Cantor's diagonalization

[WM]

[snip]
)OK. But the identification step is essential in the Cantor
)proof. There is a step in the proof that says, "for if you
)claim that my number is at position x, then I will assert that it
)differs from your number in the x'th digit."
)
)So what if I simply claim that the number is somewhere in the
)list but that if I were to tell you where it was that would lead
)to a contradiction? It seems that the Cantor proof, at least in
)that form, might only be proving that the position on the list
)cannot be made definite.
You cannot claim to have made a list without being able to specify where
it is. Or to put it another way - you are saying that the number is in
there, without there being a way to find it. But there -is- a way to
find it. It must be in -one- of the slots. So we just start scanning the
list. If it is -in- the list, then we will encounter it. The only
numbers which result in ifinitely scanning the list are numbers -not- in
the list. So if it is in the list, we -can- find it.

That is the common conclusion.
It has lead to over 100 years of set theory.
It is wrong.
When scanning the list we will always remain in a domain that is preceded by a finite number of slots but is succeeded by an infinite number of slots.
This does never change - in infinity.

Regards, WM

[WM]

Remember, if all your reals are represented as infinite decimal
expansions,

which is impossible because representations are used to convey information.

then the act of comparing two numbers for equality is a
process that terminates iff the numbers are unequal.

Wrong. Simply compare number x and all other real numbers y. For every pair (x, y), there is a finite index n where they differ, but since the set N of indices is infinite, the process will never terminate, i.e., there will be, for every n, many undecided cases. "For every n undecided" however simply means "undecided in infinity".

It is impossible to distinguish real x by decimal representations from all other reals y. Therefore it is impossible to define x by its decimal representation.

Regards, WM

[Virgil]

Post by WM

Remember, if all your reals are represented as infinite decimal
expansions,

which is impossible because representations are used to convey information.

then the act of comparing two numbers for equality is a
process that terminates iff the numbers are unequal.

Wrong.

It is WM who is TOTALLY wrong here, as usual!
Given any two real numbers, if they are NOT equal, there will be a
finite decimal place at which they differ, which can be fond in finite
number of steps if it exists at all.

Post by WM
It is impossible to distinguish real x by decimal representations from all
other reals y.

Then there would not be representation of different reals by different
decimals.

Outside of WM's witless worthless wacky world of WMytheology,
any two different reals have sufficiently different decimal
representations to allow one to tell them apart.

Decimals would be useless otherwise.

Whereas is is only WMytheology that is useless. Decimals work!

[WM]

So what if I simply claim that the number is somewhere in the
list but that if I were to tell you where it was that would lead
to a contradiction? It seems that the Cantor proof, at least in
that form, might only be proving that the position on the list
cannot be made definite.

Given a map f: N -> R from the natural numbers to the reals, the
diagonal argument shows that there exists a real number x such that
f(k) != x for each k in N.

Indeed. The way I feel about it is that there is a difference between
the existence of an enumeration and the ability to produce one (or
construct or whatever you wish to name it). The diagonal argument that
R is not enumerable merely depends on the fact that even the existence
of an enumeration leads to a contradiction. For the constructable
(recursive, or whatever you wish to call it) numbers the ability to
*produce* one would lead to a contradiction, not the mere existence.
(Which proves that, while an enumeration exists it is impossible to
produce one.)
While many mathematicians would like it that mere existence would lead
to the showing of an occurrence of whatever, alas, this is not so.

Hopefully he is now at a place where he has better insights.

Regards, WM

[Virgil]

(Which proves that, while an enumeration exists it is impossible to
produce one.)

WM again claims the impossible possible!

Unless WM can PRODUCE an enumeration of the reals,
a surjection from |N to |R. which he can't,
his claim that one must exist is
just another lie.

[WM]

Consider the halting problem. Each well-formed program can be
encoded into an integer in N. Let the set of all such codes be
S. The codes of S may be partitioned into three parts S1, S2,
and S3, respectively those codes that represent programs that are
known to terminate, those that represent programs that are known
to not terminate, and those codes that represent programs for
which the halting problem is undecidable.

You forgot the set S4, consisting of programs that are decidable, but
whose status is not yet known. In principle, each member of S4 can be
moved to either S1 or S2 by simply enumerating proofs until the right
one is found; however, you never can reach a point where S4 is empty.

