Discussion:
Is WM allowed to teach math
(too old to reply)
b***@gmail.com
2018-04-21 19:39:26 UTC
Permalink
WM observed that a function on the natural numbers can have two extensions that disagree on omega. now he thinks set theory is inconsistent. can we not follow that he is completly nuts?
b***@gmail.com
2018-04-21 19:46:38 UTC
Permalink
here are the details how he draws the contradiction
1) McDuck's wealth can only change with n.
2) For every n, McDuck's wealth is positive and increasing. (This can be checked for every n.)
3) McDuck's wealth is zero after all n (since before it is not zero).
Contradiction!
WM
2018-04-21 20:13:06 UTC
Permalink
Post by b***@gmail.com
here are the details how he draws the contradiction
1) McDuck's wealth can only change with n.
2) For every n, McDuck's wealth is positive and increasing. (This can be checked for every n.)
3) McDuck's wealth is zero after all n (since before it is not zero).
Contradiction!
Simple enough to understand?

Regards, WM
b***@gmail.com
2018-04-21 21:23:38 UTC
Permalink
pretry crazy indeed. you expect from f(n) > 0 that also one of your extensions g2(omega) > 0. but g(omega) can have any value. omega is not a natural number.

if you know the values f(n) this says nothing about an extension value g(omega). thats just plain crazy crankdoom from Augsburg Crank institute of Volkwagen Omlette.
WM
2018-04-22 09:28:21 UTC
Permalink
Post by b***@gmail.com
you expect from f(n) > 0 that also one of your extensions g2(omega) > 0.
No! I prove that for every n, f(n) > 0. That means all n are insufficient to enumerate all rationals. What is difficult to iunderstand in this simple logic expression?
Post by b***@gmail.com
omega is not a natural number.
Nobody said so. omega is nothing, it is nonsense, it is the delusion of cranks.

BUT IF omega is assumed to exist and if f(omega) is the empty set, then infinitely many rational numbers not yet indexed in any finite step, must have become indexed between all n and omega. So the existence of omega and of an empty set at omega does not help to convince a thinking being of the equinumerosity of |N and |Q.
Post by b***@gmail.com
if you know the values f(n) this says nothing about an extension value g(omega).
As explained above this extension does not help your case.

Regards, WM
Zelos Malum
2018-04-23 05:42:33 UTC
Permalink
Post by WM
No! I prove that for every n, f(n) > 0. That means all n are insufficient to enumerate all rationals. What is difficult to iunderstand in this simple logic expression?
No it doesn't, cause you are talking about elements IN a set, not the set itself.
Post by WM
Nobody said so. omega is nothing, it is nonsense, it is the delusion of cranks.
Actually, you are the delusional crank here.
Post by WM
BUT IF omega is assumed to exist and if f(omega) is the empty set, then infinitely many rational numbers not yet indexed in any finite step, must have become indexed between all n and omega. So the existence of omega and of an empty set at omega does not help to convince a thinking being of the equinumerosity of |N and |Q.
It is trivial to prove that they are used and all of Q+ is covered.
It isn't non-sense cause we are given the infinite set N by ZFC.

For your arguement

1: We can use any index set we want, Natural numbers is just the base one for countability

2: This is an arguement from finite cases, this point cannot be used to say anything about N itself

2a: Counterintuitive isn't it?

2b: You are argying for 1 SPECIFIC ennumeration but the definition says there has to be AN ennumeration.

3: Does not follow from 1 through 2.
WM
2018-04-23 11:52:50 UTC
Permalink
Post by Zelos Malum
No! I prove that for every n, f(n) > 0. That means all n are insufficient to enumerate all rationals. What is difficult to understand in this simple logic expression?
No it doesn't, cause you are talking about elements IN a set, not the set itself.
The set is one entity. I does not index anything. Indexing can only be done by natural numbers.
Post by Zelos Malum
It is trivial to prove that they are used and all of Q+ is covered.
It is trivial to prove the contrary too. Contradiction.

For all n in |N: There are oo many not enumerated fractions.

Regards, WM
Alan Mackenzie
2018-04-23 15:27:54 UTC
Permalink
Post by WM
Post by Zelos Malum
Post by WM
No! I prove that for every n, f(n) > 0. That means all n are
insufficient to enumerate all rationals. What is difficult to
understand in this simple logic expression?
No it doesn't, cause you are talking about elements IN a set, not the set itself.
The set is one entity. I does not index anything. Indexing can only be
done by natural numbers.
Post by Zelos Malum
It is trivial to prove that they are used and all of Q+ is covered.
It is trivial to prove the contrary too. Contradiction.
For all n in |N: There are oo many not enumerated fractions.
Is that right? Please tell us the "smallest" such fraction, where
"smallest" here is taken to mean the lowest sum of the moduli of the
numerator and of the denominator of the fraction, when it is expressed
in its lowest terms.
Post by WM
Regards, WM
--
Alan Mackenzie (Nuremberg, Germany).
WM
2018-04-23 15:45:38 UTC
Permalink
Post by Alan Mackenzie
Post by WM
For all n in |N: There are oo many not enumerated fractions.
Is that right? Please tell us the "smallest" such fraction,
Such a fraction exists for every n. For details see https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 254.

where
Post by Alan Mackenzie
"smallest" here is taken to mean the lowest sum of the moduli of the
numerator and of the denominator of the fraction, when it is expressed
in its lowest terms.
The function of all rational numbers below n that are not enumerated by a natural less than n (in Cantor's approach) is infinite. This holds for every n.

For a formal proof see https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 256.

Regards, WM
Alan Mackenzie
2018-04-23 16:00:00 UTC
Permalink
Post by WM
Post by Alan Mackenzie
Post by WM
For all n in |N: There are oo many not enumerated fractions.
Is that right? Please tell us the "smallest" such fraction,
Such a fraction exists for every n. For details see
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p.
254.
Then write one of them down, here, on this newsgroup. Its numerator
followed by a slash followed by its denominator. If so many of them
exist, surely you can identify just one of them.
Post by WM
Post by Alan Mackenzie
where "smallest" here is taken to mean the lowest sum of the moduli of
the numerator and of the denominator of the fraction, when it is
expressed in its lowest terms.
The function of all rational numbers below n that are not enumerated by
a natural less than n (in Cantor's approach) is infinite. This holds
for every n.
For a formal proof see
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p.
256.
Regards, WM
--
Alan Mackenzie (Nuremberg, Germany).
WM
2018-04-23 16:17:56 UTC
Permalink
Post by Alan Mackenzie
Post by WM
Post by Alan Mackenzie
Post by WM
For all n in |N: There are oo many not enumerated fractions.
Is that right? Please tell us the "smallest" such fraction,
Such a fraction exists for every n. For details see
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p.
254.
Then write one of them down, here, on this newsgroup. Its numerator
followed by a slash followed by its denominator. If so many of them
exist, surely you can identify just one of them.
The smallest fractions are given on p. 255 https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf:
1/1,
1/2, 2/1,
1/3, 3/1,
1/4, 2/3, 3/2, 4/1,
1/5, 5/1,
1/6, 2/5, 3/4, 4/3, 5/2, 6/1,

Of course the sequence can be continued, but what would that be good for?

REgardes, WM
b***@gmail.com
2018-04-23 16:41:51 UTC
Permalink
you might have stepped over one of the many bijections f: N -> Q. luckily there are also definable bijections around.

a lot of stuff is definable already in ZF, like the natural ordering of the real numbers. well ordering of the reals

is not definable. how would you proof that the reals well ordering is not definable?
b***@gmail.com
2018-04-23 16:59:23 UTC
Permalink
oh you dont like proofs, you call it matheology. so what is your nonsense PDF all about? a wrong axiom of infinity on page 45.

Volkswagen Omlette instead of proofs in your PDF?
b***@gmail.com
2018-04-24 15:41:03 UTC
Permalink
Am ok here is a function f : N+ -> Q+, and its inverse. can you tell us which rational number isnt mapped?

the function f(n) = r:

r = 1;
while (n <> 1) {
if (n mod 2 == 0) {
r = r/(r+1);
} else {
r = r+1;
}
n = n div 2;
}

and for the inverse the following definition, f^(-1)(r) = n:

n = 1;
while (r <> 1) {
n = n*2;
if (r < 1) {
r = r/(1-r);
} else {
r = r-1;
n = n+1;
}
}

some example runs:

f^(-1)(115/265) = 5144575

f(5144575) = 115/265

which rational number is not bijected?
Post by WM
Post by Alan Mackenzie
Post by WM
Post by Alan Mackenzie
Post by WM
For all n in |N: There are oo many not enumerated fractions.
Is that right? Please tell us the "smallest" such fraction,
Such a fraction exists for every n. For details see
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p.
254.
Then write one of them down, here, on this newsgroup. Its numerator
followed by a slash followed by its denominator. If so many of them
exist, surely you can identify just one of them.
1/1,
1/2, 2/1,
1/3, 3/1,
1/4, 2/3, 3/2, 4/1,
1/5, 5/1,
1/6, 2/5, 3/4, 4/3, 5/2, 6/1,
Of course the sequence can be continued, but what would that be good for?
REgardes, WM
b***@gmail.com
2018-04-24 15:50:46 UTC
Permalink
credits for this algorithm: cut the knot, stern brocot

here is a Prolog rendering of the algorithm, the used library provides rational number arithmetic and comparison:

?- listing.
% user
:- use_module(library(groebner/generic)).

f(N, R) :-
f(N, 1, R).

f(1, R, S) :- !,
S = R.
f(N, R, S) :-
user:(N mod 2 =:= 0), !,
user:(M is N//2),
H is R/(R+1),
f(M, H, S).
f(N, R, S) :-
user:(M is N//2),
H is R+1,
f(M, H, S).

finv(R, N) :-
finv(R, 1, N).

finv(1, N, M) :- !,
M = N.
finv(R, N, M) :-
R < 1, !,
S is R/(1-R),
H is 2*N,
finv(S, H, M).
finv(R, N, M) :-
S is R-1,
H is 2*N+1,
finv(S, H, M).

Yes

here are the sample runs again:

?- R is 345/789, finv(R, N).
R is 115/263,
N is 5144575
?- f(5144575, R).
R is 115/263

everything done while at the beach with an Android tablet. LoL
Post by b***@gmail.com
Am ok here is a function f : N+ -> Q+, and its inverse. can you tell us which rational number isnt mapped?
r = 1;
while (n <> 1) {
if (n mod 2 == 0) {
r = r/(r+1);
} else {
r = r+1;
}
n = n div 2;
}
n = 1;
while (r <> 1) {
n = n*2;
if (r < 1) {
r = r/(1-r);
} else {
r = r-1;
n = n+1;
}
}
f^(-1)(115/265) = 5144575
f(5144575) = 115/265
which rational number is not bijected?
Post by WM
Post by Alan Mackenzie
Post by WM
Post by Alan Mackenzie
Post by WM
For all n in |N: There are oo many not enumerated fractions.
Is that right? Please tell us the "smallest" such fraction,
Such a fraction exists for every n. For details see
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p.
254.
Then write one of them down, here, on this newsgroup. Its numerator
followed by a slash followed by its denominator. If so many of them
exist, surely you can identify just one of them.
1/1,
1/2, 2/1,
1/3, 3/1,
1/4, 2/3, 3/2, 4/1,
1/5, 5/1,
1/6, 2/5, 3/4, 4/3, 5/2, 6/1,
Of course the sequence can be continued, but what would that be good for?
REgardes, WM
b***@gmail.com
2018-05-21 15:11:25 UTC
Permalink
It seems to me that WM is constantly confused about
the notation f : A -> B. He doesn't understand that
it only means that elements x of A, are mapped to

elements f(x) of B. It doesn't mean that f(A)=B.
There is no claim in the notation that the domain
itself is sent to the range by f itself.

