Author 
Message 
erc science forum Guru Wannabe
Joined: 25 Jul 2005
Posts: 269

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: sketch of PROOF OF COUNTABLE REALS



<rupertmccallum@yahoo.com> wrote in
Quote:  I thought you were putting forth the list of computable reals as your
countable list of all real numbers. What list are you putting forward?
What does "representable" mean?

Given (*), all well defined formula can be expressed in a single finite alphabet
and are countable.
Computable reals are a subset of this list, as
"UTM( 123, digit) mod 10" is some sentence
"UTM( 124, digit) mod 10" is some sentence
Not only are all the computable reals covered UTM(n e N, ..) but
"antidiag(UTM(n e N, digit) mod 10)" is also a real that it counts.
Herc 

Back to top 


Guest

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: how to find out the closed formula for this sequence?



brilliant! a thousand thanks to you! 

Back to top 


erc science forum Guru Wannabe
Joined: 25 Jul 2005
Posts: 269

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: Idiocy of Muckenheim was Re: countability of reals



"Piotr Sawuk" <piotr5@unet.univie.ac.at> wrote in >
Quote:  Not only have I formalised Cantor's proof, I knocked off about 50 steps off others' formalisms!
L(x, y) = the yth digit of the xth real.
this is not really the language of logic, but since you so believe in
such a function's definability...
1. L(a,a) = L(a,a) (obvious)
2. exists b, L(a,b) = L(b,b) (provable from 1, with b=a)
3. forall a, exists b, L(a,b) = L(b,b) (generalization of 2)
4a. not(exists a, forall b, L(a,b) != L(b,b) (negation of 3)
4b. not(exists a, forall b, L(a,b) = !L(b,b) (! is some suitable digit change function)
5. exists r, not(exits a, forall b, L(a,b) = r(b)) (r(b)= !L(b,b))
neither you did define the set in which the function r is contained.
So, suppose R:=set of enumerations of each subset of N, such that an
element r of that set R is a function with r(a) in N and forall b:
(b!=a > r(b)!=r(a)), then L(0):=R,L(a):=L(a1)C(L(a1)),L(a,b):=C(L(a))(b)
where C is the choicefunction we use for the set R and its subsets.
then your !(r(a)):=r(C(Na,r(a))) where C is a choicefunction on the
set N which takes a natural number as a hint. I think these definitions
can easily be understood as equivalent to a definition of reals between
0 and 1 and their enumeration if you restrict R to contain only functions
with r(a):=10^a times some number between 0 and 9.
"There exists a real that no member of list L matches at every digit."
Of course the free b in (r(b)= !L(b,b)) puts doubt on such an algebraic derivation.
why? it is just a definition of an r in R which is not in the list.
if you indeed can create a list L, and if you really can create a
function "!", then there does exist an r in R with r(b):=!L(b,b).
what I fail to understand is the step from (13) to 4, I just fail
to see how to argue here in the language of logic. from 1 we know
that b:=a in 2, and we can see that this does work for all a, but
even though the negation 4a is true, it is only so because one could
choose b:=a and thereby the function "!" is not guaranteed to exist.
don't you know any formulazation of this proof where the function
"!" is defined explicitely such that it does yield a value some r
would yield, or where it at least is proven to exist? in whole this
proof doesn't seem formal enough for mee! it does attempt to prove
that no surjection from N to R does exist in a constructive way
(i.e. that some r in R is never equal to some q(b):=L(a,b) forall
a in N), but as Herc said, in the 1st and 2nd step the variable
"a" is a free variable, and in the last step b is a free variable,
and in logic free variables are not even allowed for defining some
function (or predicate), or am I wrong? how does one define functions
within the language of logic? for example 2nd order logic with ZFC?
how does one define a choicefunction for finite sets? Herc, since
you had 3 years of logic, what did you learn on this topic?
You obviously know jack s**t since ! is trivial to define and its undefinability is
so paramount to you.
of course I know jack s**t, if I knew everything, then I wouldn't
have asked. it seems that for you it is indeed trivial to define
"!", while I do have some problems to understand this little step
you omitted in which it got defined. you know something which I
do not know, you understand something I do not understand. therefore
I ask you, please explain it to me! you spent 3 years with the
innyards of logic, I didn't even spend one, therefore your greater
knowledge is highly valuable for me. please show me how you would
define "!" based on the insights gained from 1,2,3 and 4a. logic
is the best language to express such trivialities, so please make
use of the language of logic as much as you need. I just wish to
understand how you did manage to understand the step from 3 to 4,
and since there is nothing you did criticize at that step, I'd have
to assume that you are able to explain it to me in the language of
logic. am I wrong with this assumption?

