10 Questions Concerning Numbers
WM
pi and only pi?
Answer: Every ISD of any other number contains a digit d_n which
differs from the digit p_n of pi, but there is no digit p_n of pi
which differs from every ISDs of every other number (which has the
same digit sequence till d_(n-1)).
Question 2: Is it possible to write an ISD which unambiguously
identifies pi?
Answer: No.
Question 3: So the ISD of pi cannot appear in a line or as the
diagonal of a Cantor's list?
Answer: That is correct.
Question 4: Is this property also true for the ISD of each other
irrational number?
Answer: Yes.
Question 5: Could this fact somehow depend on the cardinal number of
the irrational numbers? Would the result change if there were only few
irrational numbers?
Answer: No.
Question 6: Why does one conclude from the absence of non-existing
ISDs on the uncountability of the set of irrational numbers?
Answer: I don't know. (Does anybody know?)
Question 7: If irrational numbers are not unambiguously identified by
ISDs, how can we identify an irrational number?
Answer: That can be done by finite definitions, equations or
descriptions like "the solution of the eq. 2 - x^2 = 0" or "the length
of the diagonal of the unit square".
Question 8: How many finite definitions can be stated?
Answer: Countably many.
Question 9: Is there any other chance to bring an uncountable set of
irrational numbers to existence?
Answer: An axiom could require that there should exist an uncountable
set of irrational numbers.
Question 10: So can an uncountable set of irrational numbers exist
other than created by an axiom of uncountability (or by counterfactual
belief, motivated by what ever)?
Answer: [To be inserted by the reader.]
Regards, WM
Peter Webb
"WM" <muec...@rz.fh-augsburg.de> wrote in message
news:ece9d3d5-2a04-4ee7...@j19g2000yqk.googlegroups.com...
> Question 1: Is there an infinite sequence of digits (ISD) representing
> pi and only pi?
>
> Answer: Every ISD of any other number contains a digit d_n which
> differs from the digit p_n of pi,
Which proves that if there is an ISD for pi, it is not also the ISD of "any
other number".
> but there is no digit p_n of pi
> which differs from every ISDs of every other number (which has the
> same digit sequence till d_(n-1)).
>
Obviously true, but proves nothing.
> Question 2: Is it possible to write an ISD which unambiguously
> identifies pi?
>
> Answer: No.
>
Pi has a unique ISD, which commences 3.1415...
This is easy to prove by using any of the bounded series which generate pi,
and considering these as a Cauchy sequence.
Everything else in your post seems based on this completely false assumption
that pi does not have a unique ISD, so I have snipped it.
William Elliot
> Question 1: Is there an infinite sequence of digits (ISD) representing
> pi and only pi?
>
Yes.
> Question 2: Is it possible to write an ISD which unambiguously
> identifies pi?
>
Yes.
> Question 3: So the ISD of pi cannot appear in a line or as the
> diagonal of a Cantor's list?
pi will appear as the diagonal of some Cantor lists and
will not appear as the diagonal of other Cantor lists.
> Question 4: Is this property also true for the ISD of each other
> irrational number?
>
An irrational number r, will appear as the diagonal of some Cantor
lists and will not appear as the diagonal of other Cantor lists.
> Question 5: Could this fact somehow depend on the cardinal number of
> the irrational numbers?
No.
> Would the result change if there were only few
> irrational numbers?
>
That's an impossibility.
> Question 6: Why does one conclude from the absence of non-existing
> ISDs on the uncountability of the set of irrational numbers?
>
> Answer: I don't know.
Tough luck.
> (Does anybody know?)
Yes, anybody who has learned college math.
> Question 7: If irrational numbers are not unambiguously identified by
> ISDs, how can we identify an irrational number?
>
As the premise is incorrect, the question is vacuous.
> That can be done by finite definitions, equations or
> descriptions like "the solution of the eq. 2 - x^2 = 0" or "the length
> of the diagonal of the unit square".
>
That will define only countably many irrationals.
All the rest of them, the majority of them will go undefined..
> Question 8: How many finite definitions can be stated?
>
> Answer: Countably many.
>
Hurray! A correct answer.
> Question 9: Is there any other chance to bring an uncountable set of
> irrational numbers to existence?
>
Yes, as the completion of the rationals.
> Answer: An axiom could require that there should exist an uncountable
> set of irrational numbers.
>
That's not needed, not preferred and not used.
> Question 10: So can an uncountable set of irrational numbers exist
> other than created by an axiom of uncountability?
>
Yes. (The deliberate bias in your question has been removed.)
----
WM
wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>
> news:ece9d3d5-2a04-4ee7...@j19g2000yqk.googlegroups.com...
>
> > Question 1: Is there an infinite sequence of digits (ISD) representing
> > pi and only pi?
>
> > Answer: Every ISD of any other number contains a digit d_n which
> > differs from the digit p_n of pi,
>
> Which proves that if there is an ISD for pi, it is not also the ISD of "any
> other number".
>
> > but there is no digit p_n of pi
> > which differs from every ISDs of every other number (which has the
> > same digit sequence till d_(n-1)).
>
> Obviously true, but proves nothing.
>
> > Question 2: Is it possible to write an ISD which unambiguously
> > identifies pi?
>
> > Answer: No.
>
> Pi has a unique ISD, which commences 3.1415...
Please continue.
>
> This is easy to prove by using any of the bounded series which generate pi,
> and considering these as a Cauchy sequence.
That proves the existence of pi as a real number, but it does not
prove the existence of a ISD of pi.
>
> Everything else in your post seems based on this completely false assumption
> that pi does not have a unique ISD, so I have snipped it.
If pi has an ISD, then give it please. Certainly it will differ from
every term of the Cauchy sequence. Certainly it will be different from
any of its finite initial segments like 31415/10000. That means,
whatever you write: It is not the ISD of pi.
Regards, WM
WM
> On Thu, 17 Dec 2009, WM wrote:
> > Question 1: Is there an infinite sequence of digits (ISD) representing
> > pi and only pi?
>
> Yes.
>
> > Question 2: Is it possible to write an ISD which unambiguously
> > identifies pi?
>
> Yes.
Please do it.
>
> > Question 3: So the ISD of pi cannot appear in a line or as the
> > diagonal of a Cantor's list?
>
> pi will appear as the diagonal of some Cantor lists and
> will not appear as the diagonal of other Cantor lists.
Let us continue after you proved your assertion.
>
> > Question 4: Is this property also true for the ISD of each other
> > irrational number?
>
> An irrational number r, will appear as the diagonal of some Cantor
> lists and will not appear as the diagonal of other Cantor lists.
Let us continue after you proved your assertion.
>
> > Question 5: Could this fact somehow depend on the cardinal number of
> > the irrational numbers?
>
> No.
>
> > Would the result change if there were only few
> > irrational numbers?
>
> That's an impossibility.
Consider the following Cantor list
0.0
0.1
0.11
0.111
...
and let 0 be exchanged by 1. The diagonal is 0.111... Does this proof
show that the set of periodic IDSs is uncountable?
>
> > Question 6: Why does one conclude from the absence of non-existing
> > ISDs on the uncountability of the set of irrational numbers?
>
> > Answer: I don't know.
>
> Tough luck.
>
> > (Does anybody know?)
>
> Yes, anybody who has learned college math.
Did you in fact learn there that the limits of Cauchy sequences must
have IDSs?
Did you also learn that there is a natural number larger than every
other natural number? Or does the ISD of pi contain a digit at index
omega?
>
> > Question 7: If irrational numbers are not unambiguously identified by
> > ISDs, how can we identify an irrational number?
>
> As the premise is incorrect, the question is vacuous.
>
> > That can be done by finite definitions, equations or
> > descriptions like "the solution of the eq. 2 - x^2 = 0" or "the length
> > of the diagonal of the unit square".
>
> That will define only countably many irrationals.
> All the rest of them, the majority of them will go undefined..
But you believe that each one has its own ISD which is completely
defined?
>
> > Question 8: How many finite definitions can be stated?
>
> > Answer: Countably many.
>
> Hurray! A correct answer.
How can you say that every ISD can be written if it is impossible to
define every number and hence every ISD? Do you write those ISDs which
haven't any definition with closed eyes?
Regards, WM
William Elliot
> On 18 Dez., 09:47, William Elliot <ma...@rdrop.remove.com> wrote:
>> On Thu, 17 Dec 2009, WM wrote:
>>> Would the result change if there were only few
>>> irrational numbers?
>>
>> That's an impossibility.
>
> Consider the following Cantor list
>
> 0.0
> 0.1
> 0.11
> 0.111
> ...
>
What's the rest of the list?
> and let 0 be exchanged by 1. The diagonal is 0.111... Does this proof
> show that the set of periodic IDSs is uncountable?
First finish the list.
> But you believe that each one has its own ISD which is completely
> defined?
You're confusing religion with math.
> How can you say that every ISD can be written if it is impossible to
> define every number and hence every ISD? Do you write those ISDs which
> haven't any definition with closed eyes?
>
To wright out an infinite sequence requires infinite ink.
Do a search for "infinite ink".
> Regards, WM
>
Marshall
>
> If pi has an ISD, then give it please. Certainly it will differ from
> every term of the Cauchy sequence. Certainly it will be different from
> any of its finite initial segments like 31415/10000. That means,
> whatever you write: It is not the ISD of pi.
I claim Asia does not exist, and will only concede otherwise
if you give it to me.
Marshall
Peter Webb
>
> If pi has an ISD, then give it please. Certainly it will differ from
> every term of the Cauchy sequence. Certainly it will be different from
> any of its finite initial segments like 31415/10000. That means,
> whatever you write: It is not the ISD of pi.
1/3 has an ISD. While I cannot write out the full decimal expansion of 1/3,
I can provide an algorithm to specify exactly the nth digit of the ISD of
1/3 - in fact for all n, the nth digit is 3.
Similarly, I can give you an algorithm to find the unique nth digit of the
ISD of pi for any n.
Do you accept that 1/3 has an ISD, even if I can't write out 0.333... in
full in a finite amount of time?
So why don't you accept that pi has an ISD?