But something else. Did we see WM define his beloved
sequences of natural numbers? I guess like N he
wants N* to be a potential set.

Lets assume potential set means class. How would
one define N* as a class in some logical framework
so that its a workable definition?

Did WM ever define it? Or is all his saying handwaving?
Post by WM
Post by Zelos Malum
No! I prove that for every n, f(n) > 0. That means all n are insufficient to enumerate all rationals. What is difficult to understand in this simple logic expression?
No it doesn't, cause you are talking about elements IN a set, not the set itself.
The set is one entity. I does not index anything. Indexing can only be done by natural numbers.
Post by Zelos Malum
It is trivial to prove that they are used and all of Q+ is covered.
It is trivial to prove the contrary too. Contradiction.
For all n in |N: There are oo many not enumerated fractions.
Regards, WM
b***@gmail.com
2018-05-20 14:40:05 UTC
Permalink
Logic and model theory is by way not simple to
unerstand. Here is a challenge for WM.

Take this theory here, only one axiom, the language L
consists of the constants {0,1,2,3,...} i.e. for each
natural number a constant:

WM_Challenge_Theory = { forall x (x = 1) }

Here are the trick questions:

1) Can WM_Challenge_Theory have uncountable models M?

2) Does the Löwenheim-Skolem Downward Theorem hold?

3) Does the Löwenheim-Skolem Upward Theorem hold?

Here are some answers:

1) It depends whether you norm (=) or not. If you do not
norm (=) then: Yes, the cardinality of the model, says
nothing about the cardinality of the = equivalence classes.

2) Löwnheim-Skolem Downward Theorem is usually without
normed (=), otherwise the theorem wouldn't work. That
means (=) is just viewed as a predicate guarded by the usual
equality theory.

3) Dito.

So here is a uncountable model:

M = <R, =_WM_Challenge_Theory , 1_WM_Challenge_Theory >

with =_WM_Challenge_Theory = < (r,r) | r in R >

1_WM_Challenge_Theory = r_0 /* any real number works */
Post by WM
Post by b***@gmail.com
here are the details how he draws the contradiction
1) McDuck's wealth can only change with n.
2) For every n, McDuck's wealth is positive and increasing. (This can be checked for every n.)
3) McDuck's wealth is zero after all n (since before it is not zero).
Contradiction!
Simple enough to understand?
Regards, WM
b***@gmail.com
2018-05-20 14:46:55 UTC
Permalink
Corr.:

=_WM_Challenge_Theory = < (r,s) | r in R, s in R >

This is an unnormed equality, and its an equality,
as one can easy verify:

x = x yes

x = y => y = x yes

x = y & y = z => x = z yes

For the Löwnheim Skolem theorem you need also unnormed
equality, where you do not require:

=_M = { (m,m) | m in M } /* not used in Löwnheim Skolem */

Otherwise you cannot make term models. What are
term models. Well term models are models where the
domain is a freely generated by the language.

Term models started already with Herbrand:
https://en.wikipedia.org/wiki/Herbrand_interpretation
Post by b***@gmail.com
Logic and model theory is by way not simple to
unerstand. Here is a challenge for WM.
Take this theory here, only one axiom, the language L
consists of the constants {0,1,2,3,...} i.e. for each
WM_Challenge_Theory = { forall x (x = 1) }
1) Can WM_Challenge_Theory have uncountable models M?
2) Does the Löwenheim-Skolem Downward Theorem hold?
3) Does the Löwenheim-Skolem Upward Theorem hold?
1) It depends whether you norm (=) or not. If you do not
norm (=) then: Yes, the cardinality of the model, says
nothing about the cardinality of the = equivalence classes.
2) Löwnheim-Skolem Downward Theorem is usually without
normed (=), otherwise the theorem wouldn't work. That
means (=) is just viewed as a predicate guarded by the usual
equality theory.
3) Dito.
M = <R, =_WM_Challenge_Theory , 1_WM_Challenge_Theory >
with =_WM_Challenge_Theory = < (r,r) | r in R >
1_WM_Challenge_Theory = r_0 /* any real number works */
Post by WM
Post by b***@gmail.com
here are the details how he draws the contradiction
1) McDuck's wealth can only change with n.
2) For every n, McDuck's wealth is positive and increasing. (This can be checked for every n.)
3) McDuck's wealth is zero after all n (since before it is not zero).
Contradiction!
Simple enough to understand?
Regards, WM
b***@gmail.com
2018-05-20 15:31:02 UTC
Permalink
So whats the gist of all this. Well
theories don't see the domain, they
only see what is mediated by the language,

like for example a theory here:

forall x (x = 1)

is completely agnostic to the size of
the domain, it only sees what the equivalence
relation (=)/2 gives.

Same for axioms, theorems in set theory,
they are formulate with the membership
relation (and equality most often):

.... in .... = ....

So you only see from the domain what is
mediated by (in)/2 and (=)/2. So its very
difficult to falsify a claim,

that there are uncountable models. You
would need to have a means for that.
But for this end there are no means

on the language level. You can even not
get hold of equality. There could be real
numbers in your model, and black swans,

and a black swan could be pi. You wouldn't
notice. This has even nothing to do with
non-standard models. Its just logic and

model theory. You would need to norm equality,
but the usual axioms for equality in a logic, or
also if you have inference rules for equality

in a logical system, how would they norm
equality? We can require this at the meta
level, when we discuss model theory, and maybe

say, ok, we look only at normed models etc...
Post by b***@gmail.com
=_WM_Challenge_Theory = < (r,s) | r in R, s in R >
This is an unnormed equality, and its an equality,
x = x yes
x = y => y = x yes
x = y & y = z => x = z yes
For the Löwnheim Skolem theorem you need also unnormed
=_M = { (m,m) | m in M } /* not used in Löwnheim Skolem */
Otherwise you cannot make term models. What are
term models. Well term models are models where the
domain is a freely generated by the language.
https://en.wikipedia.org/wiki/Herbrand_interpretation
Post by b***@gmail.com
Logic and model theory is by way not simple to
unerstand. Here is a challenge for WM.
Take this theory here, only one axiom, the language L
consists of the constants {0,1,2,3,...} i.e. for each
WM_Challenge_Theory = { forall x (x = 1) }
1) Can WM_Challenge_Theory have uncountable models M?
2) Does the Löwenheim-Skolem Downward Theorem hold?
3) Does the Löwenheim-Skolem Upward Theorem hold?
1) It depends whether you norm (=) or not. If you do not
norm (=) then: Yes, the cardinality of the model, says
nothing about the cardinality of the = equivalence classes.
2) Löwnheim-Skolem Downward Theorem is usually without
normed (=), otherwise the theorem wouldn't work. That
means (=) is just viewed as a predicate guarded by the usual
equality theory.
3) Dito.
M = <R, =_WM_Challenge_Theory , 1_WM_Challenge_Theory >
with =_WM_Challenge_Theory = < (r,r) | r in R >
1_WM_Challenge_Theory = r_0 /* any real number works */
Post by WM
Post by b***@gmail.com
here are the details how he draws the contradiction
1) McDuck's wealth can only change with n.
2) For every n, McDuck's wealth is positive and increasing. (This can be checked for every n.)
3) McDuck's wealth is zero after all n (since before it is not zero).
Contradiction!
Simple enough to understand?
Regards, WM
WM
2018-04-21 20:12:38 UTC
Permalink
Post by b***@gmail.com
WM observed that a function on the natural numbers can have two extensions that disagree on omega.
In fact, omega is not at all required.

The question is whether all rational numbers can be indexed by natural numbers. The answer is no. For every natural number there remain infinitely many rational numbers without index.

Regards, WM
Python
2018-04-21 21:27:26 UTC
Permalink
Post by WM
Post by b***@gmail.com
WM observed that a function on the natural numbers can have two extensions that disagree on omega.
In fact, omega is not at all required.
The question is whether all rational numbers can be indexed by natural numbers. The answer is no. For every natural number there remain infinitely many rational numbers without index.
Then (by your silly argument), even natural numbers are not countable.

For anyone with a little for mental sanity this should ring a bell about
the whole argument being invalide. For YOU, Herr Mueckenheim, from
Augsburg, it didn't. Conclusion: YOU ARE MENTALLY INSANE.
Ross A. Finlayson
2018-04-21 21:43:10 UTC
Permalink
Post by Python
Post by WM
Post by b***@gmail.com
WM observed that a function on the natural numbers can have two extensions that disagree on omega.
In fact, omega is not at all required.
The question is whether all rational numbers can be indexed by natural numbers. The answer is no. For every natural number there remain infinitely many rational numbers without index.
Then (by your silly argument), even natural numbers are not countable.
For anyone with a little for mental sanity this should ring a bell about
the whole argument being invalide. For YOU, Herr Mueckenheim, from
Augsburg, it didn't. Conclusion: YOU ARE MENTALLY INSANE.
Troll!
Python
2018-04-21 21:59:23 UTC
Permalink
Post by Ross A. Finlayson
Post by Python
Post by WM
Post by b***@gmail.com
WM observed that a function on the natural numbers can have two extensions that disagree on omega.
In fact, omega is not at all required.
The question is whether all rational numbers can be indexed by natural numbers. The answer is no. For every natural number there remain infinitely many rational numbers without index.
Then (by your silly argument), even natural numbers are not countable.
For anyone with a little for mental sanity this should ring a bell about
the whole argument being invalide. For YOU, Herr Mueckenheim, from
Augsburg, it didn't. Conclusion: YOU ARE MENTALLY INSANE.
Troll!
You do think, Ross, than f:f(n)=n does not define a bijection from
N to N? Really Ross?
Ross A. Finlayson
2018-04-26 04:23:37 UTC
Permalink
I think f(n) -> n + 1
has N go to N+.
WM
2018-04-22 09:41:51 UTC
Permalink
Post by WM
In fact, omega is not at all required.
The question is whether all rational numbers can be indexed by natural numbers. The answer is no. For every natural number there remain infinitely many rational numbers without index.
Then (by your argument), even natural numbers are not countable.
Of course they are not countable. Countability requires finished infinity. That is nonsense. But that is not easy to see for an average matheologians who confuse the identity mapping f(n) = n up to every n with a mapping for all n. The simple logic used in the following however, should be comprehensible:

1) Only a natural number n can index a fraction.

2) For all n in |N: there are infinitely many not indexed fractions.

2a) The number of intevals (n, n+1) without any fraction indexed by k < n increases with n/2.

2b) The first interval (0, 1) contains most indexed fractions, but for every n it contains less than 1 % indexed fractions.

3) Never all fractions are indexed.
For anyone with a little for mental sanity this should ring a bell about
the whole argument being invalide.
If we can state without being contradicted: "FOR ALL n in |N", then this is a firm and logical conclusion whereon we can errect a valid mathematical theorem. In the present case the theorem reads:

Never all fractions are indexed by natural numbers.

