Finite Models of the Complex Numbers

[RussellE]

Several people have conjectured the
complex numbers are a model of
Modular Arithmetic (MA).

MA has finite models and I have been
wondering what a finite model of the
complex numbers looks like.

The defining feature of the complex
numbers is i, the square root of -1.
There are finite models of MA where
-1 is a square.

U={0,1,2,3,4}
U={0,1,2,3,4,5,6,7,8,9}
...

Notice there are both even and odd
models of MA where -1 is a square.
We can use compactness to prove
there are odd infinite models of MA
where -1 is a square and even infinite
models where -1 is a square.

If the standard complex numbers are
a model of MA, they are an odd
infinite model.


Russell
- Integers are an illusion

[RussellE]

Some more strange things that are provable
if the complex numbers are a model of MA:

x*x = (Sum of odd numbers from 0 to 2*x)
2*x*x = (Sum of all numbers from 0 to 2*x)-x

Let -x = (-1)*x
Let i be the square root of -1.

-1 = Sum of all odd numbers from 0 to 2*i

2*i*i = (Sum of all numbers from 0 to 2*i)-i
-2 = (Sum of all numbers from 0 to 2*i)+i*i*i
0 = (Sum of all numbers from 0 to 2*i)+i*i*i+2

The sum of all numbers in an odd model
of MA is 0.

-1 = -1/2 + -1/2

(-1/2)*(-1/2) = Sum of all odd numbers from 0 to -1
1/4 = Sum of all odd complex numbers

2*(1/4) = 1/2 = Sum of all complex numbers - (-1/2) = 0-(-1/2)

[RussellE]

Let a "complex" model of MA be a model
where Ex(S(x*x)=0) is true.

I think this is a theorem of MA:

(-1 = x^2) -> (x = x^5 mod(x^2+1))

x=1
1=1^5 mod 2

x=2
2=2^5=32 mod 5

x=3
3=3^5=243 mod 10


Russell
- The integers are an illusion

[David Libert]

Probably no good finite analogue to C
=====================================

Author: David Libert <dave@aptosidbox>
Date: 2012-01-29 01:52:26 EST


Table of Contents
=================
1 This is followup and response to Russell's base of thread
2 comments on article
3 C can be axiomatized as being an algebarically closed field of characteristic 0
4 Any field of charactersitc 0 is infinite
5 Any algebraically closed field is infinite
5.1 Proofs all algebraically closed fields are infinite
5.1.1 by Lang above
5.1.2 by Hungerford above
6 How can any finite field be considered a resonable analogue of C?


1 This is followup and response to Russell's base of thread
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[1]
RussellE "Finite Models of the Complex Numbers"
sci.logic, sci.math
Jan 28, 2012
[http://groups.google.com/group/sci.logic/msg/5b48dd7e4eb02149]