Owen Jacobson
<webbf...@DIESPAMDIEoptusnet.com.au> said:
>
>
>>
>> If pi has an ISD, then give it please. Certainly it will differ from
>> every term of the Cauchy sequence. Certainly it will be different from
>> any of its finite initial segments like 31415/10000. That means,
>> whatever you write: It is not the ISD of pi.
>
> 1/3 has an ISD. While I cannot write out the full decimal expansion of
> 1/3, I can provide an algorithm to specify exactly the nth digit of the
> ISD of 1/3 - in fact for all n, the nth digit is 3.
You can formalise this even further:
Given some real number r, there exists an ordered pair (f_r, m_r) such that:
1. f_r : N -> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
That is, f is a function that, given a natural number, produces a digit.
2. m_r ∈ Z
m_r represents the "magnitude" of the number, which allows us to
distinguish between pairs for distinct reals (a ≠ b) with the same
digit sequence (f_a = f_b).
3. (∑ (i in N) f_r (i) * (10 ^ -i)) * (10 ^ m_r) = r
That is, if you string the digits given by f_r together into a number
using the usual powers-of-ten sum of digits and shift the decimal place
to the right m_r places, you get r.
Obviously, for your example above, for r = 1 / 3, f_r(x) = 3 and m_r =
-1 (or 0 if N does not contain 0).
This works in other bases, too, which can be convenient: using the BBP
formula, you can determine the n'th base-16 digit of pi without first
determining the digits from 0 through n - 1. A similar formula exists
for e and others will likely be developed. Therefore we can construct
f_pi and f_e, and from there m_pi and m_e are trivial to determine.
(Actually, for many reals, there exist many pairs as above that produce
them. Consider the obvious case of a real that begins with a finite
sequence of zeros: do you use f_d => (0, 0, 1, ...) and m_d = 0 or f_d
=> (0, 1, ...) and m_d = -1? However, the important part is that each
pair (f_r, m_r) identifies one and only one real number r. Proof left
as an exercise.)
This technique (formalising some infinite list as a function from N to
the enumerated items) is fairly standard, and /noticably absent/ from
WM's treatment of lists of paths. Given how fundamentally related it is
to the concept of countably infinite sets, that's a telling omission.
-o
Peter Webb
"Owen Jacobson" <angryb...@gmail.com> wrote in message
news:2009121903240516807-angrybaldguy@gmailcom...
Of course, not that I care that I have to generate digits 0 to n-1 to
generate the nth digit.
>Therefore we can construct f_pi and f_e, and from there m_pi and m_e are
>trivial to determine.
>
> (Actually, for many reals, there exist many pairs as above that produce
> them. Consider the obvious case of a real that begins with a finite
> sequence of zeros: do you use f_d => (0, 0, 1, ...) and m_d = 0 or f_d =>
> (0, 1, ...) and m_d = -1? However, the important part is that each pair
> (f_r, m_r) identifies one and only one real number r. Proof left as an
> exercise.)
>
> This technique (formalising some infinite list as a function from N to the
> enumerated items) is fairly standard, and /noticably absent/ from WM's
> treatment of lists of paths. Given how fundamentally related it is to the
> concept of countably infinite sets, that's a telling omission.
>
> -o
>
Reals, and pi in particular, can be defined in lots of other ways as well.
I think his underlying argument is one about using decimal expansion, and
not about the property of pi at all. It is true that you cannot show pi as a
finite decimal, but you can't show 1/3 as a finite decimal either. And this
is purely an argument about decimal notation, because in other ways of
specifying Reals (such as pi) the problem doesn't exist. But he thinks that
a property of decimal notation is a property of the Real itself; its not. In
base pi, pi=1. That's a pretty finite length string.
Virgil
"Peter Webb" <webbf...@DIESPAMDIEoptusnet.com.au> wrote:
> "Owen Jacobson" <angryb...@gmail.com> wrote in message
> news:2009121903240516807-angrybaldguy@gmailcom...
> > On 2009-12-18 19:51:29 -0500, "Peter Webb"
> > <webbf...@DIESPAMDIEoptusnet.com.au> said:
> >
> >>
> >>
> >>>
> >>> If pi has an ISD, then give it please. Certainly it will differ from
> >>> every term of the Cauchy sequence. Certainly it will be different from
> >>> any of its finite initial segments like 31415/10000. That means,
> >>> whatever you write: It is not the ISD of pi.
> >>
> >> 1/3 has an ISD. While I cannot write out the full decimal expansion of
> >> 1/3, I can provide an algorithm to specify exactly the nth digit of the
> >> ISD of 1/3 - in fact for all n, the nth digit is 3.
> >
> > You can formalise this even further:
> >
> > Given some real number r, there exists an ordered pair (f_r, m_r) such
> > that:
> >
> > 1. f_r : N -> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
> >
> > That is, f is a function that, given a natural number, produces a digit.
> >
> > 2. m_r Å∏ Z
> >
> > m_r represents the "magnitude" of the number, which allows us to
> > distinguish between pairs for distinct reals (a ÅÇ b) with the same digit
> > sequence (f_a = f_b).
> >
> > 3. (É∞ (i in N) f_r (i) * (10 ^ -i)) * (10 ^ m_r) = r
I think that much of WM's confusion about numbers is that he cannot
distinguish them from numerals, which are merely names for numbers, and
not the numbers themselves.
In some contexts, the distinction between a number and a numeral may be
ignored without harm, but whenever one starts talking about the digits
in something, as WM so often does, that something is necessarily a
numeral and not a number, and that distinction is often essential.
WM
> On Fri, 18 Dec 2009, WM wrote:
> > On 18 Dez., 09:47, William Elliot <ma...@rdrop.remove.com> wrote:
> >> On Thu, 17 Dec 2009, WM wrote:
> >>> Would the result change if there were only few
> >>> irrational numbers?
>
> >> That's an impossibility.
>
> > Consider the following Cantor list
>
> > 0.0
> > 0.1
> > 0.11
> > 0.111
> > ...
>
> What's the rest of the list?
>
> > and let 0 be exchanged by 1. The diagonal is 0.111... Does this proof
> > show that the set of periodic IDSs is uncountable?
>
> First finish the list.
Such lists cannot be finished. That is the cardinal error of Cantor.
For a countable set of numbers and lists, however, I can give you the
blueprint for construction. Aboce a line contains as much 1's as its
ordinal number requires.
>
> > But you believe that each one has its own ISD which is completely
> > defined?
>
> You're confusing religion with math.
Not me. I dont believe that there is a complete ISD, i.e. an ISD which
belongs to one and only one number and which can be recognized by its
digits at finite positions. That means, there are digits (at finite
positions) which allow to identify the ISD but, alas, these digits
cannot be found at finite positions. That is religion.
>
> > How can you say that every ISD can be written if it is impossible to
> > define every number and hence every ISD? Do you write those ISDs which
> > haven't any definition with closed eyes?
>
> To wright out an infinite sequence requires infinite ink.
> Do a search for "infinite ink".
The problem is not the ink alone. The main problem is your logic. You
say that digits at finite positions are sufficient to identify the
number but you cannot determine any finite position where these digits
belong to.
I say: For every finite position, there is a ISD which is exactly
identical to the ISD of a certain number, but defining a different
number.
Therefore there is no ISD for any number but there are only finite
approximations for every number. The set of finite approximations
however is countable.
Regards, WM
WM
wrote:
> > If pi has an ISD, then give it please. Certainly it will differ from
> > every term of the Cauchy sequence. Certainly it will be different from
> > any of its finite initial segments like 31415/10000. That means,
> > whatever you write: It is not the ISD of pi.
>
> 1/3 has an ISD.
No. 1/3 has a Cauchy sequence, in fact many Cauchy sequences. It has
not an ISD. But contrary to the transzendental number 5.75376... we
know how to determine every term of the sequence 0.3, 0.33, 0.333, ...
So it apperas as if 1/3 would have an ISD.
> While I cannot write out the full decimal expansion of 1/3,
> I can provide an algorithm to specify exactly the nth digit of the ISD of
> 1/3 - in fact for all n, the nth digit is 3.
Yes, that is true. But this is true only for a countable set of
numbers.
>
> Similarly, I can give you an algorithm to find the unique nth digit of the
> ISD of pi for any n.
True.
>
> Do you accept that 1/3 has an ISD, even if I can't write out 0.333... in
> full in a finite amount of time?
No.
Look here for an easy example: Consider the unit fractions 1/1, 1/2,
1/3, ...
Do you think that 0 can be written as a unit fraction?
That is the same case! You can write every approximation by means of
unit fractions. But you cannot write 0 as such.
it sounds silly, but even 0 in decimal has no ISD. But if we define,
that every missing didgit should be zero, then we can write 0,000...
in an unambiguous way.
That is possible for a countable set of numbers.
Have you figured out what transcendental number I refered to above?
Neither do I.
>
> So why don't you accept that pi has an ISD?
Because all you can do is to determine any finite approximation.
Regards, WM
WM
> This technique (formalising some infinite list as a function from N to
> the enumerated items) is fairly standard, and /noticably absent/ from
> WM's treatment of lists of paths. Given how fundamentally related it is
> to the concept of countably infinite sets, that's a telling omission.
Consider the set of unit fractions 1/1, 1/2, 1/3, ...
Would you say that 0 can be written as a unit fraction? That is as
impossible as writing 1/3 in binary or decimal.
Regards, WM
WM
<webbfam...@DIESPAMDIEoptusnet.com.au>
> Reals, and pi in particular, can be defined in lots of other ways as well.
Yes. Continued fractions, Cauch-sequences, series, ...
>
> I think his underlying argument is one about using decimal expansion, and
> not about the property of pi at all. It is true that you cannot show pi as a
> finite decimal, but you can't show 1/3 as a finite decimal either. And this
> is purely an argument about decimal notation, because in other ways of
> specifying Reals (such as pi) the problem doesn't exist. But he thinks that
> a property of decimal notation is a property of the Real itself;
No. But it is a property required when using Cantor's list in its
common form.
