specific function Halt()
That is different to PROVING antidiag exists.

constructivism : the denial of existence resulting in contradiction does
not prove existence.

======================
e.g. "this statement has no proof"

Assume it is FALSE.
then it has a proof, so it is true
CONTRADICTION

therefore "this statement has no proof:" is true.
=======================

Now, compare that to the constructivist proof.

true(facts) = proven(facts) U provable(facts).

factX <=> true(factX) = ~proven(factX)
factX <=> proven(factX) U provable(factX) = ~proven(factX)
factX <=> t U p = ~t
factX <=> t = ~t (extension of case p={})
factX <=> CONTRADICTION

Sorry, no incompleteness theorem just because you got stuck working out
"YOU CANT PROVE ME"

Herc

From:	Kent Paul Dolan
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 10:16:40 +0000 (UTC)

"|-|erc" wrote:

> constructivism :
> the denial of existence resulting in
> contradiction does not prove existence.

But Barb's proof of the proposition "Herc is a
nutcase", reproduced here in full:

> "Barb Knox" wrote:

:- Search Google Groups on
:- 'group:sci.logic author:|-|erc (god OR adam)'
:- and you'll have your answer...

is both sufficient (anyone reviewing that evidence
will agree with her implied conclusion), and is not
dependent on a "proof by contradiction", one that,
for example, you've never demonstrated any sane
behaviors, and _therefore_ you are a nutcase, but
instead, it directly demonstrates your nutcase
behaviors, a constructive proof not only showing
that something exists, but also showing how to
reify an instance, as required by constructivists.

So, this being a proof in a style you approve, why
did you snip it?

Of course, like many constructive proofs, it only
constructs a _single_ sufficient instance, omitting
your Truman gig, your numerology beliefs, your
paranoia that antennas are beaming messages toward
your head, your claims of generating vast wealth
via "click to read ads for pay" sites, and on and
on.

You are at least a fully flowered, multifunctional,
multitasking nutcase, if that helps at all.

The problem is that whatever newsgroup you touch, you
turn to trash, as for example comp.theory, suddenly
awash in postings with uppercased subject lines, of
your devising, subject lines insulting saner posters
by name, of your devising, subject lines full of
obscenities and rude language, of your devising.

You are not merely insane, you are screamingly insane.

That others insist on screaming back does them no
credit, but it doesn't make you less the nutcase who
is causing the deluge of dreck.

Go away Herc. You have nothing of value to
contribute, and you contribute it in carload lots,
to the exclusion of useful discussion. This may
gratify your ego, but it does not glorify your name.

xanthian.

--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG

From:	Ken Johnson
Subject:	Re: How many digits is pi computable to?
Date:	Thu, 20 Jan 2005 15:30:17 GMT

> How many digits is pi computable to?

One. Everyone knows that pi is 3. Says so in the Barble.

Ken Johnson

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 20:34:33 +1000

"Kent Paul Dolan" wrote in >
> Go away Herc. You have nothing of value to
> contribute, and you contribute it in carload lots,
> to the exclusion of useful discussion. This may
> gratify your ego, but it does not glorify your name.
>
> xanthian.

no pipsqueak. i've designed the simplest system of computation, a proof
of an infallible halting function, polynomial search space, 3d intersection rendering,
simplest formalism of cantors proof, meta-godel statements, ....
I've even helped the crippled walk and the blind see

>yes I access a PC and the net through my Braille Display
>it is called a Powerbraille 80 Cell Display
>there are many makes of Braille Displays and in many sizes.
>With out my braille Display I could never be able to use a Computer in any way.
>I can not use speech software like many other Blind people
>because I can not hear speech.
>But your letter to the newsgroup is being read on a Braille Display
>and typed on a keyboard that has braille dots on it
>so one can use it just like you.

