See Hot Sexy Star Aishwarya Rai Videos In All Angles.
at http://ukcitygirls.co.cc
Due to high sex content,i have hidden the videos in an image. in that
website on left side below search box click on image and watch
videos in all angles.please dont tell to anyone.
————————SCI.MATH—————————–
Take any list of reals
123
456
789
Diag = 159
AntiDiag = 260
It’s a NEW DIGIT SEQUENCE and it works on EVERY LIST.
—————————HERC——————————
defn(herc_cant_3)
The list of computable reals contains every digit (in order) of all possible infinite sequences.
..as a result of containing ALL (infinitely many) finite prefixes.
THEREFORE YOU CANNOT CONSTRUCT A NEW DIGIT SEQUENCE
————————–SCI.MATH————————–
BUT:
0.0
0.1
0.2
…
0.01
0.02
0.03
…
0.99
0.101
0.102
…
ALSO contains every finite prefix
AND 0.111… is not on that list.
THEREFORE ANTI-DIAG STILL *IS* A NEW DIGIT SEQUENCE.
—————————–HERC——————————
A correction to a correction does not prove the original assertion.
You STILL have not come up with a NEW DIGIT SEQUENCE.
You use the term NEW DIGIT SEQUENCE for the finite example 260
then you BAIT AND SWITCH and call it NEW NUMBER because
An AD(n) =/= L(n,n).
Is it a *NEW DIGIT SEQUENCE* or not?
Herc
In
http://math.stanford.edu/~feferman/papers/unfolding.pdf
Feferman gives a sense in which Mahlo cardinals can be seen as part of
an "unfolding" of ZF in a similar sense to that in which predicative
analysis is the "unfolding" of PA.
However he only permits reflection formulas in which the class
variables are universally quantified, and so draws the line at weakly
compact cardinals.
I find this quite a nice justification of the small part of the large-
cardinal spectrum. I would be interested if anyone could offer me any
reasons why I should accept weakly compact cardinals.
As it says, this may be a stupid query. Be gentle.
1st-order PA has many models. But it seems to me that there is
a very clear "minimal model", in that the standard model is
(isomorphic to) a subset of any other model. And I gather
that this fact can be proved with a fairly trivial extension
of PA itself, extended into extremely basic model theory
(i.e. set theory).
Assuming I am not yet too haywire:- Does this notion also
apply to ZF? Is there in some sense "a minimal model"?
Is this easy to prove? As easy as for PA?
I’m pretty sure I’ve read at some time a reference to
a "minimal model" for ZF, probably in Cohen ’66.
But I don’t know if this is the same thing.
TIA.
– Baffled Bill
Hypothesis: a real number contains a finite sequence that is not computable.
Contradiction
Therefore: all digits of every real are contained in the list of computable reals.
_________________________________________________________________
This may not IMPLY that all infinite digit sequences are computable, but
it trivially defeats this argument:
123
456
789
Diag = 159
AntiDiag = 260
A new digit sequence can be found on all real lists.
Herc
—
If you ever rob someone, even to get your own stuff back, don’t use the phrase
"Nobody leave the room!" ~ OJ Simpson
In his thesis "The Search for New Axioms" Peter Koellner writes
"There is no known example of a natural sentence phi of first-order
arithmetic such that (1) phi is known to be independent of ZF and (2)
it is not known whether phi is true."
Is there any example at all of a "natural" sentence of first-order
arithmetic which is known to be independent of ZF?
See Hot Sexy Star Aishwarya Rai Videos In All Angles
at http://lifeisbeatiful.co.cc
Due to high sex content,i have hidden the videos in an image. in that
website on left side below search box click on image and watch
videos in all angles.please dont tell to anyone.
The antidiagonal is too general to meaningfully define a number.
It’s not just based on all digits in
forall n, L(n,n)
The antidiagonal argument also has to work on EVERY PERMUTATION of a list.
That means, you can construct an anti-diagonal using the 1st digit of *any* listed real,
the second digit of *any* other listed real, and so on.
As long as you choose ANY increasing unique position digit of EVERY real, and eventually the selected reals
fillas out from the top, you can create a real from any ‘diagonal’, like so:
0. _ _ _ x _ _
0. x _ _ _ _ _
0. _ _ x _ _ _
0. _ x _ _ _ _
0. _ _ _ _ _ x
0. _ _ _ _ x _
A valid diagonal from the leftmost x to the rightmost x.
You can select ANY digit {0, .. 9} except change the x value and that should be a NEW real
according to Cantor.
If you designed an algorithm that could alter the x positions so that any digit of the
diagonal could be changed to a different digit, then that would prove the diagonal
argument doesn’t work!
Herc
—
If you ever rob someone, even to get your own stuff back, don’t use the phrase
"Nobody leave the room!" ~ OJ Simpson
The KBH Word Coordinate Application assigns 17,636 most frequent words
to the KBH Word Coordinate System. The purpose of the KBH Word
Coordinates are shorter text messages and easier input of text
messages.
The KBH Word Coordinate Application is a reference source for text
messaging that installs on a Win32 PC computer.
Here are a few KBH Word Coordinates:
ATAB, above-average
BDB, baby-sitter
EO, especially
ILB, idealistic
IH, important
JRB, jealous
OQ, opportunity
UD, understand
UZ, usually
Here is a link to the KBH Word Coordinate Application:
http://www.kbhscape.com/texting.htm
:
:
The origin of this question comes from a colleague, who asked "Can you
have two nonisomorphic rings R and S, such that the additive group of
R is isomorphic to the additive group of S, and the multiplicative
semigroup of R is isomorphic to the multiplicative semigroup of S"?
My answer, after some thought, was "yes", with the following example.
Let F be a field, and let R=F[x], S=F[x,y]. The isomorphism of (R,+)
and (S,+) can be done explicitly, as they are vector spaces of F of
dimension aleph_0, and one can give an explicit bijection by ordering
the monomials in R by degree, and the monomials in S by total degree
and lexicographically among those of the same degree.
To prove the isomorphism between (R,*) and (S,*), I argued as follows:
the cardinality of the set of monic irreducible polynomials in F[x] is
|F|*aleph_0: for infinite F, there are at least |F| monic irreducibles
(the linear polynomials), and since |F[x]|=|F|, there are at most |F|
of them. For finite F, there are at least aleph_0 monic irreducibles
(at least one for each positive integer n, given by the finite
extensions and the primitive element theorem), and |F[x]|=aleph_0, so
you get aleph_0 again. Since |F[x,y]|=|F|*aleph_0 as well, this gives
that the cardinality of the set of irreducibles in S is the same. If M
is the free commutative monoid on a set with |F|*aleph_0 elements,
then (R,*) and (S,*) are both isomorphic to F x M, since they are both
UFDs with isomorphic (in fact, identical) group of units.
That R and S are not isomorphic follows from any number of arguments
(R is a PID, S is not; the Krull dimension of S is one more than that
of R, etc).
Now, the isomorphism between additive structures can be given
constructively/explicitly, without having to invoke the Axiom of
Choice. Can the same be done with the isomorphism of the
multiplicative structures? I’m not sure if I’m invoking AC above; I
may be in arguing some of the cardinalities or the inequalities
between cardinalities.
If not in general, can it be done for some specific fields, say F=Q,
where we have explicit well-orderings of Q; or C, where the set of
monic irreducibles of F[x] is easy?
(I think that replacing F with any UFD will also give counterexamples
to the original question, but then we may need choice to select
representatives from the equivalence classes of irreducible elements,
so let’s stick to fields where there is an obvious choice of
representative).
–
Arturo Magidin