Regards, WM
Peter Webb
> On 19 Dez., 09:24, Owen Jacobson <angrybald...@gmail.com> wrote:
>
>> This technique (formalising some infinite list as a function from N to
>> the enumerated items) is fairly standard, and /noticably absent/ from
>> WM's treatment of lists of paths. Given how fundamentally related it is
>> to the concept of countably infinite sets, that's a telling omission.
>
> Consider the set of unit fractions 1/1, 1/2, 1/3, ...
>
> Would you say that 0 can be written as a unit fraction?
No.
> That is as
> impossible as writing 1/3 in binary or decimal.
>
They aren't the same thing at all.
But so what?
> Regards, WM
Peter Webb
> On 19 Dez., 01:51, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
> wrote:
>> > If pi has an ISD, then give it please. Certainly it will differ from
>> > every term of the Cauchy sequence. Certainly it will be different from
>> > any of its finite initial segments like 31415/10000. That means,
>> > whatever you write: It is not the ISD of pi.
>>
>> 1/3 has an ISD.
>
> No. 1/3 has a Cauchy sequence, in fact many Cauchy sequences. It has
> not an ISD. But contrary to the transzendental number 5.75376... we
> know how to determine every term of the sequence 0.3, 0.33, 0.333, ...
> So it apperas as if 1/3 would have an ISD.
>
So does 1/3 have an ISD or not?
WM
wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
The series
0*10^-1 + 1*10^-2 + 0*10^-3 + 1*10^-4 + ...
converges (to 1/99) because the powers of 10 force it to do so.
The sequence of digits
0, 1, 0, 1, 0, 1, ...
does not converge.
You cannot estimate from seing this sequence without its last term
(which does not exist) what number it determines.
Regards, WM
A Nony Mouse
<af0a087b-7220-44fd...@a21g2000yqc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 18 Dez., 16:10, William Elliot <ma...@rdrop.remove.com> wrote:
> > On Fri, 18 Dec 2009, WM wrote:
> > > On 18 Dez., 09:47, William Elliot <ma...@rdrop.remove.com> wrote:
> > >> On Thu, 17 Dec 2009, WM wrote:
> > >>> Would the result change if there were only few
> > >>> irrational numbers?
> >
> > >> That's an impossibility.
> >
> > > Consider the following Cantor list
> >
> > > 0.0
> > > 0.1
> > > 0.11
> > > 0.111
> > > ...
> >
> > What's the rest of the list?
> >
> > > and let 0 be exchanged by 1. The diagonal is 0.111... Does this proof
> > > show that the set of periodic IDSs is uncountable?
> >
> > First finish the list.
>
> Such lists cannot be finished. That is the cardinal error of Cantor.
A list CAN be "finished" by producing a first member and then creating
an algorithm for producing the successor of any given member.
That very method creates the list N.
> For a countable set of numbers and lists, however, I can give you the
> blueprint for construction. Aboce a line contains as much 1's as its
> ordinal number requires.
Even with the misspelling corrected, that last sentence is nonsense.
> >
> > > But you believe that each one has its own ISD which is completely
> > > defined?
> >
> > You're confusing religion with math.
>
> Not me.
Yes you!
> I dont believe that there is a complete ISD, i.e. an ISD which
> belongs to one and only one number and which can be recognized by its
> digits at finite positions.
WM is assuming, wrongly, that a number can be defined ONLY by decimal
(or other base) expansions, and cannot exist otherwise.
He again is conflating numbers with numerals, which are merely names for
numbers, but are not numbers themselves.
The (positive) square root of two, for example, is not definable that
way but is nevertheless quite adequately defined.
Virgil
<f699d159-34a1-4c95...@h9g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
>
> No. 1/3 has a Cauchy sequence, in fact many Cauchy sequences. It has
> not an ISD. But contrary to the transzendental number 5.75376... we
> know how to determine every term of the sequence 0.3, 0.33, 0.333, ...
> So it apperas as if 1/3 would have an ISD.
It is quite possible for a number to be perfectly well defined and exist
as a number without its necessarily having any particular form of
numerical representation. For example the positive square root of two
has no decimal representataion, but evertheless exists.
A number is not merely a numeral. Various forms of numerals are merely
different form of names for numbers, and there is not even any inherent
reason that a number must have a name in order to exist. The existence f
such a name only indicates our awareness of that particular number.
Numbers which do not have any form of name are, of course, hard to talk
about, and inaccessible, but their inaccessibility does not necessarily
imply their non-existence.
Except in Wolkenmuekenheim.
WM
> > I dont believe that there is a complete ISD, i.e. an ISD which
> > belongs to one and only one number and which can be recognized by its
> > digits at finite positions.
>
> WM is assuming, wrongly, that a number can be defined ONLY by decimal
> (or other base) expansions, and cannot exist otherwise.
You are wrong. A number can be defined by many means. But what Cantor
needs in his diagonal argument is not the number but the sequence of
digits.
> The (positive) square root of two,
the square root of two is positive anyhow
> for example, is not definable that
> way but is nevertheless quite adequately defined
it cannot be defined by a sequence of digits. It cannot be an infinite
path in the binary tree. It cannot enter a Cantor list, and, that's
overlooked usually, it cannot result from a Cantor-list as the
antidiagonal. The antidiagonal of a Cantor-list is usually a diverging
sequence. It has nothing to do with a real number and, therefore, it
does not show anything about the cardinality of real numbers.
Regards, WM
WM
wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
No. You see it best, when looking for the ISD of 1/99:
0.010101...
Regards, WM
WM
> In article
> <f699d159-34a1-4c95-b673-9d77cd1f8...@h9g2000yqa.googlegroups.com>,
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > No. 1/3 has a Cauchy sequence, in fact many Cauchy sequences. It has
> > not an ISD. But contrary to the transzendental number 5.75376... we
> > know how to determine every term of the sequence 0.3, 0.33, 0.333, ...
> > So it apperas as if 1/3 would have an ISD.
>
> It is quite possible for a number to be perfectly well defined and exist
> as a number without its necessarily having any particular form of
> numerical representation. For example the positive square root of two
Yes.
>
> A number is not merely a numeral. Various forms of numerals are merely
> different form of names for numbers, and there is not even any inherent
> reason that a number must have a name in order to exist. The existence of
> such a name only indicates our awareness of that particular number.
Yes.
>
> Numbers which do not have any form of name are, of course, hard to talk
> about, and inaccessible, but their inaccessibility does not necessarily
> imply their non-existence.
But the lack of a sequential representation prevents a number from
being the result of a sequential operation like becoming the anti
diagonal of a Cantor list.
Regards, WM
Virgil
<a44fc25e-30ef-4df3...@d20g2000yqh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
That a particular number may not have a name of a certain form does not
mean that the number itself cannot exist.
WM concedes the existence of the number named by 1/3 as a number by
including its name among the unit fractions.
Peter Webb
On 19 Dez., 15:17, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>
> news:a44fc25e-30ef-4df3...@d20g2000yqh.googlegroups.com...
>
> > On 19 Dez., 09:24, Owen Jacobson <angrybald...@gmail.com> wrote:
>
> >> This technique (formalising some infinite list as a function from N to
> >> the enumerated items) is fairly standard, and /noticably absent/ from
> >> WM's treatment of lists of paths. Given how fundamentally related it is
> >> to the concept of countably infinite sets, that's a telling omission.
>
> > Consider the set of unit fractions 1/1, 1/2, 1/3, ...
>
> > Would you say that 0 can be written as a unit fraction?
>
> No.
>
> > That is as
> > impossible as writing 1/3 in binary or decimal.
>
> They aren't the same thing at all.
>
> But so what?
The series
0*10^-1 + 1*10^-2 + 0*10^-3 + 1*10^-4 + ...
converges (to 1/99) because the powers of 10 force it to do so.
___________________________________
Well, the series converges, for a number of reasons.
The sequence of digits
0, 1, 0, 1, 0, 1, ...
does not converge.
_______________________________
If you say so. Without knowing what your "..." is supposed to mean, it is
impossible to say.
You cannot estimate from seing this sequence without its last term
(which does not exist) what number it determines.
_________________________
If you are saying you cannot determine if a limit exists or what it is, that
is perfectly correct, but its only because you haven't specified the
sequence. If 0,1,0,1,0,1... is supposed to mean the pattern continues
0,1,0,1,0,1,0,1 ... then there is no limit.
But so what?
Peter Webb
So 1/3 doesn't have an ISD. So why not develop your argument using 1/3
instead of pi ?
Are there any numbers at all with infinite ISDs, according to you?
Virgil
<1b036f17-4246-4498...@h9g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Cantor's original list, of functions from N to {m,w}, does not require
any such function to correspond to any number at all.
But the proof of the incompleteness of any such list of functions can
easily be modified to apply to any alleged complete listing of any
non-trivial interval of real numbers in any base from 2 on up.
Peter Webb
But the lack of a sequential representation prevents a number from
being the result of a sequential operation like becoming the anti
diagonal of a Cantor list.
_____________________________________
If you believe there are Reals that cannot be represented as decimals, then
Cantors list which contains only decimals cannot include all Reals, so the
proof is trivial.
Unfortunately it has been known and proved for at least 200 years that all
Reals have a decimal representation, so you cannot prove Cantor's theorem
this simply. That is why Cantor had to use his digit substitution proof.
Virgil
<7bc92828-8e04-4729...@26g2000yqo.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
If its finite initial subsequences have understood numerical values
which converge to a number in the reals, how does it differ from
0*10^-1 + 1*10^-2 + 0*10^-3 + 1*10^-4 + ... ?
Virgil
<bcd13a35-85d7-40d5...@m3g2000yqf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 19 Dez., 21:55, A Nony Mouse <T...@mouse.hole> wrote:
>
> > > I dont believe that there is a complete ISD, i.e. an ISD which
> > > belongs to one and only one number and which can be recognized by its
> > > digits at finite positions.
> >
> > WM is assuming, wrongly, that a number can be defined ONLY by decimal
> > (or other base) expansions, and cannot exist otherwise.
>
> You are wrong. A number can be defined by many means. But what Cantor
> needs in his diagonal argument is not the number but the sequence of
> digits.