In principle every entry of the Cantor-list can be checked: however, you never can reach a point where all entries have been checked.

Regards, WM

[Virgil]

Post by WM

Consider the halting problem. Each well-formed program can be
encoded into an integer in N. Let the set of all such codes be
S. The codes of S may be partitioned into three parts S1, S2,
and S3, respectively those codes that represent programs that are
known to terminate, those that represent programs that are known
to not terminate, and those codes that represent programs for
which the halting problem is undecidable.

You forgot the set S4, consisting of programs that are decidable, but
whose status is not yet known. In principle, each member of S4 can be
moved to either S1 or S2 by simply enumerating proofs until the right
one is found; however, you never can reach a point where S4 is empty.

In principle every entry of the Cantor-list can be checked: however, you
never can reach a point where all entries have been checked.

A general rule for such checkings can cover
all infinitely many cases in one step.
And Cantor has such a rule!

So WM is
LYING !
AGAIN ! !
AS USUAL ! ! !

[Virgil]

Post by WM

[snip]
)OK. But the identification step is essential in the Cantor
)proof. There is a step in the proof that says, "for if you
)claim that my number is at position x, then I will assert that it
)differs from your number in the x'th digit."
)
)So what if I simply claim that the number is somewhere in the
)list but that if I were to tell you where it was that would lead
)to a contradiction? It seems that the Cantor proof, at least in
)that form, might only be proving that the position on the list
)cannot be made definite.
You cannot claim to have made a list without being able to specify where
it is. Or to put it another way - you are saying that the number is in
there, without there being a way to find it. But there -is- a way to
find it. It must be in -one- of the slots. So we just start scanning the
list. If it is -in- the list, then we will encounter it. The only
numbers which result in ifinitely scanning the list are numbers -not- in
the list. So if it is in the list, we -can- find it.

That is the common conclusion.
It has lead to over 100 years of set theory.
It is wrong.

WM's Claiming it wrong without proof does not make it wrong,
and all WM's such claims are without proof.
Cantor managed to find two quite independent proofs that he was right,
and neither WM, nor anyone else, has ever managed to show either of
Cantor's two proofs wrong.

In order for the reals to be provably countable, one must prove, for
instance, that there can be a list which includes all of them.

WM has not done this and cannot do this.

A rational has no more than two non-terminating decimal expansions and
an irrational has only one, but in any supposed list of ALL such
expansions one can easily prove, a la Cantor, that there are some
missing.

So WM is, as usual, totally, arrogantly and stupidly wrong.

[WM]

Post by Virgil

So if it is in the list, we -can- find it.
That is the common conclusion.
It has lead to over 100 years of set theory.
It is wrong.

WM's Claiming it wrong without proof does not make it wrong,
and all WM's such claims are without proof.

Look, if a certain real number is not in the list, then we will never know that.
If all rational numbers are in the list, then we will never have checked them all.
When checking an arbitrary list, then we will always stay in finite distance from the first row with infinitely many rows remaining to check.

What else remains? Unprovable (counterfactual) belief in the possibility to finish the list.

Regards, WM

[Virgil]

Post by WM

Post by Virgil

So if it is in the list, we -can- find it.
That is the common conclusion.
It has lead to over 100 years of set theory.
It is wrong.

WM's Claiming it wrong without proof does not make it wrong,
and all WM's such claims are without proof.

Look, if a certain real number is not in the list, then we will never know that.

That depends on the number and the list.

If every number on the list is given in one of its possible decimal
forms, Cantor showed how to construct any number of numbers not
themselves listed.

Post by WM
If all rational numbers are in the list, then we will never have checked them all.

Cantors diagonal method checks all!

Post by WM
When checking an arbitrary list, then we will always stay in finite distance
from the first row with infinitely many rows remaining to check.

Not when using Cantor's diagonal method!

Post by WM
What else remains?

Cantors first proof is still totally valid and Cantor's second proof has
not actually been invalidated.

Those who are familiar with Cantor's two proofs of the
uncountability of |R, which WM apparently is not,
at least not well enough to really understand them,
accept that |R is not countable.

NOTE ALSO: if the reals really are countable, they must also be
well-orderable, but no one has ever been able to demonstrate a
well-ordering of them.