If the chess tournament sites don't have a text option then this stops you
playing realtime over the web, you can still get email opponents for
slower games. I hope while you are learning that some tournament sites
add a text feature, maybe the board will look something like this
~~~
RNBQKBNR
PPPPPPPP
.........
.........
.........
.........
pppppppp
rnbqkbnr

r is rook, n is for knight though I call them horses, and so on
the most popular first move is

RNBQKBNR
PPPPPPPP
.........
.........
.....p...
.........
pppp.ppp
rnbqkbnr

that's moving your pawn in front of your king up 2 squares, hope I haven't
complicated things, only 6 pieces to learn how to move.

A bipedal robot hip is disclosed. The hip contains rotational bearings with
respective axis being coplanar but not collinear. As the hip pivots via
actuation of the ground engaged leg with respect to the hip, opposing
actuation of the raised leg results in lateral movement of the raised leg.
Using two actuators in the hip allows two degrees of freedom for each leg
not ground engaged.

maybe comp.theory could try contributing to theory instead of sucking up to
the 2 geriatrics Barb and xanthian who pray to their text books all day. I
told you xan, keep annoying me and I"ll SUBSCRIBE to comp.theory, take
a look at sci.logic, the top 100 subjects are 80% Herc, you'll know what
computer theory is all about then.

Herc

From:	nikolai kingsley
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 01:28:14 +1100

|-|erc wrote:

> "Kent Paul Dolan" wrote in >
>
>>Go away Herc. You have nothing of value to
>>contribute...
>
>
> no pipsqueak. i've designed the simplest

if i could find you i would hurt you.

and you, Kent! why the fuck do you have to keep dragging your fucking
arguments into talk.bizarre?

From:	Kent Paul Dolan
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 00:24:06 +0000 (UTC)

"nikolai kingsley" wrote:

> and you, Kent! why the fuck do you have to keep
> dragging your fucking arguments into talk.bizarre?

0) I'm not "dragging in my arguments", I'm
crossposting my more creative flames.

1) I consider Herc, like Julian, an interesting
specimen, hugely insane, bizarrely deranged.

[Notice his complete inability to remain on
topic, or to avoid deistic claims, even when
trying to lend credence to pretensions of
utility or sanity? This is every bit the
equivalent of Julian on a manic spree: Herc is
completely out of his own control. Notice
Herc's threats to "subscribe" to comp.theory?
Who on earth cares, and besides, he's already
trashing the place. Probably, if the stats from
the days of Arbitron still hold true, there are
hundreds of thousands to perhaps millions of
people "subscribed", whatever that is, to
comp.theory. Herc is not only not dealing with
the real world, he isn't even in contact with a
reasonable mock-up of a real galaxy. He fails
to grasp that his "subscription" is something
local to his news server, having no outside
effect.]

2) Thus, his spew is funny to read, though not in
any way he intends, to one in an appropriate frame
of mind.

3) Reading Herc's ravings balances out the drearily
dull stuff shoveled in by imbeciles posting from the
alt groups, and in response to Polewka's trolling.

I answered your question; now your turn. Once more,
why do you find it impossible for you to give proper
attribution to material you quote? THat makes it
tremendously hard to keep context in threads where
you participate.

xanthian.

--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 00:34:09 +1000

>
> if i could find you i would hurt you.

the trillion dollar US media empire might not like you interfering with their investment The Truman,
already 10GB of your thoughts have been uplinked to the pentagon

Herc

From:	Kent Paul Dolan
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 05:15:12 +0000 (UTC)

"Herc" wrote:

> the trillion dollar US media empire might not like
> you interfering with their investment The Truman,
> already 10GB of your thoughts have been uplinked
> to the pentagon

I couldn't _pay_ a sane person to so quickly convict
himself of insanity when accused, yet you fall into
the self-same trap every time you get led by your
nose into the same hemisphere where it sits.

You really spoil the sport, Herc.

One could only wish those wasting Net resources by
paying you the attention appropriate to a sane
correspondent would catch the hint and let you
yammer your nonsense on totally ignored.

xanthian.

--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Wed, 19 Jan 2005 15:24:41 +1000

"Kent Paul Dolan" wrote in
> "Herc" wrote:
>
> > the trillion dollar US media empire might not like
> > you interfering with their investment The Truman,
> > already 10GB of your thoughts have been uplinked
> > to the pentagon
>
> I couldn't _pay_ a sane person to so quickly convict
> himself of insanity when accused, yet you fall into
> the self-same trap every time you get led by your
> nose into the same hemisphere where it sits.
>
> You really spoil the sport, Herc.
>
> One could only wish those wasting Net resources by
> paying you the attention appropriate to a sane
> correspondent would catch the hint and let you
> yammer your nonsense on totally ignored.
>

yada yada yada yada yada.... yeah we know you've seen the inside
of mental asylms but can you stop the 20,000 word obsession with my psyche?

Herc

From:	Scott Dorsey
Subject:	Re: How many digits is pi computable to?
Date:	19 Jan 2005 13:02:50 -0500

Using the numerical method of Leibniz, I can only do one decimal place
before I lose track of everything and have to write the intermediate steps
down.

Using graphical methods, I think I could get three digits if there is enough
space in the parking lot. I already have the chalk and a long enough tape
measure.
--scott

--
"C'est un Nagra. C'est suisse, et tres, tres precis."

From:	Don Del Grande
Subject:	Re: How many digits is pi computable to?
Date:	Mon, 17 Jan 2005 20:59:13 GMT

Michael Mendelsohn wrote:

> Bill Smythe wrote:

>> "|-|erc" wrote:

>>> An infinite amount of people each flip coins infinite times each.
>>> Can you come up with a new sequence of flips?

>> Or, instead of using a countability argument (but still assuming the number