Regards, WM
Python
2018-04-22 12:18:23 UTC
Permalink
Post by WM
Post by WM
In fact, omega is not at all required.
The question is whether all rational numbers can be indexed by natural numbers. The answer is no. For every natural number there remain infinitely many rational numbers without index.
Then (by your argument), even natural numbers are not countable.
1) Only a natural number n can index a fraction.
Trivial assertion: usual Mueckenheim sophistry
Post by WM
2) For all n in |N: there are infinitely many not indexed fractions.
Meaningless statement.
Post by WM
2a) The number of intevals (n, n+1) without any fraction indexed by k < n increases with n/2.
Irrelevant.
Post by WM
2b) The first interval (0, 1) contains most indexed fractions, but for every n it contains less than 1 % indexed fractions.
Irrelevant.
Post by WM
3) Never all fractions are indexed.
All fraction are indexed, "never" is a misleading (usual Mueckenheim
sophistry) term here.
Post by WM
For anyone with a little for mental sanity this should ring a bell about
the whole argument being invalide.
It is neither firm nor logical: this is insanity from Mueckenheim's
silly mind.
WM
2018-04-22 12:46:44 UTC
Permalink
Am Sonntag, 22. April 2018 14:18:28 UTC+2 schrieb Python:
#
Post by Python
Post by WM
1) Only a natural number n can index a fraction.
Important when arguing with matheologians.
Post by Python
Post by WM
2) For all n in |N: there are infinitely many not indexed fractions.
Meaningless statement.
Incomprehensible for you? Try this: Not enumerating all positive rational numbers (formal proof), https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 256
Post by Python
Post by WM
2a) The number of intevals (n, n+1) without any fraction indexed by k < n increases with n/2.
Irrelevant.
For believers in nonsensical delusions everthing else is irrelevant.
Post by Python
Post by WM
2b) The first interval (0, 1) contains most indexed fractions, but for every n it contains less than 1 % indexed fractions.
Irrelevant.
"Forall n: P(n)" is usually accepted in logic (if it does not contradict set theory).
Post by Python
Post by WM
3) Never all fractions are indexed.
All fraction are indexed, "never" is a misleading term here.
"Never" is leading to the statement: Forall n: not all fractions are indexed.
Nice try. I hope that some readers are not yet too insane to understand your point of view and its relevance.

Regards, WM
b***@gmail.com
2018-04-22 14:40:53 UTC
Permalink
i dont see any proof that a bijection N ~ Q is impossible. only the usual halucination and confusion by WM. lets introduce some terminology to highlight the fallacy by WM.

a set A of natural numbers is called cofinite, if there is a set B finite, such that A = N \ B. There is the following trivial lemma:

A cofinite set is infinite

thats always true and in no way a contradiction. We can also apply it to any bijection f : N -> C . i do not specify what C is at the moment.

the obviously for a segment S = {0,1,2,...,n-1} we have this set here finite, the already enumerated elements of C by 0,1,2,...,n-1:

B = { f(k) | k in S } is finite

but then this set of the not yet enumerated elements:

A = C \ B is cofinite and hence infinite

its simply isomorphic, since f is a bijection, to its natural number pendant. So what WM tells us about some not yet enumerated elements is anyway alwas true, also when C = Q, i.e. the rational numbers.

this happens for any bijection and does it no way contradict anything. So we can conclude we is completely nuts.
b***@gmail.com
2018-04-22 14:43:22 UTC
Permalink
i dont see any proof that a bijection N ~ Q is impossible. only the usual halucination and confusion by WM. lets introduce some terminology to highlight the fallacy by WM.

a set A of natural numbers is called cofinite, if there is a set B finite, such that A = N \ B. There is the following trivial lemma:

A cofinite set is infinite

thats always true and in no way a contradiction. We can also apply it to any bijection f : N -> C . i do not specify what C is at the moment.

then obviously for a segment S = {0,1,2,...,n-1} we have this set here finite, the already enumerated elements of C by 0,1,2,...,n-1:

B = { f(k) | k in S } is finite

but then this set of the not yet enumerated elements:

A = C \ B is cofinite and hence infinite

its simply isomorphic, since f is a bijection, to its natural number pendant. So what WM tells us about some not yet enumerated elements is anyway alwas true, also when C = Q, i.e. the rational numbers.

this happens for any bijection and does it no way contradict anything. So we can conclude WM is completely nuts.
WM
2018-04-22 15:06:26 UTC
Permalink
Post by b***@gmail.com
lets introduce some terminology to highlight the fallacy by WM.
No necessity to introduce the terminology of matheology. Simply apply logic:

Forall n in |N: f(n) is not empty.
f can change only at an indexed step.
==> f annot be empty in the limit.
Post by b***@gmail.com
this happens for any bijection and does it no way contradict anything.
It contradicts logic.

Regards, WM
b***@gmail.com
2018-04-22 16:39:24 UTC
Permalink
you see, you still introduce some limit, like omega. but a bijection is based on "every" and not your "all".

you are definitively crazy. lets recap, what a function f : A -> B is. its just mapping such that:

forall x (x in A => f(x) in B)

not your "all", but an "every". Got it? Possibly no, it seems you got only Volkswagen Omlette for brains. total bird brain.
WM
2018-04-22 18:48:17 UTC
Permalink
Post by b***@gmail.com
you see, you still introduce some limit, like omega.
If all rationals shall be indexed, this cannot happen at a finite step. Without limit no completeness.
Post by b***@gmail.com
but a bijection is based on "every" and not your "all".
A bijection determining equinumerosity is based on "all". Who would believe it if only every number (that is followed by infinitely many) is in the bijection?

Regards, WM
b***@gmail.com
2018-04-22 23:59:02 UTC
Permalink
what else do you want in the bijection. some elephants?

you are completly crazy WM. more crazy than bird brain JG.
Python
2018-04-22 20:36:55 UTC
Permalink
Post by WM
Post by b***@gmail.com
lets introduce some terminology to highlight the fallacy by WM.
Sure, let's not forget what f(n) is, f(n) is the set of banknotes in
McDuck safe at the end of day n.
Post by WM
Forall n in |N: f(n) is not empty.
f can change only at an indexed step.
==> f annot be empty in the limit.
There is no "f in the limit", there is f(n) for any n. Period.

There is, nevertheless, set limit of f(n), which is not f(n)
for any n, but the set of banknotes given and kept indefinitely.
This is the empty set (which didn't ask Wolfgang Mueckenheim
permission to exist), and is never a set of the sequence f(n).

(There is nothing surprising, btw, that the limit of a sequence
is not a member of the sequence, it is quite usual)
Post by WM
Post by b***@gmail.com
this happens for any bijection and does it no way contradict anything.
It contradicts logic.
During the last course AWP 0397 "Geschichte des Unendlichen" at
Hochschule Augsburg, Rolph und Gisela, two student of Hochschule
Augsburg were attending while Wolfgang Mueckenheim was "teaching"
the kind of nonsense you may read in his posts here. I got the
record:

Rolph - Zzzzz

Gisela - Rolph! Rolph! Wake up!

Rolph - Zzzzz

Gisela - Rolph! Rolph! Wake up! We have to leave this place, this
is insane!

Rolph - Hmm? What? Is the silly course finished yet?

Gisela - What this old buzzard is saying does not make any sense!
Out of a few sentences that are short enough, which are wrong!

Rolph - Don't worry Gisela, all students here know for ages that
Mueckenheim is fit for the loony bin. But the course is
easy: no need to study, just use the same silly words in approximate
order at the exam, confirm his obsession and you'll get very good
grades...

Gisela - Then, Rolph, how come you failed last year?

Rolph - I tried the same trick in a real math exam...
WM
2018-04-23 11:48:33 UTC
Permalink
Post by Python
Post by WM
Post by b***@gmail.com
lets introduce some terminology to highlight the fallacy by WM.
Sure, let's not forget what f(n) is, f(n) is the set of banknotes in
McDuck safe at the end of day n.
Post by WM
Forall n in |N: f(n) is not empty.
f can change only at an indexed step.
==> f annot be empty in the limit.
There is no "f in the limit", there is f(n) for any n. Period.
Accordingn to set theory there is a limit. The sequence 9, 18, 27, ... has limit omega.
Post by Python
There is, nevertheless, set limit of f(n), which is not f(n)
for any n, but the set of banknotes given and kept indefinitely.
If this set is empty, then all must have been issued at finite steps, no?

Regards, WM
Python
2018-04-23 13:10:18 UTC
Permalink
Post by WM
Post by Python
Post by WM
Post by b***@gmail.com
lets introduce some terminology to highlight the fallacy by WM.
Sure, let's not forget what f(n) is, f(n) is the set of banknotes in
McDuck safe at the end of day n.
Post by WM
Forall n in |N: f(n) is not empty.
f can change only at an indexed step.
==> f annot be empty in the limit.
There is no "f in the limit", there is f(n) for any n. Period.
Accordingn to set theory there is a limit. The sequence 9, 18, 27, ... has limit omega.
Irrelevant, omega is not involved in any way in McDuck case. In
McDuck case you have a sequence and a set limit. PERIOD.
Post by WM
Post by Python
There is, nevertheless, set limit of f(n), which is not f(n)
for any n, but the set of banknotes given and kept indefinitely.
If this set is empty, then all must have been issued at finite steps, no?
This is your quantifiers dyslexia (or dirty trick) again:

From the trivial fact that:

for all n: there is step N so that from step N banknote_n has been
issued (and will not be given back)

you suggest (by voluntary sloppy misleading words) that set theory
would say:

there exist N so that from step N: for all n: banknote_n has been
issued.

It doesn't say that. Only you pretend so.

Neither Rolph nor Gisela, from Hochschule Augsburg will be fooled.
WM
2018-04-23 14:01:38 UTC
Permalink
Post by Python
Post by WM
Post by Python
Post by WM
Forall n in |N: f(n) is not empty.
f can change only at an indexed step.
==> f annot be empty in the limit.
There is no "f in the limit", there is f(n) for any n. Period.
You are wrong. Proof: f(1) = 9.
Post by Python
Post by WM
Accordingn to set theory there is a limit. The sequence 9, 18, 27, ... has limit omega.
Irrelevant, omega is not involved in any way in McDuck case.
Every step before omega is insufficient. Proof: f(n) = 9n.
Post by Python
In
McDuck case you have a sequence and a set limit. PERIOD.
The sequence is (f(n)) and the set limit is empty set. The limit results from the sets of the sequence by losing all elements. How?
Post by Python
Post by WM
Post by Python
There is, nevertheless, set limit of f(n), which is not f(n)
for any n, but the set of banknotes given and kept indefinitely.
If this set is empty, then all must have been issued at finite steps, no?
It is a logical necessity.
Post by Python
for all n: there is step N so that from step N banknote_n has been
issued (and will not be given back)
you suggest (by voluntary sloppy misleading words) that set theory
there exist N so that from step N: for all n: banknote_n has been
issued.
If this is never the case, then never all banknotes have been returned.
Post by Python
It doesn't say that. Only you pretend so.
If it doesn't say that, then there are banknotes remaining, no?

Regards, WM
Python
2018-04-23 14:15:38 UTC
Permalink
Post by WM
Post by Python
Post by WM
Post by Python
Post by WM
Forall n in |N: f(n) is not empty.
f can change only at an indexed step.
==> f annot be empty in the limit.
There is no "f in the limit", there is f(n) for any n. Period.
You are wrong. Proof: f(1) = 9.
Is it wrong that "there is f(n) for any n" because "f(1)=9" ?

f(n) is a SET, Muecke, not a number. f(1) = {1,2,3,4,5,6,7,8,9}

Stop drinking Schnaps, Muecke.
Post by WM
Post by Python
Post by WM
Accordingn to set theory there is a limit. The sequence 9, 18, 27, ... has limit omega.
Irrelevant, omega is not involved in any way in McDuck case.
Every step before omega is insufficient. Proof: f(n) = 9n.
There you should stop drinking vodka.

f(n) is a SET, Mucke, not a number. You confuse f(n) with Card f(n).