Cantor's original diagonal argument did not even need that. Cantor
proved that for ANY list of functions from N to {m,w}, or any other set
of more than one element, there were other such function not appearing
in that list.
>
> > The (positive) square root of two,
>
> the square root of two is positive anyhow
There are actually two square roots for any positive real,
as x^2 = (-x)^2 > 0, for any non-zero real x number, and one of them is
definitely NOT postive.
> > for example, is not definable that
> > way but is nevertheless quite adequately defined
>
> it cannot be defined by a sequence of digits. It cannot be an infinite
> path in the binary tree. It cannot enter a Cantor list, and, that's
> overlooked usually, it cannot result from a Cantor-list as the
> antidiagonal. The antidiagonal of a Cantor-list is usually a diverging
> sequence.
If the members of the list are bounded by a converging seqeunce, as they
usually are represented to be, then every one of the uncountably many
possible anti-diagonals is also convergent.
In all such sequences, the maximal potential value of each term is no
more than 1/2 that of the maximal potential value of the previous term,
so that all such seqeunces, including all so called anti-diagonals,
converge.
WM is losing even the faint touch with reality that he once had.
Virgil
<8161a31f-c8aa-4bf6...@m16g2000yqc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Thus, according to WM, 1/3 is NAN (not a number).
Virgil
<eea43479-617b-4a4a...@g7g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
For lists of binary sequences, the only way to exclude the binary
sequence for 1/3 from ever being an anti-diagonal is to require that it
be a member of EVERY such list.
and it is fairly easy to create a list of binaries for which the binary
sequence for 1/3 IS the Cantor anti-diagonal.
At least if one is a bit more competent than WM.
Virgil
<eea43479-617b-4a4a...@g7g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
That WM does not know what a sequence is does not prevent it from
existing.
Owen Jacobson
<webbf...@DIESPAMDIEoptusnet.com.au> said:
>
> "WM" <muec...@rz.fh-augsburg.de> wrote in message
> news:8161a31f-c8aa-4bf6...@m16g2000yqc.googlegroups.com...
On
>
>> 19 Dez., 17:16, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
>> wrote:
>>> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>>>
>>> news:f699d159-34a1-4c95...@h9g2000yqa.googlegroups.com...
>>>
>>> > On 19 Dez., 01:51, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
>>> > wrote:
>>> >> > If pi has an ISD, then give it please. Certainly it will differ from
>>> >> > every term of the Cauchy sequence. Certainly it will be different
>>> >> > from
>>> >> > any of its finite initial segments like 31415/10000. That means,
>>> >> > whatever you write: It is not the ISD of pi.
>>>
>>> >> 1/3 has an ISD.
>>>
>>> > No. 1/3 has a Cauchy sequence, in fact many Cauchy sequences. It has
>>> > not an ISD. But contrary to the transzendental number 5.75376... we
>>> > know how to determine every term of the sequence 0.3, 0.33, 0.333, ...
>>> > So it apperas as if 1/3 would have an ISD.
>>>
>>> So does 1/3 have an ISD or not?
>>
>> No. You see it best, when looking for the ISD of 1/99:
>> 0.010101...
>
> So 1/3 doesn't have an ISD. So why not develop your argument using 1/3
> instead of pi ?
>
> Are there any numbers at all with infinite ISDs, according to you?
The risks of using a crank's own terminology instead of the accepted
nomenclature: you just asked about the existence of "infinite infinite
sequences of digits".
-o
Peter Webb
Lets be fair here. Real cranks use accepted nomenclemature but with
different meanings. WM has defined his own term "ISD", which is perfectly
legitimate and happens all the time.
However he does have to define it exactly, which he has not yet done. Based
upon him excluding 1/3 from having an ISD, I have a horrible suspicion that
only those rational fractions with finite decimal expansions are ISDs, and
ISDs would be better named FSDs, for Finite Strings of Digits. In other
words the term is defined to mean the opposite of what you would think from
its name.
Hence my question "Are there any numbers at all with infinite ISDs?".
Hopefully he will tell us, and we will have a better idea of what an ISD is
supposed to mean.
Or, more likely, he will wander off on some tangent, or not reply at all.
William Elliot
> Such lists cannot be finished. That is the cardinal error of Cantor.
You're in the extreme minority of finite constructionism.
Though constructionism is one of the philosophical notions
of math, finite constructionism doesn't apply. I suggest
you switch to computeratics which is limited to finite
thoughts that fit well with finite constructionism.
WM
> In article
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 19 Dez., 09:24, Owen Jacobson <angrybald...@gmail.com> wrote:
>
> > > This technique (formalising some infinite list as a function from N to
> > > the enumerated items) is fairly standard, and /noticably absent/ from
> > > WM's treatment of lists of paths. Given how fundamentally related it is
> > > to the concept of countably infinite sets, that's a telling omission.
>
> > Consider the set of unit fractions 1/1, 1/2, 1/3, ...
>
> > Would you say that 0 can be written as a unit fraction? That is as
> > impossible as writing 1/3 in binary or decimal.
>
> That a particular number may not have a name of a certain form does not
> mean that the number itself cannot exist.
Of course not.
>
> WM concedes the existence of the number named by 1/3 as a number by
> including its name among the unit fractions.
I concede the existence of the number 1.
I see that there are many ways to express this number, for instance by
the *finite* formule "1.000...". But I do not see a way to express it
by an infinite sequence of zeros following the 1. Without knowing
beforhand, that there will follow only zeros you cannot be sure that
there will follow only zeros.
That means: A finite formula defines an infinite sequence of digits.
An infinite sequence of digits does not define a finite formula.
Regards, WM
WM
<webbfam...@DIESPAMDIEoptusnet.com.au>
> The sequence of digits
> 0, 1, 0, 1, 0, 1, ...
> does not converge.
>
> _______________________________
> If you say so. Without knowing what your "..." is supposed to mean, it is
> impossible to say.
It is meaning that for every finite initial segment of this sequence
ones and zeros are following one another.
>
> You cannot estimate from seing this sequence without its last term
> (which does not exist) what number it determines.
>
> _________________________
> If you are saying you cannot determine if a limit exists or what it is, that
> is perfectly correct, but its only because you haven't specified the
> sequence. If 0,1,0,1,0,1... is supposed to mean the pattern continues
> 0,1,0,1,0,1,0,1 ... then there is no limit.
>
> But so what?
That is important for the following reason: There are only countably
many ways to specify a sequence. Unspecified sequences are said to be
more frequent. But they are not capable of specifying numbers.
Regards, WM
WM
An ISD is infinite sequence of digits representing a special number
like pi or 1/3. As it is impossible to obtain, without knowíng the
rule of the sequence, how it will further develop (and as it is
impossible to check its end), it is impossible to obatin a number from
an ISD. In fact, there is no ISD. What exists is the infinite set of
finite initial segments like
0.3
0.33
0.333
...
such that for every number z that is not representing 1/3 there is a
digit where z differs from a better approximation of 1/3. But it is
meaningless to talk about the ISD of 1/3, i.e. a sequence that
represents 1/3 and only 1/3 exactly.
Regards, WM
Herbert Newman
> I suggest you switch to computeratics ...
A psychiatric approach might also be appropriate.
Herb
WM
wrote:
> But the lack of a sequential representation prevents a number from
> being the result of a sequential operation like becoming the anti
> diagonal of a Cantor list.
>
> _____________________________________
> If you believe there are Reals that cannot be represented as decimals, then
> Cantors list which contains only decimals cannot include all Reals, so the
> proof is trivial.
Every real can be represented by a Cauchy-sequence. Every real can be
represented by a sequence of approximations. But there is simply no
sequence of digits that alone, i.e., without knowing its generating
rule, can represent a real number.
How long would it take you to obtain from a very long seqeunce of
zeros that the sequence consists of zeors only? Do you think that
there is a digit at some finitely indexed place where you could
decide: All digits are zero? No. You can never be sure.
It is not my personal belief that reals cannot be represented as
decimals. It is objectively the case.
>
> Unfortunately it has been known and proved for at least 200 years that all
> Reals have a decimal representation,
Unfortunately this has not been proved but is an axiom of Cantor-
Dedekind: Every point on the real line can be represented as a real
number.
More unfortunately this is wrong, as just proved. No sequence of
digits will be sufficient to determine a number. Only finite
definitions are sufficient. But there are only countably many.
In general a mathematician has no problems with: A ==> B does not impy
B ==> A.
(A law determines a sequence but a sequence does not determine a law.)
Regards, WM
Peter Webb
An ISD is infinite sequence of digits representing a special number
like pi or 1/3. As it is impossible to obtain, without know�ng the
rule of the sequence, how it will further develop (and as it is
impossible to check its end), it is impossible to obatin a number from
an ISD. In fact, there is no ISD.
__________________________________
"There is no ISD". Thank God that's cleared up. You are talking about the
empty set. That explains a lot.
WM
wrote:
> <SNIP>
>
> An ISD is infinite sequence of digits representing a special number
> rule of the sequence, how it will further develop (and as it is
> an ISD. In fact, there is no ISD.
> __________________________________
> "There is no ISD". Thank God that's cleared up. You are talking about the
> empty set. That explains a lot.
There is no ISD. There are only finite rules.
But in what form do uncountably many real numbers exist?
Regards, WM
WM
> On Sat, 19 Dec 2009, WM wrote:
> > Such lists cannot be finished. That is the cardinal error of Cantor.
>
> You're in the extreme minority of finite constructionism.
No, you have misunderstood me. I do not object to infinite sequences.
I only asked whether one can obtain from an infinite sequence, when
looking at all digits at finite places only, what numbers they
represent.
The answer is: a number cannot be obtained. (That is also the argument
why the construction of all finite paths of the binary tree is said to
miss the construction of numbers like 1/3.)
And there are no digits at infinite places.
Therefore there is no infinite sequence of digits that unambiguously
identifies a real number, unless its finite rule of construction is
known.