2 comments on article
~~~~~~~~~~~~~~~~~~~~~~

On Jan 28, 6:11 pm, RussellE <reaste...@gmail.com> wrote:
> Several people have conjectured the
> complex numbers are a model of
> Modular Arithmetic (MA).

See also

[2]
David Libert
"Re: Even and Odd Infinities"
sci.logic, sci.math
Jan 16. 2012
[http://groups.google.com/group/sci.logic/msg/6e6aa8d2331d9518]


which is claiming to have a proof. [2] also references previous
articles conjecturing this as you wrote, so giving exact references.

> MA has finite models and I have been
> wondering what a finite model of the
> complex numbers looks like.

That question of a finite model of the complex numerbs will be a
main topic below.

> The defining feature of the complex
> numbers is i, the square root of -1.
> There are finite models of MA where
> -1 is a square.
>
> U={0,1,2,3,4}
> U={0,1,2,3,4,5,6,7,8,9}
> ...
>
> Notice there are both even and odd
> models of MA where -1 is a square.
> We can use compactness to prove
> there are odd infinite models of MA
> where -1 is a square and even infinite
> models where -1 is a square.
>
> If the standard complex numbers are
> a model of MA, they are an odd
> infinite model.

I agree with all of above. In particular, here an easy way to
remember which models are odd, and so to see your claim above that C
is odd.

A finite MA model Z/n has n many elements (0, ... , n-1},
(More acurately [3]_n, ... [n-1]_n).

Just referring to Z/n elements as i instead of [i]_n, we have in
Z/n n-1 + 2 = 1, so 1 is even <-> n-1 is even.

n-1 is even in Z/n <-> n-1 is even in the usual sense in Z.

So 1 is even in Z/n <-> n-1 is even in Z/n <-> n is odd in Z.

So for finite MA models, the model is odd <-> 1 is even.

Most of your definitions odd1, even1, etc were like this.

In C, 1 is even since 1 = 1/2 + 1/2.

So C is odd, as you wrote.

> Russell
> - Integers are an illusion

3 C can be axiomatized as being an algebarically closed field of characteristic 0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Characteristic 0 just means no finite sum of 1's = 0.

It is obvious C is a field of characterisitic 0.

C being algebraically closed is the Fundamental Theorem of Algebra:
[http://en.wikipedia.org/wiki/Fundamental\_theorem\_of\_algebra]

Tarski published his monograph on the decidability of the theory of
the reals. That was the first this theory was known to be
decidable.

Tarski mentioned in there he was only using properties of the reals
being an ordered field, with all positive elements having square
roots, and all odd degree polynomials having at least one root.
Tarski remarked his proof also showed this was a complete theory.

Tarski mentioned, in this theory you could intepret the theory of
the complex numbers by mimicing the usual construction of the
complex numbers from the reals. Ie intepret an ordered pair of real
numbers as two real vaiables, so a quantofier over complex numbers
expands as two nest real quantifiers.

I recall Tarksi remarked there that his completeness result for the
reals thereby shows the theory of the complexes interepreted as
above intot he reals is thereby shown complete. And that this in
turn can be reworked to show that the above theory of complexes as
algebraically closed filed of characterisit 0 is thereby complete.

Tarksi's monograph shows the theory of reals complete by an
elimination of quatntifiers argument.

(In fact, I seem to recall it was Tarski who invented the method of
elimination of quantifiers.)

Or else [2] above directly argued elimination of quantifiers in C.
All [2] used was the C was infinite and was algebraically closed.

So from C being characterisitc 0 and alegbraically closed, [2] could
apply to get elimination of quantifiers and show the resulting
theory is complete.

So by Tarski or by [2], being an algebraically closed field of
characterisitc 0 axiomatizes all the the theory of C in field
language.

Being of characteristic 0 can be formalized in field langaueg with a
schematum. And being algebraically closed can be formalized in
field language with a schematum.

4 Any field of charactersitc 0 is infinite
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

5 Any algebraically closed field is infinite
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Serge Lang _Algebra_ my revised printing January 1971
Chapter VIII Galois Theory
Section 9 The equation X^n - a = 0
page 223

Corollary 2: Let k be a field and assume the algebraic closure
k^bar (Lang write it with a br above the k) of k is of finite
degree > 1 over k. Then k_bar = k(i) where i^2 = -1
and k has characteristic 0.


Thomas W. Hungerford _Algebra_ copyright 1974
Chapter V Fields and Galois Theory
Section 5 Finite Fields
page 281

Corolarry 5.9 If K is a finite field and n >= 1 an integer, then
there exists an irreducible polynomial of degree n in K[x].


5.1 Proofs all algebraically closed fields are infinite
========================================================

5.1.1 by Lang above
--------------------
Given an algerically closed filed F, let k be the prime subsfield
of F, then F = k^bar as in Lang's corollary 2.

If F were finite, then k^bar would be of finite degree over k, so
Lang's Corollary 2 applies, and k is of characteristic 0.

Hence k is infinite (being characteristic 0), so F extending k
is infinite, contraryt to F assumed finite.

QED proof using Lang

5.1.2 by Hungerford above
--------------------------
If F is finite, then it has irreducible polynomals of degrees > 1
by Hungerford, and hence is not algebraically closed

QED proof using Hungerford

6 How can any finite field be considered a resonable analogue of C?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Among fields C is characterized as being algebraically closed and
characteric 0.

But either one of those 2 properties is individually ruled out by
above for finite fields.

I bought you an expensive red car.

Only it wasn't expensive and it wasn't red and it wasn't a car and I
didn't buy it for you.



--
David Libert ah...@FreeNet.Carleton.CA

[Frederick Williams]

RussellE wrote:
>
> Several people have conjectured the
> complex numbers are a model of
> Modular Arithmetic (MA).
>
> MA has finite models and I have been
> wondering what a finite model of the
> complex numbers looks like.

I understand "model of a set of formulae", but what does "model of the
complex numbers" mean?

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

[RussellE]

> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~­~~~~~~~

I had to look up algebraically closed field.
http://en.wikipedia.org/wiki/Algebraically_closed_field

Wikipedia says x^2+1=0 has no solution in the reals.
I define a "complex" model of MA to be any model
where Ex(S(x*x)=0) is true. x^2+1=0 has a solution
in any such model.

Wikipedia goes on to say no finite field
is algebraically closed because the
polynomial (x-a0)(x-a1)...(x-an)+1 = 0
has no root. I think this is true for finite
models of MA except the 0-model.

(x-0)+1=0. Let x=0.

Since I don't think infinite sets exist,
I would argue the 0-model is the
only algebraically closed field.


Russell
- Never never means never in set theory

[Frederick Williams]

RussellE wrote:

>
> Since I don't think infinite sets exist,

Usually it is the axiom of infinity that "makes" them exist. What is
your objection to it?

[RussellE]

On Jan 29, 2:40 pm, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> RussellE wrote:
>
> > Since I don't think infinite sets exist,
>
> Usually it is the axiom of infinity that "makes" them exist.  What is
> your objection to it?

Neither PA nor MA have an axiom of infinity.
I think infinite sets are provably inconsistent.
This is one of the reasons I am interested in MA.

[Transfer Principle]

On Jan 29, 2:40 pm, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> RussellE wrote:
> > Since I don't think infinite sets exist,
> Usually it is the axiom of infinity that "makes" them exist.  What is
> your objection to it?

Maybe it's because some people are _finitists_. And based on this
post, I'd consider RE to be a _finitist_.

Disclaimer: I'm aware that RE hasn't given an epistemologically
relevant proof of the inconsistency of any standard theory.

So now we ask, why do finitists object to the Axiom of Infinity
and the existence of infinite sets?

Well, let's turn it around. I now ask ZFC users (and here I
include Williams), do you think that Tarski-infinite yet
Dedekind-finite sets exist? Of course, ZFC proves that such
sets don't exist. So what is your _objection_ to the existence
of T-infinite D-finite sets?

Then finitists object to infinite sets for the same reason that
ZFC users object to infinite D-finite sets.

In summary:

RE: objects to existence of T-infinite sets.
ZFC: objects to existence of T-infinite D-finite sets.

The exact same objection, except that the finitists' objection
extends from certain infinite sets to all of them.

[Jesse F. Hughes]

Transfer Principle <david.l...@lausd.net> writes:

> On Jan 29, 2:40 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
>> RussellE wrote:
>> > Since I don't think infinite sets exist,
>> Usually it is the axiom of infinity that "makes" them exist.  What is
>> your objection to it?
>
> Maybe it's because some people are _finitists_. And based on this
> post, I'd consider RE to be a _finitist_.
>
> Disclaimer: I'm aware that RE hasn't given an epistemologically
> relevant proof of the inconsistency of any standard theory.
>
> So now we ask, why do finitists object to the Axiom of Infinity
> and the existence of infinite sets?
>
> Well, let's turn it around. I now ask ZFC users (and here I
> include Williams), do you think that Tarski-infinite yet
> Dedekind-finite sets exist? Of course, ZFC proves that such
> sets don't exist. So what is your _objection_ to the existence
> of T-infinite D-finite sets?

I have no objection to them at all. It's a simple fact that ZFC proves
they don't exist. If some other interesting theory shows otherwise,
that is also a simple fact. The only adjustment to be made in that case
is that one must be a bit more explicit in mentioning the relevant
theory when discussing the (non-)existence of such sets.

Honestly, it is not much like RE's (baby) finitism at all. ZFC is a
theory of interest for some of us, and we notice that it proves there
are no T-infinite D-finite sets. Russell, on the other hand, has a
prior notion that there are no infinite sets and spends his time vainly
trying to show that ZF is inconsistent. Not really all that similar, is
it?

>
> Then finitists object to infinite sets for the same reason that
> ZFC users object to infinite D-finite sets.

Name a ZFC user who "objects" to such sets.

> In summary:
>
> RE: objects to existence of T-infinite sets.
> ZFC: objects to existence of T-infinite D-finite sets.
>
> The exact same objection, except that the finitists' objection
> extends from certain infinite sets to all of them.

Not very similar at all, for reasons given above.

--
Jesse F. Hughes

"Had you told it like it was, it wouldn't be like it is."
-- Albert King

[RussellE]

On Jan 29, 8:30 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Transfer Principle <david.l.wal...@lausd.net> writes:
> > On Jan 29, 2:40 pm, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> >> RussellE wrote:
>
>  Russell, on the other hand, has a
> prior notion that there are no infinite sets and spends his time vainly
> trying to show that ZF is inconsistent.

It is more accurate to say you have a prior notion
infinite sets exist and will call anyone who suggests
otherwise a crank.

You make it clear you will ignore anything I do prove.
I can prove omega is both even and odd.

Do you really believe a lamp can be both on and off?
Or do you consider Thompson's Lamp to be just
another "paradox".
http://en.wikipedia.org/wiki/Thomson's_lamp

Since the complex numbers are an odd model of MA,
turning a lamp on and off |C| times means the lamp
will be on if it started off.

[Jesse F. Hughes]

RussellE <reas...@gmail.com> writes:

> On Jan 29, 8:30 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Transfer Principle <david.l.wal...@lausd.net> writes:
>> > On Jan 29, 2:40 pm, Frederick Williams <freddywilli...@btinternet.com>
>> > wrote:
>> >> RussellE wrote:
>>
>>  Russell, on the other hand, has a
>> prior notion that there are no infinite sets and spends his time vainly
>> trying to show that ZF is inconsistent.
>
> It is more accurate to say you have a prior notion
> infinite sets exist and will call anyone who suggests
> otherwise a crank.

Not at all! I hate to sound cliche, but one of my best friends is a
perfectly competent finitist[1]. I don't understand his opinion, but he's
capable of distinguishing mathematically valid reasoning from invalid
reasoning (and he doesn't pretend to have a mathematical argument that
there are no infinite sets, but rather a philosophical argument).

> You make it clear you will ignore anything I do prove.
> I can prove omega is both even and odd.

Keen! I assume that you mean omega is both a multiple of 2 and not a
multiple of 2. Show me what you multiply 2 by to get omega, and then I
will believe it is even.

> Do you really believe a lamp can be both on and off?
> Or do you consider Thompson's Lamp to be just
> another "paradox".
> http://en.wikipedia.org/wiki/Thomson's_lamp

Certainly, that paradox is not proof of any contradiction in ZFC. It's
a fanciful word problem that is most obviously modeled as a
non-convergent sequence. Since no one ever claimed that every sequence
converges, I don't see how this spells doom for ZFC.

> Since the complex numbers are an odd model of MA,
> turning a lamp on and off |C| times means the lamp
> will be on if it started off.

Er. Hmm... I have no idea why you think this, but let it go. I didn't
say anything about your work on MA. In fact, I didn't call you a crank
in the post you're responding to, nor say that you are foolish for
believing there are no infinite sets. To be sure, I *do* think that
you're a crank, and I do think that you're on a fool's errand trying to
prove ZFC is inconsistent, given your incompetence in producing and
evaluating mathematical arguments.

But I'm not volunteering to read your entire thread on MA just to point
out its errors.

Sorry.

Footnotes:
[1] At least, he was the last time we spoke on the subject, which was
years ago. We do not see each other often at all these days.

--
"Other times it seems like we're just monkeys in a cage,
Throwing pieces of our feces on the crowd.
Better look out! Monkeys getting out!"
-- Randy Crouch, "Rock 'n Roll Delusion"

[Transfer Principle]

On Jan 30, 4:10 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> RussellE <reaste...@gmail.com> writes:
> > It is more accurate to say you have a prior notion
> > infinite sets exist and will call anyone who suggests
> > otherwise a crank.
> Not at all!  I hate to sound cliche, but one of my best friends is a
> perfectly competent finitist[1].

So Hughes _claims_, that is, but what's actually _posted_ tells a
much different story. For _every_ time a finitist posts to sci.math,
what sort of reception does he get? Of course, he receives insults,
if not from Hughes per se, then from some other ZFC user.

Furthermore, what about Doron Zeilberger? Hughes was prepared to
accept Zeilberger as reasonable (even if he didn't agree with him),
until he found out that the finitist had a high opinion of WM. Then
Hughes's opinion of Zeilberger instantly plummeted.

> I don't understand his opinion, but he's
> capable of distinguishing mathematically valid reasoning from invalid
> reasoning

I hope this isn't the claim that posters who agree with you have
mathematically valid reasoning, while posters who disagree with you
have mathematically invalid reasoning (a typical majority claim).

> > You make it clear you will ignore anything I do prove.
> > I can prove omega is both even and odd.
> Keen!  I assume that you mean omega is both a multiple of 2 and not a

> multiple of 2.  Show me what you multiply 2 by to get omega [...]

The short answer is -- omega.

For a more detailed answer, we must first determine whether this is
_ordinal_ or _cardinal_ multiplication. Based on the context, as
most of RE's posts refer to _cardinals_, I will assume that this is
cardinal multiplication. From this point on, I'll write "aleph_0"
in order to emphasize that this is cardinal multiplication.

So what does it mean for a cardinal to be even or odd? Once again,
based on context, RE probably intends these to be defined in terms
of models of MA. Specifically:

-- a cardinal kappa is even iff there exists a even model of MA
(i.e., a model of MA+"Ex (x+x=0 & ~x=0)") of cardinality kappa
-- a cardinal kappa is odd iff there exists an odd model of MA
(i.e., a model of MA+"Ax (x+x=0 -> x=0)") of cardinality kappa

So then we have:

Theorem:
aleph_0 is even

Proof:
There exist arbitrarily large models of MA+"Ex (x+x=0 & ~x=0)",
for example, Z/2nZ for each natural number n. So, by L-S, we
have a countably infinite even model. Thus aleph_0 is even. QED

Theorem:
aleph_0 is odd

Proof:
There exist arbitrarily large models of MA+"Ax (x+x=0 -> x=0)",
for example, Z/(2n+1)Z for each natural number n. So, by L-S, we
have a countably infinite even model. Thus aleph_0 is even. QED

Indeed, note that by full L-S, ZFC proves that every infinite
cardinal is both even and odd -- i.e., the only cardinals not
both even and odd are the _finite_ cardinals.

But based on what Hughes wrote:

> Keen! I assume that you mean omega is both a multiple of 2 and not a

> multiple of 2. Show me what you multiply 2 by to get omega [...]

apparently Hughes's definition of "even" and "odd" has nothing
to do with models of MA. Instead, he appears to hold that:

-- a cardinal kappa is even iff there exists a cardinal lambda
such that 2*lambda = kappa
-- a cardinal kappa is odd iff there exists a cardinal lambda
such that 2*lambda+1 = kappa

And of course, cardinal multiplication is defined in terms of
the Cartesian product, so we really have:

-- a cardinal kappa is even iff there exists a cardinal lambda
such that {0,1}xlambda has cardinality kappa
-- a cardinal kappa is odd iff there exists a cardinal lambda
such that ({0,1}xlambda) u {0} has cardinality kappa

And we can check that for kappa = aleph_0, lambda = aleph_0
works for both evenness and oddness. In fact, ZFC proves that
for any infinite cardinal kappa, we have that:

2*kappa = 2*kappa+1 = kappa

I suspect that AC is used in the proof somewhere, and that
2*kappa and 2*kappa+1 aren't necessarily equal to each other
or to kappa if kappa were, say, a T-infinite D-finite cardinal.

But what if _ordinal_ multiplication, rather than _cardinal_
multiplication, were intended? Then it still follows that
omega is even, but not that omega is odd. After all, for
ordinals, we have the Division Algorithm:

Given ordinals alpha, beta, there uniquely exist ordinals
rho (< beta) and gamma such that beta*gamma+rho = alpha.

We see that for alpha = omega and beta = 2, we determine that
gamma = omega and rho = 0, since 2*omega = omega. And so we
conclude that omega is even. But since there is no ordinal
gamma such that 2*gamma+1 = omega, so omega is not odd.

Note if we write the ordinal alpha in Cantor normal form:

alpha = omega*delta+n

then alpha is even iff the finite ordinal n is even. So every
ordinal alpha is exactly one of even, odd.

> To be sure, I *do* think that you're [RE] a crank
[emphasis Hughes's]

Insulter count for this thread: 1. Hughes is the first poster
to call RE the c-word in this thread.

Actually, as I explained in another thread, I can safely use an
insult as long as I'm only referring to the word, and that I'm
not the first poster to use the word in the thread. So I can
call Hughes the first poster to call RE a "_crank_" without
being a hypocrite myself.

> I do think that you're [RE] on a fool's errand trying to
> prove ZFC is inconsistent

Disclaimer: I'm aware that RE hasn't given an epistemologically
relevant proof of the inconsistency of any standard theory.

Still, I participate in this thread in order to discuss not
whether we _ought_ to forbid cardinals that are both even and
odd, not whether ZFC is inconsistent.

[Jesse F. Hughes]

Transfer Principle <david.l...@lausd.net> writes:

> On Jan 30, 4:10 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > It is more accurate to say you have a prior notion
>> > infinite sets exist and will call anyone who suggests
>> > otherwise a crank.
>> Not at all!  I hate to sound cliche, but one of my best friends is a
>> perfectly competent finitist[1].
>
> So Hughes _claims_, that is, but what's actually _posted_ tells a
> much different story. For _every_ time a finitist posts to sci.math,
> what sort of reception does he get? Of course, he receives insults,
> if not from Hughes per se, then from some other ZFC user.

I'm not responsible for anyone else's reaction.

I do regard the overwhelming majority of local, self-proclaimed
finitists as cranks, no doubt. Not because they express a philosophical
opinion that finite sets do not exist, however.

> Furthermore, what about Doron Zeilberger? Hughes was prepared to
> accept Zeilberger as reasonable (even if he didn't agree with him),
> until he found out that the finitist had a high opinion of WM. Then
> Hughes's opinion of Zeilberger instantly plummeted.

Yes, for good reason. Doron apparently endorsed a fallacious argument.
He was either careless or incapable (I assume the former).

>> I don't understand his opinion, but he's
>> capable of distinguishing mathematically valid reasoning from invalid
>> reasoning
>
> I hope this isn't the claim that posters who agree with you have
> mathematically valid reasoning, while posters who disagree with you
> have mathematically invalid reasoning (a typical majority claim).

Obviously not, and it is your prejudice on display here, not mine.

>> > You make it clear you will ignore anything I do prove.
>> > I can prove omega is both even and odd.
>> Keen!  I assume that you mean omega is both a multiple of 2 and not a
>> multiple of 2.  Show me what you multiply 2 by to get omega [...]
>
> The short answer is -- omega.
>
> For a more detailed answer, we must first determine whether this is
> _ordinal_ or _cardinal_ multiplication. Based on the context, as
> most of RE's posts refer to _cardinals_, I will assume that this is
> cardinal multiplication. From this point on, I'll write "aleph_0"
> in order to emphasize that this is cardinal multiplication.

His claim was about omega, so I took it to be about ordinals, not
cardinals. But no matter. I was being silly and thinking that
2 * omega = omega + omega, which is mistaken. Evidently, omega
certainly *is* even, and my question should have been, "How do you
conclude that omega is *not* a multiple of 2?"

(We should, of course, be clear on what kind of multiple we mean. TP
has chosen left side multiples below, i.e. 2 * alpha for some ordinal
alpha, and I see no reason not to choose the same. I don't know if
there's a conventional choice, but I was obviously being ambiguous
above.)

> So what does it mean for a cardinal to be even or odd? Once again,
> based on context, RE probably intends these to be defined in terms
> of models of MA. Specifically:
>
> -- a cardinal kappa is even iff there exists a even model of MA
> (i.e., a model of MA+"Ex (x+x=0 & ~x=0)") of cardinality kappa
> -- a cardinal kappa is odd iff there exists an odd model of MA
> (i.e., a model of MA+"Ax (x+x=0 -> x=0)") of cardinality kappa

Why would he mean that? Certainly not what I mean when I discuss
parity.

But, okay, let's suppose this is what he means. Then perhaps omega is
both even and odd. As far as I can tell, this is not any more a
contradiction than the fact that certain sets are both open and closed
(topologically speaking).

Unless Russell can prove that no cardinality is both even and odd in his
sense, then I see nothing particularly interesting here.


[...]

>
> But based on what Hughes wrote:
>
>> Keen! I assume that you mean omega is both a multiple of 2 and not a
>> multiple of 2. Show me what you multiply 2 by to get omega [...]
>
> apparently Hughes's definition of "even" and "odd" has nothing
> to do with models of MA. Instead, he appears to hold that:
>
> -- a cardinal kappa is even iff there exists a cardinal lambda
> such that 2*lambda = kappa

Quite right!

> -- a cardinal kappa is odd iff there exists a cardinal lambda
> such that 2*lambda+1 = kappa

Er, no. Simpler: A cardinal kappa is odd iff it is not even.

Under this definition, a cardinal which is both even and odd would
indeed be a puzzle. Hence, this is why I thought it was Russell's
meaning. (I also have not read much of his MA thread, so was unaware of
the meaning you suggest he had.)

> And of course, cardinal multiplication is defined in terms of
> the Cartesian product, so we really have:
>
> -- a cardinal kappa is even iff there exists a cardinal lambda
> such that {0,1}xlambda has cardinality kappa
> -- a cardinal kappa is odd iff there exists a cardinal lambda
> such that ({0,1}xlambda) u {0} has cardinality kappa
>
> And we can check that for kappa = aleph_0, lambda = aleph_0
> works for both evenness and oddness. In fact, ZFC proves that
> for any infinite cardinal kappa, we have that:
>
> 2*kappa = 2*kappa+1 = kappa

Well, yes, if you define odd as "having the form 2*kappa + 1", then
there are cardinals that are both even and odd. So what? I thought
Russell was trying to prove that ZFC was inconsistent, and this is
evidently not an inconsistency.

> I suspect that AC is used in the proof somewhere, and that
> 2*kappa and 2*kappa+1 aren't necessarily equal to each other
> or to kappa if kappa were, say, a T-infinite D-finite cardinal.
>
> But what if _ordinal_ multiplication, rather than _cardinal_
> multiplication, were intended? Then it still follows that
> omega is even, but not that omega is odd. After all, for
> ordinals, we have the Division Algorithm:
>
> Given ordinals alpha, beta, there uniquely exist ordinals
> rho (< beta) and gamma such that beta*gamma+rho = alpha.
>
> We see that for alpha = omega and beta = 2, we determine that
> gamma = omega and rho = 0, since 2*omega = omega. And so we
> conclude that omega is even.

Right. I was wrong about that. Thanks for the correction.

> But since there is no ordinal gamma such that 2*gamma+1 = omega, so
> omega is not odd.

More obvious: since I defined odd as "not even" (which is, I think,
traditional and the equivalent definition in terms of 2n + 1 a theorem),
it is pretty clear that no even ordinal is odd.

> Note if we write the ordinal alpha in Cantor normal form:
>
> alpha = omega*delta+n
>
> then alpha is even iff the finite ordinal n is even. So every
> ordinal alpha is exactly one of even, odd.
>
>> To be sure, I *do* think that you're [RE] a crank
> [emphasis Hughes's]
>
> Insulter count for this thread: 1. Hughes is the first poster
> to call RE the c-word in this thread.

Yes, that's nice. Thanks for keeping us informed, you pathetic twat.

> Actually, as I explained in another thread, I can safely use an
> insult as long as I'm only referring to the word, and that I'm
> not the first poster to use the word in the thread. So I can
> call Hughes the first poster to call RE a "_crank_" without
> being a hypocrite myself.

Be my guest.

>> I do think that you're [RE] on a fool's errand trying to
>> prove ZFC is inconsistent
>
> Disclaimer: I'm aware that RE hasn't given an epistemologically
> relevant proof of the inconsistency of any standard theory.
>
> Still, I participate in this thread in order to discuss not
> whether we _ought_ to forbid cardinals that are both even and
> odd, not whether ZFC is inconsistent.

I express no opinions on what one "ought" to forbid in mathematics!

--
"'Every man who has ever lived holds tight to the belief that for him
alone the laws of probability are canceled out by love[...] Therefore,
you will marry Guinevere. You do not want advice --- only agreement.'
Merlin sighed..." -- John Steinbeck

[RussellE]

On Jan 30, 4:10 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> RussellE <reaste...@gmail.com> writes:
> > On Jan 29, 8:30 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Transfer Principle <david.l.wal...@lausd.net> writes:
> >> > On Jan 29, 2:40 pm, Frederick Williams <freddywilli...@btinternet.com>
> >> > wrote:
> >> >> RussellE wrote:
>
> >>  Russell, on the other hand, has a
> >> prior notion that there are no infinite sets and spends his time vainly
> >> trying to show that ZF is inconsistent.
>
> > It is more accurate to say you have a prior notion
> > infinite sets exist and will call anyone who suggests
> > otherwise a crank.
>
> Not at all!  I hate to sound cliche, but one of my best friends is a
> perfectly competent finitist[1].

I'm not prejudiced. Some of my best friends are ...

>  I don't understand his opinion, but he's
> capable of distinguishing mathematically valid reasoning from invalid
> reasoning (and he doesn't pretend to have a mathematical argument that
> there are no infinite sets, but rather a philosophical argument).

I don't want a philosophical argument. I want to prove
infinite sets can't exist.

> > You make it clear you will ignore anything I do prove.
> > I can prove omega is both even and odd.
>
> Keen!  I assume that you mean omega is both a multiple of 2 and not a
> multiple of 2.  Show me what you multiply 2 by to get omega, and then I
> will believe it is even.

You mean other than 2*w = w?

Even and Odd Infinities
Russell Easterly Dec 27 2011
http://groups.google.com/group/sci.logic/browse_thread/thread/aab9ea7a2d5372e2/ca7f04676bd4835f

I give four definitions for even vs odd models of MA.
Every finite model of MA satisfies all of the odd
definitions or all of the even definitions.
PA satisfies two of the defintions for even and
two of the definitions for odd.
The complex numbers satisfy all four definitions for odd.

PA satisfies ExAy(x ~= y+y) which is an even definition
and Ax(x=0 or x+x ~= 0) which is an odd definition.

> > Do you really believe a lamp can be both on and off?
> > Or do you consider Thompson's Lamp to be just
> > another "paradox".
> >http://en.wikipedia.org/wiki/Thomson's_lamp
>
> Certainly, that paradox is not proof of any contradiction in ZFC.

I like how "paradox" doesn't mean "contradiction" in set theory.
I didn't ask if you thought ZFC was consistent.
Obviously, you do.

I asked if you believe a lamp can be both on and off.

>  It's
> a fanciful word problem

I don't think British philosopher James F. Thomson
considered it a "fanciful word problem".

> that is most obviously modeled as a
> non-convergent sequence.

When did lamps become non-convergent?
I assume you are referring to Grandi's series.
http://en.wikipedia.org/wiki/Grandi%27s_series

>  Since no one ever claimed that every sequence
> converges, I don't see how this spells doom for ZFC.

The question is whether the lamp is on or off.
I will let Thomson speak for himself:

"The impossibility of a super-task does not depend at all on
whether some vaguely-felt-to-be-associated arithmetical
sequence is convergent or divergent."

Since we can prove omega is both even and odd,
we can solve Thomson's "Paradox". After flipping
the switch an omega number of times, the lamp
will be both on and off.

Wikipedia says Grandi's series converges to 1/2
in some systems. If you believe ZFC is consistent,
you must also believe a lamp can be both on and off.

> > Since the complex numbers are an odd model of MA,
> > turning a lamp on and off |C| times means the lamp
> > will be on if it started off.
>
> Er.  Hmm...  I have no idea why you think this, but let it go.  I didn't
> say anything about your work on MA.  In fact, I didn't call you a crank
> in the post you're responding to, nor say that you are foolish for
> believing there are no infinite sets.  To be sure, I *do* think that
> you're a crank, and I do think that you're on a fool's errand trying to
> prove ZFC is inconsistent, given your incompetence in producing and
> evaluating mathematical arguments.

You didn't call me a crack earlier, but you
want everyone to know you think I am a crank.
And only a fool would try to prove ZFC is inconsistent.

> But I'm not volunteering to read your entire thread on MA just to point
> out its errors.

Thanks. I wouldn't want you pointing out any errors.
And, so far, you haven't.

> Sorry.
>
> Footnotes:
> [1]  At least, he was the last time we spoke on the subject, which was
> years ago.  We do not see each other often at all these days.

If your friend is a professional mathematician, they
best keep their finitist opinions to themselves if
they want to be published.

Fields medalist Vladimir Voevodsky gave a lecture
last year on what would happen if PA was found
to be inconsistent. These are some of the remarks
made on FOM by prominent theorists:
http://cs.nyu.edu/pipermail/fom/2011-June/015534.html

"If we screen out all the mistakes and misleading statements"
"Vladimir's provocative speculations regarding the consistency of PA"
"Voevodsky's speculations are neither "interesting" nor
"provocative", but merely misinformed"
"even "none of us know whether PA is consistent" is, I
believe, indefensible "

This last statement is indefensible.
There is no proof infinite sets exist.

At best, we can assume infinite sets exist
and claim, again without proof, that it
is impossible to derive a contradiction.

I think Godel's 2nd Incompleteness theorem
proves PA is inconsistent.
I think Zeno's Paradoxes prove a number of
contradictions. Thomson's Lamp is another
contradiction. I think proving omega is
both even and odd is like proving 0=1.

I won't be surprised when someone tries
to convince me "0=1" is a non-intuitive
property of infinite sets. just like sets
having bijections with proper subsets.

Believers in the consistency of PA and ZFC
have become a cult in the mathematical
community and the study of a legitimate
question has become almost impossible
to even talk about.

[Marshall]

On Jan 30, 6:40 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Transfer Principle <david.l.wal...@lausd.net> writes:
> > On Jan 30, 4:10 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> RussellE <reaste...@gmail.com> writes:
> >> > It is more accurate to say you have a prior notion
> >> > infinite sets exist and will call anyone who suggests
> >> > otherwise a crank.
> >> Not at all!  I hate to sound cliche, but one of my best friends is a
> >> perfectly competent finitist[1].
>
> > So Hughes _claims_, that is, but what's actually _posted_ tells a
> > much different story. For _every_ time a finitist posts to sci.math,
> > what sort of reception does he get? Of course, he receives insults,
> > if not from Hughes per se, then from some other ZFC user.
>
> I'm not responsible for anyone else's reaction.

I'm not sure you correctly followed Transfer's insinuation here.
What he's saying is that although you say you don't call all
finitists cranks, whenever a finitist posts in sci.math, someone
(not necessarily you) insults them, therefore you are lying.


Marshall

[Marshall]

On Jan 30, 7:13 pm, RussellE <reaste...@gmail.com> wrote:
>
> There is no proof infinite sets exist.

The counting numbers are an example of an infinite set.