Even with apple juice, vodka is bad for you.
Post by WM
Post by Python
In
McDuck case you have a sequence and a set limit. PERIOD.
The sequence is (f(n)) and the set limit is empty set. The limit results from the sets
of the sequence by losing all elements. How?
The sets of sequence are not losing all elements (so now, you remember
that f(n) is a set, not a number, thanks to Alka-Seltzer).

Union_n Intersect_(k>n) (f(k)) = Intersect_n Union_(k>n) (f(k)) = {}

There is nothing there claiming that "the sets are losing all
elements", the claim is only "there is no element kept forever".
Post by WM
Post by Python
Post by WM
Post by Python
There is, nevertheless, set limit of f(n), which is not f(n)
for any n, but the set of banknotes given and kept indefinitely.
If this set is empty, then all must have been issued at finite steps, no?
It is a logical necessity.
Post by Python
for all n: there is step N so that from step N banknote_n has been
issued (and will not be given back)
you suggest (by voluntary sloppy misleading words) that set theory
there exist N so that from step N: for all n: banknote_n has been
issued.
If this is never the case, then never all banknotes have been returned.
Switching to wine won't help you, Herr Mueckenheim.
Post by WM
Post by Python
It doesn't say that. Only you pretend so.
If it doesn't say that, then there are banknotes remaining, no?
When you will have identified at least ONE banknote McDuck will
keep forever, can you post which banknote it is? Thanks.

Try not to drink a shot of slivoviça for any banknote McDuck is
issuing, this is bad for you, Herr Mueckenheim.
WM
2018-04-23 14:35:07 UTC
Permalink
Post by Python
You confuse f(n) with Card f(n).
Take s_n as a set, f(n) as a number, namely its cardinality.
Post by Python
There is nothing there claiming that "the sets are losing all
elements", the claim is only "there is no element kept forever".
If there is no element kept forever, then every element must be lost. This can only happen at a finite step.
Post by Python
Post by WM
If it doesn't say that, then there are banknotes remaining, no?
When you will have identified at least ONE banknote McDuck will
keep forever, can you post which banknote it is?
I do not claim that any banknote is kept forever!
I do not claim that the complete finished infinite set exists.
I do not claim that all rationals can be enumerated or all dollars can be returned.

But if you claim that all rationals can be enumerated, then this must happen somewhere. Then all banknotes must have been returned at some existing stage.

Or do your claims concern not existing issues only?

Regards, WM
Python
2018-04-23 14:42:36 UTC
Permalink
Post by WM
Post by Python
You confuse f(n) with Card f(n).
Take s_n as a set, f(n) as a number, namely its cardinality.
You made the confusion as f(n) was defined as a set from the very
first line... Stop wine, after so much vodka it is not good.
Post by WM
Post by Python
There is nothing there claiming that "the sets are losing all
elements", the claim is only "there is no element kept forever".
If there is no element kept forever, then every element must be lost. This can only happen at a finite step.
Post by Python
Post by WM
If it doesn't say that, then there are banknotes remaining, no?
When you will have identified at least ONE banknote McDuck will
keep forever, can you post which banknote it is?
I do not claim that any banknote is kept forever!
"there are banknotes remaining" is your words, not mine.

Rolph und Gisela won't be fooled, you know?
WM
2018-04-23 15:31:35 UTC
Permalink
Post by Python
Post by WM
I do not claim that any banknote is kept forever!
"there are banknotes remaining" is your words, not mine.
For every step n, 9n banknotes are remaining.
That does not contradict anything in potential infinity.

But if the completeness of infinite sets is assumed, then all banknotes must be issued. This is not the case in every step where this could happen. Therefore actual infinity is contradicted.

For every banknote we can say when it is returned. No problem. Every banknote belongs to a finite initial segment which is followed by infinitely many. But since there is never completeness, we cannot prove equinumosity. (Cantor knew why he demanded actual infinity.)

Regards, WM
Python
2018-04-23 15:37:54 UTC
Permalink
Post by WM
Post by Python
Post by WM
I do not claim that any banknote is kept forever!
"there are banknotes remaining" is your words, not mine.
For every step n, 9n banknotes are remaining.
That does not contradict anything in potential infinity.
But if the completeness of infinite sets is assumed, then all banknotes
must be issued.
No. You're confused. "limit" is not a "final step". I told you not
to drink a shot of slivoviça for every issued note.
WM
2018-04-23 15:52:27 UTC
Permalink
Post by Python
Post by WM
Post by Python
Post by WM
I do not claim that any banknote is kept forever!
"there are banknotes remaining" is your words, not mine.
For every step n, 9n banknotes are remaining.
That does not contradict anything in potential infinity.
But if the completeness of infinite sets is assumed, then all banknotes
must be issued.
No.
Yes.
Post by Python
"limit" is not a "final step".
Nothing must remain when all naturals have been applied. Is that possible? Necessarily. If some element remains, then the bijection is not complete.

Regards, WM
Python
2018-04-23 15:55:29 UTC
Permalink
Post by WM
Post by Python
Post by WM
Post by Python
Post by WM
I do not claim that any banknote is kept forever!
"there are banknotes remaining" is your words, not mine.
For every step n, 9n banknotes are remaining.
That does not contradict anything in potential infinity.
But if the completeness of infinite sets is assumed, then all banknotes
must be issued.
No.
Yes.
Post by Python
"limit" is not a "final step".
Nothing must remain when all naturals have been applied. Is that possible?
It is not what "set limit = {}" means. It means EXACTLY that there is
not a single banknote kept forever. This can be proven from the
definition of set limit.

"after all naturals have been applied" is meaningless in such context
Post by WM
Necessarily. If some element remains, then the bijection is not complete.
There is no bijection involved in McDuck case. Stop drinking on Mondays.
WM
2018-04-23 16:11:30 UTC
Permalink
Post by Python
Post by WM
Nothing must remain when all naturals have been applied. Is that possible?
It is not what "set limit = {}" means.
Set limit means every buck is returned. That can only happen at natural steps.
Post by Python
It means EXACTLY that there is
not a single banknote kept forever.
Then each one must be returned at a step n. Then none must remain after all steps n. The set limit says exatly this: Look at the formlism: n = 1 to oo.
Post by Python
This can be proven from the
definition of set limit.
have you seen it? n = k to oo.
Post by Python
"after all naturals have been applied" is meaningless in such context
n = 1 to oo covers all finite steps.
Post by Python
Post by WM
Necessarily. If some element remains, then the bijection is not complete.
There is no bijection involved in McDuck case.
Of course it is. But we can switch to the enumeration of fractions as well.

If all are enumerated, then it is after all finite numbers have been applid. This is expressed formally as n = 1 to oo.

Look it up, for instance here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 55. Or Wiki: LimSup.

Regards, WM
Python
2018-04-23 16:17:27 UTC
Permalink
Post by WM
Post by Python
Post by WM
Nothing must remain when all naturals have been applied. Is that possible?
It is not what "set limit = {}" means.
Set limit means every buck is returned. That can only happen at natural steps.
Post by Python
It means EXACTLY that there is
not a single banknote kept forever.
Then each one must be returned at a step n. Then none must remain after all steps n. The set limit says exatly this: Look at the formlism: n = 1 to oo.
Post by Python
This can be proven from the
definition of set limit.
have you seen it? n = k to oo.
Post by Python
"after all naturals have been applied" is meaningless in such context
n = 1 to oo covers all finite steps.
Post by Python
Post by WM
Necessarily. If some element remains, then the bijection is not complete.
There is no bijection involved in McDuck case.
Of course it is. But we can switch to the enumeration of fractions as well.
If all are enumerated, then it is after all finite numbers have been applid. This is expressed formally as n = 1 to oo.
Look it up, for instance here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 55. Or Wiki: LimSup.
I told you not to do so: you've emptyed the bottle of slivoviça and you
have read at the bottom of it "n = 1 to oo" (old Serbian joke).

Time to go to bed, Herr Mueckenheim. Have a good hangover!
b***@gmail.com
2018-04-23 15:42:23 UTC
Permalink
LoL if all else fails, WM either starts hallucinating matheology or potential/actual infinite. What a creep and loser.
b***@gmail.com
2018-04-23 14:55:46 UTC
Permalink
all rational numbers can be enumerated because there is a bijection f : N -> Q. that for a cofinite ( not cofinal ) set M , the result:

P = { f(n) | n in M }

is also cofinite, is very easy to see. if now WM thinks that a cofinite image of a cofinite set, is a contradiction to a bijection,

then the existence of natural numbers would be also a contradiction, since they can be bijected to them selves by id : N -> N.

but the natural numbers exist by the axiom of inifinity. but WM is so stupid, he not only believes that cofinity contradicts bijections,

he can also not write down the axiom of infinity. his PDF page 45 is still wrong.
b***@gmail.com
2018-04-23 15:07:28 UTC
Permalink
so what is the error of WM. he says for every n, there are infinitely many not yet enumerated rational numbers.

so he takes the cofinite set M = {n, n+1, n+2, ...}. and say they are not enumerated. but this is nonsense, they are all enumerated,

its just the other cofinite set

P = {f(n), f(n+1), f(n+2), ...}

how crazy is WM? answer augsburg crank crazy.

appendix: technical detail for an infinite domain C, call a set A cofinite, if A = C \ B for some finite set B. both M and P are cofinite.
WM
2018-04-23 15:45:53 UTC
Permalink
Post by b***@gmail.com
all rational numbers can be enumerated because there is a bijection f : N -> Q. that
has been brought down to us by the Lord.

Regards, WM
Alan Smaill
2018-04-23 14:48:26 UTC
Permalink
Post by WM
Post by Python
Post by Python
Post by WM
Forall n in |N: f(n) is not empty.
f can change only at an indexed step.
==> f annot be empty in the limit.
There is no "f in the limit", there is f(n) for any n. Period.
You are wrong. Proof: f(1) = 9.
[...]
Post by WM
Every step before omega is insufficient. Proof: f(n) = 9n.
Post by Python
In
McDuck case you have a sequence and a set limit. PERIOD.
The sequence is (f(n)) and the set limit is empty set.
But that is not the sequence of sets at issue.
That's a sequence of numbers.
Different problem.

Guess what, different problems have different answers!!

(sheesh)
Post by WM
Regards, WM
--
Alan Smaill
WM
2018-04-23 15:46:00 UTC
Permalink
Post by Alan Smaill
Post by WM
Every step before omega is insufficient. Proof: f(n) = 9n.
Post by Python
In
McDuck case you have a sequence and a set limit. PERIOD.
The sequence is (f(n)) and the set limit is empty set.
But that is not the sequence of sets at issue.
That's a sequence of numbers.
Different problem.
In mathematics we can reason as follows: The sequence is a sequence of cardinal numbers greater than 0. That means the corresponding sets have elements. So they are not empty for ev. But according to set theory the set limit is empty. So the elements must disappear. Alas they do not where it is possible, namely at n.
Post by Alan Smaill
Guess what, different problems have different answers!!
Apply mathematical reasoning: A set with card 3 cannot be empty.
Post by Alan Smaill
(sheesh)
sheesh.

Regards, WM
Zelos Malum
2018-04-23 05:34:13 UTC
Permalink
Post by WM
Post by b***@gmail.com
WM observed that a function on the natural numbers can have two extensions that disagree on omega.
In fact, omega is not at all required.
The question is whether all rational numbers can be indexed by natural numbers. The answer is no. For every natural number there remain infinitely many rational numbers without index.
Regards, WM
Moron, that is only true if you think of it as an algorithm where you do a step by step process which NO ONE IS FUCKING DOING!
WM
2018-04-23 11:48:24 UTC
Permalink
Post by Zelos Malum
Post by WM
The question is whether all rational numbers can be indexed by natural numbers. The answer is no. For every natural number there remain infinitely many rational numbers without index.
that is only true if you think of it as an algorithm where you do a step by step process which NO ONE IS FUCKING DOING!
The sequence of natural numbers can be interrupted at every point. If matheologians claim their delusions, it is even obligatory for true mathematicians to interrupt the sequence at every point and to analyze it. The result is: Fot all n in |N: There remain infinitely many fractions without index.