And I, for one, will not accept that the reals are countable until I see
a specific well-ordering of them.
Does WM have such a well-ordering of |R? I doubt it!

[WM]

Post by Virgil

Post by WM
If all rational numbers are in the list, then we will never have checked them all.

Cantors diagonal method checks all!

Believing without proving leads to strange results, for instance: Cantor's list contains all rational numbers.

In mathematics these claims would have to proved. Both are nonsense.

Since every diagonl digit is followed by infinitely many diagonal digits, you will never reach even half of all. How can you reasonably claim to reach all?

Regards, WM

[i***@xlog.ch]

Believing without proving leads to strange results, for instance.

You conflate assumptions with believes. Nobody said
that the assumptions in any proofs of

Cantor are true, it can still happend that ZFC or
so turns out inconsistent.

You have no clue how math proofs work.

[i***@xlog.ch]

Post by i***@xlog.ch

Believing without proving leads to strange results, for instance.

You conflate assumptions with believes. Nobody said
that the assumptions in any proofs of
Cantor are true, it can still happend that ZFC or
so turns out inconsistent.
You have no clue how math proofs work.

Please put the bottle down.

Post by i***@xlog.ch
Cantor's list contains all rational numbers.

Its a list of real numbers, not rational numbers.

And the list assumption leads anyway to
a contradiction, so what are you bragging
about. Do you think it matters how quick
you get a conradiction in a mathematical proof?

Say [] is the contradiction. If we can proof

A -> []

Or

A -> B, B -> C, C -> []

How does it matter. By the "syllogistic rule",
already known to the Greek, A->B,B->C gives A->C,
in both cases we have still:

A -> []

and

A -> []

So it is irrelevant whether you think the list
cannot be constructed or we assume the list can
be constructed and show a contradiction via
diagonalization

in both cases we have show that |A| < |P(A)|,
only you WM didnt wanted to show that.

[i***@xlog.ch]

Post by i***@xlog.ch

Post by WM
Cantor's list contains all rational numbers.

Its a list of real numbers, not rational numbers.

Corr.:
Its supposed to be a countable list of real
numbers (from R), not rational numbers (from Q).

Countable lists of rational nummbers (from Q)
are trivial to construct, and exist easily.

Elements of Q are p/q. Besides that NxN is easily
countable, there are special enumerations for p/q,

so that only p/q with gcd(p,q)=1 are enumerated,
see for example Stern–Brocot tree
https://en.wikipedia.org/wiki/Stern%E2%80%93Brocot_tree

If you want I can give you a function f, that
enumerates all rationals in the interval [0,1].

[WM]

Post by i***@xlog.ch

Post by WM
Cantor's list contains all rational numbers.

Its a list of real numbers, not rational numbers.

Rational numbers are real numbers.

Post by i***@xlog.ch
And the list assumption leads anyway to
a contradiction, so what are you bragging
about. Do you think it matters how quick
you get a conradiction in a mathematical proof?

Yes, that is important if you never get to the desired contradiction. In case of an infinite set this is unavoidable.

Post by i***@xlog.ch
Say [] is the contradiction. If we can proof
A -> []
Or
A -> B, B -> C, C -> []

But you cannot prove []. Before havind the diagonal number you must have half of it. But you never have half of it.

Post by i***@xlog.ch
So it is irrelevant whether you think the list
cannot be constructed or we assume the list can
be constructed and show a contradiction via
diagonalization
in both cases we have show that |A| < |P(A)|,
only you WM didnt wanted to show that.

In order to show that, you have to present the whole diagonal number. But since I can prove that you never get half of it, your "proof" turns out nonsense.

Regards, WM

[i***@xlog.ch]

Post by WM
(1)
The union of the sequence of intervals [0, (n-1)/n] is [0, 1).
(2)
The union of the sequence of finite initial segments of the sequence of >
(3)
The sequence of partial sums 3.1, 3.14, 3.141, ... = 3.141... is not

What is this?
(1), (2), (3) hol die Polizei, WM is stupid as hell?

What is this claim like set theory blurs the
distinction between sets and classes? Why, where?

The distinction between sets and classes is not
strictly tied to infinity. If you want you can

make this distinction also for finite sets only,
similarly actual and potential are also not strictly

tied to infinity. *Actual* only means that you have a
"bag" for this set. *Potential* means you don't have this "bag".