So you have not merely proof but an example to boot.


Marshall

[Jesse F. Hughes]

RussellE <reas...@gmail.com> writes:

> On Jan 30, 4:10 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > On Jan 29, 8:30 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> >> Transfer Principle <david.l.wal...@lausd.net> writes:
>> >> > On Jan 29, 2:40 pm, Frederick Williams <freddywilli...@btinternet.com>
>> >> > wrote:
>> >> >> RussellE wrote:
>>
>> >>  Russell, on the other hand, has a
>> >> prior notion that there are no infinite sets and spends his time vainly
>> >> trying to show that ZF is inconsistent.
>>
>> > It is more accurate to say you have a prior notion
>> > infinite sets exist and will call anyone who suggests
>> > otherwise a crank.
>>
>> Not at all!  I hate to sound cliche, but one of my best friends is a
>> perfectly competent finitist[1].
>
> I'm not prejudiced. Some of my best friends are ...

Yes, thanks, Russell, that is the cliche I was explicitly referring to.

>>  I don't understand his opinion, but he's
>> capable of distinguishing mathematically valid reasoning from invalid
>> reasoning (and he doesn't pretend to have a mathematical argument that
>> there are no infinite sets, but rather a philosophical argument).
>
> I don't want a philosophical argument. I want to prove
> infinite sets can't exist.

Yes, well, good luck with that.

>
>> > You make it clear you will ignore anything I do prove.
>> > I can prove omega is both even and odd.
>>
>> Keen!  I assume that you mean omega is both a multiple of 2 and not a
>> multiple of 2.  Show me what you multiply 2 by to get omega, and then I
>> will believe it is even.
>
> You mean other than 2*w = w?

Yeah, I was being stupid there. I mistakenly thought 2*w = w+w.

So, let me ask the question I should have asked: Show me how you
conclude that w is *not* a multiple of 2.

>
> Even and Odd Infinities
> Russell Easterly Dec 27 2011
> http://groups.google.com/group/sci.logic/browse_thread/thread/aab9ea7a2d5372e2/ca7f04676bd4835f
>
> I give four definitions for even vs odd models of MA.
> Every finite model of MA satisfies all of the odd
> definitions or all of the even definitions.
> PA satisfies two of the defintions for even and
> two of the definitions for odd.
> The complex numbers satisfy all four definitions for odd.
>
> PA satisfies ExAy(x ~= y+y) which is an even definition
> and Ax(x=0 or x+x ~= 0) which is an odd definition.

Okay, so you have an idiosyncratic definition of even/odd and you
conclude that w is both even and odd. Let me presume for now that you
are correct. So what? Do you have a proof that no ordinal is both even
and odd? If not, this is not a contradiction.

>> > Do you really believe a lamp can be both on and off?
>> > Or do you consider Thompson's Lamp to be just
>> > another "paradox".
>> >http://en.wikipedia.org/wiki/Thomson's_lamp
>>
>> Certainly, that paradox is not proof of any contradiction in ZFC.
>
> I like how "paradox" doesn't mean "contradiction" in set theory.

If it were a set theoretic proof of P & ~P for some formula P, then it
would be a contradiction. But Thompson's Lamp is not a proof in ZFC at
all. It is a word problem that may be interpreted in ZFC, but with
unsatisfactory results.

> I didn't ask if you thought ZFC was consistent.
> Obviously, you do.
>
> I asked if you believe a lamp can be both on and off.

No, I don't, in the usual sense of the term and ignoring various dull
tricks of language.

>>  It's
>> a fanciful word problem
>
> I don't think British philosopher James F. Thomson
> considered it a "fanciful word problem".

Er, okay. Not sure why you think that's relevant.

>
>> that is most obviously modeled as a
>> non-convergent sequence.
>
> When did lamps become non-convergent?
> I assume you are referring to Grandi's series.
> http://en.wikipedia.org/wiki/Grandi%27s_series

You're babbling, son. I never suggested that lamps are non-convergent.

Look, you don't assert that Thomson's lamp yields a proof that ZFC is
inconsistent, right? Fine. I'm not interested in it, then. I'm not
about to defend any claim that we can flick a light switch infinitely
often in a finite length of time, nor do I think that any of my opinions
regarding mathematics depends on this remarkable skill.

So, let's move on.

>
>>  Since no one ever claimed that every sequence
>> converges, I don't see how this spells doom for ZFC.
>
> The question is whether the lamp is on or off.
> I will let Thomson speak for himself:
>
> "The impossibility of a super-task does not depend at all on
> whether some vaguely-felt-to-be-associated arithmetical
> sequence is convergent or divergent."
>
> Since we can prove omega is both even and odd,
> we can solve Thomson's "Paradox". After flipping
> the switch an omega number of times, the lamp
> will be both on and off.
>
> Wikipedia says Grandi's series converges to 1/2
> in some systems. If you believe ZFC is consistent,
> you must also believe a lamp can be both on and off.

I have no commitment to the claim that we should take supertasks
seriously. I have never suggested that the axiom of infinity
presupposes that every supertask yield a sensible result.

I am therefore underwhelmed by your presentation of Thomson's lamp.

Let's move on.

>
>> > Since the complex numbers are an odd model of MA,
>> > turning a lamp on and off |C| times means the lamp
>> > will be on if it started off.
>>
>> Er.  Hmm...  I have no idea why you think this, but let it go.  I didn't
>> say anything about your work on MA.  In fact, I didn't call you a crank
>> in the post you're responding to, nor say that you are foolish for
>> believing there are no infinite sets.  To be sure, I *do* think that
>> you're a crank, and I do think that you're on a fool's errand trying to
>> prove ZFC is inconsistent, given your incompetence in producing and
>> evaluating mathematical arguments.
>
> You didn't call me a crack earlier, but you
> want everyone to know you think I am a crank.

No, I just didn't want to give a false impression. I *didn't* call you
a crank in the previous post, but I certainly don't deny having that
opinion and sharing it previously.

> And only a fool would try to prove ZFC is inconsistent.

Well, I'm not sure about that. Seems pretty foolish to me, but perhaps
others differ.

It is undeniably foolish for *you* to try to prove ZFC is inconsistent,
given your mathematical ignorance. No doubt about that! Now, whether
others, more skilled than you, would be ignorant to attempt the same? I
suppose humility requires I admit that someone like, oh, whats-his-name[1]
surely knows better than I.

>> But I'm not volunteering to read your entire thread on MA just to point
>> out its errors.
>
> Thanks. I wouldn't want you pointing out any errors.
> And, so far, you haven't.
>
>> Sorry.
>>
>> Footnotes:
>> [1]  At least, he was the last time we spoke on the subject, which was
>> years ago.  We do not see each other often at all these days.
>
> If your friend is a professional mathematician, they
> best keep their finitist opinions to themselves if
> they want to be published.

A philosopher of mathematics, I'd say.

> Fields medalist Vladimir Voevodsky gave a lecture
> last year on what would happen if PA was found
> to be inconsistent. These are some of the remarks
> made on FOM by prominent theorists:
> http://cs.nyu.edu/pipermail/fom/2011-June/015534.html
>
> "If we screen out all the mistakes and misleading statements"
> "Vladimir's provocative speculations regarding the consistency of PA"
> "Voevodsky's speculations are neither "interesting" nor
> "provocative", but merely misinformed"
> "even "none of us know whether PA is consistent" is, I
> believe, indefensible "

Well, again, I'm not responsible for anyone's opinions or expressions
but my own, so I regard this particular rant as irrelevant to me.

>
> This last statement is indefensible.
> There is no proof infinite sets exist.
>
> At best, we can assume infinite sets exist
> and claim, again without proof, that it
> is impossible to derive a contradiction.
>
> I think Godel's 2nd Incompleteness theorem
> proves PA is inconsistent.
> I think Zeno's Paradoxes prove a number of
> contradictions. Thomson's Lamp is another
> contradiction. I think proving omega is
> both even and odd is like proving 0=1.
>
> I won't be surprised when someone tries
> to convince me "0=1" is a non-intuitive
> property of infinite sets. just like sets
> having bijections with proper subsets.
>
> Believers in the consistency of PA and ZFC
> have become a cult in the mathematical
> community and the study of a legitimate
> question has become almost impossible
> to even talk about.

Footnotes:
[1] You know, that guy trying to prove PA is inconsistent or what-not.
I'm blanking on the name right now. Nelson, is it?

--
"Just because you're ... in a Ph.d program it does not mean that
you're up to the challenge of being a real mathematician. Only those
who have a purity of mind and dedication to the truth as the highest
ideal have a chance." --James Harris, as Sir Galahad the Pure.

[RussellE]

On Jan 29, 12:35 pm, RussellE <reaste...@gmail.com> wrote:
> On Jan 28, 10:57 pm, ah...@FreeNet.Carleton.CA (David Libert) wrote:
>

> I had to look up algebraically closed field.http://en.wikipedia.org/wiki/Algebraically_closed_field

>
> Wikipedia says x^2+1=0 has no solution in the reals.
> I define a "complex" model of MA to be any model
> where Ex(S(x*x)=0) is true. x^2+1=0 has a solution
> in any such model.
>
> Wikipedia goes on to say no finite field
> is algebraically closed because the
> polynomial (x-a0)(x-a1)...(x-an)+1 = 0
> has no root. I think this is true for finite
> models of MA except the 0-model.
>
> (x-0)+1=0. Let x=0.

I think we can show there exists a model of PA
where the complex numbers are not an
algebraically closed field.

As usual, the definition of algebraically closed field
has been rigged to leave a loophole for infinite sets.

Assume the complex numbers are a model of MA.

Just like we did for the finite models, I want to
define a polynomial using every complex number.

(x-0)(x-1)...(x+1/2)(x-1/2)...(x+2)(x+1)+1 = 0

This polynomial has no solution in the complex numbers.
Now, you are going to complain we can't have
infinite exponents in a polynomial.

Actually, we can if we choose an appropriate
non-standard model of PA. Our infinite exponents
then become hyper-finite exponents.

We can easily define a "complex" non-standard
model of PA. We have the initial copy of C and
our successor function S(x) = x+1.
(I still wonder how you get to i using this definition.
The sum of two pure real numbers is always a
pure real where pure real means the complex
part is 0.)

Make another copy of C:

C' = {0', 1', ..., -1/2', 1/2', ..., -2', -1'}
Let S(-1) = 0'.

We then define addition and multiplcation using
methods similar to those Dave Libert used in
previous threads where he defines finite "patches"
of addition and mutiplication over the universe.

Of course, we have to keep adding new
copies, C'', C''', etc. Since the complex numbers
are uncountable, lets assume we add an
uncountable number of copies of C.

In this non-standard model of PA the complex
numbers are a hyper-finite set and are not an
algebraically closed field.


Russell
- Integers are an illusion

[quasi]

On Mon, 30 Jan 2012 19:13:52 -0800 (PST), RussellE
<reas...@gmail.com> wrote:

>I don't want a philosophical argument. I want to prove
>infinite sets can't exist.

In other words, as you've said in the past, you want to
prove ZFC is inconsistent within itself.

TP knowingly tries to distort the picture, by suggesting
that RE is implicitly working in some alternate theory,
and that based on the axioms and rules of inference of
RE's unspecified alternate theory, RE can justly claim
that infinite sets don't exist.

But note that RE has never claimed that he's working in
some alternate theory. Rather he's asserted numerous times
that he believes that he can prove that ZFC is
_self-contradictory_.

And that's why RE gets called a crank.

And TP, by his deliberate distortion of the actual
situation, once again demonstrates that he's a troll.

Sure, ZFC _might_ be inconsistent, but RE's posts to sci.math
make it clear that RE has no ability to prove anything
nontrivial, so I think we can safely say that RE's dream of
proving ZFC inconsistent is just a pipe dream.

RE makes wild claims, produces nothing, and doesn't respond
intelligently to objections.

So is RE a crank?

In my opinion, yes, definitely.

quasi

[Michael Stemper]

In article <78069bef-027a-4d79...@kg1g2000pbb.googlegroups.com>, RussellE <reas...@gmail.com> writes:

>On Jan 29, 2:40=A0pm, Frederick Williams <freddywilli...@btinternet.com> wrote:
>> RussellE wrote:

>> > Since I don't think infinite sets exist,
>>

>> Usually it is the axiom of infinity that "makes" them exist. =A0What is

>> your objection to it?
>
>Neither PA nor MA have an axiom of infinity.

Peano Arithmetic is about natural numbers, not sets. Since the Axiom
of Infinity is about sets, your statement is as sensible as stating
that your fish doesn't have roller skates.

Modular Arithmetic doesn't have axioms at all. It's about structures
that have certain properties.

>- Integers are an illusion

Tea-time doubly so.

--
Michael F. Stemper
#include <Standard_Disclaimer>
Indians scattered on dawn's highway bleeding;
Ghosts crowd the young child's fragile eggshell mind.

[Transfer Principle]

On Jan 30, 7:36 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> Footnotes:
> [1]  You know, that guy trying to prove PA is inconsistent or what-not.
> I'm blanking on the name right now.  Nelson, is it?

Yes. It is Ed Nelson.

[Transfer Principle]

On Jan 30, 6:40 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Transfer Principle <david.l.wal...@lausd.net> writes:
> > -- a cardinal kappa is even iff there exists a even model of MA
> > (i.e., a model of MA+"Ex (x+x=0 & ~x=0)") of cardinality kappa
> > -- a cardinal kappa is odd iff there exists an odd model of MA
> > (i.e., a model of MA+"Ax (x+x=0 -> x=0)") of cardinality kappa
> Why would he mean that?  Certainly not what I mean when I discuss
> parity.

In every single MA thread, when RE mentions parity, he explicitly
mentions even and odd models of MA. Therefore, it's best that we
use _RE's_ definition of parity in this _RE_ thread.

[Transfer Principle]

On Jan 30, 7:13 pm, RussellE <reaste...@gmail.com> wrote:

> Believers in the consistency of PA and ZFC
> have become a cult in the mathematical
> community and the study of a legitimate
> question has become almost impossible
> to even talk about.

Ah, there's that religious analogy again! But in my
opinion, the PA and ZFC users are more like an
establishment religion than a "cult" since they are,
after all, in the majority.

Still, this post does go to show how pervasive the
religious analogy is. Members of both sides of the
debate compare the other side to a debate.

[Jesse F. Hughes]

I agree that, if that's what he meant when he claimed omega was both
even and odd, then we should use that definition.

Of course, it's not at all obvious why we should think that being both
even and odd is paradoxical or troublesome, using this definition.

--
Jesse F. Hughes
"Such behaviour is exclusively confined to functions invented by
mathematicians for the sake of causing trouble."
-Albert Eagle's _A Practical Treatise on Fourier's Theorem_

[MoeBlee]

On Jan 29, 9:50 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:

> do you think that Tarski-infinite yet
> Dedekind-finite sets exist?

In some models of ZF there are infinite sets that have no bijection
with a proper subset and in some models there are no such sets.

Have you considered the notion that a model corresponds to an abstract
"state of affairs"? Statements that are true in some states of affairs
are not true in certain other states of affairs.

If I wish to consider abstract state of affairs in which the axiom of
choice is true, then in such states of affairs there are no infinite
sets that don't have a bijection with a proper subset. If I wish to
consider abstract states of affairs in which the axiom of choice is
false, then there infinite sets that do not have a bijection with any
proper subset.

For that matter, if I wish to consider states of affairs in which the
axioms of (ZF\I)+~I are true, then there are no infinite sets in such
states of affairs. But if I wish to consider states of affairs in
which the axioms of ZF are true, then in such states of affairs there
do exist infinite sets.

I am free to take existence claims as being true of certain states of
affairs and false of other states of affairs. And I am free to
consider that these are abstract states of affairs, as I and anyone
else has freedom to study whatever abstract states of affairs are of
interest to the person.

I don't adopt one axiom set over another as the one that is true of
the "real" state of affairs. Rather, I understand that each
consistent theory is true of certain states of affairs and (if a
contingent theory) false of other states of affairs. Then, also, one
is free to adopt whatever criteria one has for studying or taking
interest in certain theories. And, to the extent that we agree on
certain criteria, we may discuss which theories better meet those
criteria.

And I don't even dispute that there is a "real" state of affairs in
the matter of these abstractions, rather, merely that I don't assert
that there is one. I am agnostic on the matter, and don't even claim
to comprehend by what means we would determine what is "real" in
matters of such abstraction. And meanwhile, I understand and respect
that there are people who do think there is a "real" state of affairs
in such abstractions and that they propose means by which we can
determine what is true or not of this "real" abstract state of
affairs.

Meanwhile, though, I do understand a notion of finitistic truth - what
is true or not about finite strings of symbols. And in such matters I
merely say that IF there is any sure and objective knowledge we can
have in mathematics, it is knowledge about finite strings (of
symbols). And I don't even require that other people share my notion
in that way. Rather, I say that IF we are to discuss objectively what
is the case in mathematics, then the LEAST we can agree on and take as
objective are matters of finite strings (of symbols), and if one does
not even have a notion of such objectivity in matters of finite
strings (of symbols) then fine, but then I am skeptical that we would
have a basis for a meaningful discussion about mathematics.

All of that allows great intellectual freedom in mathematics. It is
quite the opposite of religious dogmatism.

MoeBlee

[Transfer Principle]

On Jan 31, 5:54 am, mstem...@walkabout.empros.com (Michael Stemper)
wrote:

> In article <78069bef-027a-4d79-8183-18c9e1aa5...@kg1g2000pbb.googlegroups.com>, RussellE <reaste...@gmail.com> writes:
> >Neither PA nor MA have an axiom of infinity.

> Modular Arithmetic doesn't have axioms at all. It's about structures
> that have certain properties.

Actually, MA, as defined by RE, does have axioms. Here are the
axioms of MA, as given by RE on the 22nd, at approx. a quarter
to 10PM Greenwich:

"Axioms of MA without induction
1) Ex (S(x)=0)
2) Ax ((S(x)=S(y)) -> (x=y))
3) Ax (x+0=x)
4) AxAy (x+S(y)=S(x+y))
5) Ax (x*0=0)
6) AxAy (x*S(y)=x*y+x)"

[Transfer Principle]

On Jan 30, 8:18 pm, RussellE <reaste...@gmail.com> wrote:
> On Jan 29, 12:35 pm, RussellE <reaste...@gmail.com> wrote:
> > Wikipedia goes on to say no finite field
> > is algebraically closed because the
> > polynomial (x-a0)(x-a1)...(x-an)+1 = 0
> > has no root. I think this is true for finite
> > models of MA except the 0-model.
> > (x-0)+1=0. Let x=0.

> Just like we did for the finite models, I want to
> define a polynomial using every complex number.
> (x-0)(x-1)...(x+1/2)(x-1/2)...(x+2)(x+1)+1 = 0
> This polynomial has no solution in the complex numbers.
> Now, you are going to complain we can't have
> infinite exponents in a polynomial.
> Actually, we can if we choose an appropriate
> non-standard model of PA. Our infinite exponents
> then become hyper-finite exponents.

Even though no ZFC user has responded to this yet
(at least as of the time I started typing), I think
I can anticipate the ZFC response.

First of all, the ZFC users will likely complain that
the left side of:

> (x-0)(x-1)...(x+1/2)(x-1/2)...(x+2)(x+1)+1 = 0

isn't even a _term_, since terms contain only finitely
many symbols.

Next, as loath as I am to quote Wikipedia, since RE
used Wikipedia I'll use it as well:

http://en.wikipedia.org/wiki/Polynomial

"In abstract algebra, one distinguishes between _polynomials_
and _polynomial_functions_. A POLYNOMIAL f in one variable X
over a ring R is defined to be a formal expression of the form:

f = a_nX^n + a_(n-1)X^(n-1) + ... + a_1X^1 + a_0X^0

where n is a natural number, the coefficients a_0,...,a_n are
elements of R, and X is a formal symbol, whose powers X^i are
just placeholders for the corresponding coefficients a_i, so
that the given formal expression is just a way to encode the
sequence (a_0,a_1,...) , where there is an n such that a_i = 0
for all i > n."
[emphasis Wikipedia's, but rendered in ASCII]

So we see that although the coefficients a_i are elements of
the ring R, the _indices_ 0,...,n are considered to be
_standard_natural_numbers_, not elements of R.

Thus, if we let R={0,1,t,t+1} be the field of four elements,
legal exponents of polynomials in R[X] are 0,1,2,3,4, etc.,
but neither t nor t+1.

Note that if the exponents were actually elements of the ring,
then Libert's proof would break down for _finite_ sets. For
example, the proof that Z/2Z isn't algebraically closed
involves the polynomial:

f(x) = x(x+1) = x^2+x

But if the exponent 2 were considered an element of Z/2Z, we
would have 2=0, and thus:

x^2+x = x^0+x = x+1

and x+1 does have a root! Indeed, there would only be two
nonconstant polynomials -- x and x+1 -- and both have roots,
which would seem to make Z/2Z algebraically closed!

But in reality, just because 2=0, we can't say x^2=x^0. The
exponents are elements of standard N, not elements of Z/2Z.

There's only one way that one could make RE's plan to have
nonstandard exponents work -- in Ed Nelson's IST, we see
that any infinite sequence (a_0,a_1,...) (since according
to Wikipedia, a polynomial is merely such a sequence) must
have elements at nonstandard indices, whether one likes it
or not!

But, as is usual for these debate threads, we see that we
can't get something for nothing. For Cantor's theorem is
still a theorem of IST, and so we can't try to fit the set
C of complex numbers into these coefficients, since there
are uncountably many complex numbers but only countably
many naturals (even including nonstandard naturals).

And so, as is usual for these debate threads, I wouldn't
mind seeing a theory in which RE's plan to allow for
nonstandard exponents would work, but none of the _known_
theories will allow it.

[Transfer Principle]

On Jan 31, 12:27 am, quasi <qu...@null.set> wrote:
> On Mon, 30 Jan 2012 19:13:52 -0800 (PST), RussellE
> <reaste...@gmail.com> wrote:
> >I don't want a philosophical argument. I want to prove
> >infinite sets can't exist.
> In other words, as you've said in the past, you want to
> prove ZFC is inconsistent within itself.

> TP knowingly tries to distort the picture [...]

But hold on a minute. In previous threads, I said that I
would try to prevent this from happening by providing a
Disclaimer with my posts in this type of thread:

Disclaimer: I'm aware that RE hasn't given an epistemologically
relevant proof of the inconsistency of any standard theory.

This is the third time that I've posted this Disclaimer in
this thread. The first time was on the 30th at just before
4AM Greenwich, the second was the same day at a little
after 10:30PM Greenwich. And this is the third time.

> [...] by suggesting

> that RE is implicitly working in some alternate theory

[...]

RE is working in an alternate theory. That theory is MA:

Axioms of MA without induction

[RE, 22nd, approx. quarter to 10PM Greenwich]

1) Ex (S(x)=0)
2) Ax ((S(x)=S(y)) -> (x=y))
3) Ax (x+0=x)
4) AxAy (x+S(y)=S(x+y))
5) Ax (x*0=0)
6) AxAy (x*S(y)=x*y+x)

> Rather he's asserted numerous times
> that he believes that he can prove that ZFC is
> _self-contradictory_.

Disclaimer: I'm aware that RE hasn't given an epistemologically
relevant proof of the inconsistency of any standard theory.

(That makes the fourth time that I've posted the Disclaimer!)

So now what? We see that RE created the theory MA with
the intent of proving ~Con(ZFC) in mind. So now I ask,
is it possible for me to discuss MA as a theory on its
own merits -- keeping in mind the Disclaimer that RE
hasn't satisfactorily proved ~Con(ZFC)? It should be
possible -- after all, quasi himself was the one who
conjectured that the set C of complex numbers is a
model of MA (the topic of the OP of this thread), and
he obviously did so without implying that RE has given
a proof of ~Con(ZFC), so it should be theoretically
possible for me to do the same.

Also, as I mentioned earlier, sometimes I want to
discuss how RE thinks math _ought_ to be -- but once
again, keeping in mind the Disclaimer that RE hasn't
satisfactorily proved ~Con(ZFC).

But if it's impossible for me to discuss MA as a
theory on its own without considering ~Con(ZFC), then
I'll have nothing left to say in this thread.

> And TP, by his deliberate distortion of the actual
> situation, once again demonstrates that he's a troll.

That word means _nothing_ to me. In this RE thread, I
only care about insults directed at RE, not at me.

> RE makes wild claims, produces nothing, and doesn't respond
> intelligently to objections.
> So is RE a crank?
> In my opinion, yes, definitely.

Above, the word "crank" is being directed at RE. Insulter
count for this thread: 2.

And before I conclude this post, let me post the Disclaimer
a fifth time to make sure everyone has seen it:

[Frederick Williams]

RussellE wrote:

>
> Just like we did for the finite models, I want to
> define a polynomial using every complex number.
>
> (x-0)(x-1)...(x+1/2)(x-1/2)...(x+2)(x+1)+1 = 0

What language is that written in? The language of MA?

[Michael Stemper]

In article <11505494-70f2-4354...@f30g2000yqh.googlegroups.com>, Transfer Principle <david.l...@lausd.net> writes:

>On Jan 31, 5:54=A0am, mstem...@walkabout.empros.com (Michael Stemper) wrote:
>> In article <78069bef-027a-4d79-8183-18c9e1aa5...@kg1g2000pbb.googlegroups.com>, RussellE <reaste...@gmail.com> writes:

>> >Neither PA nor MA have an axiom of infinity.
>> Modular Arithmetic doesn't have axioms at all. It's about structures
>> that have certain properties.
>
>Actually, MA, as defined by RE, does have axioms.

I stand corrected. His definition of modular arithmetic has axioms.

>axioms of MA, as given by RE on the 22nd, at approx. a quarter
>to 10PM Greenwich:
>
>"Axioms of MA without induction

>1) Ex (S(x)=3D0)
>2) Ax ((S(x)=3DS(y)) -> (x=3Dy))
>3) Ax (x+0=3Dx)
>4) AxAy (x+S(y)=3DS(x+y))
>5) Ax (x*0=3D0)
>6) AxAy (x*S(y)=3Dx*y+x)"

Peculiar. If I was to define modular arithmetic, I'd call it "the
study of structures that have those six properties." I wouldn't
call those axioms, I'd call them the properties that a atructure
would need to have be be studied by modular arithmetic.

Of course, I'd no more say "modular arithmetic without induction"
than I'd say "dachshund without typewriter".

--
Michael F. Stemper
#include <Standard_Disclaimer>

2 + 2 = 5, for sufficiently large values of 2

[Transfer Principle]

But that presumes that the set of counting numbers exists. The
standard proof requires the Axiom of Infinity, but we already
know that RE doesn't accept the Axiom of Infinity.

Try proving infinite sets exist in PA, MA, or ZF-Infinity.

Without the Axiom of Infinity, it could be the case that the
natural numbers are too numerous to be a set (just as the
ordinals are in standard theory).

Now that the insulter counter for this thread is at 2, this
is a good time for me to step back and decide, is it worth
it for me to continue to defend RE?

In previous threads, I said that I want to focus my efforts
on borderline cases. But is this really a borderline case?

So far, I have already considered those who post purely
religious messages (and not about the religious _analogy_,
but about religion _per_se_) to be indefensible, and maybe
even insultable (like Musatov).

We also have posters like Herc and AP. The problem with
these two posters is that both have made extreme claims
outside of mathematics, namely the Adam Herc theory, as
well as the Atom Totality theory. I'd prefer keeping these
fringe theories separate from their mathematical theories,
but other posters don't -- they hold the theories about
Adams and atoms against them when considering their _math_
as well. To me, the religious and scientific theories of
Herc and AP are indefensible - perhaps even insultable --
and while I consider their math to be defensible, their
non-mathematical theories are too extreme for others to
consider them borderline cases. And so I left the recent
Herc and AP threads to pursue this one.

So what about RE? Well, RE doesn't claim to be Adam or
that the Universe is an atom, so that helps. He has come
up with a theory, MA, that's more rigorous than anything
posted by Herc, AP, or Musatov. But as we've already seen,
RE's problem is that he claims to have proved ~Con(ZFC).

Let's compare RE, with his claim of a proof of ~Con(ZFC),
with the borderline case from last week for whom a
successful defense has already occurred -- Hovdan, with
his claim of a proof of Goldbach's Conjecture. Both has
made grandiose claims about proofs, but there is one
major difference here: the majority of mathematicians
believe that GC is probably true, and provable as well,
but they surely _don't_ believe that ZFC is inconsistent.

And so anyone who claims a proof of ~Con(ZFC) is a priori
further away from borderline than anyone who merely claims
a proof of GC (or twin primes, RH, etc.). If RE could've
introduced MA without tying it to ~Con(ZFC), then I could
called RE borderline, but his claims about ~Con(ZFC) is
enough to convince posters that he's not borderline.

One thing that remains, though -- in previous threads, I
allowed for the possibility that I might have to insult a
poster myself. Insulting RE certainly won't help RE out,
but it would help a future, borderline case out (because
I'd be able to compare him favorably to RE).

The insult counter right now is at 2. We already know that:

-- if I'm the first poster to insult, I'd be a hypocrite.
-- if I'm the fourth poster to insult, I'd be a bully.

but there's a loophole here. I can be the second or third
poster to insult without being a hypocrite or bully. With
the insult counter at 2, if I were to insult RE right now,
I'd be the third insulter, which is OK. If it would help
posters out in future threads, I'd be willing to insult RE.

And so I await responses to this and other recent posts to
see whether my becoming the third insulter in this thread
is appropriate or not.

[quasi]

On Tue, 31 Jan 2012 14:40:09 -0800 (PST), Transfer Principle
<david.l...@lausd.net> wrote:

>Disclaimer: I'm aware that RE hasn't given an
>epistemologically relevant proof of the inconsistency of
>any standard theory.

Hence, given his repeated claims that ZFC _is_ inconsistent,
and that his theories (MA, odd & even models, etc) will
_prove_ that inconsistency -- those claims together with
not _one_ proof by RE of anything except trivialities (and
even those he mangles badly), says what about RE?

It says he's a _crank_, what else?

And when you, TP, whine about the insults directed at RE even
though it's clear (and it should be clear to you ) that those
insults are well deserved, but yet you close your eyes to that
and start your idiotic "insult counter", what does that say
about you?

It says you're a _troll_, what else?

quasi

[RussellE]

On Jan 31, 2:42 pm, mstem...@walkabout.empros.com (Michael Stemper)
wrote:

> In article <11505494-70f2-4354-ba59-5fd3caf9e...@f30g2000yqh.googlegroups.com>, Transfer Principle <david.l.wal...@lausd.net> writes:
>
> >On Jan 31, 5:54=A0am, mstem...@walkabout.empros.com (Michael Stemper) wrote:
> >> In article <78069bef-027a-4d79-8183-18c9e1aa5...@kg1g2000pbb.googlegroups.com>, RussellE <reaste...@gmail.com> writes:
> >> >Neither PA nor MA have an axiom of infinity.
> >> Modular Arithmetic doesn't have axioms at all. It's about structures
> >> that have certain properties.
>
> >Actually, MA, as defined by RE, does have axioms.
>
> I stand corrected. His definition of modular arithmetic has axioms.

I didn't give the axioms in this thread.
I have posted a number of threads about MA.

> >axioms of MA, as given by RE on the 22nd, at approx. a quarter
> >to 10PM Greenwich:
>
> >"Axioms of MA without induction
> >1) Ex (S(x)=3D0)
> >2) Ax ((S(x)=3DS(y)) -> (x=3Dy))
> >3) Ax (x+0=3Dx)
> >4) AxAy (x+S(y)=3DS(x+y))
> >5) Ax (x*0=3D0)
> >6) AxAy (x*S(y)=3Dx*y+x)"

MA also has the first order axiom schema of induction.
TP quoted from a post comparing MA to ring theory.
Ring theory doesn't have induction, so I didn't
include induction in that post.

> Peculiar. If I was to define modular arithmetic, I'd call it "the
> study of structures that have those six properties." I wouldn't
> call those axioms, I'd call them the properties that a atructure
> would need to have be be studied by modular arithmetic.
>
> Of course, I'd no more say "modular arithmetic without induction"
> than I'd say "dachshund without typewriter".

I came up with MA as a way to prove PA is inconsistent.
The idea was to prove MA is consistent and turn
such a proof into a "finitistic" proof for PA.
PA is inconsistent if PA can prove its own consistency.

Both PA and MA can prove MA has a model
using FOL and the axioms of MA.
In other threads we argue over whether
Godel's Incompleteness theorem applies
to MA. If so, then MA is inconsistent which
means PA is inconsistent since PA proves
MA is consistent.

Inconsistency proofs can get very weird.

MA seems to be interesting enough to study
in its own right. If PA is inconsistent then
MA might be the only other option for
defining natural numbers.


Russell

[RussellE]

On Jan 31, 3:32 pm, quasi <qu...@null.set> wrote:
> On Tue, 31 Jan 2012 14:40:09 -0800 (PST), Transfer Principle
>

> <david.l.wal...@lausd.net> wrote:
> >Disclaimer: I'm aware that RE hasn't given an
> >epistemologically relevant proof of the inconsistency of
> >any standard theory.
>
> Hence, given his repeated claims that ZFC _is_ inconsistent,
> and that his theories (MA, odd & even models, etc) will
> _prove_ that inconsistency -- those claims together with
> not _one_ proof by RE of anything except trivialities (and
> even those he mangles badly), says what about RE?

It means I work harder than people who claim ZFC
in consistent and don't even bother trying to prove it.

> It says he's a _crank_, what else?

Does this mean people who claim ZFC is
consistent and don't prove it are also cranks?
I would really like to see a consistency
proof for ZFC in ZFC.

[RussellE]

On Jan 28, 10:57 pm, ah...@FreeNet.Carleton.CA (David Libert) wrote:
>

>   Characteristic 0 just means no finite sum of 1's  = 0.

Isn't this Robinson's principle?
Wikipedia says compactness implies Robinson's principle.
http://en.wikipedia.org/wiki/Compactness_theorem

I couldn't find much when I did a quick search for
Robinson's principle.

1+1+...+1=0 is true is all models of MA.
Let U be the universe of the model.
Adding 1 |U| times is always equal to 0.

>   It is obvious C is a field of characterisitic 0.

Normally, yes.
But, we are talking about a model of MA.
Lots of strange things can happen in MA.

Why can't a hyper-finite sum of 1's equal 0?

>   C being algebraically closed is the Fundamental Theorem of Algebra:
>   [http://en.wikipedia.org/wiki/Fundamental\_theorem\_of\_algebra]

I like the part about how the Fundamental Theorem of Algebra
is neither fundamental nor a theorem of algebra.


Russell
- The integers are an illusion

[RussellE]

I should also explain the axioms of MA are
the same as the axioms of Peano Arithmetic (PA)
except A(Sx ~= 0) is replaced with Ex(Sx=0).

The axioms of PA are independent whch means
we can't derive any given axiom from the other axioms.
It also means, if PA is consistent, I can negate
an axiom and still have a consistent theory.

I don't really like induction. Originally,
I replaced induction with the Omega rule.
http://en.wikipedia.org/wiki/%CE%A9-consistent_theory

P(0) & P(S0) & P(SS0) & ... -> Ax(P(x))

I was able to prove MA has no infinite models
with the omega rule, but, no one found that
interesting.

The Modular Arithmetic vs Ring Theory thread
made it obvious I need induction to prove
basic things about MA, like MA is commutative.
http://groups.google.com/group/sci.math/browse_thread/thread/672eaf335c8e9f54/14d6146eb254c970

I am thinking of negating the induction axiom of PA
to see if it is really independent.

[RussellE]

On Jan 31, 2:40 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:

> On Jan 31, 12:27 am, quasi <qu...@null.set> wrote:
>
> Disclaimer: I'm aware that RE hasn't given an epistemologically
> relevant proof of the inconsistency of any standard theory.

Disclaimer: I'm aware that no one hasn't given an epistemologically
relevant proof of the consistency of any standard theory.

[RussellE]

On Jan 31, 2:10 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:

> On Jan 30, 8:18 pm, RussellE <reaste...@gmail.com> wrote:
>
> Even though no ZFC user has responded to this yet
> (at least as of the time I started typing), I think
> I can anticipate the ZFC response.
>
> First of all, the ZFC users will likely complain that
> the left side of:
>
> > (x-0)(x-1)...(x+1/2)(x-1/2)...(x+2)(x+1)+1 = 0
>
> isn't even a _term_, since terms contain only finitely
> many symbols.

Please define what you mean by finite.
Do you mean finite in a model of MA?

> f(x) = x(x+1) = x^2+x

This should be 0 = x(x-1)+1
There is no solution in the binary model of MA.

> But if the exponent 2 were considered an element of Z/2Z, we
> would have 2=0, and thus:
>
> x^2+x = x^0+x = x+1

This is just false, even in MA.

I do think MA allows us to reduce
polynomials a lot more than PA allows.

These statements are true in models of MA
where Ex(S(x*x)=0)), even the finite models:

Ex(x = x^5)
Ex(0 = x^3+x)

Let x*x = -1
x = x^5 = (-1)(-1)x
0 = x^3 + x = (-1)x + x

>
> But, as is usual for these debate threads, we see that we
> can't get something for nothing. For Cantor's theorem is
> still a theorem of IST, and so we can't try to fit the set
> C of complex numbers into these coefficients, since there
> are uncountably many complex numbers but only countably
> many naturals (even including nonstandard naturals).

There are uncouintable models of PA.
A number of people have claimed the complex numbers
are a model of MA and the complex numbers are
uncountable. So far, no one has come up with a countable
model of MA. Compactness says there must be one.

[RussellE]

I am sorry quasi. Of course, I meant:

[quasi]

On Tue, 31 Jan 2012 19:05:05 -0800 (PST), RussellE
<reas...@gmail.com> wrote:
>
>On Jan 31, 3:32 pm, quasi <qu...@null.set> wrote:
>>
>> Hence, given his repeated claims that ZFC _is_ inconsistent,
>> and that his theories (MA, odd & even models, etc) will
>> _prove_ that inconsistency -- those claims together with
>> not _one_ proof by RE of anything except trivialities (and
>> even those he mangles badly), says what about RE?
>
>It means I work harder than people who claim ZFC

>inconsistent and don't even bother trying to prove it.

>
>> It says he's a _crank_, what else?
>
>Does this mean people who claim ZFC is
>consistent and don't prove it are also cranks?

If someone publicly and repeatedly claims that they can
prove something but never provide a proof, and worse, show
no ability to prove anything nontrivial, then it doesn't
matter whether they are claiming a proof that ZFC is
consistent or a proof that ZFC is inconsistent --
either way, that person is a crank.

quasi

[Nam Nguyen]

On 31/01/2012 3:09 PM, MoeBlee wrote:
> On Jan 29, 9:50 pm, Transfer Principle<david.l.wal...@lausd.net>
> wrote:
>> do you think that Tarski-infinite yet
>> Dedekind-finite sets exist?
>
> In some models of ZF there are infinite sets that have no bijection
> with a proper subset and in some models there are no such sets.
>
> Have you considered the notion that a model corresponds to an abstract
> "state of affairs"? Statements that are true in some states of affairs
> are not true in certain other states of affairs.

Your position sounds like mine: relativity of arithmetic truths
(in general at least): statements that are true in some kind of
naturals numbers are not true in certain other kinds of natural
numbers.

Note that the naturals are just a (language) model.

> All of that allows great intellectual freedom in mathematics. It is
> quite the opposite of religious dogmatism.

Agree. Post Godel, a lot of FOL reasoning is just "religious dogmatism".

--
----------------------------------------------------
There is no remainder in the mathematics of infinity.

NYOGEN SENZAKI
----------------------------------------------------

[quasi]

On Tue, 31 Jan 2012 15:11:38 -0800 (PST), Transfer Principle
<david.l...@lausd.net> wrote:
>
>Let's compare RE, with his claim of a proof of ~Con(ZFC),
>with the borderline case from last week for whom a
>successful defense has already occurred -- Hovdan, with
>his claim of a proof of Goldbach's Conjecture.

The comparison isn't even close.

Hovdan clearly understands Number Theory and Combinatorics, at
least at the level required for the proof he presented, and was
able to construct a rigorous 7-page mathematical argument. When
his argument was shown to be flawed, he retracted his claim.

RE understands very little of the math he talks about. He has
admitted that he doesn't know much about the complex numbers,
didn't know the formal definition of a ring until recently, and
was unaware of the concept of an algebraically closed field
until a few days ago. He makes many claims, produces few proofs.
The proofs he produces are mostly flawed, and the ones that are
actually correct are trivialities (and even those are presented
in a style which is almost incoherent).

My sense of RE is that he has far less knowledge and
understanding of mathematics than the average undergraduate
math major. Perhaps he once actually was a math major, but if
so, I suspect he was a failed math major.

In fact, I suspect that many of the sci.math cranks were
failed math majors, and are bitter about it.

Bottom line -- Hovdan and RE are levels apart, and
grouping them together insults Hovdan yet again.

>Both has made grandiose claims about proofs,

You are such a spin doctor.

Hovdan claimed a proof, posted it, and asked politely for
comments.

RE makes wild assertions, often claiming that he can prove
those claims, but produces only hand-waving gibberish, or
else nothing at all.

>but there is one major difference here: the majority of
>mathematicians believe that GC is probably true, and provable
>as well, but they surely _don't_ believe that ZFC is
>inconsistent.
>
>And so anyone who claims a proof of ~Con(ZFC) is a priori
>further away from borderline than anyone who merely claims
>a proof of GC (or twin primes, RH, etc.).

What proof of ~Con(ZFC) are you talking about?

RE's imagined proof?

>If RE could've introduced MA without tying it to ~Con(ZFC),
>then I could called RE borderline,

Borderline what? And why would his introduction of MA make
him borderline anything?

>but his claims about ~Con(ZFC) is enough to convince posters
>that he's not borderline.

Believing ~Con(ZFC) is not the issue. Rather, it's his repeated
claim that he can _prove_ ~Con(ZFC), without ever presenting a
valid proof -- that's what gets him called a crank.

>One thing that remains, though -- in previous threads, I
>allowed for the possibility that I might have to insult a
>poster myself. Insulting RE certainly won't help RE out,
>but it would help a future, borderline case out (because
>I'd be able to compare him favorably to RE).
>
>The insult counter right now is at 2. We already know that:

But RE really _is_ a crank.

If you can't defend RE regarding his empty claims that he can
prove ~Con(ZFC), what justifies your complaints about insults
directed to him?

>-- if I'm the first poster to insult, I'd be a hypocrite.
>-- if I'm the fourth poster to insult, I'd be a bully.
>
>but there's a loophole here. I can be the second or third
>poster to insult without being a hypocrite or bully. With
>the insult counter at 2, if I were to insult RE right now,
>I'd be the third insulter, which is OK. If it would help
>posters out in future threads, I'd be willing to insult RE.
>
>And so I await responses to this and other recent posts to
>see whether my becoming the third insulter in this thread
>is appropriate or not.

The above is so ridiculous, I don't know what to say.

Maybe this ...

TP proves himself an idiot yet again!

quasi

[Alan Smaill]

RussellE <reas...@gmail.com> writes:

> There are uncouintable models of PA.
> A number of people have claimed the complex numbers
> are a model of MA and the complex numbers are
> uncountable. So far, no one has come up with a countable
> model of MA. Compactness says there must be one.

What about the algebraic numbers?

> Russell
> - Integers are an illusion

--
Alan Smaill

[quasi]

On Wed, 01 Feb 2012 14:10:25 +0000, Alan Smaill
<sma...@SPAMinf.ed.ac.uk> wrote:

>RussellE <reas...@gmail.com> writes:
>
>> There are uncouintable models of PA.
>> A number of people have claimed the complex numbers
>> are a model of MA and the complex numbers are
>> uncountable. So far, no one has come up with a countable
>> model of MA. Compactness says there must be one.
>
>What about the algebraic numbers?

I'm pretty sure the algebraic numbers _are_ a model of PA.

And for the benefit of Russell, in case he doesn't know,
the set of algebraic numbers is countable.

quasi

[quasi]

On Wed, 01 Feb 2012 09:38:04 -0500, quasi <qu...@null.set> wrote:

>On Wed, 01 Feb 2012 14:10:25 +0000, Alan Smaill
><sma...@SPAMinf.ed.ac.uk> wrote:
>
>>RussellE <reas...@gmail.com> writes:
>>
>>> There are uncouintable models of PA.
>>> A number of people have claimed the complex numbers
>>> are a model of MA and the complex numbers are
>>> uncountable. So far, no one has come up with a countable
>>> model of MA. Compactness says there must be one.
>>
>>What about the algebraic numbers?
>
>I'm pretty sure the algebraic numbers _are_ a model of PA.

I meant MA, not PA (typo).

[Frederick Williams]

RussellE wrote:

> I would really like to see a consistency
> proof for ZFC in ZFC.

Such a thing would show that ZFC is inconsistent. The best you'll get
is such things as: if ZF is consistent then ZFC is consistent.

[Frederick Williams]

Is there an unintended double negation there?

[FredJeffries]

On Jan 31, 8:20 pm, RussellE <reaste...@gmail.com> wrote:
>
> There are uncouintable models of PA.
> A number of people have claimed the complex numbers
> are a model of MA and the complex numbers are
> uncountable. So far, no one has come up with a countable
> model of MA. Compactness says there must be one.

A pseudo-finite prime field.

http://en.wikipedia.org/wiki/Field_%28mathematics%29#Ultraproducts
http://www.jstor.org/stable/1970573
http://arxiv.org/abs/0909.2189
http://books.google.com/books?id=b0Fvrw9tBcMC&pg=PA120&lpg=PA120
http://www.mymathforum.com/viewtopic.php?f=21&t=13

[MoeBlee]

On Feb 1, 1:21 am, Nam Nguyen <namducngu...@shaw.ca> wrote:

> Your position sounds like mine: relativity of arithmetic truths
> (in general at least): statements that are true in some kind of
> naturals numbers are not true in certain other kinds of natural
> numbers.

I do not, and would not, state my view in that way. I don't use a
rubric "kind of natural numbers", though I don't disallow that you may
have your own notions and defintions as to "kind of natural numbers".

> Note that the naturals are just a (language) model.

That's not my own view, but I don't disallow that one may have such a
view.

> > All of that allows great intellectual freedom in mathematics. It is
> > quite the opposite of religious dogmatism.
>
> Agree. Post Godel, a lot of FOL reasoning is just "religious dogmatism".

To be clear, I'm not saying anything like, "a lot of FOL reasoning is
just "religious dogmatism"."

MoeBlee

[Tim Golden BandTech.com]

On Jan 28, 10:46 pm, RussellE <reaste...@gmail.com> wrote:
> On Jan 28, 5:10 pm, RussellE <reaste...@gmail.com> wrote:
> > Some more strange things that are provable
> > if the complex numbers are a model of MA:
>
> > x*x = (Sum of odd numbers from 0 to 2*x)
> > 2*x*x = (Sum of all numbers from 0 to 2*x)-x

Let z = - 1 - 2 i .
then z * z = + 1 + 2 i + 2 i + 4 i i
so z * z = - 3 + 4 i .

Instancing the complex number as general z does not yield clean
reductions.
However RussellE, there is a clean way to get MA(3) as the complex
numbers, when this MA(3) is treated as sign.
It is unfortunate that you tend to ignore me here, since I happen to
take great interest in the care that you give to the modular
arithmetic system, for it is deeply imprortant. As soon as we realize
that the sign of the real number is MA(2) then we see the option to
build not just MA(3), MA(4), ..., but also to build MA(1), which is
the most significant perhaps, since it poses a unidirectional system
which matches the physical properties of time.

MA(n) alone cannot resolve the complex numbers, for they are
continuous. Still, by marrying discrete sign to continuous magnitude
as
s x
where s is sign and x is magnitude then the complex numbers can
alight, where s is an MA(3) system; say the symbols
( -, +, * )
whose quantity of strokes mnemonically map. But this is not a complete
rendition, for we must also extend the symmetry of the real numbers by
generalizing the behavior on R (P2):
- 1 + 1 = 0, - x + x = 0
to (for the case of P3 (the complex numbers though in a new format):
- 1 + 1 * 1 = 0, - x + x * x = 0, - z + z * z = 0 .
Just thinking in the terms of the first cancellation instance here is
sufficient and simplistic, but the latter two can likewise be
employed, so that x may be any magnitude, and z may likewise be any
value in P3. Of course this law is extensible to P4 and so forth, but
here I will stick with just the P3 or MA(3) sign instance.

We now have behaviors which expose the 2D nature of expressions in P3
without too much work to do from the usual real valued (P2) instances:
Let z1 = - 1 + 2 , z2 = + 2 * 3
Sum( z1, z2 ) = - 1 + 2 + 2 * 3 = - 1 + 4 * 3 = + 3 * 2 .
Product( z1, z2 ) = ( - 1 + 2 )( + 2 * 3 ) = * 2 - 3 - 4 + 6 = - 7
+ 6 * 2 = - 5 + 4 .
These expressions may seem cryptic, but the pattern analyst will
likely see his way through to the core which is presented more
gradually at
http://bandtech.com/polysigned
As well the graphical interpretation of P3 as a vector algebra is
covered, and what we witness in those balanced vectors is a clean
geometrical coordinate system called (by me) the simplex coordinate
system. These systems are naturally nonorthogonal, and they do not
need any more rules than the real numbers started out with. All that
they need is that we grant the generalization of sign, and mull over
the details. Polysign numbers are fundamental to physics and
mathematics yet few can even appreciate them yet. You RussellE are one
of the few people who should come to appreciate them, yet you don't
seem to bother with the information that I present to you. It is not
so terribly advanced. I suspect that I could teach a gradeschool child
to perform polysign computations sooner that I could a college
graduate. This is because by that time the real number as fundamental
has been ironed in; soldered fast; hard wired. The discrepancy starts
roughly in the 1600's and carries on to this day. Welcome to polysign
numbers. They should be a blast from the past, and it would not
surprise me if locked away in someones vault of archaic documents
there is a mathematical constructivist who likewise broke through to
them. They are more of a discovery than they are an invention, and the
details which support this claim go ignored by those programmed in the
modern curriculum. Even Bucky Fuller overlooked the purity of the
coordinate system which he should have developed for P4:
- 1 + 1 * 1 # 1 = 0 , - z + z * z # z = 0
but instead he reached back for the orthogonal basis as a means of
reliance and posed a need for the perpendicular angle. It need not be
so, and what we have is a means to dimensionality within the modulus
that you are so keyed on to.

- Tim

>
> > Let -x = (-1)*x
> > Let i be the square root of -1.
>
> > -1 = Sum of all odd numbers from 0 to 2*i
>
> > 2*i*i = (Sum of all numbers from 0 to 2*i)-i
> > -2 = (Sum of all numbers from 0 to 2*i)+i*i*i
> > 0 = (Sum of all numbers from 0 to 2*i)+i*i*i+2
>
> > The sum of all numbers in an odd model
> > of MA is 0.
>
> > -1 = -1/2 + -1/2
>
> > (-1/2)*(-1/2) = Sum of all odd numbers from 0 to -1
> > 1/4 = Sum of all odd complex numbers
>
> > 2*(1/4) = 1/2 = Sum of all complex numbers - (-1/2) = 0-(-1/2)

>
> > Russell
> > - Integers are an illusion
>

> Let a "complex" model of MA be a model
> where Ex(S(x*x)=0) is true.
>
> I think this is a theorem of MA:
>
> (-1 = x^2) -> (x = x^5 mod(x^2+1))
>
> x=1
> 1=1^5 mod 2
>
> x=2
> 2=2^5=32 mod 5
>
> x=3
> 3=3^5=243 mod 10

[FredJeffries]

On Jan 31, 2:10 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:
>
>

> And so, as is usual for these debate threads, I wouldn't
> mind seeing a theory in which RE's plan to allow for
> nonstandard exponents would work, but none of the _known_
> theories will allow it.

https://profiles.google.com/114134834346472219368/buzz/1dvqMdm1S8G
by Terrence Tao.
Scroll down to Definition 8: "A good polynomial is a nonstandard
polynomial...of two
nonstandard variables of some (nonstandard) degree at most D (in each
variable),
where D is a nonstandard natural number"

See also
http://books.google.com/books?id=iYtSrOqxnHAC&pg=PA23
for a use of the binomial theorem with nonstandard exponents

[Transfer Principle]

On Feb 1, 12:55 pm, FredJeffries <fredjeffr...@gmail.com> wrote:
> On Jan 31, 2:10 pm, Transfer Principle <david.l.wal...@lausd.net>
> wrote:
> > And so, as is usual for these debate threads, I wouldn't
> > mind seeing a theory in which RE's plan to allow for
> > nonstandard exponents would work, but none of the _known_
> > theories will allow it.
> https://profiles.google.com/114134834346472219368/buzz/1dvqMdm1S8G
> by Terrence Tao.
> Scroll down to Definition 8: "A good polynomial is a nonstandard
> polynomial...of two
> nonstandard variables of some (nonstandard) degree at most D (in each
> variable), where D is a nonstandard natural number"

Thanks for the link!

OK, so RE's nonstandard polynomials are possible. Still, it appears
doubtful that they do everything that RE wants. In particular, I
doubt that:

product(x-c)

where the product ranges over all _complex_ numbers c, is a good
polynomial in the Tao sense. Too bad...

[Transfer Principle]

On Jan 31, 2:09 pm, MoeBlee <modem...@gmail.com> wrote:
> On Jan 29, 9:50 pm, Transfer Principle <david.l.wal...@lausd.net>

> wrote:
> > do you think that Tarski-infinite yet Dedekind-finite sets exist?
> In some models of ZF there are infinite sets that have no bijection
> with a proper subset and in some models there are no such sets.
> Have you considered the notion that a model corresponds to an abstract

> "state of affairs"? Statements that are true in some states of affairs
> are not true in certain other states of affairs.

OK, but still, I see lots of criticism directed at posters who try
to discuss these alternate "states of affairs."

Consider, for example, the current TO thread. (Note: that thread
was originally a Herc thread, then a DC thread. I would've posted
the following in the TO thread, but I already said that I'd leave
the thread after declaring Herc and DC to be indefensible.) In
that thread, TO proposed an alternate "state of affairs" in which
quantifiers are eliminated, and of course, he was criticized. And
then he proposed an alternate "state of affairs" in which all
cardinals and ordinals inherit certain properties of their finite
counterparts (which, BTW, sounds like Bigulosity again, except
that TO never actually uses that word), and of course, he was
again criticized.

> I don't adopt one axiom set over another as the one that is true of
> the "real" state of affairs.

...except that TO is continually asked to read books, and what
else would these books discuss but the "real" state of affairs --
that is, the state that doesn't draw criticism?

> Meanwhile, though, I do understand a notion of finitistic truth - what
> is true or not about finite strings of symbols.

And if I'm convinced that a poster is refuting such finitistic
truth, then I'll declare the poster indefensible. Such appears to
have happened to RE in this thread.

> All of that allows great intellectual freedom in mathematics. It is
> quite the opposite of religious dogmatism.

Freedom for oneself, yes, but what about one's opponents?

[Transfer Principle]

On Jan 31, 2:42 pm, mstem...@walkabout.empros.com (Michael Stemper)
wrote:

> In article <11505494-70f2-4354-ba59-5fd3caf9e...@f30g2000yqh.googlegroups.com>, Transfer Principle <david.l.wal...@lausd.net> writes:

> >"Axioms of MA without induction

> >1) Ex (S(x) = 0)
> >2) Ax ((S(x) = S(y)) -> (x = y))
> >3) Ax (x+0 = x)
> >4) AxAy (x+S(y) = S(x+y))
> >5) Ax (x*0 = 0)
> >6) AxAy (x*S(y) = x*y+x)"

> Peculiar. If I was to define modular arithmetic, I'd call it "the
> study of structures that have those six properties." I wouldn't

> call those axioms, I'd call them the properties that a [s]tructure

> would need to have be be studied by modular arithmetic.

Actually, this was the big debate over in the Dan Christensen
thread (which has since become a Tony Orlow thread), except with
regards to group theory.

In this case, DC and Stemper would in fact agree. To DC, the
group properties are those which a structure (i.e., a _group_)
would need to have to be studied by group theory. DC wants to
refer to the elements of that structure, in order to express
the closure property.

RE, on the other hand, would agree with the majority in the
other DC thread. To RE and the majority, these are axioms of a
first-order theory, either of groups (or mod arithmetic). They
would say that having first-order axioms is helpful because
every model of the axioms is a group (or modular ring).

> Of course, I'd no more say "modular arithmetic without induction"
> than I'd say "dachshund without typewriter".

RE explains his induction elsewhere in this and other grades.

[Transfer Principle]

On Feb 1, 12:13 am, quasi <qu...@null.set> wrote:
> On Tue, 31 Jan 2012 15:11:38 -0800 (PST), Transfer Principle

> >And so anyone who claims a proof of ~Con(ZFC) is a priori
> >further away from borderline than anyone who merely claims
> >a proof of GC (or twin primes, RH, etc.).
> What proof of ~Con(ZFC) are you talking about?
> RE's imagined proof?

Yes -- his _claim_ of a proof.

> >If RE could've introduced MA without tying it to ~Con(ZFC),
> >then I could called RE borderline,
> Borderline what?

Borderline "crank."

> And why would his introduction of MA make
> him borderline anything?

Simply introducing MA didn't make RE a full-fledged "crank,"
but tying it to ~Con(ZFC) did.

> If you can't defend RE regarding his empty claims that he can
> prove ~Con(ZFC), what justifies your complaints about insults
> directed to him?

This is enough to convince me that I can't discuss MA in this
thread without running into the ~Con(ZFC) problem.

So now what? I already agreed to become the third poster to
insult RE if I can't discuss MA without ~Con(ZFC). And so,
here goes nothing:


RE, you are a _crank_.


Note: I only want to help outnumbered posters. I insult RE in
the hopes that it will help an outnumbered poster in some
future thread.

But even though I insulted RE, I still want to help him. And
there's a way I can help RE out -- by leaving this thread and
going to his Presburger arithmetic thread. In that thread, I
see _no_ mention of ZFC or its inconsistency -- only about the
_consistency_ of Presburger and whether or not it can have a
modular analog. Since there is no mention of ~Con(ZFC) in that
thread, I should be able to discuss MPresA without insults
(but I hope that RE won't turn around and say, now here's how
he'd use MPresA to prove ~Con(ZFC)).

And so it's time for me to leave this thread. (I still can't
believe I let a majority trick me into insulting a poster who
is outnumbered -- I feel so dirty!)

[Nam Nguyen]

On 01/02/2012 10:15 AM, MoeBlee wrote:
> On Feb 1, 1:21 am, Nam Nguyen<namducngu...@shaw.ca> wrote:

>
>> Note that the naturals are just a (language) model.
>
> That's not my own view, but I don't disallow that one may have such a
> view.

So are the natural numbers a model of PA?

If the natural numbers aren't a language model then what is it to you?

[RussellE]

Thanks for the links!
I read we need compactness to prove pseudo-finite fields exist.
What is finite about pseudo-finite filed?


Russell
- Disclaimer: I'm aware no one has given an epistemologically

[RussellE]

On Feb 1, 12:13 am, quasi <qu...@null.set> wrote:

> On Tue, 31 Jan 2012 15:11:38 -0800 (PST), Transfer Principle
>

> RE understands very little of the math he talks about. He has
> admitted that he doesn't know much about the complex numbers,
> didn't know the formal definition of a ring until recently, and
> was unaware of the concept of an algebraically closed field
> until a few days ago.

I have never had any interest in rings or complex numbers.

Recently, I have been reading about rings because of
the similarity between ring theory and MA.
My only interest in the complex numbers is proving
they are not a model of MA.

> He makes many claims, produces few proofs.
> The proofs he produces are mostly flawed, and the ones that are
> actually correct are trivialities (and even those are presented
> in a style which is almost incoherent).

I think you are being dishonest here.
I have proven lots of things.
You may think my proofs are trivial.
So what. Most proofs in math are trivial.

These are things I and others have proven about MA.

Negating one axiom of Peano Arithmetic
gives us a theory of arithmetic with finite models.

MA can prove MA has a model.

Any infinite model of MA must be non-standard.

MA has no infinite models if first order induction
is replaced with the omega rule.

MA is omega inconsistent.

MA can't be ordered without additional axioms.

MA isn't well founded.

The integers are not a model of MA.

I defined numerous definitions of even vs odd
and proved all of these statements are
independent of the axioms of MA.
None of these statements are independent of PA.

All finite models of MA satisfy all of my
definitions of even or all of my definitions of odd.

If MA has infinite models, there are infinite
models satisfying all of my definitions of even
and infinite models satisfying all of my
definitions of odd.

The complex numbers satisfy all of my
definitions of odd.

There are finite models of MA where -1 is square.

MA can prove these congruences:

x = x^5 mod(x^2+1)
0 = x^3+x mod(x^2+1)

Maybe these congruences are well known.
I wouldn't know. But, you can't honestly say
I haven't proven anything.

If Dave Libert or others want to add to this list,
please do so. MoeBlee has been asking for a
list of things that have definitely been proven about MA.

I almost gave up on MA.
I didn't think MA would be able
to help me prove PA is inconsistent.
Just because MA has no infinite models
doesn't mean PA has no models.

It wasn't until MoeBlee and I got into
an argument about infinite models of MA
that I learned about compactness.

Now, I am trying to prove compactness is false.
I know little about compactness, but, I suspect
compactness being false would be bad news
for induction. I am not that fond of induction.

> My sense of RE is that he has far less knowledge and
> understanding of mathematics than the average undergraduate
> math major. Perhaps he once actually was a math major, but if
> so, I suspect he was a failed math major.
>
> In fact, I suspect that many of the sci.math cranks were
> failed math majors, and are bitter about it.

The validity of a proof depends on the status
of the person giving the proof?
Its sad how true this is in modern mathemetics.

Many years ago, I co-authored a few papers
about low Reynolds number hydrodynamics.
http://onlinelibrary.wiley.com/doi/10.1002/bip.360220409/abstract

I realized I wasn't cut out to be an academic.
Now, I am a web developer and I make a lot more money.

The nice thing about being an amatuer is I don't
have to care about people thinking I am a crank.

[quasi]

RussellE wrote:

>
>quasi wrote:
>>
>> RE understands very little of the math he talks about. He
>> has admitted that he doesn't know much about the complex
>> numbers, didn't know the formal definition of a ring until
>> recently, and was unaware of the concept of an algebraically
>> closed field until a few days ago.
>
>I have never had any interest in rings or complex numbers.
>
>Recently, I have been reading about rings because of the
>similarity between ring theory and MA.

But the fact is, _all_ models of MA are commutative rings and
many results about models of MA are essentially trivial when
interpreted as statements about commutative rings.

>My only interest in the complex numbers is proving they are
>not a model of MA.

How do you propose to do that?

And what about David Libert's posted proof that the complex
_are_ a model of MA?

>> He makes many claims, produces few proofs. The proofs he
>> produces are mostly flawed, and the ones that are actually
>> correct are trivialities (and even those are presented in a
>> style which is almost incoherent).
>
>I think you are being dishonest here.

Not really -- at least, there was no _intent_ on my part to
distort the truth. But I'll stand by my claims until I see the
evidence showing that I was mistaken.

>I have proven lots of things.

Reviewed by who?

>You may think my proofs are trivial.

That's my impression so far, yes.

>So what. Most proofs in math are trivial.

Sure, many proofs in math are trivial, perhaps even most, but
there are certainly many results that have depth and subtlety.

But it's just too far-fetched to believe that you can prove ZFC
inconsistent with a trivial proof, and as far as I can see,
deep, careful, rigorous thinking is beyond you.

>These are things I and others have proven about MA.

You and _others_?

We were talking about _your_ proofs, no?

>Negating one axiom of Peano Arithmetic gives us a theory of
>arithmetic with finite models.

A triviality -- Z/nZ is a model of MA for all positive
integers n.

>MA can prove MA has a model.

I don't even know what that means.

>Any infinite model of MA must be non-standard.

What does it even mean to say that a model of MA is
nonstandard? Nonstandard what?

Is this something you understand well enough to define
rigorously, or is it just some concept you have a vague notion
about? I suspect the latter.

>MA has no infinite models if first order induction
>is replaced with the omega rule.

I'm not familiar with the omega rule, but is the proof of
above claim difficult, and if so, was it a proof given by you?

>MA is omega inconsistent.

Presumably this follows immediately from the previous claim by
The Compactness Theorem.

>MA can't be ordered without additional axioms.

But that's obvious since for the rings Z/nZ, the inequality
0 < 1 would imply (after adding it to itself finitely many
times) that 0 < 0.

Thus, another triviality.

>MA isn't well founded.

I'm not familiar with the term "well founded". But once again,
I'll ask -- is the proof difficult? For example, is it just an
immediate corollary to omega-inconsistency? And the proof that
MA is not well founded -- was it _your_ proof?

>The integers are not a model of MA.

Ok, that's not so trivial, but again, was it _you_ who proved
it, or was it proved by someone else?

>I defined numerous definitions of even vs odd and proved all
>of these statements are independent of the axioms of MA.

Your various proposed definitions were of some interest, but
for all the versions, the rings Z/nZ are immediately seen to
be even for some values of n and odd for others, hence the
claimed independence is obvious.

>None of these statements are independent of PA.

Another triviality.

>All finite models of MA satisfy all of my definitions of even
>or all of my definitions of odd.

You didn't say that very well, but presumably you are only
claiming what I said above about the rings Z/nZ, and as I noted,
the truths are obvious.

>If MA has infinite models, there are infinite models
>satisfying all of my definitions of even and infinite models
>satisfying all of my definitions of odd.

An immediate consequence of The Compactness theorem, which goes
further and asserts that there actually _are_ infinite models.

>The complex numbers satisfy all of my definitions of odd.

A triviality.

>There are finite models of MA where -1 is square.

An immediate consequence of the well known result from
Elementary Number Theory, namely that for any prime p with
p = 1 (mod 4), -1 is a quadratic residue mod p.

>MA can prove these congruences:
>
>x = x^5 mod(x^2+1)
>0 = x^3+x mod(x^2+1)
>
>Maybe these congruences are well known.

They are trivially true in any ring.

>I wouldn't know. But, you can't honestly say I haven't proven
>anything.

I said you hadn't proved anything _nontrivial_, and as far as I
can see, that assertion still stands.

You certainly haven't proved your claim that ZFC is
inconsistent, and it was that claim, repeated many times and
backed up by nothing, which earned you the crank label.

>If Dave Libert or others want to add to this list, please
>do so.

David Libert claims to have proved that C is a model of MA, and
while I don't fully understand his proof, that proof does _not_
appear to be a triviality.

On the other hand, given that you make the claim that there
_aren't_ any infinite models of MA, why is that you haven't
challenged David Libert's proof?

>MoeBlee has been asking for a list of things that have