Regards, WM
Virgil
<6c684586-a378-4a00...@n35g2000yqm.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 20 Dez., 02:24, Virgil <Vir...@home.esc> wrote:
> > In article
> > <a44fc25e-30ef-4df3-b47e-76b4f2c66...@d20g2000yqh.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 19 Dez., 09:24, Owen Jacobson <angrybald...@gmail.com> wrote:
> >
> > > > This technique (formalising some infinite list as a function from N to
> > > > the enumerated items) is fairly standard, and /noticably absent/ from
> > > > WM's treatment of lists of paths. Given how fundamentally related it is
> > > > to the concept of countably infinite sets, that's a telling omission.
> >
> > > Consider the set of unit fractions 1/1, 1/2, 1/3, ...
> >
> > > Would you say that 0 can be written as a unit fraction? That is as
> > > impossible �as writing 1/3 in binary or decimal.
> >
> > That a particular number may not have a name of a certain form does not
> > mean that the number itself cannot exist.
>
> Of course not.
Then why do you argue to the contrary?
> >
> > WM concedes the existence of the number named by 1/3 as a number by
> > including its name among the unit fractions.
>
> I concede the existence of the number 1.
Reluctantly, I'm sure.
>
> I see that there are many ways to express this number, for instance by
> the *finite* formule "1.000...". But I do not see a way to express it
> by an infinite sequence of zeros following the 1. Without knowing
> beforhand, that there will follow only zeros you cannot be sure that
> there will follow only zeros.
Do you mean that you must know something before you can learn it?
Virgil
<3e126ac8-4d5c-456f...@j19g2000yqk.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
That, at most, merely limits the number of numerals, not the number of
numbers.
Virgil
<178a307b-f538-4640...@v30g2000yqm.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> like pi or 1/3. As it is impossible to obtain, without know�ng the
> rule of the sequence, how it will further develop (and as it is
> impossible to check its end), it is impossible to obatin a number from
> an ISD. In fact, there is no ISD. What exists is the infinite set of
> finite initial segments like
> 0.3
> 0.33
> 0.333
> ...
> such that for every number z that is not representing 1/3 there is a
> digit where z differs from a better approximation of 1/3. But it is
> meaningless to talk about the ISD of 1/3, i.e. a sequence that
> represents 1/3 and only 1/3 exactly.
If there can be an infinite set containing each of those infinitely many
finite sequences of digits, what prevents an infinite sequence of digits
from existing?
WM seems to accept some infinite sequences but not others even though
both can be equally well and equally finitely defined.
Virgil
<9ed20e50-406e-464c...@26g2000yqo.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 20 Dez., 02:34, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
> wrote:
> > But the lack of a sequential representation prevents a number from
> > being the result of a sequential operation like becoming the anti
> > diagonal of a Cantor list.
> >
> > _____________________________________
> > If you believe there are Reals that cannot be represented as decimals, then
> > Cantors list which contains only decimals cannot include all Reals, so the
> > proof is trivial.
>
> Every real can be represented by a Cauchy-sequence. Every real can be
> represented by a sequence of approximations. But there is simply no
> sequence of digits that alone, i.e., without knowing its generating
> rule, can represent a real number.
One such rule, at least for non-negatives is that each next digit is the
largest that does not overstep the number itself.
>
> How long would it take you to obtain from a very long seqeunce of
> zeros that the sequence consists of zeors only?
I can tell right away that it does not contain a single one of those
ZEORS [sic].
> > Unfortunately it has been known and proved for at least 200 years that all
> > Reals have a decimal representation,
>
> No sequence of digits will be sufficient to determine a number.
False!
> In general a mathematician has no problems with: A ==> B does not impy
> B ==> A.
> (A law determines a sequence but a sequence does not determine a law.)
Only in the sense that the law-to-sequence correspondence is
many-to-one, so does not have an inverse.
Virgil
<3a1994c6-3c2d-4cb8...@v25g2000yqk.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 20 Dez., 16:19, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
> wrote:
> > <SNIP>
> >
> > An ISD is infinite sequence of digits representing a special number
> > like pi or 1/3. As it is impossible to obtain, without know�ng the
> > rule of the sequence, how it will further develop (and as it is
> > impossible to check its end), it is impossible to obtain a number from
> > an ISD. In fact, there is no ISD.
> > __________________________________
> > "There is no ISD". Thank God that's cleared up. You are talking about the
> > empty set. That explains a lot.
>
> There is no ISD. There are only finite rules.
> But in what form do uncountably many real numbers exist?
WM is actually asking about numerals, not about numbers.
And there is nothing in the structure of the system of real numbers that
requires any real numbers, with the possible exception of 0 and 1, to be
represented by numerals.
Virgil
<b1e29cb3-0c27-48f3...@d21g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 20 Dez., 09:11, William Elliot <ma...@rdrop.remove.com> wrote:
> > On Sat, 19 Dec 2009, WM wrote:
> > > Such lists cannot be finished. That is the cardinal error of Cantor.
> >
> > You're in the extreme minority of finite constructionism.
>
>
> No, you have misunderstood me. I do not object to infinite sequences.
WM has been objecting vociferously to anything being infinite for years.
> I only asked whether one can obtain from an infinite sequence, when
> looking at all digits at finite places only, what numbers they
> represent.
Which digits in an infinite sequence of them does WM suggest do not
occur at finite places?
>
> The answer is: a number cannot be obtained. (That is also the argument
> why the construction of all finite paths of the binary tree is said to
> miss the construction of numbers like 1/3.)
There is much of mathematics that WM cannot obtain, but not everyone is
so handicapped.
>
> And there are no digits at infinite places.
No one claims there are any "infinite places" in an infinite sequence,
of digits, or anything else, except possibly WM himself.
WM
> In article
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 20 Dez., 02:24, Virgil <Vir...@home.esc> wrote:
> > > In article
> > > <a44fc25e-30ef-4df3-b47e-76b4f2c66...@d20g2000yqh.googlegroups.com>,
>
> > > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > On 19 Dez., 09:24, Owen Jacobson <angrybald...@gmail.com> wrote:
>
> > > > > This technique (formalising some infinite list as a function from N to
> > > > > the enumerated items) is fairly standard, and /noticably absent/ from
> > > > > WM's treatment of lists of paths. Given how fundamentally related it is
> > > > > to the concept of countably infinite sets, that's a telling omission.
>
> > > > Consider the set of unit fractions 1/1, 1/2, 1/3, ...
>
> > > > Would you say that 0 can be written as a unit fraction? That is as
> > > > impossible as writing 1/3 in binary or decimal.
>
> > > That a particular number may not have a name of a certain form does not
> > > mean that the number itself cannot exist.
>
> > Of course not.
>
> Then why do you argue to the contrary?
I do not. I only recognized that the number cannot be learned from an
infinite sequence of digits.
>
> > I see that there are many ways to express this number, for instance by
> > the *finite* formule "1.000...". But I do not see a way to express it
> > by an infinite sequence of zeros following the 1. Without knowing
> > beforhand, that there will follow only zeros you cannot be sure that
> > there will follow only zeros.
>
> Do you mean that you must know something before you can learn it?
Yes, if infinite sequences are concerned. You cannot learn the meaning
of an infinite sequence from its digits.
Regards, WM
WM
> > That is important for the following reason: There are only countably
> > many ways to specify a sequence. Unspecified sequences are said to be
> > more frequent. But they are not capable of specifying numbers.
>
> That, at most, merely limits the number of numerals, not the number of
> numbers
Cantors argument does not concern numbers but the meaning of infinite
sequences.
Regards, WM
Nam Nguyen
>
> cannot
> distinguish them [numbers] from numerals, which are merely names for numbers, and
> not the numbers themselves.
Well, that isn't true about the naturals. Each natural number is represented
exactly by 1 numeral and vice versa, and there's no natural number than can't
be represented by a numeral. In such case it's impossible to distinguish
between a number and its numeral, other than the fact that a numeral is
a syntactical object while a number isn't.
Virgil
<53ee8f23-1552-4e44...@g7g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
There are all sorts of numbers that can be learned from all sorts of
infinite sequences, by those with the wits to learn them.
> >
>
> > > I see that there are many ways to express this number, for instance by
> > > the *finite* formule "1.000...". But I do not see a way to express it
> > > by an infinite sequence of zeros following the 1. Without knowing
> > > beforhand, that there will follow only zeros you cannot be sure that
> > > there will follow only zeros.
> >
> > Do you mean that you must know something before you can learn it?
>
> Yes, if infinite sequences are concerned. You cannot learn the meaning
> of an infinite sequence from its digits.
Speak only for yourself!
Many of us CAN learn the meaning of infinite sequences of digits from
those digits.
Virgil
<16a5c9a4-37e5-4f3e...@r5g2000yqb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Cantor's arguments have nothing to do with the MEANING of such
sequences, only about how many of them there are.
William Elliot
> On 20 Dez., 09:11, William Elliot <ma...@rdrop.remove.com> wrote:
>> On Sat, 19 Dec 2009, WM wrote:
>>> Such lists cannot be finished. That is the cardinal error of Cantor.
>>
>> You're in the extreme minority of finite constructionism.
>
> No, you have misunderstood me. I do not object to infinite sequences.
> I only asked whether one can obtain from an infinite sequence, when
That sentence is incomplete: can obtain what?
> looking at all digits at finite places only, what numbers they
> represent.
>
All the digits of a sequence are at finite places
unless per chance, you mean transfinite sequences.
> The answer is: a number cannot be obtained. (That is also the argument
> why the construction of all finite paths of the binary tree is said to
> miss the construction of numbers like 1/3.)
>
That's because they're the dyatic rationals from 0 to 1.
> And there are no digits at infinite places.
>
So what?
> Therefore there is no infinite sequence of digits that unambiguously
> identifies a real number, unless its finite rule of construction is
> known.
>
So you're a constructionist. Most mathematicians aren't.
Peter Webb
> On 20 Dez., 09:11, William Elliot <ma...@rdrop.remove.com> wrote:
>> On Sat, 19 Dec 2009, WM wrote:
>> > Such lists cannot be finished. That is the cardinal error of Cantor.
>>
>> You're in the extreme minority of finite constructionism.
>
>
> No, you have misunderstood me. I do not object to infinite sequences.