Regards, WM
Zelos Malum
2018-04-25 05:47:36 UTC
Permalink
Post by WM
The sequence of natural numbers can be interrupted at every point. If matheologians claim their delusions, it is even obligatory for true mathematicians to interrupt the sequence at every point and to analyze it. The result is: Fot all n in |N: There remain infinitely many fractions without index.
Which is irrelevant, you don't need to stpo or anything. You are again thinking in terms of algorithms which is not how things work. We are talking about sets in their entirety and there, the bijection exists.
Post by WM
The set is one entity. I does not index anything. Indexing can only be done by natural numbers.
You can have any index set you want, including real numbers =)
Post by WM
It is trivial to prove the contrary too. Contradiction.
For all n in |N: There are oo many not enumerated fractions.
Which is not a proof that there is no bijection, that is a proof that step by step thinking like yorus don't grant it.
WM
2018-04-25 10:03:03 UTC
Permalink
Post by Zelos Malum
Post by WM
The sequence of natural numbers can be interrupted at every point. If matheologians claim their delusions, it is even obligatory for true mathematicians to interrupt the sequence at every point and to analyze it. The result is: Fot all n in |N: There remain infinitely many fractions without index.
Which is irrelevant
No, you have to suppress it in order to maintain your matheologial delusions.
Post by Zelos Malum
You are again thinking in terms of algorithms which is not how things work.
It is the only way to proceed in mathematics. That distinguishes matheology from mathematics.
Post by Zelos Malum
We are talking about sets in their entirety and there, the bijection exists.
That is belief, not mathematics. Mathematics is the formal proof that the bijection does not exist, for instance here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 256.
Post by Zelos Malum
Post by WM
For all n in |N: There are oo many not enumerated fractions.
Which is not a proof that there is no bijection, that is a proof that step by step thinking like yorus don't grant it.
The sequence of natural numbers including its construction *is* step by step. See also Hausdorff's counting to omega and beyond --- step by step. Of course he is not a modern author. But your great present leaders endorse it:

Emil Jeřábek counterfactually claimed: "This does not violate any Peano axioms. It is a perfectly valid and commonly used construction. [...] Peano axioms are axioms of natural numbers. The sequence here is not indexed by natural numbers, but by ordinals, so Peano axioms are irrelevant."

No comment necessary, I hope.

And Joel David Hamkins boasted: "I endorse this method."

Look, if you think mathematically, then set theory is recognized as nonsense. If you try to avoid to think mathematically, but to believe in "the whole set", you will lose many newbies.

Regards, WM
Zelos Malum
2018-04-25 11:52:54 UTC
Permalink
Post by WM
No, you have to suppress it in order to maintain your matheologial delusions.
No one has to surpress anything because it is wholy irrelevant. The fact that if you go like an algorithm and stop and has done finite steps is in no way relevant to set theory because it isn't doing thigns in a step by step fashion.
Post by WM
It is the only way to proceed in mathematics. That distinguishes matheology from mathematics.
That is where your assumption is, in mathematics that is not the case. Algorithms is the standard norm in computer science, but in mathematics thigns either exist or dosen't exist and they are not constructed in a step by step, algorithmical fashion, they simply either are or are not from the axioms.
It is a provable fact that the bijection exist, it is not a belief. It is pure mathematics, somethign you do not understand.

Your proof is fundamentally flawed on so many levels. Just because the limit is all of those intervals, and therefore Q, doesn't that mean a truth about any single of them is true for the limit itself.

You are again doing the usual erroneous extrapoulation of finite to infinite which doesn't work.
Post by WM
The sequence of natural numbers including its construction *is* step by step.
No it isn't, the set N has a specific property but the set itself exist as a whole isntantly, it is not constructed step by step.
WM
2018-04-25 16:53:11 UTC
Permalink
Post by Zelos Malum
Post by WM
No, you have to suppress it in order to maintain your matheologial delusions.
No one has to surpress anything because it is wholy irrelevant. The fact that if you go like an algorithm and stop and has done finite steps is in no way relevant to set theory because it isn't doing thigns in a step by step fashion.
Not after having revealed the fallacies perhaps. But Cantor understood that soem sets are countable, i.e., that all elements can be counted. Why do you think has he used that term, contrary to uncountable?
Post by Zelos Malum
Post by WM
It is the only way to proceed in mathematics. That distinguishes matheology from mathematics.
That is where your assumption is, in mathematics that is not the case. Algorithms is the standard norm in computer science, but in mathematics thigns either exist or dosen't exist and they are not constructed in a step by step, algorithmical fashion, they simply either are or are not from the axioms.
What axiom says that the rational numbers are countable? What axiom says that the algebraic numbers are a complete set and in bijection with the naturals? Why is a bijection based on individual pairs?
Post by Zelos Malum
It is a provable fact that the bijection exist, it is not a belief. It is pure mathematics,
No.
Post by Zelos Malum
somethign you do not understand.
Yes, I can't understand that abusing newbies like you is not yet punished.
Post by Zelos Malum
Your proof is fundamentally flawed on so many levels. Just because the limit is all of those intervals, and therefore Q, doesn't that mean a truth about any single of them is true for the limit itself.
A proof for all rationals proves something for all rationals, no?
Post by Zelos Malum
You are again doing the usual erroneous extrapoulation of finite to infinite which doesn't work.
It does not work, but it is applied in set theory: Remember: Countable sets.
Post by Zelos Malum
Post by WM
The sequence of natural numbers including its construction *is* step by step.
No it isn't, the set N has a specific property but the set itself exist as a whole isntantly, it is not constructed step by step.
Not by "if A then {A}, if n then n + 1?

Regards, WM
Python
2018-04-26 00:14:36 UTC
Permalink
Doktor Frankenheim wrote:
...
Post by WM
Yes, I can't understand that abusing newbies like you is not yet punished.
This is the reason you could "teach" you nonsense for years, Herr
Mueckenheim, from Augsburg: you knew you won't be punished.
Zelos Malum
2018-04-26 06:04:45 UTC
Permalink
Post by WM
Not after having revealed the fallacies perhaps. But Cantor understood that soem sets are countable, i.e., that all elements can be counted. Why do you think has he used that term, contrary to uncountable?
Because it is a useful one as the natural numbers can you cunt through to any point in finite time so you can "count" them, but for example real numbers you cannot, you won't get anywhere trying to count them.
Post by WM
What axiom says that the rational numbers are countable? What axiom says that the algebraic numbers are a complete set and in bijection with the naturals? Why is a bijection based on individual pairs?
It is not an axiom, it is a theorem, it is proven they are countable because there exists a bijection from them to the natural numbers.

Why a bijection is based on pairing 1 to another? that is a definition you moron.
Post by WM
No.
Yes, you can cry all you want but it is so.
Post by WM
Yes, I can't understand that abusing newbies like you is not yet punished.
No, you cannot understand logic at all, that is your whole issue.

No one is abusing anyone, they teach proper mathematics, which you cannot understand.
Post by WM
A proof for all rationals proves something for all rationals, no?
A proof for all parts is not a proof for the whole.
Post by WM
It does not work, but it is applied in set theory: Remember: Countable sets.
It doesn't work, so use the entire set rather than parts.
Post by WM
Not by "if A then {A}, if n then n + 1?
That is a descriptive property of the set, not constructing it. Get it you moron?
WM
2018-04-26 09:23:43 UTC
Permalink
Post by Zelos Malum
Not after having revealed the fallacies perhaps. But Cantor understood that some sets are countable, i.e., that all elements can be counted. Why do you think has he used that term, contrary to uncountable?
Because it is a useful one as the natural numbers can you cunt through to any point in finite time so you can "count" them
or construct them all.
Post by Zelos Malum
, but for example real numbers you cannot, you won't get anywhere trying to count them.
Same with rationals. You don't get anywhere. Even with naturals you don't get anywhere.
Post by Zelos Malum
What axiom says that the rational numbers are countable? What axiom says that the algebraic numbers are a complete set and in bijection with the naturals? Why is a bijection based on individual pairs?
It is not an axiom, it is a theorem, it is proven they are countable because there exists a bijection from them to the natural numbers.
What axionm says that this bijection exists if it cannot be constructed but only believed?
Post by Zelos Malum
Why a bijection is based on pairing 1 to another? that is a definition
But you refuse to consider individual pairs.
Post by Zelos Malum
No one is abusing anyone, they teach proper mathematics, which you cannot understand.
A proof for all rationals proves something for all rationals, no?
A proof for all parts is not a proof for the whole.
But only all parts are important. "The whole" is simply an abbreviation. It does not change any of the parts. It does not change mathematics.
Post by Zelos Malum
It does not work, but it is applied in set theory: Remember: Countable sets.
It doesn't work, so use the entire set rather than parts.
The "entire set" does not contain anything more than all its parts.
Post by Zelos Malum
Not by "if A then {A}, if n then n + 1?
That is a descriptive property of the set, not constructing it.
What is in the set that cannot be constructed by these steps?

Regards, WM
Zelos Malum
2018-04-26 10:28:17 UTC
Permalink
Post by WM
or construct them all.
if you follow ZFC work on it, they aren't constructed.
Post by WM
Same with rationals. You don't get anywhere. Even with naturals you don't get anywhere.
You can with the bijection, it is a strange form of counting but it works in the same way it does for naturals.
Post by WM
What axionm says that this bijection exists if it cannot be constructed but only believed?
It is constructed using ZFC tools.
Post by WM
But you refuse to consider individual pairs.
The bijection as a whole is what matters, that it has the property of being injective, surjective with the domain being Naturals and the image being Q+, that is all that matters.
Post by WM
But only all parts are important. "The whole" is simply an abbreviation. It does not change any of the parts. It does not change mathematics.
It depends on the question, if I talk about some thigns the parts don't matter but the properties of the whole does.

It doesn't change mathematics the fact that you are too stupid to get it.
Post by WM
The "entire set" does not contain anything more than all its parts.
The set contains what it contains but that doesn't mean the set has the same properties as what it contains. A set can contain only finite sets but itself being infinite.
Post by WM
What is in the set that cannot be constructed by these steps?
That which is in it. That set contains some elements that have that specific property, it isn't constructed through it, it just describes a property of the set.
WM
2018-04-26 11:05:59 UTC
Permalink
Post by Zelos Malum
The bijection as a whole is what matters, that it has the property of being injective, surjective
That can only be proved by considering the elements.
Post by Zelos Malum
Post by WM
But only all parts are important. "The whole" is simply an abbreviation. It does not change any of the parts. It does not change mathematics.
It depends on the question,
Bijection requires to show all elements are mapped. McDuck shows that never all elements are mapped.

Regards, WM
Zelos Malum
2018-04-26 11:55:30 UTC
Permalink
Post by WM
That can only be proved by considering the elements.
You need to check that the predicate of them is true, which can be done through many means.
Post by WM
Bijection requires to show all elements are mapped. McDuck shows that never all elements are mapped.
Not at all, thats a limit thing, not a bijection between rationals and naturals.
s***@googlemail.com
2018-04-26 22:12:00 UTC
Permalink
And in fact all elements ARE mapped.
WM
2018-04-27 08:33:18 UTC
Permalink
Post by s***@googlemail.com
And in fact all elements ARE mapped.
There are no "facts" in set theory but belief.