Better send the eMails to netscape.net, as to
evade useless burthening of my provider's /dev/null...
P

!a = a+1 mod 3
we generalised a in step 3, its no longer equal to b
3. forall a, exists b, L(a,b) = L(b,b) (generalization of 2)
this captures the notion of the diagonal of any list,
"all members of a list have a matching digit to the diagonal at some point".
The 1st 3 steps are Daryls, my original was cumbersome, this is my usual syntax..
Assume a real number list L where
L(a) is the ath real
L(a, b) is the bth digit of the ath real.
Required To Prove :
the list is incomplete
=
Er e R, Aa e N, L(a) =/= r
=
Er e R, Aa e N, Eb e N, L(a, b) =/= r(b) ...r(b) is the bth digit of r
Assume r is on the list
Ea e N, L(a) = r [1]
=
Ea e N, Ab e N, L(a, b) = r(b)
Let
r(b) = L(b,b)+1 (mod 9) [2]
=>
r(b) = !L(b,b) ...! is some digit change function
Therefore
Ea e N, Ab e N, L(a,b) = !L(b,b)
When a=b
Ea e N, L(a,a) =!L(a,a)
=>
Ec e N, c=!c
CONTRADICTION
CONCLUSION
NOT (Ea e N, L(a) = r AND r(b) = L(b,b)+1) ...[1] AND [2]
=
NOT (Ea e N, Ab e N, L(a, b) = L(b, b) + 1)
=
NOT (Ea e N, Ab e N, L(a, b) != L(b, b))
=
Aa e N, Eb e N, L(a, b) = L(b, b)
AXIOM
all reals in the list have a digit on the diagonal that equals itself
Herc

Quote:  then it is uniquely defined AFTER the flippers have flipped
enough (given that every flipper is only a finite number of
flippers from the beginning, and every flip is only a finite
number of flips from the beginning, no individual flip or anti
flip needs infinite inputs to compute). GEORGE GREENE sci.logic 


Back to top 


Munsey science forum beginner
Joined: 24 Mar 2005
Posts: 5

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: 0 * X = null?



lim(x>0) f(x)=x/x is considered indeterminate, not undefined. To
find 0/0, think what number must be multiplied by 0 to get 0. The
answer is not "no number," the answer is "any number."
But you are right, my logic does have some flaws. 

Back to top 


Jason science forum addict
Joined: 24 Mar 2005
Posts: 72

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: Epistemology 201: The Science of Science



Quote:  Most of maths is built from sets, so the basic assumptions of maths
are the axioms of set theory, in particular ZFC set theory. [Chapter Zero 
Fundamental Notions of Abstract Mathematics, Carol Schumacher]
However, after Chapter Zero, nobody cares about the axioms of ZFC,
let alone about the rules of predicate logic.
This is not to say that ZFC does not play an important role in
contemporary mathematics.

Mathematical proofs are in the form of logical deduction. Does nobody care
about proofs all of a sudden? 

Back to top 


r.e.s. science forum beginner
Joined: 29 Apr 2005
Posts: 27

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: My claim on Omega's defn



"Daryl McCullough" <stevendaryl3016@yahoo.com> wrote ...
Quote:  Finally, define Omega to be the limit as n>infinity of S(n).
This is the same as
Omega = sum over all valid bit strings p of 2^{length(p)}

Letting A be the prefixfree set {s: s is a valid bitstring},
with some measure theory I can show pretty directly that
Omega(A) := SUM[s in A] 1/2^s = Pr(X has a prefix in A), [1]
where X is a random bitstring of *infinite* length.
OTOH, the type of approach you've used here shows that
Omega(A) = LIM[n > oo] Pr(X_n has a prefix in A), [2]
where X_n is a random bitstring of *finite* length n.
(Your S(n) = Pr(X_n has a prefix in A).)
The second approach has avoided a measure on an uncountable
set, but *hasn't* established convergence of the infinite
sequence [2]  and I'm only sure the limit exists because
of its connection to [1]. In the first approach, convergence
of the sum in [1] is assured by the countable additivity of a
probability measure, but the measure is on an uncountable set.
Is it possible to establish convergence of [2] without
resorting to a measure on an uncountable set?
r.e.s. 