For example set theory delegates many many potential
sets to classes. For example a set that you can draw

by a finite diagram such as:

* = { * }

Does not exist in ZFC, since it would violate the axiom
of regularity. Although in physics such "sets" exist:

http://www.scienceelf.com/121-2/

[Virgil]

Post by WM

Post by i***@xlog.ch

Post by WM
Cantor's list contains all rational numbers.

But cannot contain all real numbers. No list can! As Cantor twice proved!

Post by WM
Rational numbers are real numbers.

Post by i***@xlog.ch
And the list assumption leads anyway to
a contradiction, so what are you bragging
about. Do you think it matters how quick
you get a conradiction in a mathematical proof?

Yes, that is important if you never get to the desired contradiction. In case

Cantor got there twice!

Post by WM

Post by i***@xlog.ch
So it is irrelevant whether you think the list
cannot be constructed or we assume the list can
be constructed and show a contradiction via
diagonalization
in both cases we have show that |A| < |P(A)|,
only you WM didnt wanted to show that.

In order to show that, you have to present the whole diagonal number.

NOt when using Cantor's FIRST proof!

But construction of the "antidiagonal" is trivially simple.

For any list of decimals, the nth decimal has an nth decimal place
decimal digit. Changing each such digit by 5, always possible, creates a
new "antidiagonal"number, different in value from every listed number,
thus unlisted!

[g***@gmail.com]

Post by Virgil

Post by WM

Post by i***@xlog.ch

Post by WM
Cantor's list contains all rational numbers.

But cannot contain all real numbers. No list can! As Cantor twice proved!

Post by WM
Rational numbers are real numbers.

Post by i***@xlog.ch
And the list assumption leads anyway to
a contradiction, so what are you bragging
about. Do you think it matters how quick
you get a conradiction in a mathematical proof?

Yes, that is important if you never get to the desired contradiction. In case

Cantor got there twice!

Post by WM

Post by i***@xlog.ch
So it is irrelevant whether you think the list
cannot be constructed or we assume the list can
be constructed and show a contradiction via
diagonalization
in both cases we have show that |A| < |P(A)|,
only you WM didnt wanted to show that.

In order to show that, you have to present the whole diagonal number.

NOt when using Cantor's FIRST proof!
But construction of the "antidiagonal" is trivially simple.
For any list of decimals, the nth decimal has an nth decimal place
decimal digit. Changing each such digit by 5, always possible, creates a
new "antidiagonal"number, different in value from every listed number,
thus unlisted!

Nobody wants to base the definition of numbers on kindergarten jargon

THIS DIGIT CHANGES TO 5 UNLESS ITS 5 AND MAKE ALL THOSE DIGITS INTO A NUMBER!



START WITH D = 0.0123456789 0123456789 01234...


NOW YOU ARE GIVEN 1 REAL NUMBER AT A TIME FOR INFINITY


D = 0.5012346789 0123456789 01234...




ADD 2ND REAL NUMBER

D = 0.5401236789 0123456789 01234...



WE CAN ADD *ANY* REAL NUMBER FOR INFINITY


YOU GET A RECURRING PERIOD





YOU CAN ANTI-DIAGONALISE THE LIST IF YOU WANT TO...
BUT

|A| < |N|

[WM]

Post by i***@xlog.ch

Believing without proving leads to strange results, for instance.

You conflate assumptions with believes. Nobody said
that the assumptions in any proofs of
Cantor are true, it can still happend that ZFC or
so turns out inconsistent.

It has happened to turn out already. The very hidden contradiction is raised by confusing potentail and actual infinity.

(1)
The union of the sequence of intervals [0, (n-1)/n] is [0, 1).
The limit of the sequence of intervals [0, (n-1)/n] is [0, 1].

(2)
The union of the sequence of finite initial segments of the sequence of natural numbers {1}, {1, 2}, {1, 2, 3}, ... has less then aleph_0 elements.
The limit of this sequence (if existing) has aleph_0 elements.

(3)
The sequence of partial sums 3.1, 3.14, 3.141, ... = 3.141... is not irrational.
The limit of this sequence is irrational.

REgards, WM

[Virgil]

Post by WM
It has happened to turn out already. The very hidden contradiction is raised
by confusing potentail and actual infinity.