To believe in the whole before having constructed it and in spite of the proof that it fails for all n in |N is not mathematics.

"To explain the unknown by the known is a logical procedure; to explain
the known by the unknown is a form of theological lunacy." ~David Brooks

Regards, WM
Zelos Malum
2018-04-27 11:15:33 UTC
Permalink
Post by WM
Post by s***@googlemail.com
And in fact all elements ARE mapped.
There are no "facts" in set theory but belief.
To believe in the whole before having constructed it and in spite of the proof that it fails for all n in |N is not mathematics.
"To explain the unknown by the known is a logical procedure; to explain
the known by the unknown is a form of theological lunacy." ~David Brooks
Regards, WM
Oh really? Well lets put that to the test shall we?

Let N be our domain and Q be our target.

by ZFC N U Q exists and P(N U Q) and P(P(N U Q)) and from there we can show that the cartesian product exist with ZFC and from tehre we can pick the subset that sates the criteria of a function, so that subset does exist. So can get functions from N to Q, and its not a step by step process, we have it all done and ready from the get go because they exist as whole sets.

And from tehre it is trivial to create a set theory predicate that gives us a bijection.

So no, it is a fact and provably so
WM
2018-04-27 18:13:27 UTC
Permalink
Post by Zelos Malum
Post by WM
Post by s***@googlemail.com
And in fact all elements ARE mapped.
There are no "facts" in set theory but belief.
To believe in the whole before having constructed it and in spite of the proof that it fails for all n in |N is not mathematics.
"To explain the unknown by the known is a logical procedure; to explain
the known by the unknown is a form of theological lunacy." ~David Brooks
Oh really? Well lets put that to the test shall we?
Let N be our domain and Q be our target.
by ZFC N U Q exists
So you start with the unknown. No reason to continue.

PS: McDuck constructs from 1 to n and from that we construct n+1 and look how far we can get with tfansfers at finite step.

Regards, WM
Zeit Geist
2018-04-28 08:15:01 UTC
Permalink
Post by WM
Post by Zelos Malum
Post by WM
Post by s***@googlemail.com
And in fact all elements ARE mapped.
There are no "facts" in set theory but belief.
To believe in the whole before having constructed it and in spite of the proof that it fails for all n in |N is not mathematics.
"To explain the unknown by the known is a logical procedure; to explain
the known by the unknown is a form of theological lunacy." ~David Brooks
Oh really? Well lets put that to the test shall we?
Let N be our domain and Q be our target.
by ZFC N U Q exists
So you start with the unknown. No reason to continue.
Just because you can’t understand it does not mean that it is unknown.
You are a crank and lost with the entire idea.
Post by WM
PS: McDuck constructs from 1 to n and from that we construct n+1 and look how far we can get with tfansfers at finite step.
PPS. You are clueless. Name a note that is not lost.
Post by WM
Regards, WM
You’re a jackass, ZG
WM
2018-04-28 09:59:07 UTC
Permalink
Post by Zeit Geist
Post by WM
PS: McDuck constructs from 1 to n and from that we construct n+1 and look how far we can get with tfansfers at finite step.
Name a note that is not lost.
Name a note that is not belonging to a tiny finite initial segment followed by infinitely many.

Regards, WM
b***@gmail.com
2018-04-28 16:17:50 UTC
Permalink
if you get two outcomes of a "function". then
either your are not using the same function:

f1(x) <> f2(x) is possible

or you are not using the same actual parameter:

f(x1) <> f(x2) is possible

but if the function and the actual parameter are
the same, then the outcome is the same:

f(x) <> f(x) is not possible

at least not for so called mathematical functions
that are neither supposed to depend whether it
rains outside nor whether your cat sneezes.

this also holds when x is a sequence and when
f is a limit operator, so you McDuck nonsense doesnt
make any sense, it doesnt make any sense since

when you reorder or regroup the actual parameter x
changes or when you compare "set theoretical limit"
with "real number limit" the function f changes,

and nobody cares that the outcome changes.
b***@gmail.com
2018-04-28 16:25:56 UTC
Permalink
somebody would care if you would name exactly
which theorem or theorems are violated, but

this would also require precise definition of
the f s and x s, and formulation of the theorems,

but so far there is only Volkswagen Omlette from
Augsburg Crank institute. very sad.
Post by b***@gmail.com
if you get two outcomes of a "function". then
f1(x) <> f2(x) is possible
f(x1) <> f(x2) is possible
but if the function and the actual parameter are
f(x) <> f(x) is not possible
at least not for so called mathematical functions
that are neither supposed to depend whether it
rains outside nor whether your cat sneezes.
this also holds when x is a sequence and when
f is a limit operator, so you McDuck nonsense doesnt
make any sense, it doesnt make any sense since
when you reorder or regroup the actual parameter x
changes or when you compare "set theoretical limit"
with "real number limit" the function f changes,
and nobody cares that the outcome changes.
Zelos Malum
2018-04-30 05:46:55 UTC
Permalink
Post by WM
Post by Zelos Malum
Post by WM
Post by s***@googlemail.com
And in fact all elements ARE mapped.
There are no "facts" in set theory but belief.
To believe in the whole before having constructed it and in spite of the proof that it fails for all n in |N is not mathematics.
"To explain the unknown by the known is a logical procedure; to explain
the known by the unknown is a form of theological lunacy." ~David Brooks
Oh really? Well lets put that to the test shall we?
Let N be our domain and Q be our target.
by ZFC N U Q exists
So you start with the unknown. No reason to continue.
PS: McDuck constructs from 1 to n and from that we construct n+1 and look how far we can get with tfansfers at finite step.
Regards, WM
I never started with unknown, the set N, as from set theory, exists by N, going all the way to Q is trivial
WM
2018-04-30 08:39:17 UTC
Permalink
Post by Zelos Malum
Post by WM
Post by Zelos Malum
by ZFC N U Q exists
So you start with the unknown. No reason to continue.
I never started with unknown, the set N, as from set theory, exists by N, going all the way to Q is trivial
Known are 1, 2, 3, and some more natural numbers, in any case a not actually infinite set. Most of |N is unknown and never will be known. You may conclude from the known naturals that |N or big portions of it exist. But instead you start with |N to explain the finite numbers. Your approach is to explain the known by the unknown, a form of theological lunacy.

Regards, WM
Zelos Malum
2018-04-30 09:18:42 UTC
Permalink
Post by WM
Known are 1, 2, 3, and some more natural numbers, in any case a not actually infinite set
Those are just as known as N, because they derive from axioms and is in set theory constructed.
WM
2018-05-01 09:42:14 UTC
Permalink
Post by Zelos Malum
Post by WM
Known are 1, 2, 3, and some more natural numbers, in any case a not actually infinite set
Those are just as known as N, because they derive from axioms and is in set theory constructed.
You believe in te complete set of natural numnbers. If the sequence of natural numbers has a limit (cf. McDuck), then the sequence of revolutions of a merry-go-round (cf. R.M. Rilke), has a limit too (since its revolutions can be enumerated). What will happen then in the Jardin du Luxembourg? Will the carousel horses jump from the carousel?

Regards, WM
b***@gmail.com
2018-05-01 10:21:19 UTC
Permalink
Well this is a nice analogy, the carousel is the
set, and the natural numbers are the horses.

Well its an infinite carousel, but you see the
distinction between set and element, something

that the Augsburg Crank institute didn't understand
so far. All that they understand is

Volkswagen Omlette.
Post by WM
Post by Zelos Malum
Post by WM
Known are 1, 2, 3, and some more natural numbers, in any case a not actually infinite set
Those are just as known as N, because they derive from axioms and is in set theory constructed.
You believe in te complete set of natural numnbers. If the sequence of natural numbers has a limit (cf. McDuck), then the sequence of revolutions of a merry-go-round (cf. R.M. Rilke), has a limit too (since its revolutions can be enumerated). What will happen then in the Jardin du Luxembourg? Will the carousel horses jump from the carousel?
Regards, WM
Zelos Malum
2018-05-03 06:01:54 UTC
Permalink
I accept the axiom as it is. Nothing more about it.
WM
2018-05-03 09:42:41 UTC
Permalink
Post by Zelos Malum
I accept the axiom as it is. Nothing more about it.
You accept the axiom in a wrong interpretation. The axiom does not create omega or a limit.

But if you accept the wrong interpretation, then you also have to accept the limit in the merry-go-round: If the sequence of natural numbers has a limit (cf. McDuck), then the sequence of revolutions of a merry-go-round (cf. R.M. Rilke), has a limit too (since its revolutions can be enumerated). What will happen then in the Jardin du Luxembourg? Will the carousel horses jump from the carousel?

Is your acquired brain damage really too severe to understand?

Regards, WM
Zelos Malum
2018-05-03 10:23:58 UTC
Permalink
Post by WM
You accept the axiom in a wrong interpretation
This comes from someone who cannot understand basic quantifiers.
Post by WM
The axiom does not create omega or a limit.
It states that a set exists with 2 properties to it, that is not an intepritation, it is simply what it says in FOL.
Post by WM
Is your acquired brain damage really too severe to understand?
Not as bad as yours, how many times do I have to tell you, you cannot extrapolate from finite to the infinite or to the limit.
b***@gmail.com
2018-05-06 13:22:54 UTC
Permalink
When will you correct your botched AOI on
page 45 of your awful PDF? Well actually its

any a huge pile of shit, so a currected AOI
wouldn't make any dent anyway. So maybe lets

leave it like that, so that everybody on the
internet can have its laugh.
Post by WM
Post by Zelos Malum
I accept the axiom as it is. Nothing more about it.
You accept the axiom in a wrong interpretation. The axiom does not create omega or a limit.
But if you accept the wrong interpretation, then you also have to accept the limit in the merry-go-round: If the sequence of natural numbers has a limit (cf. McDuck), then the sequence of revolutions of a merry-go-round (cf. R.M. Rilke), has a limit too (since its revolutions can be enumerated). What will happen then in the Jardin du Luxembourg? Will the carousel horses jump from the carousel?
Is your acquired brain damage really too severe to understand?
Regards, WM
b***@gmail.com
2018-05-06 13:26:48 UTC
Permalink
This nonsense here, with the missing inner
forall quantifier, when will it be corrected?

WMs blunder
https://gist.github.com/jburse/fb43afd01048feac7028b5642817af0a#gistcomment-2383428
Post by b***@gmail.com
When will you correct your botched AOI on
page 45 of your awful PDF? Well actually its
any a huge pile of shit, so a currected AOI
wouldn't make any dent anyway. So maybe lets
leave it like that, so that everybody on the
internet can have its laugh.
Post by WM
Post by Zelos Malum
I accept the axiom as it is. Nothing more about it.
You accept the axiom in a wrong interpretation. The axiom does not create omega or a limit.
But if you accept the wrong interpretation, then you also have to accept the limit in the merry-go-round: If the sequence of natural numbers has a limit (cf. McDuck), then the sequence of revolutions of a merry-go-round (cf. R.M. Rilke), has a limit too (since its revolutions can be enumerated). What will happen then in the Jardin du Luxembourg? Will the carousel horses jump from the carousel?
Is your acquired brain damage really too severe to understand?
Regards, WM
Zeit Geist
2018-04-28 08:12:08 UTC
Permalink
Post by WM
Post by s***@googlemail.com
And in fact all elements ARE mapped.
There are no "facts" in set theory but belief.
This just shows the deep level of your misunderstanding.
Post by WM
To believe in the whole before having constructed it and in spite of the proof that it fails for all n in |N is not mathematics.
"To explain the unknown by the known is a logical procedure; to explain
the known by the unknown is a form of theological lunacy." ~David Brooks
Regards, WM
Sma, ZG
WM
2018-04-27 08:33:13 UTC
Permalink
Post by Zelos Malum
The bijection as a whole is what matters, that it has the property of being injective, surjective with the domain being Naturals and the image being Q+, that is all that matters.
To believe in the whole before having constructed it and in spite of the proof that it fails for all n in |N is not mathematics.