>definitely been proven about MA.

Thus, would you include David Libert's proof that C is model
of MA as one of the "definitely proven" things?

>I almost gave up on MA. I didn't think MA would be able
>to help me prove PA is inconsistent.

And in fact, I'm 100% certain that MA _won't_ help you

prove PA is inconsistent.

>Just because MA has no infinite models doesn't mean PA
>has no models.
>
>It wasn't until MoeBlee and I got into an argument about
>infinite models of MA that I learned about compactness.

It makes sense to learn the basics of Model Theory before
making claims about the non-existence of models, no?

>Now, I am trying to prove compactness is false. I know
>little about compactness, but, I suspect compactness being
>false would be bad news for induction. I am not that fond
>of induction.

You have about as much chance of proving compactness false as
you have of understanding why it's true.

>> My sense of RE is that he has far less knowledge and
>> understanding of mathematics than the average undergraduate
>> math major. Perhaps he once actually was a math major, but
>> if so, I suspect he was a failed math major.
>>
>> In fact, I suspect that many of the sci.math cranks were
>> failed math majors, and are bitter about it.
>
>The validity of a proof depends on the status of the person
>giving the proof?

No definitely not, at least not in my book.

A proof can stand for itself, regardless of the source.

But when you talk about "the validity of a proof", you are
referring to the validity of _what_ proof?

Your imagined proof of the inconsistency of ZFC?

>Its sad how true this is in modern mathemetics.

There's some validity to that complaint -- there _are_
strong aristocratic tendencies throughout academia, not
just in math.

>Many years ago, I co-authored a few papers about low
>Reynolds number hydrodynamics.
>
><http://onlinelibrary.wiley.com/doi/10.1002/bip.360220409/abstract>
>
>I realized I wasn't cut out to be an academic. Now, I am a web
>developer and I make a lot more money.

I hope your programming skills are more existent than your
capacity for rigorous mathematical thinking.

>The nice thing about being an amatuer is I don't have to care
>about people thinking I am a crank.

Right, but the fact remains -- your claim that ZFC is
inconsistent, and that you can prove it, but with no proof
forthcoming -- that's what gets you called a "crank".

quasi

[Tim Golden BandTech.com]

On Feb 1, 8:37 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:

> RE, you are a _crank_.
>
> Note: I only want to help outnumbered posters. I insult RE in
> the hopes that it will help an outnumbered poster in some
> future thread.

OK T. P. Still, RE's commitment to the modular arithmetic is
authentic; he just applies it wrongly. It's discrete mathematics and
deserves to be applied to discrete systems, such as the sign of the
real number.

I do occasionally attempt a conversation and never have gotten much
back from RE.
I guess I will continue to post my own opinion to his threads as
rarely as I show up.
Any who want a serious application of the modular numbers please see
polysign numbers:
http://bandtechnology.com/polysigned
It's a thick presentation, yet the rules are no different than those
of the real number (but generalized).
Magnitude is more fundamental than the real number!
Structured thinking has come with the advent of programming languages.
The mathematician never did have a compiler to test his work against,
and we are mere humans.
Mathematicians cannot grant themselves immunity.

In relation to this thread the complex numbers are constructed
naturally as a three-signed arithmetic.
This arithmetic already carries an implied geometry in the balance
- 1.0 + 1.0 * 1.0 = 0.0
no different than the real line was constructed by the balance
- 1.0 + 1.0 = 0.0 .

- Tim

[Frederick Williams]

RussellE wrote:

> MA can prove MA has a model.

You cannot even express "MA has a model" in the language of MA. I've
pointed this out to you many times.

[Frederick Williams]

Transfer Principle wrote:

>
> ...except that TO is continually asked to read books, and what
> else would these books discuss but the "real" state of affairs --
> that is, the state that doesn't draw criticism?

The reason for that is that he discuses the usual approach from a
position of ignorance.

[Frederick Williams]

Frederick Williams wrote:
>
> Transfer Principle wrote:
>
> >
> > ...except that TO is continually asked to read books, and what
> > else would these books discuss but the "real" state of affairs --
> > that is, the state that doesn't draw criticism?
>
> The reason for that is that he discuses the usual approach from a
> position of ignorance.

... as is usual with cranks.

[Aatu Koskensilta]

MoeBlee <mode...@gmail.com> writes:

> In some models of ZF there are infinite sets that have no bijection
> with a proper subset and in some models there are no such sets.

In some models of ZF there is a proof of contradiction in ZF.

> Have you considered the notion that a model corresponds to an abstract
> "state of affairs"? Statements that are true in some states of affairs
> are not true in certain other states of affairs.

The problem with this idea is that there are models of ZF in which
there exist no models of ZF, there are models of ZF in which there are
models of ZF but they're all non-standard, etc. So just how are we to
understand talk about these states of affairs? When are we entitled to
assert there is (or is not) a state of affairs of this or that sort?

> For that matter, if I wish to consider states of affairs in which the
> axioms of (ZF\I)+~I are true, then there are no infinite sets in such
> states of affairs. But if I wish to consider states of affairs in
> which the axioms of ZF are true, then in such states of affairs there
> do exist infinite sets.

But how do you know there is any such state of affairs? What if there
is, but "there is no model of ZF" is true in all of them? Recall once
again that consistency -- or the existence of a model -- is a piddling
correctness condition.

> I am free to take existence claims as being true of certain states of
> affairs and false of other states of affairs.

What of claims about existence of states of affairs? And so on, and so
forth. The idea your describe is certainly natural and appealing, in the
sense that it occurs to many people pondering these matters. Alas, it is
ultimately incoherent.

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, darüber muss man schweigen."
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

[MoeBlee]

On Feb 1, 6:53 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:

> On Jan 31, 2:09 pm, MoeBlee <modem...@gmail.com> wrote:

You respond as if you read some other very different post from the one
I wrote. THAT is insulting:

> TO proposed an alternate "state of affairs" in which
> quantifiers are eliminated,

I said I was talking about MODELS as abstract states of affairs. That
is different from what you mention. But anyway, I have no objection to
someone specifying whatever syntax and semantics they want, including
a language without quantifiers that still expresses quantification.
But in Orlow's case, his recent proposal is incoherent and does not
provide a semantics for expressing quantification.

> And
> then he proposed an alternate "state of affairs" in which all
> cardinals and ordinals inherit certain properties of their finite
> counterparts

I have not objected to the mere idea of specifying a theory and
definitions in which cardinals and ordinals have properties unlike in
ordinary set theory. However, again, Orlow has never given a coherent
formulation of an alternative theory. Also, I do note that for purpose
of communication, it would be better for him to use special terms
(such as 't-ordinals' and 't-cardinals') so that it is easier to
discuss his notions without having to qualify each time that his
ordinals are not defined as usually.

But notice that Orlow, for example, claims that his notion is the
correct one and that other people who work in ordinary set theory are
"deluded". In contrast, I allow that if one has a coherently formed
syntax and then a theory that is true of abstract states of affairs
different from the class of models of ordinary set theory, then there
is nothing in itself wrong or "deluded" with that. So it is Orlow who
is dogmatic while my view is most tolerant.

But, yes, I do ask that one's proposal be coherent - starting at least
with the syntax (which, ideally, at least eventually, should be
recursive so that it is objectively verifiable whether a given
sequence is or is not a formal proof in the system). I don't wish to
restrict the liberty of a person to stipulate a syntax that is
incoherent, but I also don't restrict my own liberty to point out that
their proposal is incoherent.

> ...except that TO is continually asked to read books, and what
> else would these books discuss but the "real" state of affairs --
> that is, the state that doesn't draw criticism?

That is a very stupid comment. I made clear I'm talking about ABSTRACT
(mathematical) states of affairs. I do NOT claim that any book in
mathematics describes a "real" abstract state of affairs as opposed to
some other book that describes an "unreal" abstract state of affairs.
(And I don't even claim that one cannot have good arguments for
viewing some mathematics as true of some real abstract state of
affairs, but rather I adopt neutrality on the whole issue.)

Meanwhile Orlow makes a number of claims about how ordinary symbolic
logic, set theory, and mathematics work. In this regard, I'm talking
about the everyday literary (not abstract) facts about what authors
have actually written. It would be silly to say that I'm imposing
aherence to a certain abstract state of affairs if I pointed out that
Christopher Hitchens did not write that India and China are two parts
of the same country. And I am not imposing adherence to a a certain
abstract state of affairs when I point out that, in the way ordinary
symbolic logic is written about, it is not the case that for all
formulas P, we have that P and AxP are equivalent. Nor am I imposing
adherence to a certain abstract state of affairs when I point out that
ordinal addition is usually defined in ordinary textbooks in such a
way that it Orlow does not grasp since he is ignorant of it.

> Freedom for oneself, yes, but what about one's opponents?

What about one's "opponents"? I WROTE "ANYONE" [emphasis added] and
"other people" and in "one" in general. You didn't see that on your
computer screen?

MoeBlee

[MoeBlee]

On Feb 1, 8:02 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:

The point of my remark to which you are now responding was merely to
make clear (in context of you starting by saying that you agree with
me) that your own way of stating things does not necessarily reflect
my own. I don't wish to dispute your mathematical characterization in
this case, but merely to make clear, without prejudice to your own
characterization, that it happens not to be the way I would state
things.

I don't have special notions about the natural numbers. I regard them
intuitively, informally, in an ordinary way such as people ordinarily
regard the counting numbers starting with 0 and with each one arrived
at by adding 1 to a predecessor.

Formally, when I'm working in formal set theory formlulated in a
certain ordinary way, I regard the set of natural numbers to be the
least set closed under successor, and each natural number is a finite
ordinal. And then the set of natural numbers, each natural number, and
certain natural numbers also have other properties too.

I don't carry such notions with a polemical intent. I don't wish to
convince you or anyone to adopt either my informal or informal notion
of the natural numbers. Rather, I'm just telling you how I happen to
ordinarily regard them. And also, I don't necessarily dispute such
conceptions of the natural numbers as the intuitionistic, realist,
structuralist, et. al conceptions.

> So are the natural numbers a model of PA?

The set of natural numbers is a universe for many models of PA. The
set of natural numbers is also universe for a model for many theories
inconsistent with PA. And there are many sets other than the set of
natural numbers that are a universe for a model of PA.

> If the natural numbers aren't a language model then what is it to you?

I take a model for a language in an ordinary sense in mathematical
logic. (For sake of definiteness, I use Enderton's definition, but
that definition is but one among several variants that differ in
certain particulars but are essentially the same.) So a model is a
certain kind of function. The set of natural numbers is not a
function. The set of natural numbers though, of course, is the
universe for certain structures. And, of course, the system of natural
numbers (the natural numbers along with the basic operations) is
corresponds to the standard model of PA.

You said, "the naturals are just a (language) model". I don't take the
set of naturals to be model (though, the set of naturals is a UNIVERSE
for a model and the SYSTEM of the naturals and the basic operations
corresponds to a certain model).

I don't wish to dispute that you might have some defensible notions in
this regard. Rather, I merely wished to make clear (in context in
which you started by saying you agree with me) that your way of
stating certain things is not the way I would state them.

I hope that satisfies you and that it does not suggest to you that we
go on and on pursuing more about my notions of the natural numbers,
since, as I mentioned, my own notions are not special and I don't have
any motivation to convince you or anyone to adhere to my own notion of
the natural numbers.

MoeBlee

[MoeBlee]

On Feb 2, 10:28 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:

> MoeBlee <modem...@gmail.com> writes:

> > Have you considered the notion that a model corresponds to an abstract
> > "state of affairs"? Statements that are true in some states of affairs
> > are not true in certain other states of affairs.
>
>   The problem with this idea is that there are models of ZF in which
> there exist no models of ZF, there are models of ZF in which there are
> models of ZF but they're all non-standard, etc. So just how are we to
> understand talk about these states of affairs? When are we entitled to
> assert there is (or is not) a state of affairs of this or that sort?

I don't know. That it may be difficult to comprehend certain odd or
"seemingly paradoxical" (whatever is the best word) state of affairs
does not perhaps disallow the notion that statements are true in some
states of affairs and false in others.

> > For that matter, if I wish to consider states of affairs in which the
> > axioms of (ZF\I)+~I are true, then there are no infinite sets in such
> > states of affairs. But if I wish to consider states of affairs in
> > which the axioms of ZF are true, then in such states of affairs there
> > do exist infinite sets.
>
>   But how do you know there is any such state of affairs?

I don't opine that we have knowledge of this or that regarding various
abstractions. I could better answer the question: For a given property
P, how do you prove that there are models having that property?

MoeBlee

[Michael Stemper]

In article <2f9ac23c-8b51-41b7...@p12g2000yqe.googlegroups.com>, Transfer Principle <david.l...@lausd.net> writes:

>On Jan 31, 2:42=A0pm, mstem...@walkabout.empros.com (Michael Stemper) wrote:
>> In article <11505494-70f2-4354-ba59-5fd3caf9e...@f30g2000yqh.googlegroups.com>, Transfer Principle <david.l.wal...@lausd.net> writes:

>> >"Axioms of MA without induction

>> >1) Ex (S(x) =3D 0)
>> >2) Ax ((S(x) =3D S(y)) -> (x =3D y))
>> >3) Ax (x+0 =3D x)
>> >4) AxAy (x+S(y) =3D S(x+y))
>> >5) Ax (x*0 =3D 0)
>> >6) AxAy (x*S(y) =3D x*y+x)"

>> Peculiar. If I was to define modular arithmetic, I'd call it "the
>> study of structures that have those six properties." I wouldn't
>> call those axioms, I'd call them the properties that a [s]tructure
>> would need to have be be studied by modular arithmetic.
>
>Actually, this was the big debate over in the Dan Christensen
>thread (which has since become a Tony Orlow thread), except with
>regards to group theory.
>
>In this case, DC and Stemper would in fact agree.

Please don't put words in my mouth. I am aware (from that thread)
that there is more than one way of discussing groups. Although I
don't follow what's behind that other way (due to having never
studied it), I do not attempt to tell others that they should not
be allowed to study the consequences of that other view of groups.

There's a lot of room for a lot of different types of mathematics,
and I'm not one for supressing those about which I'm not knowledgeable.

Can we call them "alternative theories", or is that trademarked?

--
Michael F. Stemper
#include <Standard_Disclaimer>
Always remember that you are unique. Just like everyone else.

[FredJeffries]

On Feb 1, 8:02 pm, RussellE <reaste...@gmail.com> wrote:
>
> What is finite about pseudo-finite filed?
>

It is a model of the theory of finite fields

[1treePetrifiedForestLane]

yeah, but you cannot prove the "last" theorem of Fermat
with your silly-signs.

> > And so it's time for me to leave this thread. (I still can't
> > believe I let a majority trick me into insulting a poster who
> > is outnumbered -- I feel so dirty!)

thus:
no; it's mentioned on the very first page of the write-up,
published in the journal in the 19th cce, although
he did the experiment, so many times, that I'm not sure, if
it's in the first paper. others improved on this anomaly,
such as Dayton C. Miller.

> the negative results of the 1887 Michelson-Morley experiment.

thus:
you cannot give the absolute limit on the speed
of sound (not its velocity,
as with lightwaves; interestingly enough,
in the field of acoustics, a "phonon" is akin
to a massless rock of sound .-)

thus:
not as far as I know (nAFAiK), but that's not hing. anyway,
as far as I can say (AFAiCS),
the only reason that the 4cc was converted
into graph-form (or
just the geometrical dual of the mapping problemma),
was that of saving the cost of hand-coloring
the notional maps.

> Are there any theorems regarding the chromatic number of planar
> regular graphs of degree 4 and 5 that do not rely on the 4CC?

[Nam Nguyen]

On 02/02/2012 10:35 AM, MoeBlee wrote:
> On Feb 1, 8:02 pm, Nam Nguyen<namducngu...@shaw.ca> wrote:
>
> The point of my remark to which you are now responding was merely to
> make clear (in context of you starting by saying that you agree with
> me) that your own way of stating things does not necessarily reflect
> my own. I don't wish to dispute your mathematical characterization in
> this case, but merely to make clear, without prejudice to your own
> characterization, that it happens not to be the way I would state
> things.
>
> I don't have special notions about the natural numbers. I regard them
> intuitively, informally, in an ordinary way such as people ordinarily
> regard the counting numbers starting with 0 and with each one arrived
> at by adding 1 to a predecessor.

So is it conceivable to you that it's impossible to know the truth
values of cGC and ~cGC, your notion about the natural numbers being
just intuitive and informal only?

>
> Formally, when I'm working in formal set theory formlulated in a
> certain ordinary way, I regard the set of natural numbers to be the
> least set closed under successor, and each natural number is a finite
> ordinal. And then the set of natural numbers, each natural number, and
> certain natural numbers also have other properties too.

So here the naturals are part of a language model. the language being
L(ZF)?

>
> I don't carry such notions with a polemical intent. I don't wish to
> convince you or anyone to adopt either my informal or informal notion
> of the natural numbers. Rather, I'm just telling you how I happen to
> ordinarily regard them. And also, I don't necessarily dispute such
> conceptions of the natural numbers as the intuitionistic, realist,
> structuralist, et. al conceptions.
>
>> So are the natural numbers a model of PA?
>
> The set of natural numbers is a universe for many models of PA. The
> set of natural numbers is also universe for a model for many theories
> inconsistent with PA. And there are many sets other than the set of
> natural numbers that are a universe for a model of PA.

But I've never said the naturals were just a set of numbers.

>
>> If the natural numbers aren't a language model then what is it to you?
>
> I take a model for a language in an ordinary sense in mathematical
> logic. (For sake of definiteness, I use Enderton's definition, but
> that definition is but one among several variants that differ in
> certain particulars but are essentially the same.) So a model is a
> certain kind of function. The set of natural numbers is not a
> function. The set of natural numbers though, of course, is the
> universe for certain structures. And, of course, the system of natural
> numbers (the natural numbers along with the basic operations) is
> corresponds to the standard model of PA.
>
> You said, "the naturals are just a (language) model". I don't take the
> set of naturals to be model (though, the set of naturals is a UNIVERSE
> for a model and the SYSTEM of the naturals and the basic operations
> corresponds to a certain model).

By "the naturals" I never mean to refer them _collectively_ as just
a universe (set), no more that a geometrical plane were just a set
of points!

>
> I don't wish to dispute that you might have some defensible notions in
> this regard. Rather, I merely wished to make clear (in context in
> which you started by saying you agree with me) that your way of
> stating certain things is not the way I would state them.
>
> I hope that satisfies you and that it does not suggest to you that we
> go on and on pursuing more about my notions of the natural numbers,
> since, as I mentioned, my own notions are not special and I don't have
> any motivation to convince you or anyone to adhere to my own notion of
> the natural numbers.

All I can say is that if you and I don't agree on the definition
of what the naturals be then it's just hard for me to convey to
you, say, what relativity in mathematical reasoning is, as well
as it'd be impossible for you to understand what I've been presenting
on this foundation issue.

[MoeBlee]

On Feb 3, 1:03 am, Nam Nguyen <namducngu...@shaw.ca> wrote:
> is it conceivable to you that it's impossible to know the truth
> values of cGC and ~cGC,

In other threads, I have done my best as to your notion of "impossible
to know the truth values". Those discussions with you landed on the
rocks. I can't do more.

> your notion about the natural numbers being
> just intuitive and informal only?

I never said my notions are intuitive and informal ONLY. Indeed, in my
very next paragraph, I mention a formal setting.

Here it is:

> > Formally, when I'm working in formal set theory formlulated in a
> > certain ordinary way, I regard the set of natural numbers to be the
> > least set closed under successor, and each natural number is a finite
> > ordinal. And then the set of natural numbers, each natural number, and
> > certain natural numbers also have other properties too.

> So here the naturals are part of a language model. the language being
> L(ZF)?

The set of natural numbers is a universe for a model (i.e. structure)
for any language whatsoever, just as ANY non-empty set is a universe
for a model for any language whatsoever. Moreover, for any theory that
has an infinite model, the set of natural numbers is a universe for
some model of that theory.

> But I've never said the naturals were just a set of numbers.

And I didn't say that you said tha the naturals are just a set of
numbers.

> > You said, "the naturals are just a (language) model". I don't take the
> > set of naturals to be model (though, the set of naturals is a UNIVERSE
> > for a model and the SYSTEM of the naturals and the basic operations
> > corresponds to a certain model).
>
> By "the naturals" I never mean to refer them _collectively_ as just
> a universe (set), no more that a geometrical plane were just a set
> of points!

Wonderful!

> All I can say is that if you and I don't agree on the definition
> of what the naturals be then it's just hard for me to convey to
> you, say, what relativity in mathematical reasoning is, as well
> as it'd be impossible for you to understand what I've been presenting
> on this foundation issue.

I quite accept that it's possible (likely, even) that I'll never
understand, for example, what you're trying to convey about
"relativity in mathematical reasoninig". I have tried. Maybe I'm just
too stupid and dishonest, or more likely, the fault lies elsewhere.

MoeBlee

[Rupert]

On Jan 29, 2:10 am, RussellE <reaste...@gmail.com> wrote:
> Some more strange things that are provable
> if the complex numbers are a model of MA:
>
> x*x = (Sum of odd numbers from 0 to 2*x)

In order to state and prove this proposition in MA you would have to
go through the procedure which we use in PA for representing a finite
sequence of numbers by a single number. I doubt that proving all the
necessary theorems can be done in MA.

The complex numbers are a model of MA, by the way. This is because the
definable subsets of n-dimensional affine space over the complex
numbers in the first-order language of field theory are precisely the
constructible subsets. If the intersection of such a subset with a
linear subvariety of dimension one was infinite, then it would be
equal to the whole linear subvariety. This shows that the induction
axioms are true in the complex numbers, if we interpret + and * by the
usual field operations, 0 by the number 0, and S by the function which
sends x to x+1.

[Frederick Williams]

RussellE wrote:
>
> On Jan 29, 2:40 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > RussellE wrote:
> >
> > > Since I don't think infinite sets exist,
> >
> > Usually it is the axiom of infinity that "makes" them exist. What is
> > your objection to it?
>
> Neither PA nor MA have an axiom of infinity.
> I think infinite sets are provably inconsistent.

Inconsistent with what?

[RussellE]

On Feb 3, 9:56 am, Rupert <rupertmccal...@yahoo.com> wrote:
> On Jan 29, 2:10 am, RussellE <reaste...@gmail.com> wrote:
>
> > Some more strange things that are provable
> > if the complex numbers are a model of MA:
>
> > x*x = (Sum of odd numbers from 0 to 2*x)
>
> In order to state and prove this proposition in MA you would have to
> go through the procedure which we use in PA for representing a finite
> sequence of numbers by a single number. I doubt that proving all the
> necessary theorems can be done in MA.

I am not sure I understand your objection.
I don't have a proof by induction, but surely
there is one. If we can prove this relation
in PA using induction, we can prove it in MA.

There are also strictly "mathematical" proofs.

2*x*x = (Sum of all numbers 0 to 2*x)-x.

We can derive this from:
(Sum of all numbers 0 to x) = x*(x+1)/2

We would have to define division.
Once division is defined the relation
above should follow immediately.

I have thought a lot on how to represent
finite sets of natural numbers in MA.
It is not as simple as in PA, but,
it can be done.

Any finite set of naturals can be
represented as a sum of powers of 2.

Let U be the universe of a model
of MA and let u = |U|. We can easily
represent any finite set of numbers
where each number is less than log2(u).
(log2 can be defined with the halving function.)

We can encode sets with larger numbers using "cycles".
Let U = {0,1,2,3,4,5,6,7}
Assume we want to encode the set {1,4}.

2^1 + 2^4 = 18

18 is too large to represent in our
model using only one variable.
We can represent 18 by adding
another variable: Let x=1 and y=8.

18 = 2*x + 2*y

We can now represent any set of
members of U using no more than
log2(u) variables.

If all else fails, I can still prove there
exists a countably infinite model of MA
where these relations are true.

> The complex numbers are a model of MA, by the way. This is because the
> definable subsets of n-dimensional affine space over the complex
> numbers in the first-order language of field theory are precisely the
> constructible subsets. If the intersection of such a subset with a
> linear subvariety of dimension one was infinite, then it would be
> equal to the whole linear subvariety. This shows that the induction
> axioms are true in the complex numbers, if we interpret + and * by the
> usual field operations, 0 by the number 0, and S by the function which
> sends x to x+1.

At one time, Dave Libert was using a restricted
form of induction that does not include the
predecessor of 0. I am not sure why he decided
to define induction this way, but, I suspect he
came across some contradiction using full
induction. As you know, I am trying to show
induction fails in MA.

I think I can show a contradiction by assuming
the successor function for the complex numbers
is S(x) = x+1. The problem is i, the square root of -1.
Exactly how does successor reach i from 0?
I hope to give a better explanation of the problem
in another post.

[RussellE]

On Feb 2, 1:38 am, quasi <qu...@null.set> wrote:
> RussellE wrote:

> >quasi wrote:
>
> >You may think my proofs are trivial.
>
> That's my impression so far, yes.
>
> >So what. Most proofs in math are trivial.
>
> Sure, many proofs in math are trivial, perhaps even most, but
> there are certainly many results that have depth and subtlety.

The important ones are often simple.

> But it's just too far-fetched to believe that you can prove ZFC
> inconsistent with a trivial proof, and as far as I can see,
> deep, careful, rigorous thinking is beyond you.

The nice thing about inconsistency proofs
is they can be very simple. Consider
Russell's paradox. Even though the
argument is simple, Russell's paradox
proves naive set theory is inconsistent.
http://en.wikipedia.org/wiki/Russell's_paradox

I will tell you a secret about inconsistency proofs.
You can't refute an inconsistency proof.

Proof by refutation assumes the theory
in question is consistent. You are
assuming a statement can't be both
true and false.

Counter examples prove nothing.
Without a counter example
it's not an inconsistency proof.

What is the "counter example" in
Russell's paradox? It is easy to find
a counter example, which is the point
of the proof.

[Rupert]

On Feb 4, 2:32 am, RussellE <reaste...@gmail.com> wrote:
> On Feb 3, 9:56 am, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > On Jan 29, 2:10 am, RussellE <reaste...@gmail.com> wrote:
>
> > > Some more strange things that are provable
> > > if the complex numbers are a model of MA:
>
> > > x*x = (Sum of odd numbers from 0 to 2*x)
>
> > In order to state and prove this proposition in MA you would have to
> > go through the procedure which we use in PA for representing a finite
> > sequence of numbers by a single number. I doubt that proving all the
> > necessary theorems can be done in MA.
>
> I am not sure I understand your objection.
> I don't have a proof by induction, but surely
> there is one. If we can prove this relation
> in PA using induction, we can prove it in MA.
>

First you have to specify how you are going to express the assertion
in the first-order language of arithmetic at all.

There is a standard way of doing this in PA which involves coding
finite sequences of natural numbers by single numbers. But the
properties of this method of coding which are provable in PA may not
be provable in MA. In fact they can't be, because for example it's a
theorem of PA that distinct ordered pairs of natural numbers are coded
for by distinct natural numbers, and this obviously can't be true in
any finite model of MA.

[RussellE]

Define a complex number as
http://en.wikipedia.org/wiki/Complex_number

(a+b*i)

where a and b are members of the
universal set and -1 = i*i.

Negation:
-x = (-1)*x

Addition:
(a+bi)+(c+di) = (a+c)+(b+d)i

Multiplication:
(a+b*i)*(c+d*i) = (ac-bd)+(bc+ad)i

Let U = {0,1,2,3,4} =
{0,1,2,-2,-1}
Let i = 2

Prove
i*i = -1

(0+1*2)*(0+1*2) =
(0*0-1*1)+(1*0+0*1)*2 =
(-1+0*2) = (4+0*2)

Prove 3+4 = 2 mod 5

We can represent 4 as 2i.

(3+0*2)+(0+2*2) =
(3+0)+(0+2)*2 =
(3+2*2) = (2+0*2)

Let's try an even model.

U = {0,1,2,3,4,5,6,7,8,9} =
{0,1,2,3,4,5,-4,-3,-2,-1}
This is base 10 where i = 3.

Prove 5*6 = 0 mod 10

Let 6 = (0+2*3)

(5+0*3)*(0+2*3) =
(5*0-0*2) + (0*0+5*2)*3 =
(0+10*3) = (0+0*3)

The following statement is
true in all finite models of MA:

AxAyEz( (x+y*i) = (z+0*i) )

If this is a theorem of MA and the

complex numbers are a model of MA

then every complex number with a
non-zero imaginary component is
equal to a complex number where
the imaginary component is 0.

Let the successor function for
the complex numbers be:

S(x) = x + (1+0*i)

Notice the imaginary part is 0.
No matter how many times we apply
the successor function, the
imaginary part will always be 0.

[RussellE]

On Feb 3, 10:09 pm, Rupert <rupertmccal...@yahoo.com> wrote:
> On Feb 4, 2:32 am, RussellE <reaste...@gmail.com> wrote:
>
>
>
>
>
> > On Feb 3, 9:56 am, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > > On Jan 29, 2:10 am, RussellE <reaste...@gmail.com> wrote:
>
> > > > Some more strange things that are provable
> > > > if the complex numbers are a model of MA:
>
> > > > x*x = (Sum of odd numbers from 0 to 2*x)
>
> > > In order to state and prove this proposition in MA you would have to
> > > go through the procedure which we use in PA for representing a finite
> > > sequence of numbers by a single number. I doubt that proving all the
> > > necessary theorems can be done in MA.
>
> > I am not sure I understand your objection.
> > I don't have a proof by induction, but surely
> > there is one. If we can prove this relation
> > in PA using induction, we can prove it in MA.
>
> First you have to specify how you are going to express the assertion
> in the first-order language of arithmetic at all.

I see what you are saying.
I don't know how to express (Sum of all numbers 0 to x) in FOL.

> There is a standard way of doing this in PA which involves coding
> finite sequences of natural numbers by single numbers. But the
> properties of this method of coding which are provable in PA may not
> be provable in MA. In fact they can't be, because for example it's a
> theorem of PA that distinct ordered pairs of natural numbers are coded
> for by distinct natural numbers, and this obviously can't be true in
> any finite model of MA.

If it is possible to prove in PA then it can be proven in MA
for sets of numbers where every number less than log2(u).
(where U is the universe and u = |U|).

In complex models of MA i =< log2(u).
(where i*i = -1).

[RussellE]

Sorry, This is wrong.

[David Libert]

Complex MA model and primitive recursion
========================================

Author: David Libert <dave@aptosidbox>
Date: 2012-02-04 16:46:24 EST


Table of Contents
=================
1 Followup and response
2 C and MA style primitive recursion
3 C as MA model


1 Followup and response
~~~~~~~~~~~~~~~~~~~~~~~~
I had intended this to be a followup and response to