Potential infiniteness is irrelevant in any mathemtics, like Peano's,
which deals directly with actual infiniteness.

Post by WM
(1)
The union and limit of the sequence of intervals [0, (n-1)/n]
is [0, 1).

WM claims that lim {[0,1-1/n): n in |N} should be [0,1]
but [0,1-1/n) is a subset of [0,1) for every n in |N
so lim {[0,1-1/n): n in |N} is necessarily a subset
of lim {[0,1): n in |N} = [0,1),
at least in any PROPER mathematics.

Post by WM
(2)
The union of the sequence of finite initial segments of the sequence of
natural numbers {1}, {1, 2}, {1, 2, 3}, ... has less then aleph_0 elements.
The limit of this sequence (if existing) has aleph_0 elements.
(3)
The sequence of partial sums 3.1, 3.14, 3.141, ... = 3.141... is not irrational.
The limit of this sequence is irrational.
REgards, WM

[Virgil]

Post by WM

Post by Virgil

Post by WM
If all rational numbers are in the list, then we will never have checked
them
all.

Cantors diagonal method checks all!

Believing without proving leads to strange results, for instance: Cantor's
list contains all rational numbers.

All sorts of people can define and and have defined ways to list
the rationals which will necesssarily contain all rational numbers
One way is to assign each rational to a natural so different rtional are
assigned to different naturals as follows:
to rational zero assign natural 1
to reduced rational m/n, assign prime 2*m to the power n
to reduced rational -m/n, assign prime 2*m+1 to the power n
then different rationals all have different naturals though most
naturals are not even needed.

Post by WM
In mathematics these claims would have to proved.

Above is a valid proof that the naturals can surject onto the rationals
Both are nonsense.

Post by WM
How can you reasonably claim to reach all?

Since every rational has been assigned a different natural
in the above, how can you claim it has not been done?
Can you name a single rational left out? NO!

[John Gabriel]

I was having a discussion on another forum re: Cantor's
diagonalization proof that the irrationals are uncountable. The
question arose as to whether the proof relies on the Axiom of
Choice.
I was directed to this group for helpful advice.
TIA,
Shack

The following video on Cantor has many hits because it is a real zinger that exposes the fallacious diagonal "argument":

[b***@gmail.com]

No thanks it contains too many errors.

Your index for pi, which you wrote as 414...,
would have infinitely many digits to the left ¨
of the decimal place. It's a divergent sum. So
your index is not a natural number.

[Python]

Post by b***@gmail.com
No thanks it contains too many errors.

Your index for pi, which you wrote as 414...,
would have infinitely many digits to the left ¨
of the decimal place. It's a divergent sum. So
your index is not a natural number.

This is old stuff anyway, Mr. John Gabriel has been refuted
for this kind of mistake years ago:

http://scienceblogs.com/goodmath/2009/12/09/another-cantor-crank-represent/
http://scienceblogs.com/goodmath/2010/02/04/so-remember-back-in/

Looks like Mr. Gabriel is not anymore able to produce new mistakes,
so he has to recycle his old ones. Too bad. He'd better stop the
math business, which is no good for him, or kill himself.

[b***@gmail.com]

This one by JG , Nr, 188 in the first link is hillarious:

So, do you now agree that one can perform a left-to right infinite traversal? Hey, let me put it to you this way: I can perform an infinite left-to-right traversal of 1/3 in finite time. How? I find the path that starts with a digit 3 and then I stop! Why? I know that every possible *permutation* is in my tree. I don’t need to continue any further.
I could apply this same process to pi or e or sqrt(2) or any other irrational number I like, provided I know the first few significant digits.

[Python]

Post by b***@gmail.com
So, do you now agree that one can perform a left-to right infinite
traversal? Hey, let me put it to you this way: I can perform an
infinite left-to-right traversal of 1/3 in finite time. How? I find
the path that starts with a digit 3 and then I stop! Why? I know that
every possible *permutation* is in my tree. I don’t need to continue
any further.
I could apply this same process to pi or e or sqrt(2) or any other
irrational number I like, provided I know the first few significant digits.

It reminds me an old joke about prime numbers:

Physicist:
2 is prime
3 is prime
5 is prime
7 is prime
9 is prime (experimental error)
11 is prime
...