Back to top 


W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: abundance of irrationals



Robin Chapman wrote:
Quote:  mueckenh@rz.fhaugsburg.de wrote:
Jesse F. Hughes wrote:
Ok, I was too hasty. Imagine all rationals D3 with three digits
in
decimal representation In the interval (10,10). Add pi to each
of
them: D3 + pi. Form the set of all these rational numbers D3 and
all
these irrational numbers D3 + pi. The intersection of this set
with
the
real interval (0,1) results in a set which also can be
considered
as an
alternating sequence of rational and irrational numbers. If you
are
not
sure, you can calculate it for all these numbers. But there is a
more
general proof: Both sets, that of the rationals D3 and that of
the
irrationals D3 + pi consist of equidistant elements. None of
them
are
equal. Hence, in the domain where they overlap, thesy must
alternate.
Okay, so the sets D3 and D3+pi alternate.
This does not entail that the set Doo and Doo+pi "alternate",
where
Doo is the union of the (nested) sets Dn.
There is no natural number oo. There is no terminating rational
with an
infinite number of digits. My proof runs over all terminating
rationals.
There is also no immediate successor to any rational number in
the set of rational numbers.

And there is no red bicycle on the moon. Who cares?
I consider only rationals with n digits. There is alway an immediate
successor. That is the point (in case you are able to grasp that).
Regards, WM 

Back to top 


W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: abundance of irrationals



David Kastrup wrote:
Quote:  There is no natural number oo. There is no terminating rational
with
an infinite number of digits. My proof runs ober all terminating
rationals.
No, it doesn't. It holds for any set of rationals terminating at a
specified position. But the set of terminating rationals itself is
no
such set. It is the smallest superset of all such sets.

This superb superset covers these minor sets like dust and rubbish
cover the ruins of Troy? Or like the icing covers the cake? So what!
Who told you so? Or did you discover it by yourself?
Regards, WM 

Back to top 


Torkel Franzen science forum Guru
Joined: 30 Apr 2005
Posts: 639

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: Epistemology 201: The Science of Science



"Jason" <jasonstevensNOSPAM@free.net.nz> writes:
Quote:  Mathematical proofs are in the form of logical deduction.

They are by no means formal derivations. 

Back to top 


Eckard Blumschein science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 128

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: Where's respect? was Re: Corrective interpretation of real numbers



On 2/1/2005 7:39 PM, Bernd Funke wrote:
Quote:  I frankly admit that I was not polite. But that does not cancel my objection
raised by 1<1/n and 2>1/n.

I only vaguely recall the matter. Perhaps I made my consideration within
IR+ while you are assuming IR.
Quote:  What does it mean "in practice"? Where do these two numbers *come*from*?

Among practice I reckon measurement in geometry too.
What about the ln, I will hopefully get in position to further elaborate
my ideas. So far I am just trying to comprehend the dilemma that zero is
the neutral element of addition but it is also not approachable by
multiplication. Therefore I am interested in the relationship between
addition/subtraction and multiplication/division:
ln(a)+ ln(b) = ln(a b)
Quote:  *nobody* comes along, asking for ln(oo+1).

This would be Cantorian nonsense.
I would like to endorse the clarification oo+1=oo but not the unjustifed
calculation oooo=1. I see it an example how formalisms may fail.
Well any, even the least qualified mathematician is in position to
explain me this in the correct language of a tortuous mathematical
excuse, I am not interested in.
Quote:  The loss happens gradually and because any number relates to the human
capacity to recognize it.
That's *your* subset of the real numbers: The subset B of numbers the
decimal representation of which can be recognized by a human in a
(necessarily finite) time. That's fine, restrict *yourself* to those
subset.