> I only asked whether one can obtain from an infinite sequence, when
> looking at all digits at finite places only, what numbers they
> represent.
>
> The answer is: a number cannot be obtained.
That is simply wrong.
If two infinite strings (using "infinite" in the same sense as in decimal
expansions) are identical at every finite position, then they are identical.
If you don't believe that, produce a counter-example.
If you specify that at every finite place in the decimal expansion the
numeral is "3", you get 0.333... = 1/3.
If you thing that 0.333.... (ie a "3" at every finite position) potentially
represents a Real other than 1/3, what is it?
> (That is also the argument
> why the construction of all finite paths of the binary tree is said to
> miss the construction of numbers like 1/3.)
>
Does miss 1/3. Its not just said to miss 1/3, it does actually miss 1/3, as
this does not terminate in base 2.
> And there are no digits at infinite places.
>
Not in Reals, no.
> Therefore there is no infinite sequence of digits that unambiguously
> identifies a real number, unless its finite rule of construction is
> known.
>
Your conclusion could not possibly derive from your premise, as you have not
mentioned a "finite rule of construction" in your premise.
It is true that there are Reals which cannot be computed, and these cannot
therefore appear on a list of all Reals. This in a sense is another proof of
Cantor's thereom, as the existence of a non-computable real guarantees that
no list of all Reals can be formed. Cantor didn't know of the existence of
such things when he produced his diagonal proof.
Note that the issue of non-computability does not directly affect Cantor's
argument. The diagonal number can always be computed if given the list of
Reals. (And nor does the Axiom of Choice affect Cantor's argument, as its
not used)
> Regards, WM
I still have not seen a list of the numbers which you consider have ISDs,
other than it does not include either pi or 1/3. Are you just giving up on
the whole idea of ISDs having any value at all as a mathematical concept? If
not, please provide some numbers that do and don't have ISDs.
WM
> In article
Only a fool can claim that.
Regards, WM
WM
> On Sun, 20 Dec 2009, WM wrote:
> > On 20 Dez., 09:11, William Elliot <ma...@rdrop.remove.com> wrote:
> >> On Sat, 19 Dec 2009, WM wrote:
> >>> Such lists cannot be finished. That is the cardinal error of Cantor.
>
> >> You're in the extreme minority of finite constructionism.
>
> > No, you have misunderstood me. I do not object to infinite sequences.
> > I only asked whether one can obtain from an infinite sequence, when
>
> That sentence is incomplete: can obtain what?
>
> > looking at all digits at finite places only, what numbers they
> > represent.
>
> All the digits of a sequence are at finite places
> unless per chance, you mean transfinite sequences.
No, I completely agree. All digits d_n are at finite places, indexed
by n in N.
Nevertheless, for every digit d_n there is a sequence d_1, ..., d_n
that belongs to infinitely many different numbers.
Hence there is no sequence that identifies a certain number.
>
> > The answer is: a number cannot be obtained. (That is also the argument
> > why the construction of all finite paths of the binary tree is said to
> > miss the construction of numbers like 1/3.)
>
> That's because they're the dyatic rationals from 0 to 1.
>
> > And there are no digits at infinite places.
>
> So what?
>
> > Therefore there is no infinite sequence of digits that unambiguously
> > identifies a real number, unless its finite rule of construction is
> > known.
>
> So you're a constructionist. Most mathematicians aren't.
That has nothing to do with constructivism but only with logic:
If there is no digit d_n at a finite place n such that the sequence
d_1, ..., d_n identifies a number, then the value of a number cannot
be obtained from a sequence of digits at finite places.
Regards, WM
WM
wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>
> news:b1e29cb3-0c27-48f3...@d21g2000yqn.googlegroups.com...
>
> > On 20 Dez., 09:11, William Elliot <ma...@rdrop.remove.com> wrote:
> >> On Sat, 19 Dec 2009, WM wrote:
> >> > Such lists cannot be finished. That is the cardinal error of Cantor.
>
> >> You're in the extreme minority of finite constructionism.
>
> > No, you have misunderstood me. I do not object to infinite sequences.
> > I only asked whether one can obtain from an infinite sequence, when
> > looking at all digits at finite places only, what numbers they
> > represent.
>
> > The answer is: a number cannot be obtained.
>
> That is simply wrong.
>
> If two infinite strings (using "infinite" in the same sense as in decimal
> expansions) are identical at every finite position, then they are identical.
> If you don't believe that, produce a counter-example.
What do you understand by "at every finite position"?
I know that at every finite position n you can find a sequence of
digits d_1, ..., d_n which is shared by infinitely many different
numbers. There is no digit d_n that is unique to a certain number.
So your "every" must imply digits d_n at positions n which are not
shared by different numbers. Please produce an example for that case.
>
> If you specify that at every finite place in the decimal expansion the
> numeral is "3", you get 0.333... = 1/3.
Of course, but that means a finite specification.
The sequence alone (without that specification) does never yield that
information (because infinite means never ending).
>
> If you think that 0.333.... (ie a "3" at every finite position) potentially
> represents a Real other than 1/3, what is it?
I do not think so. I talked about a sequence that should give the
information, not e finite rule like your (ie a "3" at every finite
position).
>
> > (That is also the argument
> > why the construction of all finite paths of the binary tree is said to
> > miss the construction of numbers like 1/3.)
>
> Does miss 1/3. Its not just said to miss 1/3, it does actually miss 1/3, as
> this does not terminate in base 2.
But above you said that the path 0.333... (which is simply the union
of all finite paths of threes (in the decimal tree) is 1/3.
Nevertheless, it does not terminate in base 10 either.
>
> > And there are no digits at infinite places.
>
> Not in Reals, no.
>
> > Therefore there is no infinite sequence of digits that unambiguously
> > identifies a real number, unless its finite rule of construction is
> > known.
>
> Your conclusion could not possibly derive from your premise, as you have not
> mentioned a "finite rule of construction" in your premise.
Look here:
Question 2: Is it possible to write an ISD which unambiguously
identifies pi?
This is impossible.
>
> It is true that there are Reals which cannot be computed,
Where are they? Where is "there".
> and these cannot
> therefore appear on a list of all Reals. This in a sense is another proof of
> Cantor's thereom,
No. Cantors theorem uses the diagonal of a list. That can be computed.
> as the existence of a non-computable real guarantees that
> no list of all Reals can be formed. Cantor didn't know of the existence of
> such things when he produced his diagonal proof.
Cantor vehemently attacked that position.
>
> I still have not seen a list of the numbers which you consider have ISDs,
> other than it does not include either pi or 1/3. Are you just giving up on
> the whole idea of ISDs having any value at all as a mathematical concept? If
> not, please provide some numbers that do and don't have ISDs.
There is no ISD. If you think so, then produce a sequence that,
without finite rule, identifies a number.
Regards, WM
WM
<webbfam...@DIESPAMDIEoptusnet.com.au>
> It is true that there are Reals which cannot be computed,
Another question. Do these reals start with digit seqeuences like
0.123? Which is the first digit that cannot be computed?
Regards, WM
Marshall
>
> There is no ISD. If you think so, then produce a sequence that,
> without finite rule, identifies a number.
There is no Asia. If you think there is, please produce it.
By the way, what do you think the definition of "sequence" is?
Note that the definition of "sequence" does not forbid a
"finite rule." That's another one of your own private strictures.
Marshall
Virgil
<f110ab79-94a1-42d3...@k19g2000yqc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Then being a proper mathematician requires one to be a fool.
Which is a better deal than WM has, being so obviously an ass.
Virgil
<c3ffee87-bd32-443c...@b32g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 21 Dez., 08:16, William Elliot <ma...@rdrop.remove.com> wrote:
> > On Sun, 20 Dec 2009, WM wrote:
> > > On 20 Dez., 09:11, William Elliot <ma...@rdrop.remove.com> wrote:
> > >> On Sat, 19 Dec 2009, WM wrote:
> > >>> Such lists cannot be finished. That is the cardinal error of Cantor.
> >
> > >> You're in the extreme minority of finite constructionism.
> >
> > > No, you have misunderstood me. I do not object to infinite sequences.
> > > I only asked whether one can obtain from an infinite sequence, when
> >
> > That sentence is incomplete: �can obtain what?
> >
> > > looking at all digits at finite places only, what numbers they
> > > represent.
> >
> > All the digits of a sequence are at finite places
> > unless per chance, you mean transfinite sequences.
>
> No, I completely agree. All digits d_n are at finite places, indexed
> by n in N.
> Nevertheless, for every digit d_n there is a sequence d_1, ..., d_n
> that belongs to infinitely many different numbers.
According to the Gospel of ST. WM, there cannot be infinitely many
anything.
>
> Hence there is no sequence that identifies a certain number,
In decimal notation, what number does an infinite sequence of 3's
following the decimal point not identify?
> >
> > > The answer is: a number cannot be obtained. (That is also the argument
> > > why the construction of all finite paths of the binary tree is said to
> > > miss the construction of numbers like 1/3.)
> >
> > That's because they're the dyatic rationals from 0 to 1.
> >
> > > And there are no digits at infinite places.
> >
> > So what?
> >
> > > Therefore there is no infinite sequence of digits that unambiguously
> > > identifies a real number, unless its finite rule of construction is
> > > known.
> >
> > So you're a constructionist. �Most mathematicians aren't.
>
> That has nothing to do with constructivism but only with logic
Wrong! Logic does not demand what WM demands.
>
> If there is no digit d_n at a finite place n such that the sequence
> d_1, ..., d_n identifies a number, then the value of a number cannot
> be obtained from a sequence of digits at finite places.
Let f:N -> {0,1} be any function, and let V(f) = sum_[k in N] f(k)/2^k
be a mapping from the set of all such functions, F2, to the set of real
numbers R.
Then EVERY such f defines a real number, though there are a few
instances of two such functions defining the same number. and the range
of V:Fr -> R is [0,1]
Similarly for functions of form f:N -> {0,1,2,...,n-1} and mappings
W(f) = sum_[k in N] f(k)/n^k.