Software engineer:
1 is prime
1 is prime
1 is prime
1 is prime
...

What would the John Gabriel's version in the
joke?
0 is prime
0.33333... is not 1/3 I PISS ON YOUR CUNT
0.999999999... is filthy MORON I'LL KILL ALL YOU ALL
1 AAARGH (KILLED MYSELF)

I guess that in the future "John Gabriel" will some kind of common
word, like in "Oh sorry, I really goofed up there, I Johngabrieled
in some way".

[b***@gmail.com]

The earlies of a great mathematician:

Comment #188, John Gabriel, February 3, 2010
"So, do you now agree that one can
perform a left-to right infinite
traversal? Hey, let me put it to
you this way: I can perform an
infinite left-to-right traversal
of 1/3 in finite time.
How? I find the path that starts
with a digit 3 and then I stop!
Why? I know that every possible
*permutation* is in my tree. I
don’t need to continue any further.
I could apply this same process to
pi or e or sqrt(2) or any other
irrational number I like, provided
I know the first few significant
digits."

Wut?

[g***@gmail.com]

Post by b***@gmail.com
Comment #188, John Gabriel, February 3, 2010
"So, do you now agree that one can
perform a left-to right infinite
traversal? Hey, let me put it to
you this way: I can perform an
infinite left-to-right traversal
of 1/3 in finite time.
How? I find the path that starts
with a digit 3 and then I stop!
Why? I know that every possible
*permutation* is in my tree. I
don’t need to continue any further.
I could apply this same process to
pi or e or sqrt(2) or any other
irrational number I like, provided
I know the first few significant
digits."
Wut?

Cantors Proof is NONSENSE!

Instead of starting with a LIST

and calculating a DIAGONAL



START WITH THE DIAGONAL 0.012345678901234567890123456789.....


NOW! WATCH AS I PRODUCE AN INFINITE LIST OF REALS

*INDEPENDANT FROM THE DIAGAONAL*





R_1 = 0.55555...


CHANGE THE DIAGONAL A SMALL FINITE DISTANCE DIGIT SWAP!



DIAGONAL = 0.50123467890123456789....




On average we can add 5 REALS into the LIST
then pad the block of 10 REALS with 0.00.. 0.11.. as required.



THE RESULT????


ALL CANTORS MISSING REALS HAVE SMALL PERIOD!


MISSING REALS ALL HAVE A PERIOD
0.1234098765 6543209871 0123456798 9876543210 ....



HAHAHAHAHAHAHAHAHA!

WHAT A PISSANT " S U P E R D U P E R D E N S E " SUBSET OF R!

[b***@gmail.com]

The only nonsense here is that those that
fight Cantor here, are in fact in desperate
need of Cantor, to straighten their thinking.

JG, BTW they need you in the cheese cake factory.

[John Gabriel]

Post by g***@gmail.com

Post by b***@gmail.com
Comment #188, John Gabriel, February 3, 2010
"So, do you now agree that one can
perform a left-to right infinite
traversal? Hey, let me put it to
you this way: I can perform an
infinite left-to-right traversal
of 1/3 in finite time.
How? I find the path that starts
with a digit 3 and then I stop!
Why? I know that every possible
*permutation* is in my tree. I
don’t need to continue any further.
I could apply this same process to
pi or e or sqrt(2) or any other
irrational number I like, provided
I know the first few significant
digits."
Wut?

Cantors Proof is NONSENSE!
Instead of starting with a LIST
and calculating a DIAGONAL
START WITH THE DIAGONAL 0.012345678901234567890123456789.....
NOW! WATCH AS I PRODUCE AN INFINITE LIST OF REALS
*INDEPENDANT FROM THE DIAGAONAL*
R_1 = 0.55555...
CHANGE THE DIAGONAL A SMALL FINITE DISTANCE DIGIT SWAP!
DIAGONAL = 0.50123467890123456789....
On average we can add 5 REALS into the LIST
then pad the block of 10 REALS with 0.00.. 0.11.. as required.
THE RESULT????
ALL CANTORS MISSING REALS HAVE SMALL PERIOD!
MISSING REALS ALL HAVE A PERIOD
0.1234098765 6543209871 0123456798 9876543210 ....
HAHAHAHAHAHAHAHAHA!
WHAT A PISSANT " S U P E R D U P E R D E N S E " SUBSET OF R!