Lets clarify: The genuine continuum as defined by Peirce evades set
theory. It might be interpreted as containing too much elements as to be
resolved into distinct elements.
One has to perform a so called "Grenzübergang" (border crossing) as to
reach the reciprocal of the so called actual infinity, in other words
Weierstrass's mathematical construct of a continuum. Be aware that
natural numbers as well as irrational numbers like pi and e are just
embedded in this somewhat vague construct. According to logics, this
constructed continuum originally belongs to IR+, not to IR. So it does
not naturally include zero and not include infinity. It includes as many
numbers as you like, of course only a finite number. Otherwise it would
not make sense. Let me declare it quite clear that Cantor was not
correct when he pretended his construct to be actually unlimited. The
limitation cannot be found at a certain value. It resides in the
possiblity to resolve the continuum into distinct sets. So I guess, all
effort to extend the reals towards the genuine continuum is doomed to fail.
Having understood the fundamental and insurmountable difference between
the genuine continuum and Weierstrass's IR+ continuum, one gets aware
that the border between the two is still not yet defined. It is merely
clear that the latter relates to the former like a subset, even if the
former cannot be splitted into "its" element, by definition. In order to
quantify a practical border Mueckenheim tried to refer to a physical
quantity. I prefer a more humancentered point of view because
mathematics has been made from people for the sake of people.
Quote:  Fine that we agree on that one. Thus, 0.[9] is formally as different from 1
as is 1+1 from 2.

No. 0.[9] is the factually distinct formal left neighbour of 1 within IR
while 1 is the left neighbour of 2 in IN. The pretended identity between
0.[9] and 1 would only exist within the genuine (Peirce) continuum.
Quote:  So what? Then use a *different* symbol for that purpose, not without saying
what you mean by a "sequence that ends at any delta < 0". This "property"
as it reads now, doesn't make any sense at all.

It is the precondition of set theory.
Quote:  You seem to be the only one having problems. You always like to refer to
other critics (sure, because you're unable to do the math on your own),
but either you didn't read their work or you didn't understand it.
Please look into what I quoted nearby.
Sorry, doubleclick on "nearby" doesn't work. Provide the msg ID, please.

It should be among these:
http://physicsforums.com/showthread.php?t=43179&page=2
http://mathforum.org/dr.math/faq/analysis_hyperreals.html#briefhistory
http://www.maa.org/pubs/Calc_articles/ma002.pdf
Quote:  Citation please (a link will do).

At the moment ....
Quote: 
So you suggest to build up a purely verbal math with no formalism? Because
you love so much being vague instead of being precise?

No. I just interprete Cauchy's hesitation as a possible hint to his
insight that things are a bit more tricky. I already explained my
suggestions.
Quote:  As I said: You utterly failed to comprehend Weierstrass' delta.
Teacher of mathematics are used to reproache their student for not
comprehend Weierstrass's delta.
Mostly justifiably so.
So what's your point? That you don't need to understand it either, even when
using or criticizing it?

Maybe, you do not understand that the most nowledgeable of your students
do not have any problem with the formalism but are just reluctant to
swallow something illogical you not even got aware of.
Quote:  Please read within what I quoted as opinion
of a mathematician on that.
*Where* did you quote that? Provide the msg ID, please.

See above.
Eckard 

Back to top 


W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: abundance of irrationals!)



Jesse F. Hughes wrote:
Quote: 
Okay, so the sets D3 and D3+pi alternate.
This does not entail that the set Doo and Doo+pi "alternate",
where
Doo is the union of the (nested) sets Dn.
There is no natural number oo. There is no terminating rational
with an
infinite number of digits. My proof runs ober all terminating
rationals.
I never said there was a natural number oo. I defined the notation
Doo above.

You abused the symbol oo because it is reserved to express infinity. In
particular we have oo = oo + 1. The nonterminating rational 1/3 for
instance, has oo many digits. oo is not a number of digits of a
nonterminating rational.
Quote:  Doo does not include any nonterminating rationals. Read the
definition.
The fact that the sets D3 and D3 + pi are alternating is utterly
irrelevant to whether the set
{ x  x is a "terminating rational" } and the set
{ x + pi  x + pi is a "terminating rational"} alternate.
But even assuming that the latter two sets alternate (which is
apparently false), that would have nothing to do with whether the set
Q and the set {q + pi  q in Q} alternate.
And even assuming that those two sets alternate, this would not show
that Q and R \ Q alternate or that Q = R or whatever you were after.

