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Author
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Topic: Supplementary Fermat Conjecture
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Rex Kerr
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Member # 632
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posted 25. June 2003 20:32
Limits aren't the appropriate way to compare cardinality. Rather, you show that aleph_0 + aleph_0 = aleph_0 by finding an embedding from Z^2 -> Z and vice versa (where Z is the integers (or natural numbers or positive integers if you prefer, since |Z| = |Z+| = |N|)).
To show that aleph_0 ^ aleph_0 = aleph_0, embed them as follows: given x^y, write each of x and y in some base B with digits x0,...,xN and y0,...,yN. Then let z = sum(xi*B^(2i)) + sum(yi*B^(2i+1)). For each x^y there therefore exists a unique z and thus |Z^Z| <= |Z|.
As an example, if B = 10, we would write 987^321 as 938271.
So, certainly x^aleph_0 <= aleph_0^aleph_0, but aleph_0^aleph_0 = aleph_0. x^aleph_0 >= aleph_0, so x^aleph_0 = aleph_0. So Fermat's Last Theorem reduces to aleph_0 + aleph_0 = aleph_0, which we already saw was true.
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Erik
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Member # 160
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posted 25. June 2003 21:32
Rex Kerr, you are right about limits not being the appropriate way to make calculation with cardinal numbers, but you are wrong about the specifics.
To prove that Aleph_0 + Aleph_0 = Aleph_0, you need to find two disjoint sets A,B such that
|A| = |B| = |A union B| = Aleph_0.
(The positive and negative integers will do.) If you find a bijection between Z^2 and Z you have instead proven that Z^2 and Z have the same cardinality, which means that
Aleph_0 * Aleph_0 = |Z| * |Z| = |Z^2| = |Z| = Aleph_0.
An appropriate bijection can be found by thinking of Z^2 as an infinite grid and drawing a spiral that passes through all grid points.
It is not true that Aleph_0^Aleph_0 = Aleph_0, because the power set always has higher cardinality than the set itself. To compute the value, you need to determine the cardinality of the set of all functions from Z to Z. (It is equal to the cardinality of the real numbers.)
Erik
[ 25. June 2003, 21:41: Message edited by: Erik ]
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Rex Kerr
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Member # 632
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posted 26. June 2003 22:57
Oi, my reply disappeared.
Anyway, in summary: ack, you're right. I'm forgetting my definitions. How embarassing! My rust is showing.
I still maintain that x^Aleph_0 + y^Aleph_0 = z^Aleph_0 if we interpret x^Aleph_0 as the cardinality of (Z_x)^omega_0.
However, I don't really feel like redoing the proof; you basically interpret (Z_x)^omega_0 as a real number in base x, and note that |[0,1]| = |R|.
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chimp
Member
Member # 333
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posted 28. June 2003 07:59
Dr. Cantor's "alephs" are different than limits in that they appear as "completed" infinities, while limits are numerical quantities that only approach infinity. These alephs appear to be types of cardinal "identities"?, possibly similar to qualia?
Interesting...
Russ
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chimp
Member
Member # 333
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posted 23. January 2004 14:17
p is a prime number
x,y, and z, are positive integers.
x^p + y^p = z^p
[x^p + y^p]^w = z^[p]^n
then
z^[wp] = z^[p]^n
w = p = n
z^[np] = z^[p]^n
np = p^n
n = p^[n-1]
n = n^[n-1]
n^[n-2] = 1
n^[0] = 1
n-2 = 0
w = p = n = 2
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chimp
Member
Member # 333
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posted 30. January 2004 02:50
Someone may ask: "What does Fermat's last theorem have to do with intelligent design?"
The answer, is, a better understanding of this numerical enigma, possibly, leads to an answer to the question: "What is space?"
Of course, if the equations are expressed as a polynomial of 2 variables, raised to odd prime p.
[x+y]^5 + [x-y]^5 =
2*x^5 + 20*x^3 * y^2 + 10*x*y^4
It appears that Fermat's last theorem is related to binomial expansions.
So the equation really becomes:
For positive integers x,y,p:
If
[[x+y]^p + [x-y]^p ]/p = x^p + y^p
then
p = 2
x^2 + (x+1)^2 = (x+2)^2 , x = 3
x^2 +(x+1)^2 = (x+9)^2 , x = 20
x^2 + (x+1)^2 = (x+50)^2 , x = 119
x^2 + (x+1)^2 = (x+289)^2 , x = 696
x^2 + (x+1)^2 = (x+1682)^2 , x = 4059
x^2 + (x+1)^2 = (x+9801)^2 , x = 23660
x^2 + (x+1)^2 = (x+57122)^2 , x = 137903
etc...
etc...
etc...