"To explain the unknown by the known is a logical procedure; to explain
the known by the unknown is a form of theological lunacy." ~David Brooks

Regards, WM
Zelos Malum
2018-04-27 11:10:01 UTC
Permalink
Post by WM
To believe in the whole before having constructed it and in spite of the proof that it fails for all n in |N is not mathematics.
It is trivially shown to exist in ZFC, it doesn't need to be constructed in the sense you want to think of it.
Zeit Geist
2018-04-28 08:11:08 UTC
Permalink
Post by WM
Post by Zelos Malum
The bijection as a whole is what matters, that it has the property of being injective, surjective with the domain being Naturals and the image being Q+, that is all that matters.
To believe in the whole before having constructed it and in spite of the proof that it fails for all n in |N is not mathematics.
First of all, you are NOT allowed to DEFINE what a “bijection” is. A “bijection” IS GIVEN by DEFINITION. This definition is simply that a function, f, is a BIJECTION if and only if f is a SURJECTION and an INJECTION. You seem to have a problem with the injection part. Since you are dense (not to mention slow), I shall define what it is for a function to be an injection. A function, f mapping X to Y, is an INJECTION iff for any y in Y, there is some x in X such that f(x)=y. If you don’t understand what s function is, and I won’t be surprised if you don’t. Then your beyond help. However, the fact is for a function to surject onto another set we only need to find some element of the domain, X, that maps to our chosen element of the range, Y. So, if you are mapping from an ordered set, such as the natural numbers, then we need only show that each number in their range, which is the rational numbers in this case, there is EVENTUALLY a some element of the domain such that f(x)=y. And we know for sure that each rational number is EVENTUALLY hit by a number in the domain.

I apologize if the above is too complex for your dense brain, but if wish to discuss mathematics you need at least a minimum of UNDERSTANDING.
Post by WM
"To explain the unknown by the known is a logical procedure; to explain
the known by the unknown is a form of theological lunacy." ~David Brooks
All that was shown was done by the know, so your quote is irrelevant. As a matter of fact, all YOU say is irrelevant.
Post by WM
Regards, WM
Suck my ass, ZG
WM
2018-04-28 09:57:00 UTC
Permalink
Post by Zeit Geist
we need only show that each number in their range, which is the rational numbers in this case, there is EVENTUALLY a some element of the domain such that f(x)=y. And we know for sure that each rational number is EVENTUALLY hit by a number in the domain.
You now with same certainty that each rational number belongs to a tiny finite initial segment which is followed by an infinity of rational numbers. But you are too blind to understande that in logic, the proof: "For all n in |N: there are infinitely many rationals not indexed", is a contradiction of what you believe (not of what you have proved in fact, namely potential infinity).

Regards, WM
b***@gmail.com
2018-04-25 13:51:33 UTC
Permalink
because your proof in your PDF is flawed. You
assume that S_n is non empty. but your S_n
are all empty.

there are no centered integral intervals with
unenumerated rational numbers. alread the
rational number 1/2 is enumerated.

your proof must be result of heavy drinking.
Post by WM
Regards, WM
WM
2018-04-25 16:53:07 UTC
Permalink
Post by b***@gmail.com
You
assume that S_n is non empty. but your S_n
are all empty.
Which n makes a first S_n empty?
Post by b***@gmail.com
there are no centered integral intervals with
unenumerated rational numbers. alread the
rational number 1/2 is enumerated.
THe first interval s_1 has most hits, about half of all hits, alas for every n in |N, less than 0.000000000000000000000000000000000000001 % of its fractions are enumerated.
Regards, WM
b***@gmail.com
2018-04-25 17:29:19 UTC
Permalink
you cannot percent fraction from a finity when
the whole is infinite. i guess you are total

out of your mind WM. one bijection f : N+ -> Q+ is
for example, no rational number will be missing:

the function f(n) = r:

r = 1;
while (n <> 1) {
if (n mod 2 == 0) {
r = r/(r+1);
} else {
r = r+1;
}
n = n div 2;
}

and for the inverse the following definition, f^(-1)(r) = n:

n = 1;
while (r <> 1) {
n = n*2;
if (r < 1) {
r = r/(1-r);
} else {
r = r-1;
n = n+1;
}
}

some example runs:

f^(-1)(115/265) = 5144575

f(5144575) = 115/265

which rational number is not bijected?
WM
2018-04-25 17:45:02 UTC
Permalink
Post by b***@gmail.com
which rational number is not bijected?
Every rational number is indexed.

This can have two explanations:

In actual infinity, there are all rationals but also all naturals.
All rationals are indexed.
But for all naturals we can prove that they do not index all rationals.
This is a contradiction, since: What else could index?

In potential infinity there is no all, neither all naturals nor all rationals.
No contradiction.

Regards, WM
Python
2018-04-26 00:16:38 UTC
Permalink
Post by WM
Post by b***@gmail.com
which rational number is not bijected?
Every rational number is indexed.
Good! Case closed! card Q = card N

Send this good news to all students, abused before Herr Mueckenheim
cleared his mind.
b***@gmail.com
2018-04-26 06:57:17 UTC
Permalink
no you cannot proof that every natural number does not index every rational number. you are halucinating and confused WM.

also "all" vs "every", "potential" vs "actual" doesnt save your blunder. thats just incomprehensible gibberish,

not related to bijection.
WM
2018-04-26 09:26:31 UTC
Permalink
Post by b***@gmail.com
no you cannot proof that every natural number does not index every rational number.
I have proved that up to every natural number infinitely many rationals are not indexed. "For all n in |N: P(n) is sufficient in logic to disprove the existence of any n with ~P(n). But there is nothing else than natural numbers that could index rationals.

Regards, WM
b***@gmail.com
2018-04-26 09:32:57 UTC
Permalink
no there are no natural numbers >0 that do not
index a rational number in the bijection i

have given. also there id no rational number >0
that doesnt have an index.

you are halucinating and confused.
Zelos Malum
2018-04-23 05:25:08 UTC
Permalink
Post by b***@gmail.com
WM observed that a function on the natural numbers can have two extensions that disagree on omega. now he thinks set theory is inconsistent. can we not follow that he is completly nuts?
Of course not, he cannot even understand basic FOL
Archimedes Plutonium
2022-07-14 17:04:20 UTC
Permalink
Why is Andrew Wiles allowed to teach, when he cannot tell apart a ellipse from a oval in slant cut of cone?? How many students have to be brainwashed by this math quack-crank???

3rd published book

AP's Proof-Ellipse was never a Conic Section // Math proof series, book 1 Kindle Edition
by Archimedes Plutonium (Author)

Ever since Ancient Greek Times it was thought the slant cut into a cone is the ellipse. That was false. For the slant cut in every cone is a Oval, never an Ellipse. This book is a proof that the slant cut is a oval, never the ellipse. A slant cut into the Cylinder is in fact a ellipse, but never in a cone.

Product details
• ASIN ‏ : ‎ B07PLSDQWC
• Publication date ‏ : ‎ March 11, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 1621 KB
• Text-to-Speech ‏ : ‎ Enabled
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 20 pages
• Lending ‏ : ‎ Enabled



Proofs Ellipse is never a Conic section, always a Cylinder section and a Well Defined Oval definition//Student teaches professor series, book 5 Kindle Edition
by Archimedes Plutonium (Author)

Last revision was 14May2022. This is AP's 68th published book of science.

Preface: A similar book on single cone cut is a oval, never a ellipse was published in 11Mar2019 as AP's 3rd published book, but Amazon Kindle converted it to pdf file, and since then, I was never able to edit this pdf file, and decided rather than struggle and waste time, decided to leave it frozen as is in pdf format. Any new news or edition of ellipse is never a conic in single cone is now done in this book. The last thing a scientist wants to do is wade and waddle through format, when all a scientist ever wants to do is science itself. So all my new news and thoughts of Conic Sections is carried out in this 68th book of AP. And believe you me, I have plenty of new news.

In the course of 2019 through 2022, I have had to explain this proof often on Usenet, sci.math and sci.physics. And one thing that constant explaining does for a mind of science, is reduce the proof to its stripped down minimum format, to bare bones skeleton proof. I can prove the slant cut in single cone is a Oval, never the ellipse in just a one sentence proof. Proof-- A single cone and oval have just one axis of symmetry, while a ellipse requires 2 axes of symmetry, hence slant cut is always a oval, never the ellipse.

Product details
• ASIN ‏ : ‎ B081TWQ1G6
• Publication date ‏ : ‎ November 21, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 827 KB
• Simultaneous device usage ‏ : ‎ Unlimited
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 51 pages
• Lending ‏ : ‎ Enabled

#11-2, 11th published book

World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
by Archimedes Plutonium (Author)

Last revision was 15Dec2021. This is AP's 11th published book of science.
Preface:
Actually my title is too modest, for the proof that lies within this book makes it the World's First Valid Proof of Fundamental Theorem of Calculus, for in my modesty, I just wanted to emphasis that calculus was geometry and needed a geometry proof. Not being modest, there has never been a valid proof of FTC until AP's 2015 proof. This also implies that only a geometry proof of FTC constitutes a valid proof of FTC.

Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis". And very surprising that most math professors cannot tell the difference between a "proving something" and that of "analyzing something". As if an analysis is the same as a proof. We often analyze various things each and every day, but few if none of us consider a analysis as a proof. Yet that is what happened in the science of mathematics where they took an analysis and elevated it to the stature of being a proof, when it was never a proof.

To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?

Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.