Please remain at the topic. Nobody talks about Q and R \ Q. We were at
D3 and then we went to Dn, but not to Doo. You may include Doo,
however, if you can explain how Cantor is able to identify and exchange
the diagonal number in line oo. Otherwise, you are incoherent.
Quote:  You're incoherent. I said nothing at all about any line number oo.

Then don't raise wrong impressions by talking about Doo either.
Quote:  It is not my fault that you are incapable of reading defined
notation. 
Stick to usual definitions, please. And in case you insist on Dn with n
= oo then explain how Cantor circumvents that same problem n = oo. (But
attention please, I do NOT propose that oo could be a natural number.)
Regards, WM 

Back to top 


Jason science forum addict
Joined: 24 Mar 2005
Posts: 72

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: Epistemology 201: The Science of Science



Quote:  Mathematical proofs are in the form of logical deduction.
They are by no means formal derivations.

I don't follow.
Do you mean that people don't go to the book of mathematical inference
and follow the semantics as if they were a machine?
Or do you mean that they are informal derivations? Or something else? 

Back to top 


W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: abundance of irrationals



David Kastrup wrote:
Quote:  mueckenh@rz.fhaugsburg.de writes:
Robin Chapman wrote:
mueckenh@rz.fhaugsburg.de wrote:
This does not entail that the set Doo and Doo+pi "alternate",
where Doo is the union of the (nested) sets Dn.
There is no natural number oo. There is no terminating rational
with an infinite number of digits. My proof runs over all
terminating rationals.
There is also no immediate successor to any rational number in the
set of rational numbers.
And there is no red bicycle on the moon. Who cares?
I consider only rationals with n digits. There is alway an
immediate
successor. That is the point (in case you are able to grasp that).
It is your same old quantor dyslexia. You think that a statement
for every n in N, for the set of all numbers terminating after n
digits ...
implies something about
for the set of all numbers terminating after n in N digits ...
It doesn't. Every single set of numbers terminating after n digits
is
a finite set. But the set of numbers terminating after a natural
number of digits is an infinite set. A statement about every first
such set does not imply a statement about the complete union of all
those sets.

Please make at least one serious attempt to understand. It is not
difficult though you are not familiar with it.
I run my proof over only one fixed n of N. For this fixed n the Dn and
Dn + Pi do alternate. I do not consider n = oo, because n cannot become
oo.
If you assert that oo came into the play at any stage, then, please,
explain how Cantor with his list could circumvent this very same
problem.
Regards, WM 

Back to top 


Torkel Franzen science forum Guru
Joined: 30 Apr 2005
Posts: 639

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: Epistemology 201: The Science of Science



Jason <jasonstevensNOSPAM@free.net.nz> writes:
Quote:  Or do you mean that they are informal derivations?

A derivation in ZFC is a mathematically defined object, a sequence
of formulas satisfying certain conditions. Actual mathematical proofs
are nothing like such sequences. 

Back to top 


Me1 science forum beginner
Joined: 28 Apr 2005
Posts: 12

Posted: Thu Mar 24, 2005 5:47 pm Post subject:
Re: how to find out the closed formula for this sequence?



Is that really correct? if r is positive big enough (I dont see any
condition for r except that its not 1) then making n big in the
equation of the original post gives some positive value and on the one
given by Virgil it gives a negative value...
Sincerely,
Jose Capco
Virgil wrote:
Quote:  In article <1107320053.143066.23070@l41g2000cwc.googlegroups.com>,
hfyan0@hotmail.com wrote:
Hello there
I would like to get a closed form for the sequence
1/(1+r) + 2/(1+r)^2 + 3/(1+r)^3 + ... + n/(1+r)^n
Could anyone out there help me out?
thank you very much!!
[r+1  (r*(n+1)+1)/(r+1)]/r^2 


Back to top 


Google


Back to top 



The time now is Sat Mar 24, 2018 1:42 pm  All times are GMT