[1]
Rupert
"Re: Finite Models of the Complex Numbers"
sci.logic, sci.math
Feb 3, 2012
[http://groups.google.com/group/sci.logic/msg/ab63146ce7dee811]


I found [1] on Google and was able to write below a response to
that. So I had intended as noted to just post this as a followup to
[1].

I discovered that my own news provider didn't recieve [1], even
though it had arrived to Google. I waited in case it would arrive
late, but as time has passed I will instead just place this as a
followup eksehwere.

Below, I quote from [1]. You can also find [1] thrugh the Google
link above.

[1] and my response was about C as MA model.

So I found another article also mentioning that C as MA model to
attach this to as followup. That article is:

[2]
quasi
"Re: Finite Models of the Complex Numbers"
sci.logic, sci.math
Feb 2, 2012
[http://groups.google.com/group/sci.logic/msg/4dc06dc9d589afbb]

The connection of [2] to C is quasi responding to Russell:

>>My only interest in the complex numbers is proving they are
>>not a model of MA.
>
>How do you propose to do that?
>And what about David Libert's posted proof that the complex
>_are_ a model of MA?

I continue with the main body of this article, responding to
Rupert's [1] :

2 C and MA style primitive recursion
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

On Feb 3, 12:56 pm, Rupert <rupertmccal...@yahoo.com> wrote:
> On Jan 29, 2:10 am, RussellE <reaste...@gmail.com> wrote:
>
> > Some more strange things that are provable
> > if the complex numbers are a model of MA:
>
> > x*x = (Sum of odd numbers from 0 to 2*x)
>
> In order to state and prove this proposition in MA you would have to
> go through the procedure which we use in PA for representing a finite
> sequence of numbers by a single number. I doubt that proving all the
> necessary theorems can be done in MA.

Some months back, I had thought it was routine to redo Godel's use
to the beta function as from PA in MA. On that basis I conjectured
Tennebaum's theorem for MA, and claimed proofs that MA proves
propositional logic is inconsistent, and MA can define a linear
ordering the universe.

I later realized a problem for trying to copy the beta function to
MA. I noted that and posted a retraction about the MA proof of ~con
and the definition of < :


[3]
David Libert
"Re: Modular Arithmetic is Uncomputable"
sci.logic, sci.math
Sep 27, 2011
[http://groups.google.com/group/sci.logic/msg/376f1f35a0996018]


[3] didn't mention the Tennenbaum conjecture. But I recalled later
my reasons for conjecturing Tennenbaum for MA were also affected by
this issue of the beta function.

My old articles conjecturing Tennenbaum were the first two which
gave and infinite MA model, as referenced early in

[4] [http://davesscribbles.wikispaces.com/modarithth]


Godel used the beta function to formalize definitions by primitive
recursion in PA.

Such definitions by primitive recursision are the obvious method to
try to get something like Russell is claiming above, ie summations up
to a bound for example.

My retraction above was about difficulites to formalize primitive
recursion in MA, and so blocking an obvious try at definitions as
Russell suggested above.

Later developments made it yet worse for primitive recursion in MA.

In

[5] [http://davesscribbles.wikispaces.com/modarithorder]

I wrote about how define a linear ordering on the MA universe using MA
style primite recursion.

In particular, [5] implies that if MA style primitive recursive
functions are definable without paramters in an MA model, then a
linear ordering of the MA universe is definable without paramters in
that MA model.

Below, you mention C being a model for MA. I agree with that and
will discuss it further below.

No linear ordering of C is definable in C in field language. I
discussed the without parameters version of this in


[6]
David Libert
"Re: Modular Arithmetic is Uncomputable"
sci.logic, sci.math
Oct 2, 2011
[http://groups.google.com/group/sci.logic/msg/dec382ea29439809]


So if C is an MA model as we think, this gives an example of an MA
model for which primitive recursive functions are not definable. Not
merely a retraction, but a counterexample.

3 C as MA model
~~~~~~~~~~~~~~~~

> The complex numbers are a model of MA, by the way. This is because the
> definable subsets of n-dimensional affine space over the complex
> numbers in the first-order language of field theory are precisely the
> constructible subsets. If the intersection of such a subset with a
> linear subvariety of dimension one was infinite, then it would be
> equal to the whole linear subvariety. This shows that the induction
> axioms are true in the complex numbers, if we interpret + and * by the
> usual field operations, 0 by the number 0, and S by the function which
> sends x to x+1.

You are affirming this as well. I had posted to news and to
Wikispaces my claimed proof:

[7] [http://davesscribbles.wikispaces.com/modarithcomplex]


You are noting similar to my claim in [7], about definable sets
being exactly finite and cofinite.

Also, after [7] I posted

[8]
David Libert
"Re: Finite Models of the Complex Numbers"
sci.logic, sci.math
Jan 29, 2012
[http://groups.google.com/group/sci.logic/msg/54ee734d10b46e2d]


where I recalled Tarski writing about elimination of quantifiers for
C. That was main part of [7].

I Googled "complex numbers elimination of quantifiers" and got hits
including

[9]
http://www.wolframalpha.com/entities/mathworld/quantifier_elimination/06/1c/h6/

--
David Libert ah...@FreeNet.Carleton.CA

[David Libert]

Algebraic numbers model MA

==========================

Author: David Libert <dave@aptosidbox>

Date: 2012-02-04 17:07:07 EST


Table of Contents
=================
1 Background
2 AC based proof A models MA from claim that C models MA
2.1 Prove this theory is consistent
2.2 Consider M a model of this consistent theory
2.3 Consider N continuum sized elementary substructre of M
2.4 C satisfies same theory as A
3 Alternate proof A models MA apply the proof for C directly
4 An internet reference


1 Background
~~~~~~~~~~~~~
This is a followup and response to quasi

[1]

quasi
"Re: Finite Models of the Complex Numbers"
sci.logic, sci.math

Feb 1, 2012
[http://groups.google.com/group/sci.logic/msg/612536fd38953223]



On Feb 1, 9:39 am, quasi <qu...@null.set> wrote:
> On Wed, 01 Feb 2012 09:38:04 -0500, quasi <qu...@null.set> wrote:
> >On Wed, 01 Feb 2012 14:10:25 +0000, Alan Smaill
> ><sma...@SPAMinf.ed.ac.uk> wrote:

>
> >>RussellE <reaste...@gmail.com> writes:
>
> >>> There are uncouintable models of PA.
> >>> A number of people have claimed the complex numbers
> >>> are a model of MA and the complex numbers are
> >>> uncountable. So far, no one has come up with a countable
> >>> model of MA. Compactness says there must be one.
>

> >>What about the algebraic numbers?
>
> >I'm pretty sure the algebraic numbers _are_ a model of PA.
>
> I meant MA, not PA (typo).
>
> >And for the benefit of Russell, in case he doesn't know,
> >the set of algebraic numbers is countable.
>
> quasi


You had previously conjectured this, that the algebraic numbers are
a model of MA.

I posted my own conjectures:

[2]
David Libert "Re: Even and Odd Infinities"
sci.logic, sci.math Jan 14. 2012
[http://groups.google.com/group/sci.logic/msg/607dfdb0cfba4261]


In [2], I noted your earlier conjectures of this, and I conjectured
it also. I noted I could prove it if the result that C models MA is
true.

I posted my claimed proof that C models MA:

[3] [http://davesscribbles.wikispaces.com/modarithcomplex]


I just posted again about C as MA model, with further citations
beyond [3]:

[4]

David Libert
"Re: Finite Models of the Complex Numbers"
sci.logic, sci.math

Feb 4, 2012
[http://groups.google.com/group/sci.logic/msg/fb33a7417f971c21]


I mention [3]-[4], because my claims about the aglebraic numbers
rest on those about C.

Let A be the set of algebraic complex numbers.

2 AC based proof A models MA from claim that C models MA
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The proof I had in mind in [2], that A models MA, was to claim that
A and C satisfy the same theory in the lanuage of fields. That is
the complete theoreis of those models in field langaueg are the same
theory.

Then if C models MA, A must also.

So to see that A and C satisfy the same theory.

Make a constant for each algebric number in A, and consider the
theory of structure A in field language with all these constants
denoting the corresponding algebraic number.

Add also continuum many new constants t_i for i in conitnuum
sized index set.

Add also the schematum of axioms which collectively say the t_i are
mutually transcendental. That is, each nonconstant polynomial in
finitely many t_i has no solution.

2.1 Prove this theory is consistent
====================================
Each finite fragment sub-theory of this consistent. Because
finitely many of these trancendental axioms only assert there are
finitely many t_i making finitely many polynomials with algebraic
coefficents non-0.

And such a finite fragment is consistent, becasue we can a take a
model of form the algebric numbers A, with all algebraic constants
interpreted by the original.

To see this, we order the finitely many t_i constants as
t_1 , ... , t_n.

In each of the finitely many polynomails in t1, ... , t_n,
We first group together like powers of t_n,
considering it as a polynomial in t_n with coefficents polynomials
in t_1 , ... , t_(n-1),

The we consider each of these polynials in t_1, ... , t_(n-1),
one polynomial for each power of t_n in each of the starting
polynomials in t_1, ... , t_n.

For each of these finitely many polynomials in t_1, ... , t_(n-1)
we group like powers of t_(n-1), regarding it as a polynomial in
t_(n-1) with coefficents polynomials in t1, ... , t(n-2).

And so we proceed through the variables, each time a sinple
polynomial is replaced by as many new polynials as its power, but
polynials in one fewer variable.

So we work down to finitely many polynomials in t_1, with
coefficients being algebraic numbers.

Each of these polynomials has only finitely many roots in A, and
there are only finitely many of these polynomials.

A is infinite, so we can find and a_1 in A so all these finitely
many polynomials with algabraic coefficients don't evaluate to 0 on
a_1.

Assign that a_1 value to t_1. Make that evaluation on the
coefficeitns of the finitely many polynomials for t_2. These were
originally polynomals in algrbric numbers and t1, but we are now
evaluti8ng t_1 as algrbric numebr a_1, so these evaluate as an
algebraic number. By our choise of a_1, all these evaluation are
to non-0.

With this evaluation of t_1 to a_1, th finately many polynomials
for t_2 become polynomials in algraic numer coeffiencts, which are
not the zero polynomial. (We made sure t_1 evaluating to a-1
wouldn't send all the polynomals up a level to 0).

So finitely many non-0 polynials in t_2 have only finitely many
roots combined, and A is infiinte. Let a_2 in A be a non-root of
all these finitely many.

So evalute t_1 as a_1 and t_2 as a_2, and consider the polymials
for t_3 with this evaluation. As before, non-0 polynomialsm so
only finitely many roots, so find a_3 to evaluate fr t_3 and keep
all results non-0.

On up to t_n and a_n.

Then with t1 , ... , t_n respectively interpreted as
a_1, ... , a_n A with the previous interpetation of algebraic
constants becomes a model of the finite fragment of our theory.

So by above, every finite fragment of our theory is consistent, so
by compactness the entire theory is consistent

2.2 Consider M a model of this consistent theory
=================================================
The original A embeds into M, by sending each A a memeber to the M
denotation of the constant for a.