2 = 1^2 + 1^2 = 2*1^2
9 = 3^2
50 = 7^2 + 1^2 = 2*5^2
289 = 17^2
1682 = 41^2 + 1^2 = 2*29^2
9801 = 99^2
57122 = 239^2 + 1^2 = 2*169^2
x^3 + (x+1)^3 = (x+2)^3
=
[x^3] + [x^3 + 3x^2 + 3x +1] - [x^3 + 6x^2 + 12x + 8] = 0
=
x^3 +3x^2 -9x - 7
=
x is not an integer...
Of course, the various polynomial symmetry "groups?" could provide some insight on the Fermat problem...
x^3 + (x+1)^3 = 2x^3 + 3x^2 + 3x + 1
is not a perfect cube, far all positive integers, x, ... no integer solutions.
x^2 + (x+1)^2 = 2x^2 + 2x + 1
has perfect square solutions for specified values of integer, x ...
Let
x^3 + (x+1)^3 = 2x^3 + y
y = 3x^2 + 3x + 1
3x^2 forms perfect cubes
3x forms perfect cubes
3x^2 + 3x forms perfect cubes
3x^2 + 1 does not
3x + 1 does not
3x^2 + 3x + 1 does not
2x^3 + y + K = z^3
Interesting...
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chimp
Member
Member # 333
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posted 31. January 2004 02:46
There is an error, with my intuition ![[Smile]](smile.gif)
3x+1 does indeed form perfect cubes.
3[333] + 1 = 10^3
3[21]+1 = 4^3
etc...
Conjecture:
2*X^X + 1
is an odd number for all positive integers X, X > 1, but 2*X^X + 1
cannot be a prime number, for all positive integers X, X > 1
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chimp
Member
Member # 333
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posted 01. February 2004 01:12
Conjecture fails at X = 12
2*(12)^(12) + 1 = 17832200896513
x^n + y^n = z^n
x^3 + y^3 = (x+y)[x^2-xy+y^2]
A more general identity for primes p, and numbers, n:
A*B means A times B
A/B means A divided by B
+ and - , we all know...
follow the order of operations:
x^p + y^p
equals
(x+y)*[(x+y)^(p-1) + [{(x^p+y^p)/(x+y)} - (x+y)^(p-1)] ]
for all p and n > 1
Question:
Let T be a metric space with distance function r(x,y) expressing the definitive predication that involves T with the real numbers, R. Therefore the juxtaposition of left and right hemispheres resonates in perfect accordance with the proposition that T and R are embedded simultaneously in the full structure of manifold M. Ergo we pass on to an enlargement *M of M, whereby the non-standard metric space is diffeomorphism invariant.
So if f(x) is a homeomorphism from T onto S, then for every point p in T, does f(u(p) = u(f(p)) ?
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chimp
Member
Member # 333
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posted 01. February 2004 23:40
A metric space is a set of points such that for every pair of points, there is a nonnegative real number called their distance that is symmetric, and satisfies the triangle inequality, which states that the sum of the measures of any two sides of any triangle is greater than the measure of the third side. Space is then a tranformation[invariant]. Two objects with relative velocity will have a relative measure that transforms into the other. In effect, the separation does not exist in an extrinsic sense. ABC = BCA = CAB
Utilizing the generalized equation:
x^3 + y^3 = (x+y)*[(x+y)^2 - 3xy]
x^5 + y^5 = (x+y)*[(x+y)^4 -5x^3 y -5(xy)^2 -5x y^3 ]
x^7 + y^7 = (x+y)*[(x+y)^6 -7yx^5 -14x^4 y^2 -21(xy)^3 -14x^2 y^4 - 7xy^5 ]
In general, x^p + y^p = (x+y)*[(x+y)^(p-1) - p*f(x,y) ]
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chimp
Member
Member # 333
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posted 02. February 2004 01:56
Utilizing the generalized equation:
x^3 + y^3 = (x+y)*[(x+y)^2 - 3xy]
x^5 + y^5 = (x+y)*[(x+y)^4 -5x^3 y -5(xy)^2 -5x y^3 ]
x^7 + y^7 = (x+y)*[(x+y)^6 -7yx^5 -14x^4 y^2 -21(xy)^3 -14x^2 y^4 - 7xy^5 ]
In general, x^p + y^p = (x+y)*[(x+y)^(p-1) - p*f(x,y) ]
If
x = 3
y = 4
then
(3+4) = 7
(3+4)^2 = 49
3*3*4 = 3xy = 36
3^3 + 4^3 = 91 = (3+4)*[ 49 - 36 ] = 7*13 = 91
(x+y)*[(x+y)^2 - 3xy ] =
(x+y)^3 - 3xy(x+y) =
(x+y)^3 - 3x^2 y - 3x y^2 =
x^3 + 3x^2 y + 3x y^2 - 3x^2 y - 3x y^2 + y^3 =
x^3 + y^3
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Rex Kerr
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Member # 632
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posted 02. February 2004 04:03
Is this supposed to have something to do with primes? It doesn't have much to do with Fermat's Last Theorem.
x^n+y^n = (x+y)*SUM[i=0..n-1]( (-1)^i * x^(n-1-i) * y^i )
(x+y)^(n-1) = SUM[i=0..n-1]( C(n-1,i)*x^(n-1-i) * y^i )
So therefore
x^n+y^n = (x+y)( (x+y)^(n-1) * SUM[i=0..n-1]( ((-1)^i - C(n-1,i))*x^(n-1-i) * y^i );
Now, it is true that if n=p is prime, then
C(p-1,i) = (-1)^i (mod p) for i=0,1,...,p-1.