Virgil
<dd73f090-c746-477e...@26g2000yqo.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 21 Dez., 13:43, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> >
> > news:b1e29cb3-0c27-48f3...@d21g2000yqn.googlegroups.com...
> >
> > > On 20 Dez., 09:11, William Elliot <ma...@rdrop.remove.com> wrote:
> > >> On Sat, 19 Dec 2009, WM wrote:
> > >> > Such lists cannot be finished. That is the cardinal error of Cantor.
> >
> > >> You're in the extreme minority of finite constructionism.
> >
> > > No, you have misunderstood me. I do not object to infinite sequences.
> > > I only asked whether one can obtain from an infinite sequence, when
> > > looking at all digits at finite places only, what numbers they
> > > represent.
> >
> > > The answer is: a number cannot be obtained.
> >
> > That is simply wrong.
> >
> > If two infinite strings (using "infinite" in the same sense as in decimal
> > expansions) are identical at every finite position, then they are identical.
> > If you don't believe that, produce a counter-example.
>
> What do you understand by "at every finite position"?
Let f and g be functions from N to some set of at least two elements
such that for all n in N, f(n) = g(n). Then f and g "are identical very
finite position".
> I know that at every finite position n you can find a sequence of
> digits d_1, ..., d_n which is shared by infinitely many different
> numbers. There is no digit d_n that is unique to a certain number.
But there are sequences of digits unique to ost given numbers, which is
all that we care about.
>
> So your "every" must imply digits d_n at positions n which are not
> shared by different numbers. Please produce an example for that case.
Given any two , say, distinct binary numbers between 0 and 1 represented
by binary sequences, their will be a first digit at which their
sequences differ.
> >
> > If you specify that at every finite place in the decimal expansion the
> > numeral is "3", you get 0.333... = 1/3.
>
> Of course, but that means a finite specification.
> The sequence alone (without that specification)
Without that specification, one does not have the sequence.
For any sequence of digits, ALL its digits must be determinate, or it is
not a sequence at all.
> >
> > If you think that 0.333.... (ie a "3" at every finite position) potentially
> > represents a Real other than 1/3, what is it?
>
> I do not think so.
> >
> > > (That is also the argument
> > > why the construction of all finite paths of the binary tree is said to
> > > miss the construction of numbers like 1/3.)
> >
> > Does miss 1/3. Its not just said to miss 1/3, it does actually miss 1/3, as
> > this does not terminate in base 2.
> Look here:
> Question 2: Is it possible to write an ISD which unambiguously
> identifies pi?
> This is impossible.
> >
> > It is true that there are Reals which cannot be computed,
>
> Where are they? Where is "there".
"There" is clearly out of WM's reach, though within the easy reach of
real mathematicians.
...
> > as the existence of a non-computable real guarantees that
> > no list of all Reals can be formed. Cantor didn't know of the existence of
> > such things when he produced his diagonal proof.
>
> Cantor vehemently attacked that position.
Actually, Cantor did know of them before his diagonal proof, since his
first proof of the uncountability of the reals was published in 1874,
which preceded the diagonal proof, not published until 1891.
Virgil
<e27c760c-3c95-4c64...@v25g2000yqk.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
For every positive, but finite, real number epsilon, however large,
there are real numbers whose values are not known to within a range of
epsilon.
Peter Webb
> On 21 Dez., 13:43, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
> wrote:
>> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>>
>> news:b1e29cb3-0c27-48f3...@d21g2000yqn.googlegroups.com...
>>
>> > On 20 Dez., 09:11, William Elliot <ma...@rdrop.remove.com> wrote:
>> >> On Sat, 19 Dec 2009, WM wrote:
>> >> > Such lists cannot be finished. That is the cardinal error of Cantor.
>>
>> >> You're in the extreme minority of finite constructionism.
>>
>> > No, you have misunderstood me. I do not object to infinite sequences.
>> > I only asked whether one can obtain from an infinite sequence, when
>> > looking at all digits at finite places only, what numbers they
>> > represent.
>>
>> > The answer is: a number cannot be obtained.
>>
>> That is simply wrong.
>>
>> If two infinite strings (using "infinite" in the same sense as in decimal
>> expansions) are identical at every finite position, then they are
>> identical.
>> If you don't believe that, produce a counter-example.
>
> What do you understand by "at every finite position"?
I understand that for every position n in the decimal expansion, the decimal
expansions are the same.
As you have already acknowledged, there are no "infinite positions" for a
decimal, therefore all positions in the expansion are finite. Therefore two
sequences being the same at every finite position means they are the same at
every position.
> I know that at every finite position n you can find a sequence of
> digits d_1, ..., d_n which is shared by infinitely many different
> numbers. There is no digit d_n that is unique to a certain number.
Sure. 0.11, 0.111, 0.112 all have d_1 = d_2 =1, but are clearly three
different Reals.
>
> So your "every" must imply digits d_n at positions n which are not
> shared by different numbers. Please produce an example for that case.
No. That does not follow.
>
>>
>> If you specify that at every finite place in the decimal expansion the
>> numeral is "3", you get 0.333... = 1/3.
>
> Of course, but that means a finite specification.
Yes.
> The sequence alone (without that specification) does never yield that
> information (because infinite means never ending).
If you are sclaiming that simply saying something like 0.386882... does not
specify a single Real number, as lots of different Reals start 0.386882 and
the rule for continuing it is not really defined by the "..." bit, then yes
that is correct and I don't think anybody would argue with you.
>>
>> If you think that 0.333.... (ie a "3" at every finite position)
>> potentially
>> represents a Real other than 1/3, what is it?
>
> I do not think so.
So this Real at least is uniquely specified by its decimal expansion.
What applies to 1/3 also exactly applies to pi, for exactly the same reason.
I can give an algorithm for specifying the nth digit of pi for all n. It is
a little more complicated than d_n = 3, but is nevertheless a finite
algorithm that complates in finite time.
> I talked about a sequence that should give the
> information, not e finite rule like your (ie a "3" at every finite
> position).
You waffled on.
>>
>> > (That is also the argument
>> > why the construction of all finite paths of the binary tree is said to
>> > miss the construction of numbers like 1/3.)
>>
>> Does miss 1/3. Its not just said to miss 1/3, it does actually miss 1/3,
>> as
>> this does not terminate in base 2.
>
> But above you said that the path 0.333... (which is simply the union
> of all finite paths of threes (in the decimal tree) is 1/3.
I said 0.333... = 1/3. I said nothing about unions of paths.
> Nevertheless, it does not terminate in base 10 either.
No.
>>
>> > And there are no digits at infinite places.
>>
>> Not in Reals, no.
>>
>> > Therefore there is no infinite sequence of digits that unambiguously
>> > identifies a real number, unless its finite rule of construction is
>> > known.
>>
>> Your conclusion could not possibly derive from your premise, as you have
>> not
>> mentioned a "finite rule of construction" in your premise.
>
> Look here:
> Question 2: Is it possible to write an ISD which unambiguously
> identifies pi?
> This is impossible.
Where do you mention a "finite rule of construction" in that, exactly?
There is of course an ISD for pi, if by the term "ISD" you mean a decimal
expansion. If by ISD you mean a *finite* decimal expansion, then no it
doesn't, any more than 1/3 does. But the news that pi does not have a finite
decimal expansion is probably well known to most readers of this group.
>>
>> It is true that there are Reals which cannot be computed,
>
> Where are they? Where is "there".
Well, if you don't believe they exist, that's fine in tyerms of what you
have already posted and Cantor's proof generally; as I previously stated
Cantor's proof does not assume whether they exist or not.
If you accept that there is even a single uncomputable Real in [0,1] then
its trivial to show that there are an infinite number of uncomptable Reals
between every two different Reals.
>
>> and these cannot
>> therefore appear on a list of all Reals. This in a sense is another proof
>> of
>> Cantor's thereom,
>
> No. Cantors theorem uses the diagonal of a list. That can be computed.
Yes. Cantor's theorem uses the diagonal of a list. That can be computed.
>
>> as the existence of a non-computable real guarantees that
>> no list of all Reals can be formed. Cantor didn't know of the existence
>> of
>> such things when he produced his diagonal proof.
>
> Cantor vehemently attacked that position.
Dunno. But that is an argument about history, not maths.
>
>>
>> I still have not seen a list of the numbers which you consider have ISDs,
>> other than it does not include either pi or 1/3. Are you just giving up
>> on
>> the whole idea of ISDs having any value at all as a mathematical concept?
>> If
>> not, please provide some numbers that do and don't have ISDs.
>
> There is no ISD.
As you have not defined what an ISD is, the statement that they don't exist
is somwwhat difficult to determine.
However, if there are no ISDs, then the set of all ISDs is identically equal
to the set of all Welsh speaking pink Unicorns that can play the violin, and
so some pretty strange properties can be vacuously true.
> If you think so, then produce a sequence that,
> without finite rule, identifies a number.
We all know that I can only write a finite rule, so obviously I can't do
that. But that is confusing two different concepts - whether a Real has a
decimal expansion (and they all do), and whether this can be expressed in a
finite rule (obviously most cannot). If they all could be expressed in a
finite rule, they would have cardinality N, but not all of them can, so get
over it.
Sure, there are uncomputable numbers (ie numbers that cannot be specified
with a finite rule), but we all already know that. And furthermore, Cantor's
proof in no way assumes the existence of uncomputable numbers or their
non-existence; the first inkling that such an category of numbers even
existed was in fact Cantor's proof. So if you think that this stuff has any
bearing on the diagonal proof, you are wrong.
>
> Regards, WM
WM
> I understand that for every position n in the decimal expansion, the decimal
> expansions are the same.
>
> As you have already acknowledged, there are no "infinite positions" for a
> decimal, therefore all positions in the expansion are finite. Therefore two
> sequences being the same at every finite position means they are the same at
> every position.
That means they have the same finite initial segments of digits d_1,
d_2, ..., d_n, and e_1, e_2, ..., e_n for every n in N. But no finite
initial segment is sufficient to prove identity. Therefore all finite
initial segments are not enough to identify a number.
>
> > I know that at every finite position n you can find a sequence of
> > digits d_1, ..., d_n which is shared by infinitely many different
> > numbers. There is no digit d_n that is unique to a certain number.