For a1 ~= a2 in A, the theory of A proves the ~= of the
constants, so M being a model of the extnesion of the theory,
satsifies the ~=. So this is an injection.

It is an elementrary embedding, since the theory included the full
theory of A.

So M is an algebraically closed field, since the thoery of A said A is
algebraically closed.

Also the extra axioms of the theory said the t_i were pairwise ~=
and jointly transcendental

2.3 Consider N continuum sized elementary substructre of M
===========================================================
There were continuum many t_i constants, and countably many
algebraic constants. So continuum many total constants.

With AC we can therefore form an elementary substructre of M of
size continuum, call this N. A still embeds into N by all those
alegraic constants. N still satifies the same theory as A.

All the t_i denotations are in N, and are still mutually
transcendental since N is an elementary substructe od M.

N is algebraically closed, it has a continuum sized set of mutually
transcendental elements, and it has overall cardinality contiuum.

So N can be charactrized up to isomophism as an agebrically closed
field of haracterisitc 0 over a continuum sized transcendence base
(using AC).

2.4 C satisfies same theory as A
=================================
C is characteristic 0 algrically closed over contuunum sized
trancendence base. (Using AC).

So C is isomorphic to N above.

But N satisfied same theory as A.

3 Alternate proof A models MA apply the proof for C directly
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This would be direclty using the [3] proof I posted

[http://davesscribbles.wikispaces.com/modarithcomplex]

All the pof used abourt C was that it extended the integers and it is
algebrically closed.

So just apply [3] directly to A.

4 An internet reference
~~~~~~~~~~~~~~~~~~~~~~~~

[5]
[http://mathoverflow.net/questions/3377/a-few-questions-on-model-theory-especially-model-theory-of-rings/3491#3491]

[5] says the complex numbers have the same theory as the theory of
algebraically closed fields.

So again, A models MA if the C claim is correct.



--
David Libert ah...@FreeNet.Carleton.CA

[Tim Golden BandTech.com]

On Feb 4, 5:16 pm, ah...@FreeNet.Carleton.CA (David Libert) wrote:
>                       Algebraic numbers model MA
>                       ==========================

Geeze, now this guy is serious. Well, let's see just how serious he
is.

The complex numbers have a continuous quality just as the real numbers
do.
These continuous principles are not covered by discrete mathematics
such as MA.

However, the rotational qualities of the complex numbers do match the
rotational qualities of MA,
as do the rotational qualities of the real number. Few can appreciate
the rotational nature of the real numbers, but they happen to contain
a discrete rotation, which is in fact the only rotational option that
they contain. We witness this in the expression
- x
where x is real. We likewise witness that the real number happens to
carry an MA(2) behavior in its sign principles, where we find the
product table:
| - +
_________
- | + -
+ | - +

which are simply modulo two sum behaved, where - represents unity and
+ represents two or zero, depending upon the standards one chooses to
construct by, neither two nor zero being exclusive of the other since
within a modulo two system they carry the same meaning. This detail
likewise explains the default usage of a blank or void sign implying
positive sign, or sign two.

These principles are extensible to signed systems of greater or lesser
modulus. It happens that the MA(3) variety does in fact match the
complex numbers, but we are overlooking the continuous portion of the
construction. In order to allow the continuous varieties we must
engage a quality which is known as magnitude, but whose fundamental
nature must be granted prior to the construction of the real number,
for we now have a construction of the real number as
s x
where s is MA(2) sign and x is this continuous magnitude of which I
write. Now, to extend the system symmetrically we must seek out a
behavior of symmetry upon the sign itself which we can extend to the
MA(3) variety, and it is here that the greatest stumbling block exists
for mankind since we have been trained upon those minus and plus signs
as fundamental and inherent to the real number, whose fundamental
nature has been preached to us since the 1600's. Yes, this is indeed
how large a discussion this is, and yet none seems capable of
registering it. I submit to you that this is a caricature of the
weakness of the human form, and our ability to be programmed just as
our computers is established. At least our computers are developing
nicely, but our own attempts to make them more like ourselves (e.g.
AI) is a misnomer.

Now, the modern ignorance has been established and is backed by the
observation that the symmetrical form
- x + x = 0
in its extensible nature is not used in the numerous constructions of
the real numbers. Instead an axiom of the existence of an inverse is
used, but such an axiom does not easily come to the same extensible
form, which for an MA(3) (P3) system reads
- x + x * x = 0
where '*' is a new sign; the third sign; and this is sufficient to
establish the complex numbers with their continuous properties intact.
No, this is not a complete rendition, but that rendition and a
substantial proof can be found at my website:
http://bandtech.com/PolySigned
http://bandtech.com/PolySigned/ThreeSignedComplexProof.html

Good luck getting past your two-signed perplexion. It is so dismal to
be an American where this logic even drives the political structure.
I'm looking forward to the adoption of a third sign.

- Tim

>   [http://mathoverflow.net/questions/3377/a-few-questions-on-model-theor...]

[RussellE]

On Feb 2, 1:38 am, quasi <qu...@null.set> wrote:
> RussellE wrote:
> >quasi wrote:
>

> >My only interest in the complex numbers is proving they are
> >not a model of MA.
>
> How do you propose to do that?
>
> And what about David Libert's posted proof that the complex
> _are_ a model of MA?

Both Dave Libert's and Rupert's proofs assume the
complex numbers are a algebraically closed field.
I've shown why the 0-model is the only model of MA
that is algebraically closed. In all other models there
exist a polynomial with no solution in the model.

x(x-1)(x-2)...(x+2)(x+1)+1 = 0

The following sentence is a theorem of MA:

AxAyAiEz ((x+y*i) = (z+0*i))

The complex "plane" is superfluous in MA.

The complex numbers are defined as (a+b*i) where
a and b are real numbers and i is the square root of -1.
If the complex numbers are a model of MA then the
square root of -1 is a real number.
This becomes even more obvious when we look
at the successor function defined for C.

S(x) = x + (1+0*i)

The complex part of all the numbers in the model
is 0 using this definition of successor.

If the complex numbers are a model of MA then
so are the real numbers. If the complex numbers
are a model why aren't the Gaussian integers a model?
If the algebraic numbers are a model of MA then why
aren't the algebraic integers a model?

When Dave Libert says he can order the universe
of MA he is adding additional axioms. He is
assuming 0 is the smallest number and the
predecessor of 0 is the largest number.
These are additional assumptions. We can
assume any element in the universe is the
"smallest" number and the predecessor of
this number is the "largest". This is like adding
the axiom:

ExEzAy( x =< y =< z)

[quasi]

RussellE wrote:
>quasi wrote:

>> RussellE wrote:
>> >
>> >My only interest in the complex numbers is proving they are
>> >not a model of MA.
>>
>> How do you propose to do that?
>>
>> And what about David Libert's posted proof that the complex
>> _are_ a model of MA?
>
>Both Dave Libert's and Rupert's proofs assume the
>complex numbers are a algebraically closed field.

The complex numbers _are_ an algebraically closed field.

>I've shown why the 0-model is the only model of MA
>that is algebraically closed.

No you haven't.

>In all other models there exist a polynomial with no
>solution in the model.
>
>x(x-1)(x-2)...(x+2)(x+1)+1 = 0

For each nonzero _finite_ model of MA, there is an equation
such as the above which has no solution in the model.

That _doesn't_ imply that for every infinite model of MA,
there exists a nonconstant univariate polynomial equation
of the form given above which has no solution in the model.

By the Compactness Theorem, there exist infinite models of
MA which are _not_ algebraically closed.

However some infinite models of MA (for example the complex
numbers) _are_ algebraically closed.

I've previously suggested (and I'm almost certain it's true)
that _every_ algebraically closed field of characteristic zero
is a model of MA.

>The following sentence is a theorem of MA:
>
>AxAyAiEz ((x+y*i) = (z+0*i))

What a trivial theorem that is!

It's true in _any_ ring.

>The complex "plane" is superfluous in MA.

If by that you mean that you can't define real and imaginary
parts in the language of MA, then yes, I agree with that.

>The complex numbers are defined as (a+b*i) where
>a and b are real numbers and i is the square root of -1.
>If the complex numbers are a model of MA then the
>square root of -1 is a real number.

Nonsense.

>This becomes even more obvious when we look
>at the successor function defined for C.
>
>S(x) = x + (1+0*i)
>
>The complex part of all the numbers in the model
>is 0 using this definition of successor.

More nonsense.

For example, the successor of i is 1 + i.

>If the complex numbers are a model of MA then
>so are the real numbers.

It doesn't follow.

The real numbers are definitely _not_ a model of MA.

>If the complex numbers are a model why aren't the
>Gaussian integers a model?

They might be.

In any case, it's not true that an arbitrary subring
of a model of MA must also be a model.

>If the algebraic numbers are a model of MA then why
>aren't the algebraic integers a model?

Again, they might be.

But again, I see no immediate reason why they _must_ be.

The fact that the Gaussian integers and the algebraic
integers are "integers" (integral over the rationals) doesn't
mean they are integers in the sense of MA. They're not
integers in the sense of PA either.

You are confusing different meanings of "integrality".

>When Dave Libert says he can order the universe of MA he
>is adding additional axioms.

We've already discussed this.

I'm fairly sure that no ordered ring (with 0 < 1) is a model
of MA.

>assuming 0 is the smallest number and the predecessor of 0
>is the largest number.

If 0 is the smallest number, then 0 < 1.

If -1 is the largest number, than -1 > 0.

Adding 1 to both sides of -1 > 0 implies 0 > 1, contradiction
(assuming the order is a ring order).

You can order any set as a _set_, but you can't order a
nonzero model of MA (assuming 0 < 1) in the ring sense.

>These are additional assumptions.

>assume any element in the universe is the
>"smallest" number and the predecessor of
>this number is the "largest". This is like adding
>the axiom:
>
>ExEzAy( x =< y =< z)

Forget adding ordering axioms.

Of course you can arbitrarily impose an order, but without
compatibility with the ring operations, such an order would not
be of much interest.

For example, you can "order" Z/2z by asserting 0 < 1, but
it's certainly not a ring order, so what's the point?

In any case, your "proof" that the complex numbers are
not a model of MA is fatally flawed, showing once again
that even though you really don't know what you're talking
about, you confidently post nonsense.

Which is why I regard you as a crank.

quasi

[Frederick Williams]

RussellE wrote:
>
> [...] In all other models there

> exist a polynomial with no solution in the model.
>
> x(x-1)(x-2)...(x+2)(x+1)+1 = 0

Is that an expression in the language of MA?

[RussellE]

On Feb 7, 5:39 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> RussellE wrote:
>
> > [...] In all other models there
> > exist a polynomial with no solution in the model.
>
> > x(x-1)(x-2)...(x+2)(x+1)+1 = 0
>
> Is that an expression in the language of MA?

I have been wondering about that myself.
How does one define polynomials in MA?

Of course, I am not the one claiming
algebraically closed fields are models of MA.

[quasi]

On Tue, 7 Feb 2012 11:44:21 -0800 (PST), RussellE
<reas...@gmail.com> wrote:

>On Feb 7, 5:39 am, Frederick Williams <freddywilli...@btinternet.com>
>wrote:
>> RussellE wrote:
>>
>> > [...] In all other models there
>> > exist a polynomial with no solution in the model.
>>
>> > x(x-1)(x-2)...(x+2)(x+1)+1 = 0
>>
>> Is that an expression in the language of MA?

As written above, it's clearly _not_ a statement in the language of
MA.

To Russell: A valid statement in MA cannot have "..." as part of the
syntax. Thus, forget about statements like

1 + 1 + ... + 1 = 0

As far as I can see, there is no statement P(x) of MA such that

(1) P(x) is provable in MA (that is P(x) is a theorem of MA)

(2) In all models of MA, P(x) is true iff x can be obtained from
zero using finitely many successor operations.

>I have been wondering about that myself. How does one define
>polynomials in MA?
>
>Of course, I am not the one claiming algebraically closed
>fields are models of MA.

Not all algebraically closed fields are models of MA -- just
the ones which have characteristic zero (and in particular,
the field of complex numbers).

One doesn't need to define polynomials "in MA" to prove that
the field of complex numbers satisfies the axioms of MA.

quasi

[RussellE]

On Feb 7, 1:17 am, quasi <qu...@null.set> wrote:
> RussellE wrote:
> >quasi wrote:
> >> RussellE wrote:
>
> >> >My only interest in the complex numbers is proving they are
> >> >not a model of MA.
>
> >> How do you propose to do that?
>
> >> And what about David Libert's posted proof that the complex
> >> _are_ a model of MA?
>
> >Both Dave Libert's and Rupert's proofs assume the
> >complex numbers are a algebraically closed field.
>
> The complex numbers _are_ an algebraically closed field.

There may be some theories where the complex
numbers are an algebraically closed field, but,
MA is not such a theory.

I am still trying to figure out how you can even
define the complex numbers in MA.

> >I've shown why the 0-model is the only model of MA
> >that is algebraically closed.
>
> No you haven't.

The axiom of deniability.
"PA is consistent" therefore PA can't be inconsistent.

> >In all other models there exist a polynomial with no
> >solution in the model.
>
> >x(x-1)(x-2)...(x+2)(x+1)+1 = 0
>
> For each nonzero _finite_ model of MA, there is an equation
> such as the above which has no solution in the model.

OK

> That _doesn't_ imply that for every infinite model of MA,
> there exists a nonconstant univariate polynomial equation
> of the form given above which has no solution in the model.

Why isn't this implied?
What theory are you using to define "finite"?
We can't define finite in MA.
Finite becomes a slippery concept in non-standard models.

> By the Compactness Theorem, there exist infinite models of
> MA which are _not_ algebraically closed.

Interesting.
If I understand Dave Libert's proof, any infinite model
of MA must be algebraically closed or induction fails.

> However some infinite models of MA (for example the complex
> numbers) _are_ algebraically closed.

This hardly matters if compactness proves there
are infinite models of MA where induction fails.

> I've previously suggested (and I'm almost certain it's true)
> that _every_ algebraically closed field of characteristic zero
> is a model of MA.

Does the 0-model have characteristic 0?

> >The following sentence is a theorem of MA:
>
> >AxAyAiEz ((x+y*i) = (z+0*i))
>
> What a trivial theorem that is!

It is, isn't it.
It is all I need to prove the complex numbers
are not a model of MA.

> It's true in _any_ ring.

Even better.

> >The complex "plane" is superfluous in MA.
>
> If by that you mean that you can't define real and imaginary
> parts in the language of MA, then yes, I agree with that.

I can define the "imaginary" part of a number in MA.
These numbers obey all of the definitions for
addition and multiplication of complex numbers.
My trivial theorem proves complex numbers are
just another way to represent an element of the
universe.

> >The complex numbers are defined as (a+b*i) where
> >a and b are real numbers and i is the square root of -1.
> >If the complex numbers are a model of MA then the
> >square root of -1 is a real number.
>
> Nonsense.

What about the trivial theorem?
Ex ( (0+1*i) = (x+0*i) )

> >This becomes even more obvious when we look
> >at the successor function defined for C.
>
> >S(x) = x + (1+0*i)
>
> >The complex part of all the numbers in the model
> >is 0 using this definition of successor.
>
> More nonsense.
>
> For example, the successor of i is 1 + i.

OK.
Ex( (1+1*i) = (x+0*i) )

If I can get from (0+0*i) to (0+1*i) using
your successor function then adding 0
exactly i times gives us 1.
And adding 1 exactly i times gives us 0.

This follows from the definition of addition
for complex numbers.

0+0+...+0 = 1 (i*0 = 1)
1+1+...+1 = 0 (i*1 = 0)

This looks like a contradiction to me.

> >If the complex numbers are a model of MA then
> >so are the real numbers.
>
> It doesn't follow.

The trivial theorem shows I can replace
every complex number having a non-zero
complex part with a complex number
having no complex part.

Once I do this, I can throw away the
complex part leaving me with a set
of real numbers.

> The real numbers are definitely _not_ a model of MA.

Oops.

> >If the complex numbers are a model why aren't the
> >Gaussian integers a model?
>
> They might be.

Then i would be an integer.

> In any case, it's not true that an arbitrary subring
> of a model of MA must also be a model.
>
> >If the algebraic numbers are a model of MA then why
> >aren't the algebraic integers a model?
>
> Again, they might be.
>
> But again, I see no immediate reason why they _must_ be.
>
> The fact that the Gaussian integers and the algebraic
> integers are "integers" (integral over the rationals) doesn't
> mean they are integers in the sense of MA. They're not
> integers in the sense of PA either.
>
> You are confusing different meanings of "integrality".

I am good at that.

> >When Dave Libert says he can order the universe of MA he
> >is adding additional axioms.
>
> We've already discussed this.
>
> I'm fairly sure that no ordered ring (with 0 < 1) is a model
> of MA.

It seems we agree on this.

> >assuming 0 is the smallest number and the predecessor of 0
> >is the largest number.
>
> If 0 is the smallest number, then 0 < 1.
>
> If -1 is the largest number, than -1 > 0.
>
> Adding 1 to both sides of -1 > 0 implies 0 > 1, contradiction
> (assuming the order is a ring order).
>
> You can order any set as a _set_, but you can't order a
> nonzero model of MA (assuming 0 < 1) in the ring sense.
>
> >These are additional assumptions.
> >assume any element in the universe is the
> >"smallest" number and the predecessor of
> >this number is the "largest". This is like adding
> >the axiom:
>
> >ExEzAy( x =< y =< z)
>
> Forget adding ordering axioms.
>
> Of course you can arbitrarily impose an order, but without
> compatibility with the ring operations, such an order would not
> be of much interest.
>
> For example, you can "order" Z/2z by asserting 0 < 1, but
> it's certainly not a ring order, so what's the point?
>
> In any case, your "proof" that the complex numbers are
> not a model of MA is fatally flawed, showing once again
> that even though you really don't know what you're talking
> about, you confidently post nonsense.

So do you.

> Which is why I regard you as a crank.

Aren't we all.

[quasi]

On Tue, 7 Feb 2012 18:39:04 -0800 (PST), RussellE
<reas...@gmail.com> wrote:

>On Feb 7, 1:17 am, quasi <qu...@null.set> wrote:
>> RussellE wrote:
>> >quasi wrote:
>> >> RussellE wrote:
>> >> >
>> >> >My only interest in the complex numbers is proving
>> >> >they are not a model of MA.
>> >>
>> >> How do you propose to do that?
>> >>
>> >> And what about David Libert's posted proof that the

>> >> complex numbers _are_ a model of MA?

>> >
>> >Both Dave Libert's and Rupert's proofs assume the
>> >complex numbers are a algebraically closed field.
>>
>> The complex numbers _are_ an algebraically closed field.
>
>There may be some theories where the complex numbers are
>an algebraically closed field, but, MA is not such a theory.

If all you're saying is that there is no set S of statements
where each statement of S is expressed in the first order
language of MA such that a model of MA is an algebraically
closed field iff all statements of S are satisfied, then
yes, I'll agree with that.

But that doesn't mean that there does not exist an
algebraically closed field satisfying the axioms of MA.

>I am still trying to figure out how you can even
>define the complex numbers in MA.

This is why you need to learn the basics of Model Theory.

You don't define the complex numbers in MA.

MA has many models of which the complex numbers is just
one example.

I don't believe there is _any_ set S of statements of MA
such that the only model of MA which satisfies S is the
set of complex numbers.

>> >I've shown why the 0-model is the only model of MA
>> >that is algebraically closed.
>>
>> No you haven't.
>
>The axiom of deniability.
>"PA is consistent" therefore PA can't be inconsistent.

I have no idea what you're talking about. Perhaps it's just
some philosophical nonsense designed to deflect attention
your mathematical nonsense.

>> >In all other models there exist a polynomial with no
>> >solution in the model.
>>
>> >x(x-1)(x-2)...(x+2)(x+1)+1 = 0
>>
>> For each nonzero _finite_ model of MA, there is an equation
>> such as the above which has no solution in the model.
>
>OK
>
>> That _doesn't_ imply that for every infinite model of MA,
>> there exists a nonconstant univariate polynomial equation
>> of the form given above which has no solution in the model.
>
>Why isn't this implied?

Because truth in some models doesn't imply truth in others --
it's that simple.

>What theory are you using to define "finite"? We can't define
>finite in MA.

We don't need to define finite in MA.

One doesn't define a model in MA. You define it in some
set theory (ZFC by default).

>Finite becomes a slippery concept in non-standard models.

There you go again with your vague "non-standard models".

In the context of MA, you should drop that terminology
unless you can rigorously define it.

>> By the Compactness Theorem, there exist infinite models
>> of MA which are _not_ algebraically closed.
>
>Interesting. If I understand Dave Libert's proof, any
>infinite model of MA must be algebraically closed or
>induction fails.

I hope he didn't claim that, since it's not true. An infinite
field which is a model of MA must have characteristic zero,
but it need not be algebraically closed.

For example by the Compactness Theorem, there must exist
infinite fields which are models of MA such that the equation
x^2 + 1 = 0 has no solution.

>> However some infinite models of MA (for example the complex
>> numbers) _are_ algebraically closed.
>
>This hardly matters if compactness proves there are infinite
>models of MA where induction fails.

Compactness proves no such thing.

>> I've previously suggested (and I'm almost certain it's true)
>> that _every_ algebraically closed field of characteristic
>> zero is a model of MA.
>
>Does the 0-model have characteristic 0?

No, it has characteristic 1 (since 1 = 0)

The characteristic of a ring is the least positive integer n
such that the sum of n ones is zero. If for a given ring there
is no such positive integer we say it has characteristic zero.

>> >The following sentence is a theorem of MA:
>>
>> >AxAyAiEz ((x+y*i) = (z+0*i))
>>
>> What a trivial theorem that is!
>
>It is, isn't it. It is all I need to prove the complex
>numbers are not a model of MA.

You're hallucinating.

That statement is just as true for the complex numbers
as it is for any other model of MA.

>> It's true in _any_ ring.
>
>Even better.
>
>> >The complex "plane" is superfluous in MA.
>>
>> If by that you mean that you can't define real and imaginary
>> parts in the language of MA, then yes, I agree with that.
>
>I can define the "imaginary" part of a number in MA.
>These numbers obey all of the definitions for
>addition and multiplication of complex numbers.
>My trivial theorem proves complex numbers are
>just another way to represent an element of the
>universe.
>
>> >The complex numbers are defined as (a+b*i) where
>> >a and b are real numbers and i is the square root of -1.
>> >If the complex numbers are a model of MA then the
>> >square root of -1 is a real number.
>>
>> Nonsense.
>
>What about the trivial theorem?
>Ex ( (0+1*i) = (x+0*i) )

That's a true statement in MA and just as true in the
complex numbers (hint -- use x = i).

>> >This becomes even more obvious when we look
>> >at the successor function defined for C.
>>
>> >S(x) = x + (1+0*i)
>>
>> >The complex part of all the numbers in the model
>> >is 0 using this definition of successor.
>>
>> More nonsense.
>>
>> For example, the successor of i is 1 + i.
>
>OK.
>Ex( (1+1*i) = (x+0*i) )

Again true in any model of MA, and once again, just as
true in the complex numbers -- just use x = 1 + i.

I'll try to help you see your rather elementary and naive
misconception ...

It's true that every complex number has a unique
representation in the form a + b*i where a and b are real.

It's not true that if z is a complex number and z = u + v*i
that u,v must be real.

Thus you seem to think that the equation

1 + 1*i = x + 0*i

forces 1 = x and 1 = 0, but that's silly, since in this
context, x is only required to be a complex number.

Thus, the equation

1 + 1*i = x + 0*i

simply implies

1 + i = x

so voila, x exists.

>If I can get from (0+0*i) to (0+1*i) using your successor
>function then adding 0 exactly i times gives us 1.

The above is complete nonsense.

>And adding 1 exactly i times gives us 0.

More nonsense.

>This follows from the definition of addition
>for complex numbers.

You're clueless.

>0+0+...+0 = 1 (i*0 = 1)
>1+1+...+1 = 0 (i*1 = 0)
>
>This looks like a contradiction to me.

Worse than clueless.

Essentially hopeless (destined to be a crank forever).

>> >If the complex numbers are a model of MA then
>> >so are the real numbers.
>>
>> It doesn't follow.
>
>The trivial theorem shows I can replace
>every complex number having a non-zero
>complex part with a complex number
>having no complex part.

No that's the same hallucination as before.

>Once I do this, I can throw away the complex part leaving
>me with a set of real numbers.

No -- you can't define real and imaginary parts in MA.

>> The real numbers are definitely _not_ a model of MA.
>
>Oops.

You're such a joke. You think the fact that

(1) the complex numbers are a model of MA

(2) the reals are not a model of MA somehow

leads to a contradiction, but forget it -- there's nothing
there. You think it's a smoking gun but it's really just
smoke (in your logic).

>> >If the complex numbers are a model why aren't the
>> >Gaussian integers a model?
>>
>> They might be.
>
>Then i would be an integer.

The word integer depends on the context.

The element i is certainly a Gaussian integer. It's
certainly an algebraic integer. It's certainly not
an ordinary integer. But none of the properties

Gaussian integer

algebraic integer

ordinary integer

are definable in the language of MA.

Do you have a definition, expressible in the language of
MA, for what it means for an element to be "an integer"?

If not, then what justifies your claim that if the field of
algebraic integers is a model of MA, then i must be
"an integer"?

I'll point out here a general kind of error you seem to be
making again and again.

Consider a first order theory T and a model X.

The definition of X need not be done from within T.

Thus X and it's elements may very well have properties
which are not expressible in the language of T.

But as long as X has the properties required by T, X
is a model of T.

As an example, try this ...

The set Z of integers with the operation of addition
is a model of the first order theory of Abelian Groups.

As an ordered set, Z has a natural order.

But there's no way to express the statement 0 < 1 in
the first order language of Abelian Groups.

Of course, we regard the elements of Z as having an order,
but those properties are outside the context of the
first order theory of Abelian Groups, and such extraneous
properties are not relevant to the question of whether
or not Z (with addition) qualifies as a model of the
first order theory of Abelian Groups.

Similarly, the complex numbers can have properties that
are not expressible in the language of MA, while still
qualifying as a model of MA.

I doubt you will understand my points above, but at least
I tried to identify a concept which is key to some of
your most blatant errors.

>> In any case, it's not true that an arbitrary subring
>> of a model of MA must also be a model.
>>
>> >If the algebraic numbers are a model of MA then why
>> >aren't the algebraic integers a model?
>>
>> Again, they might be.
>>
>> But again, I see no immediate reason why they _must_ be.
>>
>> The fact that the Gaussian integers and the algebraic
>> integers are "integers" (integral over the rationals) doesn't
>> mean they are integers in the sense of MA. They're not
>> integers in the sense of PA either.
>>
>> You are confusing different meanings of "integrality".
>
>I am good at that.

And that's your basic problem.

By blurring meanings which are context dependent, you get
seeming contradictions which fall apart when the correct
meanings are restored.

>> >When Dave Libert says he can order the universe of MA he
>> >is adding additional axioms.
>>
>> We've already discussed this.
>>
>> I'm fairly sure that no ordered ring (with 0 < 1) is a model
>> of MA.
>
>It seems we agree on this.

Yes.

Sometimes, yes.

But at least I try to learn from my mistakes.

>> Which is why I regard you as a crank.
>
>Aren't we all.

No, not all of us.

We all make mistakes, sometimes blatant ones.

But a crank is one who continually posts nonsense while
ignoring or failing to comprehend valid arguments exposing
the flaws in their reasoning.

Based on that, you qualify easily as a crank.

As far as I can tell, the flaws in your logic are based on
ignorance, thus your nonsense claims are not deliberate
nonsense. If you _knew_ that your claims were nonsense, then
you would be a troll, but I think you actually believe your
own nonsense, hence you're just a crank.

quasi

[FredJeffries]

On Feb 5, 12:15 pm, RussellE <reaste...@gmail.com> wrote:
>
> Both Dave Libert's and Rupert's proofs assume the
> complex numbers are a algebraically closed field.
> I've shown why the 0-model is the only model of MA
> that is algebraically closed.

Your 0-model is not a field. It is not clear that it makes sense to
say that it is algebraically closed.

[RussellE]

Let's agree to use ZFC as our meta-theory.

> >Finite becomes a slippery concept in non-standard models.
>
> There you go again with your vague "non-standard models".
>
> In the context of MA, you should drop that terminology
> unless you can rigorously define it.

http://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic

The standard natural numbers are the universe of the
standard model of PA. If the universe has an element
that is not a standard natural number then the model
is non-standard. The complex numbers are obviously
a non-standard model because -1, 1/2, i, etc are not
standard natural numbers.

I have previously shown all infinite models of MA
must be non-standard.

> >> By the Compactness Theorem, there exist infinite models
> >> of MA which are _not_ algebraically closed.
>
> >Interesting. If I understand Dave Libert's proof, any
> >infinite model of MA must be algebraically closed or
> >induction fails.
>
> I hope he didn't claim that, since it's not true. An infinite
> field which is a model of MA must have characteristic zero,
> but it need not be algebraically closed.

I will let Dave and Rupert speak for themselves, but,
both proofs seemed to depend on the complex numbers
being algebraically closed.

> For example by the Compactness Theorem, there must exist
> infinite fields which are models of MA such that the equation
> x^2 + 1 = 0 has no solution.

I agree.

> >> However some infinite models of MA (for example the complex
> >> numbers) _are_ algebraically closed.

I don't agree. See below.

> >This hardly matters if compactness proves there are infinite
> >models of MA where induction fails.
>
> Compactness proves no such thing.

I think this remains to be seen.
I have hardly scratched the surface of things
I can prove about infinite models of MA using
compactness. I have already shown there
must be countably infinite, even, complex
models of MA. Such a model satisfies:

Ex(x~=0 & x+x=0) & Ex(-x=x) & Ax(x+x~=1) & Ex(x*x+1=0)

No one has even suggested an infinite even model of MA.

> >> I've previously suggested (and I'm almost certain it's true)
> >> that _every_ algebraically closed field of characteristic
> >> zero is a model of MA.
>
> >Does the 0-model have characteristic 0?
>
> No, it has characteristic 1 (since 1 = 0)
>
> The characteristic of a ring is the least positive integer n
> such that the sum of n ones is zero. If for a given ring there
> is no such positive integer we say it has characteristic zero.

Hmmm.
Rupert has pointed out how large sums are
difficult to express in FOL. I will define repeated
addition as multiplication.

The characteristic of a ring is the least positive

integer, n, such that n*1 = 0.

How are you defining "positive integer"?

If we define integer in MA then
Ax(x=0 or x*1 ~= 0) would prove every
model of MA has characteristic 0.

If we define positive integer in the meta-theory,
ZFC, then we have the possibility of hyperfinite
integers:

http://en.wikipedia.org/wiki/Hyperinteger

It is easy to see the characteristic of any
model of MA is a positive hyperfinite integer.
Let U be the universe of the model and u=|U|.
u*1 = 0 in every model of MA.

You seem to be defining positive integer
using the standard model of arithmetic.
I don't think it makes much sense to use
definitions that only work in standard models
when we know we are working in a
non-standard model.

You have defined the successor function as
S(x) = x + (1 + 0*i)

I want to apply successor i times to 0.
Dave Libert has pointed out that applying
successor y times to x is equivalent to x+y.

(0+0*i) + (1+0*i) + (1+0*i) + ... + (1+0*i) = (0+1*i)
0+i = i

I define repeated addition as multiplication.

Consider just the real part of the sum above.

0+1+1+...+1 = 0
i*1 = 0

The complex numbers have characteristic i
using your definition of successor.

Now, consider just the imaginary part of the sum.

0+0+0+...+0 = 1
i*0 = 1

This is flat out inconsistent.
I am using the standard definition for
addition of complex numbers.

I know ring theory doesn't have a
successor function. Successor
is an important part of both PA and MA.

You haven't defined a model of MA
until you define a consistent
successor function.

[quasi]

Fine with me.

Ex(-x=x) is trivial since x = 0 always satisfies that.

If you intended Ex(x~=0 & -x=x) then that's redundant
since it's equivalent to Ex(x~=0 & x+x=0).

Either way there's no need for that sub-statement.

Also, I think it's confusing and basically silly to call
a model of MA "complex" just because it satisfies Ex(x^2+1=0).
By that convention, the ring Z/5Z would be "complex" since
in that ring, x^2+1=0 is satisfied using x=2. Similarly,
lots of other finite rings of the form Z/nZ would also be
"complex", while missing a critical property of the complex
numbers -- the algebraic closedness.

>No one has even suggested an infinite even model of MA.

But it's just a trivial consequence of compactness.

There are lots of properties that are inherent in arbitrary
large finite models of MA, and hence, you can make lots more
claims that by compactness, there must also be infinite models
of MA with the same properties, but so what? Compactness
makes that automatic.

>> >> I've previously suggested (and I'm almost certain it's true)
>> >> that _every_ algebraically closed field of characteristic
>> >> zero is a model of MA.
>>
>> >Does the 0-model have characteristic 0?
>>
>> No, it has characteristic 1 (since 1 = 0)
>>
>> The characteristic of a ring is the least positive integer n
>> such that the sum of n ones is zero. If for a given ring there
>> is no such positive integer we say it has characteristic zero.
>
>Hmmm.
>Rupert has pointed out how large sums are difficult to express
>in FOL. I will define repeated addition as multiplication.

That's problematic.

I don't think you get to define that since multiplication
is already defined axiomatically in MA (or am I mistaken
about that?)

In any case, your use of the phrase "repeated addition"
is certainly problematic unless you mean repeated n times
where n is a positive integer (in the meta-theory),

>The characteristic of a ring is the least positive

>integer n such that n*1 = 0.

>
>How are you defining "positive integer"?

I am using it in the sense of any ordinary positive integer
in the meta theory.

>If we define integer in MA then

You _don't_ define integer in MA (unless you are trying
for maximum confusion).

>Ax(x=0 or x*1 ~= 0) would prove every model of MA has
>characteristic 0.

No, once again you blur contexts.

If n is a positive integer in the meta-theory, a model of
MA has characteristic n if the sum of n copies of 1 is 0,
and n is the least such positive integer.

If there is no such positive integer then we say the
characteristic is zero,

But the n in this definition is not being regarded as an
element of the model but rather an _ordinary_ positive integer.

Thus the ring Z/3z with the three elements 0,1,2 has
characteristic 3 since 1 + 1 + 1 = 0, but 3 is not an
element of the model -- it simply counts the minimum
number of ones that must be added to get to zero.

So while the sum of 3 ones

1 + 1 + 1

is a legal expression in MA (since 3 is an ordinary positive
integer) there no syntax in the first order language of MA
which allows the sum of x ones

1 + 1 + ... + 1

where x is an element of the model.

>If we define positive integer in the meta-theory,
>ZFC, then we have the possibility of hyperfinite
>integers:
>
>http://en.wikipedia.org/wiki/Hyperinteger
>
>It is easy to see the characteristic of any
>model of MA is a positive hyperfinite integer.
>Let U be the universe of the model and u=|U|.
>u*1 = 0 in every model of MA.

The above makes no sense.

In MA, u*1 = 0 implies u = 0.

>You seem to be defining positive integer using the standard
>model of arithmetic. I don't think it makes much sense to use
>definitions that only work in standard models

That's a disconnect for me.

When I say "n times", I mean n times where n is an ordinary
positive integer, nothing more.

>when we know we are working in a non-standard model.

You keep saying that, but to me, your vague "non-standard
model" claim is just handwaving nonsense. There's nothing
nonstandard about the Gaussian integers or the complex
numbers.

>> >> >The following sentence is a theorem of MA:
>> >> >
>> >> >AxAyAiEz ((x+y*i) = (z+0*i))
>> >> >
>> >> What a trivial theorem that is!
>> >
>> >It is, isn't it. It is all I need to prove the complex
>> >numbers are not a model of MA.
>>
>> You're hallucinating.
>>
>> That statement is just as true for the complex numbers
>> as it is for any other model of MA.
>>
>> It's true in _any_ ring.
>>
>> >Even better.
>> >
>> >> >
>> >> >The complex "plane" is superfluous in MA.
>>
>> >> If by that you mean that you can't define real and imaginary
>> >> parts in the language of MA, then yes, I agree with that.
>>
>> >I can define the "imaginary" part of a number in MA.
>> >These numbers obey all of the definitions for
>> >addition and multiplication of complex numbers.
>> >My trivial theorem proves complex numbers are
>> >just another way to represent an element of the
>> >universe.

The above is so ridiculous, and so reeking of ignorance, I
don't think there's much point continuing.

Sorry, I stopped reading a while back.

You exceeded my limit for responding to gibberish.

quasi

[David Libert]

Algebraically closed char 0 fields are MA model
===============================================

Author: David Libert <dave@aptosidbox>
Date: 2012-02-09 08:05:03 EST


Table of Contents
=================
1 I will note this in Wikispaces
2 this is followup and response to quasi
3 conjecture algebraically closed char 0 fields are MA models
4 I will post here proofs conj alg closed char 0 are MA models
5 related previous background
6 define algebraically closed ring
7 every algebraically closed ring is a field
8 Proofs every algebraically closed field of char 0 is MA model
8.1 AC based proof based on AC proof algebraic numbers model MA
8.1.1 Claim 1
8.1.2 Proof of Claim 1
8.1.2.1 Special case of Claim 1 for F1, F2 with #F1 < #F2
8.1.2.2 Proof of special case of CLaim 1
8.1.2.3 Proof full case of Claim 1
8.1.2.4 QED CLaim 1
8.1.3 side note claim 1 for each characteristic individually
8.1.4 finish AC proof alg closed char 0 F is MA model
8.1.5 discuss characteristic 0 in above proof
8.2 Metamathematical elimination of AC assumption
8.3 Direct proof without AC


1 I will note this in Wikispaces
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
[http://davesscribbles.wikispaces.com/modarithalgclsdma]

2 this is followup and response to quasi
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

[1]
quasi
"Re: Finite Models of the Complex Numbers"
sci.logic, sci.math

Feb 7, 2012
[http://groups.google.com/group/sci.logic/msg/0cdcdeb41bf79423]

3 conjecture algebraically closed char 0 fields are MA models
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

On Feb 7, 3:52 pm, quasi <qu...@null.set> wrote:
> On Tue, 7 Feb 2012 11:44:21 -0800 (PST), RussellE

[...]

> >Of course, I am not the one claiming algebraically closed
> >fields are models of MA.
>
> Not all algebraically closed fields are models of MA -- just
> the ones which have characteristic zero (and in particular,
> the field of complex numbers).

4 I will post here proofs conj alg closed char 0 are MA models
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

5 related previous background
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
I posted that I may have a proof that C the complex numbers are an
MA model. Later independently, quasi also posted the same
conjecture about C. Soon after, quasi also conjectured that the
algebraic numbers are also an MA model. Then quasi also soon after
conjectured every algebraically closed field is an MA model.

In

[2] [http://davesscribbles.wikispaces.com/modarithcomplex] I gave
references for quasi's conjecture aricles above, and I also
conjectured algebraically closed fields of characteristic 0 are MA
models, and noted I could prove that from C being an MA model.

[2] went on to post my claimed proof that C is an MA model.

For references about C being an MA model, [2] above and another
article see

[3] [http://davesscribbles.wikispaces.com/modarithcomplexmain]

I also later posted proving the algebraic numbers are an MA
model,based on the corresponing result [3] for C. That is
referenced in

[4] [http://davesscribbles.wikispaces.com/modarithalg]

6 define algebraically closed ring
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
It could make sense to define what it means for any ring to be
algebraically closed: every non-constant polymial with coeeficients
in the ring has at leas one root.

7 every algebraically closed ring is a field
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
for a in the ring a^-1 is a root of poly a*x - 1 .

8 Proofs every algebraically closed field of char 0 is MA model
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

8.1 AC based proof based on AC proof algebraic numbers model MA
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
A is defined to be the set of algebraic complex numbers.

The new proof here for all algebraically closed fields of
characteristic 0 will be based on the AC based proof that A models
MA from [4] above.

8.1.1 Claim 1
=============
all algebraically closed characteristic 0 fields have same
complete theory of their models

8.1.2 Proof of Claim 1
======================

8.1.2.1 Special case of Claim 1 for F1, F2 with #F1 < #F2
-----------------------------------------------------------
That is, for F1 and F2 algebraically closed fields of
characteristic 0, with #F1 < #F2 then they have the same
complete theory in field language

8.1.2.2 Proof of special case of CLaim 1
----------------------------------------
Let F1 and F2 be as assumed. We seek to show they have the sasme
theory.

By recent discussions, F1, F2 being algebrically closed must both
be infinite. So F2 being strictly larger must be uncountable.

The [4] proof started with the theory of field A, with contants
for all members and added continuum many new constants with axiom
schemata saying these are mutually transcendental over A.

We similarly form the theory of F1 with constants for all F1
members, and add #F2 many new constants and axioms saying there
are mutually transcendental over F1.

As in [4], argue this theory is consistent. As in [4], find M a
model of it, extending F1 and with the same thoery. As in [4],
take N a elementary substructure of M, now of size #F2.

N is an algebrically closed field with #F2 many mutually
tanscendental elements over F1, and with N of total
cardinality #F2.

With AC, N has a #F2 sized transcenednce basis.

[http://en.wikipedia.org/wiki/Transcendence\_basis] (http://en.wikipedia.org/wiki/Transcendence_basis)

F2 was algebrically closed with #F2 uncountable, so F2 also has
an #F2 cardinality trancendence basis.

So F2 and N must be isomorphic, so they have the same theory. But
N had the same theory as F1

QED Special case of Claim 1

8.1.2.3 Proof full case of Claim 1
----------------------------------
Suppose F1 and F2 are algebraically closed characteristic 0
fields. We seek to show F1 and F2 have the same theory.

By upward Lowenhiem Skolem, there is F3 an algebraically closed
field of characteristic 0 with #F3 > #F1 and #F3 > F2.

By 2 applications of the special case of Claim 1, F1 has the same
full theory as F3 and F2 has trhe same full theory as F3.

So F1 and F2 have the same full theory

QED full case CLaim 1

8.1.2.4 QED CLaim 1
-------------------

8.1.3 side note claim 1 for each characteristic individually
=============================================================
Fields of different characteristic from each other have different
full theories.

But for each characteristic p, p = 0 or p a prime, all
algebraically closed fields of characteristic p have the same
theory as each other.

This was proven above for case p = 0.

The same proof works for each prime p. Namely the characteric 0
proof above used that F1, F2 are both infinite. But this is also
true in characteristic p since both are algebraically closed.

The proof from [4] used the field was infinite, but as noted we
have that still.

Below though, I will just need the original Claim 1 for
characteristic 0.

8.1.4 finish AC proof alg closed char 0 F is MA model
======================================================
Suppose F is algebrically closed and characterisic 0. We want to
show F is an MA model.

C is complex numbers are algebrasiclaly closed and
characterisit 0. So by Claim 1, F and C have the same full theory
of their models in field language.

But C is an MA model, from [2] above.

So F is an MA model

QED AC basied proof

8.1.5 discuss characteristic 0 in above proof
=============================================
The proof above used Claim 1. As from the disccusion above, Claim
1 holds individually in each other charateristic.

But the proof went on to use Claim 1 to relate an arbitrary
algebraially closed charactertic 0 field to C.

That comparison using Claim 1 would have worked in other
characteristics p to cpmaore an arbitrary charateristic p field F to some other
field of characteristic p.

The proof above used the [2] result about C. And that is where
characteristic 0 was used. The [2] proof used C has characteristic
0, namely to get some finite succssor chain above 0 and below x
are each outside of fionte set F.

For a field F of characteristic p =~ 0, we can't used the p
version of Calim 1 to compare F to C. Instead we have to compare
F to some other field D of characteristic p. And for such D the
[2] proof would have a gap to copy to D.

8.2 Metamathematical elimination of AC assumption
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Above proved in ZFC that all algebraically closed fields of
charactirstic 0 model MA.

So by Godel's completeness, it shows the theory of algebrically
closed fields of characteristic 0, a recursivelyt axiomatized
theory, proves the shchmeata of MA axioms.

But this claim about a reucrisely axiomized first otrder theoy
proving a recursive set of axioms, is itself a purely arithmetical
claim.

So by Godel's L construction, if ZFC proves then then ZF does.

Paul Cohen discusses this in
_Set Theory and the Continuum Hypothesis_.

There is also Shoenfield's Absolutness Lemma, a stronger form of this.

8.3 Direct proof without AC
~~~~~~~~~~~~~~~~~~~~~~~~~~~
This all traces back to the elimination of quantifiers argument for
C from [2]. As I remarked before, all this argument used about C
was that C extends the inegers and is algebraically closed.