>> of sequences is countable and each sequence is countable), you can construct
>> a new sequence of coin flips as follows:
>>
>> If the first flip by the first person was heads, define the first flip in
>> the new sequence to be tails, and vice versa.
>>
>> If the second flip by the second person was heads, define the second flip in
>> the new sequence to be tails, and vice versa.
>>
>> Et cetera.
>>
>> That way, the new sequence will be different from the Nth previous sequence
>> at the Nth flip. Hence, the new sequence will be different from all the
>> previous sequences.

> IIRC Herc rejects any kind of diagonalization argument.
> (Maybe that has changed, though.)

Is there a proof that the real numbers in [0,1] are uncountable that
does not use diagonalization? (Use binary digits instead of decimal
ones, with each 0 to the right of the decimal point (er, "binary
point"?) as "heads" and 1 as "tails", and you reduce the original
problem to this one.)

-- Don

From:	r.e.s.
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 00:37:15 GMT

"Don Del Grande" wrote ...

> Is there a proof that the real numbers in [0,1] are uncountable that
> does not use diagonalization?

See
http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof

From:	Matthew Russotto
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 08:59:45 -0600

In article <%SYGd.354$Ju1.54@newsread3.news.pas.earthlink.net>,
r.e.s. wrote:
>"Don Del Grande" wrote ...
>
>> Is there a proof that the real numbers in [0,1] are uncountable that
>> does not use diagonalization?
>
>See
>http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof

Doesn't seem like a very good proof, as it demonstrates that the
rationals are uncountable.

From:	Dave Seaman
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 15:16:05 +0000 (UTC)

On Tue, 18 Jan 2005 08:59:45 -0600, Matthew Russotto wrote:
> In article <%SYGd.354$Ju1.54@newsread3.news.pas.earthlink.net>,
> r.e.s. wrote:
>>"Don Del Grande" wrote ...
>>
>>> Is there a proof that the real numbers in [0,1] are uncountable that
>>> does not use diagonalization?
>>
>>See
>>http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof

> Doesn't seem like a very good proof, as it demonstrates that the
> rationals are uncountable.

No, because the proof depends on the "has no gaps" hypothesis, which does
not hold for the rationals.

--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.

From:	Gareth Owen
Subject:	Re: How many digits is pi computable to?
Date:	18 Jan 2005 15:06:44 +0000

russotto@grace.speakeasy.net (Matthew Russotto) writes:

> Doesn't seem like a very good proof, as it demonstrates that the
> rationals are uncountable.

No it doesn't.
Rationals do not satisfy the fourth property of the theorem.

"if it is partitioned into two nonempty sets A and B in such a way that every
member of A is less than every member of B, then there is a boundary point c,
so that every point less than c is in A and every point greater than c is in B."

although it would be more clearly stated
... there is a boundary point c *in R* ...

So {z sqrt(2)} is a partition of Q without a boundary in Q.
--
Gareth Owen
If I were to ask you a hypothetical question, would you answer it?

From:	Matthew Russotto
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 10:28:36 -0600

In article ,
Gareth Owen wrote:
>russotto@grace.speakeasy.net (Matthew Russotto) writes:
>
>> Doesn't seem like a very good proof, as it demonstrates that the
>> rationals are uncountable.
>
>No it doesn't.
>Rationals do not satisfy the fourth property of the theorem.
>
>"if it is partitioned into two nonempty sets A and B in such a way that every
> member of A is less than every member of B, then there is a boundary point c,
> so that every point less than c is in A and every point greater than c is in B."
>
> although it would be more clearly stated
> ... there is a boundary point c *in R* ...
>
>So {z sqrt(2)} is a partition of Q without a boundary in Q.

Ah, OK. As stated it didn't look any different from density.

From:	r.e.s.
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 16:30:02 GMT

"r.e.s." wrote ...
> "Don Del Grande" wrote ...
>
>> Is there a proof that the real numbers in [0,1] are uncountable that
>> does not use diagonalization?
>
> See
> http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof

Given the various objections to the quality of the presentation
on that page, perhaps someone would post a link to one that's
more faithful to Cantor's original?

My purpose in giving the wikipedia link was to confirm that,
yes, there is such a proof, and that in fact Cantor's "first
proof" did not use diagonalisation -- not to vouch for the
quality of the proof as presented there.

--r.e.s.

From:	\|-\|erc
Subject:	Re: How many digits is pi computable to?
Date:	Tue, 18 Jan 2005 11:59:30 +1000

"Bill Smythe" wrote in
> "|-|erc" wrote:
> > The question (5 months ago) was.
> > An infinite amount of people each flip coins infinite times each.
> Can you
> > come up with a new sequence of flips?
>
> If both instances of "infinite" above mean "countably infinite", then I'd
> say yes. The total number of coin flips so far is countable times
> countable, which is still countable. The number of possible countably
> infinite coin flips is 2 to the countable, which is uncountable.
>
> Bill Smythe
>

Consise argument, but does it contradict John's proposition?

"if you have the list of computables,
a random real number will be on it to an infinite number of digits"

i.e. every possible coin sequence is on the list of computables to an infinite number of flips.

> > How many numbers are in this sequence?
> > <1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ....>
>
> aleph_0
>
> >
> > How many numbers are in this sequence? (duplicates allowed)
> > <3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 6, ....>
>
> aleph_0

Right

How many flips of this random sequence

make an appearance after all their predecessors in (members of) this list?

UTM(row, col) mod 2
1 <0101101000..>
2 <1110101000..>
3 <0000000000..>
4 <1111100000..>
...

You can use an alphabet substitution you deem appropriate.
Herc