Product details
ASIN ‏ : ‎ B07PQTNHMY
Publication date ‏ : ‎ March 14, 2019
Language ‏ : ‎ English
File size ‏ : ‎ 1309 KB
Text-to-Speech ‏ : ‎ Enabled
Screen Reader ‏ : ‎ Supported
Enhanced typesetting ‏ : ‎ Enabled
X-Ray ‏ : ‎ Not Enabled
Word Wise ‏ : ‎ Not Enabled
Print length ‏ : ‎ 154 pages
Lending ‏ : ‎ Enabled
Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#2 in 45-Minute Science & Math Short Reads
#134 in Calculus (Books)
#20 in Calculus (Kindle Store)
Archimedes Plutonium
2022-07-14 21:47:20 UTC
Permalink
Terence Tao. Why is Terence Tao allowed to teach, when he cannot tell apart a ellipse from a oval in slant cut of cone?? How many students have to be brainwashed by this math quack-crank???
Post by Archimedes Plutonium
3rd published book
AP's Proof-Ellipse was never a Conic Section // Math proof series, book 1 Kindle Edition
by Archimedes Plutonium (Author)
Ever since Ancient Greek Times it was thought the slant cut into a cone is the ellipse. That was false. For the slant cut in every cone is a Oval, never an Ellipse. This book is a proof that the slant cut is a oval, never the ellipse. A slant cut into the Cylinder is in fact a ellipse, but never in a cone.
Product details
• ASIN ‏ : ‎ B07PLSDQWC
• Publication date ‏ : ‎ March 11, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 1621 KB
• Text-to-Speech ‏ : ‎ Enabled
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 20 pages
• Lending ‏ : ‎ Enabled


Proofs Ellipse is never a Conic section, always a Cylinder section and a Well Defined Oval definition//Student teaches professor series, book 5 Kindle Edition
by Archimedes Plutonium (Author)
Last revision was 14May2022. This is AP's 68th published book of science.
Preface: A similar book on single cone cut is a oval, never a ellipse was published in 11Mar2019 as AP's 3rd published book, but Amazon Kindle converted it to pdf file, and since then, I was never able to edit this pdf file, and decided rather than struggle and waste time, decided to leave it frozen as is in pdf format. Any new news or edition of ellipse is never a conic in single cone is now done in this book. The last thing a scientist wants to do is wade and waddle through format, when all a scientist ever wants to do is science itself. So all my new news and thoughts of Conic Sections is carried out in this 68th book of AP. And believe you me, I have plenty of new news.
In the course of 2019 through 2022, I have had to explain this proof often on Usenet, sci.math and sci.physics. And one thing that constant explaining does for a mind of science, is reduce the proof to its stripped down minimum format, to bare bones skeleton proof. I can prove the slant cut in single cone is a Oval, never the ellipse in just a one sentence proof. Proof-- A single cone and oval have just one axis of symmetry, while a ellipse requires 2 axes of symmetry, hence slant cut is always a oval, never the ellipse.
Product details
• ASIN ‏ : ‎ B081TWQ1G6
• Publication date ‏ : ‎ November 21, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 827 KB
• Simultaneous device usage ‏ : ‎ Unlimited
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 51 pages
• Lending ‏ : ‎ Enabled
#11-2, 11th published book
World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
by Archimedes Plutonium (Author)
Last revision was 15Dec2021. This is AP's 11th published book of science.
Actually my title is too modest, for the proof that lies within this book makes it the World's First Valid Proof of Fundamental Theorem of Calculus, for in my modesty, I just wanted to emphasis that calculus was geometry and needed a geometry proof. Not being modest, there has never been a valid proof of FTC until AP's 2015 proof. This also implies that only a geometry proof of FTC constitutes a valid proof of FTC.
Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis". And very surprising that most math professors cannot tell the difference between a "proving something" and that of "analyzing something". As if an analysis is the same as a proof. We often analyze various things each and every day, but few if none of us consider a analysis as a proof. Yet that is what happened in the science of mathematics where they took an analysis and elevated it to the stature of being a proof, when it was never a proof.
To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?
Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.
Product details
ASIN ‏ : ‎ B07PQTNHMY
Publication date ‏ : ‎ March 14, 2019
Language ‏ : ‎ English
File size ‏ : ‎ 1309 KB
Text-to-Speech ‏ : ‎ Enabled
Screen Reader ‏ : ‎ Supported
Enhanced typesetting ‏ : ‎ Enabled
X-Ray ‏ : ‎ Not Enabled
Word Wise ‏ : ‎ Not Enabled
Print length ‏ : ‎ 154 pages
Lending ‏ : ‎ Enabled
Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#2 in 45-Minute Science & Math Short Reads
#134 in Calculus (Books)
#20 in Calculus (Kindle Store)
Archimedes Plutonium
2022-07-15 19:26:42 UTC
Permalink
Thomas Hales. Why is Thomas Hales allowed to teach, when he cannot tell apart a ellipse from a oval in slant cut of cone?? How many students have to be brainwashed by this math quack-crank???

AP asks: Jan, is this why Thomas Hales could never understand that Calculus Fundamental Theorem required a geometry proof, for Calculus is geometry, and not the idiot Hales "limit analysis", for if Hales is so stupid in not recognizing the slant cut in single right circular cone is a Oval never Ellipse, he is too stupid to do right by calculus.
Post by Archimedes Plutonium
3rd published book
AP's Proof-Ellipse was never a Conic Section // Math proof series, book 1 Kindle Edition
by Archimedes Plutonium (Author)
Ever since Ancient Greek Times it was thought the slant cut into a cone is the ellipse. That was false. For the slant cut in every cone is a Oval, never an Ellipse. This book is a proof that the slant cut is a oval, never the ellipse. A slant cut into the Cylinder is in fact a ellipse, but never in a cone.
Product details
• ASIN ‏ : ‎ B07PLSDQWC
• Publication date ‏ : ‎ March 11, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 1621 KB
• Text-to-Speech ‏ : ‎ Enabled
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 20 pages
• Lending ‏ : ‎ Enabled


Proofs Ellipse is never a Conic section, always a Cylinder section and a Well Defined Oval definition//Student teaches professor series, book 5 Kindle Edition
by Archimedes Plutonium (Author)
Last revision was 14May2022. This is AP's 68th published book of science.
Preface: A similar book on single cone cut is a oval, never a ellipse was published in 11Mar2019 as AP's 3rd published book, but Amazon Kindle converted it to pdf file, and since then, I was never able to edit this pdf file, and decided rather than struggle and waste time, decided to leave it frozen as is in pdf format. Any new news or edition of ellipse is never a conic in single cone is now done in this book. The last thing a scientist wants to do is wade and waddle through format, when all a scientist ever wants to do is science itself. So all my new news and thoughts of Conic Sections is carried out in this 68th book of AP. And believe you me, I have plenty of new news.
In the course of 2019 through 2022, I have had to explain this proof often on Usenet, sci.math and sci.physics. And one thing that constant explaining does for a mind of science, is reduce the proof to its stripped down minimum format, to bare bones skeleton proof. I can prove the slant cut in single cone is a Oval, never the ellipse in just a one sentence proof. Proof-- A single cone and oval have just one axis of symmetry, while a ellipse requires 2 axes of symmetry, hence slant cut is always a oval, never the ellipse.
Product details
• ASIN ‏ : ‎ B081TWQ1G6
• Publication date ‏ : ‎ November 21, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 827 KB
• Simultaneous device usage ‏ : ‎ Unlimited
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 51 pages
• Lending ‏ : ‎ Enabled
#11-2, 11th published book
World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
by Archimedes Plutonium (Author)
Last revision was 15Dec2021. This is AP's 11th published book of science.
Actually my title is too modest, for the proof that lies within this book makes it the World's First Valid Proof of Fundamental Theorem of Calculus, for in my modesty, I just wanted to emphasis that calculus was geometry and needed a geometry proof. Not being modest, there has never been a valid proof of FTC until AP's 2015 proof. This also implies that only a geometry proof of FTC constitutes a valid proof of FTC.
Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis". And very surprising that most math professors cannot tell the difference between a "proving something" and that of "analyzing something". As if an analysis is the same as a proof. We often analyze various things each and every day, but few if none of us consider a analysis as a proof. Yet that is what happened in the science of mathematics where they took an analysis and elevated it to the stature of being a proof, when it was never a proof.
To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?
Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.
Product details
ASIN ‏ : ‎ B07PQTNHMY
Publication date ‏ : ‎ March 14, 2019
Language ‏ : ‎ English
File size ‏ : ‎ 1309 KB
Text-to-Speech ‏ : ‎ Enabled
Screen Reader ‏ : ‎ Supported
Enhanced typesetting ‏ : ‎ Enabled
X-Ray ‏ : ‎ Not Enabled
Word Wise ‏ : ‎ Not Enabled
Print length ‏ : ‎ 154 pages
Lending ‏ : ‎ Enabled
Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#2 in 45-Minute Science & Math Short Reads
#134 in Calculus (Books)
#20 in Calculus (Kindle Store)
Archimedes Plutonium
2022-07-16 01:49:12 UTC
Permalink
John Stillwell. Honestly why is John Stillwell allowed to teach, when he cannot tell apart a ellipse from a oval in slant cut of cone?? How many students have to be brainwashed by this math quack-crank???
AP asks: Jan, is this why John Stillwell could never understand that Calculus Fundamental Theorem required a geometry proof, for Calculus is geometry, and not the idiot Hales "limit analysis", for if Hales is so stupid in not recognizing the slant cut in single right circular cone is a Oval never Ellipse, he is too stupid to do right by calculus.
Post by Archimedes Plutonium
3rd published book
AP's Proof-Ellipse was never a Conic Section // Math proof series, book 1 Kindle Edition
by Archimedes Plutonium (Author)
Ever since Ancient Greek Times it was thought the slant cut into a cone is the ellipse. That was false. For the slant cut in every cone is a Oval, never an Ellipse. This book is a proof that the slant cut is a oval, never the ellipse. A slant cut into the Cylinder is in fact a ellipse, but never in a cone.
Product details
• ASIN ‏ : ‎ B07PLSDQWC
• Publication date ‏ : ‎ March 11, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 1621 KB
• Text-to-Speech ‏ : ‎ Enabled
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 20 pages
• Lending ‏ : ‎ Enabled


Proofs Ellipse is never a Conic section, always a Cylinder section and a Well Defined Oval definition//Student teaches professor series, book 5 Kindle Edition
by Archimedes Plutonium (Author)
Last revision was 14May2022. This is AP's 68th published book of science.
Preface: A similar book on single cone cut is a oval, never a ellipse was published in 11Mar2019 as AP's 3rd published book, but Amazon Kindle converted it to pdf file, and since then, I was never able to edit this pdf file, and decided rather than struggle and waste time, decided to leave it frozen as is in pdf format. Any new news or edition of ellipse is never a conic in single cone is now done in this book. The last thing a scientist wants to do is wade and waddle through format, when all a scientist ever wants to do is science itself. So all my new news and thoughts of Conic Sections is carried out in this 68th book of AP. And believe you me, I have plenty of new news.
In the course of 2019 through 2022, I have had to explain this proof often on Usenet, sci.math and sci.physics. And one thing that constant explaining does for a mind of science, is reduce the proof to its stripped down minimum format, to bare bones skeleton proof. I can prove the slant cut in single cone is a Oval, never the ellipse in just a one sentence proof. Proof-- A single cone and oval have just one axis of symmetry, while a ellipse requires 2 axes of symmetry, hence slant cut is always a oval, never the ellipse.
Product details
• ASIN ‏ : ‎ B081TWQ1G6
• Publication date ‏ : ‎ November 21, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 827 KB
• Simultaneous device usage ‏ : ‎ Unlimited
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 51 pages
• Lending ‏ : ‎ Enabled
#11-2, 11th published book
World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
by Archimedes Plutonium (Author)
Last revision was 15Dec2021. This is AP's 11th published book of science.
Actually my title is too modest, for the proof that lies within this book makes it the World's First Valid Proof of Fundamental Theorem of Calculus, for in my modesty, I just wanted to emphasis that calculus was geometry and needed a geometry proof. Not being modest, there has never been a valid proof of FTC until AP's 2015 proof. This also implies that only a geometry proof of FTC constitutes a valid proof of FTC.
Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis". And very surprising that most math professors cannot tell the difference between a "proving something" and that of "analyzing something". As if an analysis is the same as a proof. We often analyze various things each and every day, but few if none of us consider a analysis as a proof. Yet that is what happened in the science of mathematics where they took an analysis and elevated it to the stature of being a proof, when it was never a proof.
To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?
Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.
Product details
ASIN ‏ : ‎ B07PQTNHMY
Publication date ‏ : ‎ March 14, 2019
Language ‏ : ‎ English
File size ‏ : ‎ 1309 KB
Text-to-Speech ‏ : ‎ Enabled
Screen Reader ‏ : ‎ Supported
Enhanced typesetting ‏ : ‎ Enabled
X-Ray ‏ : ‎ Not Enabled
Word Wise ‏ : ‎ Not Enabled
Print length ‏ : ‎ 154 pages
Lending ‏ : ‎ Enabled
Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#2 in 45-Minute Science & Math Short Reads
#134 in Calculus (Books)
#20 in Calculus (Kindle Store)
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