So directly run the argument from [2] with F algebrally closed of
charactoeristic 0 (hence exnteding integers) replacing C, to
directly get F satisfies induction. That argument from [2] did not
use AC.


--
David Libert ah...@FreeNet.Carleton.CA

[Frederick Williams]

RussellE wrote:

>
> I have previously shown all infinite models of MA
> must be non-standard.

Since MA is your invention, you get to specify what the standard model
(or models) is (or are).

[RussellE]

On Feb 9, 6:49 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> RussellE wrote:
>
> > I have previously shown all infinite models of MA
> > must be non-standard.
>
> Since MA is your invention, you get to specify what the standard model
> (or models) is (or are).

I am surprised this has become an issue.
I am even more surprised I am the one having to explain it.

Before we can define the complex numbers,
we have to define the reals. To define the reals,
we have to define the rationals. To define
rationals, we need to define the integers.
To define integers, we first need to define
the natural numbers.

A commonly used definition of the natural
numbers is Peano's Axioms.
http://en.wikipedia.org/wiki/Peano_axioms

Peano Arithmetic (PA) is an extension of
Peano's axioms that also defines addition
and multiplication. Modular Arithmetic (MA)
is an alternate definition for natural numbers.

The language of PA is S, 0, +, and *.
A term of the form SS..S0 is called a
numeral. A term is finite in length, so
a numeral represents applying the
successor function to 0 some finite
number of times.

A standard natural number is a natural
number that can be represented with a
numeral. A non-standard natural number
is an element in the model that can not
be represented with a numeral.
Non-standard natural numbers are
sometimes called infinite natural numbers.
We need to apply successor to 0 some
infinite number of times to get a
non-standard natural number.

If PA has a model then the set of
all standard natural numbers must
be the universe of one such model.
With the standard definitions of
addition and multiplication this is
called the standard model of PA.

PA must also have non-standard models.
In these models there is an element
in the universe that can not be represented
by a numeral.
http://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic

The finite models of MA are the standard
models of MA. Every element in the
universe of these models can be represented
by a numeral.

MA does not have an infinite standard model.

The proof is "trivial". Assume some standard
natural number, z, is the predecessor of 0.
Obviously, z is the largest natural number in
the universe. SInce every element of the
universe is less than or equal to some
finite number, the universe must be finite.

In any inifinite model of MA the predecessor
of 0 must be a non-standard natural number.

[quasi]

On Thu, 9 Feb 2012 18:55:20 -0800 (PST), RussellE
<reas...@gmail.com> wrote:

>In any inifinite model of MA the predecessor of 0 must be a
>non-standard natural number.

No, you can define what you mean by nonstandard elements
of MA, but you don't get to define "non-standard natural
number" since that already has a meaning in the context
of PA.

In the complex numbers, the element -1 is _not_ a
non-standard natural number -- it's not any kind of
natural number.

This is how you blur things, by applying terminology such as
"non-standard natural number" in ways that don't match the
required context, yielding illusory contradictions.

quasi

[FredJeffries]

On Feb 7, 6:39 pm, RussellE <reaste...@gmail.com> wrote:
>
> > I've previously suggested (and I'm almost certain it's true)
> > that _every_ algebraically closed field of characteristic zero
> > is a model of MA.
>
> Does the 0-model have characteristic 0?

The 0-model is not a field

[RussellE]

OK.