>
> Sure. 0.11, 0.111, 0.112 all have d_1 = d_2 =1, but are clearly three
> different Reals.
>
>
>
> > So your "every" must imply digits d_n at positions n which are not
> > shared by different numbers. Please produce an example for that case.
>
> No. That does not follow.
It would follow, if digit sequences did actually exist, not only
potentially. For two different numbers d and e, we have two different
digit sequences
d_1, d_2, ..., d_n
and
e_1, e_2, ..., e_n
that must differ at a finite place n. If a digit sequence should
identify a number uniquely, then it must differ at a finite place from
every other number's digit sequence. As that is impossible, there is
no digit sequence that uniquely defines a number.
>
>
>
> >> If you specify that at every finite place in the decimal expansion the
> >> numeral is "3", you get 0.333... = 1/3.
>
> > Of course, but that means a finite specification.
>
> Yes.
And there are only countably man finite specifications.
>
> > The sequence alone (without that specification) does never yield that
> > information (because infinite means never ending).
>
> If you are claiming that simply saying something like 0.386882... does not
> specify a single Real number, as lots of different Reals start 0.386882 and
> the rule for continuing it is not really defined by the "..." bit, then yes
> that is correct and I don't think anybody would argue with you.
The arguing is usually as follows: There is a unique digit sequence
consisting only of finite initial segments of digits and identifying a
real number, but there is not a finite initial segment of digits
identifying a real number. Quantifyer magic at its best - but not
suitable for actually existing infinity.
>
>
>
> >> If you think that 0.333.... (ie a "3" at every finite position)
> >> potentially
> >> represents a Real other than 1/3, what is it?
>
> > I do not think so.
>
> So this Real at least is uniquely specified by its decimal expansion.
No, it specified by a finite rule. You could never obtain the number
1/3 from seeing only the sequence (as this is never finished).
>
> What applies to 1/3 also exactly applies to pi, for exactly the same reason.
> I can give an algorithm for specifying the nth digit of pi for all n. It is
> a little more complicated than d_n = 3, but is nevertheless a finite
> algorithm that complates in finite time.
Yes. A finite algorithm or number specifies an infinite sequence, but
an infinite sequence cannot specify a finite algorithm or number.
>
> > I talked about a sequence that should give the
> > information, not e finite rule like your (ie a "3" at every finite
> > position).
>
> You waffled on.
Is it that hard for you to understand, that A ==> B does not imply B
==> A ?
>
>
>
> >> > (That is also the argument
> >> > why the construction of all finite paths of the binary tree is said to
> >> > miss the construction of numbers like 1/3.)
>
> >> Does miss 1/3. Its not just said to miss 1/3, it does actually miss 1/3,
> >> as
> >> this does not terminate in base 2.
>
> > But above you said that the path 0.333... (which is simply the union
> > of all finite paths of threes (in the decimal tree) is 1/3.
>
> I said 0.333... = 1/3. I said nothing about unions of paths.
In the binary tree we can form the union of paths.
>
> > Nevertheless, it does not terminate in base 10 either.
>
> No.
>
>
>
>
>
>
>
> >> > And there are no digits at infinite places.
>
> >> Not in Reals, no.
>
> >> > Therefore there is no infinite sequence of digits that unambiguously
> >> > identifies a real number, unless its finite rule of construction is
> >> > known.
>
> >> Your conclusion could not possibly derive from your premise, as you have
> >> not
> >> mentioned a "finite rule of construction" in your premise.
>
> > Look here:
> > Question 2: Is it possible to write an ISD which unambiguously
> > identifies pi?
> > This is impossible.
>
> Where do you mention a "finite rule of construction" in that, exactly?
I do *not* mention it. *Therefore* the ISD cannot specify a number.
>
> There is of course an ISD for pi, if by the term "ISD" you mean a decimal
> expansion. If by ISD you mean a *finite* decimal expansion,
No. I mean an infinite (ISD means infinite sequence of digits).
> >> It is true that there are Reals which cannot be computed,
>
> > Where are they? Where is "there".
>
> Well, if you don't believe they exist, that's fine in tyerms of what you
> have already posted and Cantor's proof generally; as I previously stated
> Cantor's proof does not assume whether they exist or not.
Cantor's proof "proves" that such numbers exist.
>
> If you accept that there is even a single uncomputable Real in [0,1] then
> its trivial to show that there are an infinite number of uncomptable Reals
> between every two different Reals.
But there is not even one of them. Numbers have to do with counting.
>
>
>
> >> and these cannot
> >> therefore appear on a list of all Reals. This in a sense is another proof
> >> of
> >> Cantor's thereom,
>
> > No. Cantors theorem uses the diagonal of a list. That can be computed.
>
> Yes. Cantor's theorem uses the diagonal of a list. That can be computed.
>
>
>
> >> as the existence of a non-computable real guarantees that
> >> no list of all Reals can be formed. Cantor didn't know of the existence
> >> of
> >> such things when he produced his diagonal proof.
>
> > Cantor vehemently attacked that position.
>
> Dunno. But that is an argument about history, not maths.
It is an argument about sober common sense.
>
>
>
> >> I still have not seen a list of the numbers which you consider have ISDs,
> >> other than it does not include either pi or 1/3. Are you just giving up
> >> on
> >> the whole idea of ISDs having any value at all as a mathematical concept?
> >> If
> >> not, please provide some numbers that do and don't have ISDs.
>
> > There is no ISD.
>
> As you have not defined what an ISD is, the statement that they don't exist
> is somwwhat difficult to determine.
See above. An ISD is an inbfinite digit sequence that defines (without
a finite rule) a real number.
>
> However, if there are no ISDs, then the set of all ISDs is identically equal
> to the set of all Welsh speaking pink Unicorns that can play the violin, and
> so some pretty strange properties can be vacuously true.
>
> > If you think so, then produce a sequence that,
> > without finite rule, identifies a number.
>
> We all know that I can only write a finite rule, so obviously I can't do
> that. But that is confusing two different concepts - whether a Real has a
> decimal expansion (and they all do), and whether this can be expressed in a
> finite rule (obviously most cannot). If they all could be expressed in a
> finite rule, they would have cardinality N, but not all of them can, so get
> over it.
All numbers that can be expressed belong to a countable set. Every
diagonal of
a definable Cantor-list belongs to a countable set.
>
> Sure, there are uncomputable numbers (ie numbers that cannot be specified
> with a finite rule), but we all already know that. And furthermore, Cantor's
> proof in no way assumes the existence of uncomputable numbers or their
> non-existence; the first inkling that such an category of numbers even
> existed was in fact Cantor's proof.
No, that was simply a misinterpretation. It is based on the assumption
that ISDs exist and define numbers. But as we have seen, they don't.
Regards, WM
Virgil
<a921b657-8678-4b2e...@b2g2000yqi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
WM her claims that whet no single initial segment can do on its own the
set of all of them also cannot do.
But this is trivially false, since any one cannot achieve the very
infiniteness that the set of all of them necessitates.
>
> All numbers that can be expressed belong to a countable set.
Only in Wolkenmuekenheim
> Every diagonal of a definable Cantor-list belongs to a countable set.
In English, that does not require that they all belong to the SAME
countable set, which is something that they do not do outside of
Wolkenmuekenheim.
> >
> > Sure, there are uncomputable numbers (ie numbers that cannot be specified
> > with a finite rule), but we all already know that. And furthermore, Cantor's
> > proof in no way assumes the existence of uncomputable numbers or their
> > non-existence; the first inkling that such an category of numbers even
> > existed was in fact Cantor's proof.
>
> No, that was simply a misinterpretation.
On the contrary, WM's is the misinterpretation. Cantor's proofs showed
for the first time that there are more real numbers than there can be
names for them.
> It is based on the assumption
> that ISDs exist and define numbers.
Any function from N to, say, {0,1} defines a number between 0 and 1
inclusive by a standard construction, the limit of an absolutely
convergent sequence partial sums, and there are too many such functions
defining different numbers to be counted by N.
But as we have seen, they don't.
WM consistently manages to see all sort of thing which are not there
while simultaneously managing not to see many things which are there.
WM
> > > as the existence of a non-computable real guarantees that
> > > no list of all Reals can be formed. Cantor didn't know of the existence of
> > > such things when he produced his diagonal proof.
>
> > Cantor vehemently attacked that position.
>
> Actually, Cantor did know of them before his diagonal proof, since his
> first proof of the uncountability of the reals was published in 1874,
> which preceded the diagonal proof, not published until 1891
Cantor knew that believing in uncomputable numbers requires one to be
a fool of matheology. He did not share this belief. He wrote (in 1906
to Hilbert): „Unendliche Definitionen" (die nicht in endlicher Zeit
verlaufen) sind Undinge.
Waere Koenigs Satz, daß alle „endlich definirbaren" reellen Zahlen
einen Inbegriff von der Maechtigkeit aleph_0 ausmachen, richtig, so
hieße dies, das ganze Zahlencontinuum sei abzaehlbar, was doch
sicherlich falsch ist.
Es fragt sich nun, welcher Irrthum liegt dem angeblichen Beweise
seines falschen Satzes zu Grunde?
If you wanna be a fool of matheology, then you should at least try to
understand German because there this foolishness took its origin (but
do not blame it to Cantor - he refused.).
Regards, WM
Virgil
<a24573d1-ab36-4b6d...@r24g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 22 Dez., 02:08, Virgil <Vir...@home.esc> wrote:
>
> > > > as the existence of a non-computable real guarantees that
> > > > no list of all Reals can be formed. Cantor didn't know of the existence
> > > > of
> > > > such things when he produced his diagonal proof.
> >
> > > Cantor vehemently attacked that position.
> >
> > Actually, Cantor did know of them before his diagonal proof, since his
> > first proof of the uncountability of the reals was published in 1874,
> > which preceded the diagonal proof, not published until 1891
>
> Cantor knew that believing in uncomputable numbers requires one to be
> a fool of matheology.
Unless WM can cite Cantor referring to "matheology", WM is a liar.