Poor Cantor was such an idiot, even he didn't know how stupid he was.

See, the set of real numbers is uncountable, but not for the reasons Cantor and his moron followers think. It is uncountable because real numbers do not exist. How can one count objects they don't know? How can one count points in an interval? This is the delusional world of Cantor.

Now here comes the rub: if one assumes that all real numbers can be represented as infinite decimal strings, then the faux set of real numbers is INDEED countable as I proved in my video and also on so many other forums:


http://youtu.be/hlqTuuhR3-4

Cantor was not placed in an asylum because he was sane. He was confined because he lost all his marbles. Chuckle.

After Hitler was defeated, a new standard of judgment was applied by the Allies:

When a Jew loses his marbles, it's called academic brilliance.

That Einstein and not Tesla was named the greatest physicist and inventor of the 20th century is proof of this. Next to Tesla, Einstein was a mental midget. But Tesla died penniless even though his contributions changed the world forever. Only problem is that Tesla was not a Jew and the Jews still control the media.

[b***@gmail.com]

Computation with real represented as sequences as rationals
is darn simple. If we have two reals a and b, as sequences
of rationals r1,r2,.. and s1,s2,..

a = (r1,r2,..)

b = (s1,s2,..)

Then we can readily define:

a+b = (r1+s1,r2+s2,..)

a*b = (r1*s1,r2*s2,..)

The only problem is when computing with decimals and not with
rationals. Since when we have a real:

a = (r1,r2,...)

And we find some digit in rk at some place we are not sure that
it flips at some time later in the sequence. And deciding whether
a digit is stable very much depends on how the sequence was constructed.

I will give an example soon.

[g***@gmail.com]

Post by b***@gmail.com
Computation with real represented as sequences as rationals
is darn simple. If we have two reals a and b, as sequences
of rationals r1,r2,.. and s1,s2,..
a = (r1,r2,..)
b = (s1,s2,..)
a+b = (r1+s1,r2+s2,..)
a*b = (r1*s1,r2*s2,..)
The only problem is when computing with decimals and not with
a = (r1,r2,...)
And we find some digit in rk at some place we are not sure that
it flips at some time later in the sequence. And deciding whether
a digit is stable very much depends on how the sequence was constructed.
I will give an example soon.

THERE IS A BIGGER SET THAN RATIONALS!


it was such a huge proof... after the Millionth Mathematician ran naked through the streets shouting 'EUREKA' they decided to do it all again!



WHAT IS YOUR 1ST MISSING REAL?

GO ON! TELL US!


OK... WHATS THE FIRST DIGIT OF YOUR FIRST MISSING REAL?

Is it

MISSING REAL 1 = 0.1...
OR
MISSING REAL 1 = 0.2...



TELL US!

FROM AN INFINITE SET OF INFINITE SEQUENCES OF DIGITS

WHAT IS THE FIRST MISSING DIGIT????




MISSING REAL = 0.X ????


WHAT IS X?


GIVEN AN INFINITE SET OF INFINITE SEQUENCES

YOU TELL US 0.X



wot-a-joke

[b***@gmail.com]

Post by g***@gmail.com
wot-a-joke

You forgot the all caps , WOT-A-JOKE .

[g***@gmail.com]

Post by b***@gmail.com

Post by g***@gmail.com
wot-a-joke

You forgot the all caps , WOT-A-JOKE .

GO DEFINE A PREDICATE!

[b***@gmail.com]

Post by g***@gmail.com

Post by b***@gmail.com

Post by g***@gmail.com
wot-a-joke

You forgot the all caps , WOT-A-JOKE .

GO DEFINE A PREDICATE!

is_all_cap(L) :<=>
forall (1=<i && i=<length(L) -> is_cap(L[i]))

[g***@gmail.com]

Post by b***@gmail.com

Post by g***@gmail.com

Post by b***@gmail.com

Post by g***@gmail.com
wot-a-joke

You forgot the all caps , WOT-A-JOKE .

GO DEFINE A PREDICATE!

is_all_cap(L) :<=>
forall (1=<i && i=<length(L) -> is_cap(L[i]))

assert( if( <(iq(MAN),99) , dumb(MAN) ).