Proof:
C(p-1,i) = (p-1) * (p-2) * ... * (p-i) / 1 * 2 * ... * i
= -1 * -2 * ... * -i / 1 * 2 * ... * i (mod p)
= (-1/1) * (-2/2) * ... * (-i/i) (mod p)
= (-1)^i (mod p)
So every term in the sum is divisible by p.
(This only works for primes since Z(mod p) is only a ring for primes; division doesn't work properly in non-rings.)
[ 02. February 2004, 04:07: Message edited by: Rex Kerr ]
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chimp
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Member # 333
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posted 08. February 2004 23:15
If space is *quantized* yet also continuous, then it too, has the property called "wave-particle" duality. If space consists of indivisible units, then a measurement of space means that Fermat's
last theorem holds, for it.
According to the Pythagorean theorem:
x^2 + y^2 = z^2
All possible integer solutions are then rerpresented as:
[a^2 - b^2]^2 + [2ab]^2 = [a^2 + b^2]^2
a^4 -2(ab)^2 + b^4 + 4(ab)^2 =
a^4 + 2(ab)^2 + b^4 = [a^2 + b^2]^2
all odd numbers can be represented as:
[a^2 - b^2] or Z^p - Y^p
if Y is an "even" natural n and Z is odd, same for a and b .
Fermat's last theorem, for integers a,b,Z,Y,p:
[a^2 - b^2]^p + Y^p = Z^p
[a^2 - b^2]^p = Z^p - Y^p
a^2 - b^2 = [Z^p - Y^p]^[1/p]
When Z^p - Y^p is a prime number, it cannot have an integer root.
[a^2 - b^2] is not an integer, for
[Z^p - Y^p]^[1/p] , for a,b,Z,Y,p,
unless p = 2.
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chimp
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Member # 333
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posted 10. February 2004 03:44
A topological space is a set X along with a happy family of subsets of X, called the open sets, requred to satisfy certain conditions, like the empty set and X itself are both open, if the subsets of X, U and V are open, so is the intersection of U and V, of course! And if the sets U_a of X are open, then so is the union of U_a. The collection of sets taken to be open is called the topology of X. An open set containing a point x, which is an element of X, is called a neighborhood of x. The complement of an open set is called "closed".
So with the use of topology it becomes possible to define continuous functions, and roughly speaking, a function is continuous if it sends nearby points to nearby points, of course. The conceptual notion of "nearby" can be made precise using open sets. Ergo a function f: X-->Y from one topological space to another is defined to be continuous because if we are given any open set U subset of Y, then the inverse image f^-1 U subset of X, is open. So the concept of manifold can be likened to that of a globe, whereby it may be covered with patches that look just like R^n.
A collectionU_a of open sets "covers" a topological space X if their union is all of X.
Well allrighty then, so, given topological spaces X and Y, there is a product X x Y, i.e. the product topology in which a set is open iff it is a union of sets of the form U x V, where U is open in X and V is open in Y .
If M is an m-dimensional manifold and N is an n-dimensional manifold, then M x N is an (m+n) dimensional manifold.
So simultanaety "S" is a spacelike hypersurface or "slice" through spacetime that cuts through event P, with a set of observers having worldlines crossing the simultanaety "simultaneously-orthogonally" having clocks that all read the same "proper" time at the instant of crossing.
The metric spaces are thus defined as being diffeomorphism invariant. Intersecting cotangent bundles[manifolds] are the set of all possible configurations of a system, i.e. they describe the phase space of the system. When the "wave-functions" intersect, and are "in phase", they are at "resonance", giving what is called the "wave-function collapse" of the Schrodinger equation.
Yes, is it possible to derive Einstein's field equation strictly in terms of quantum mechanical operators? using n-dimensional cross sections of cotangent vector spaces? Near a massive object M, the *isobar* cross sections increase in density as wavefunction density gradients, a possible solution? to Hartle and Hawking's "wavefunction of the universe"?
There is the Schrodinger equation: H(psi) = E(psi), where H is the Hamiltonian operator, the sum of potential and kinetic energies, and "psi" is the wavefunction. E is the energy of the system. The square of the wavefunction, is the probability of the position and momentum for the system.
The Wheeler DeWitt equation is the Schrodinger equation applied to the whole universe. Since the total energy of the universe is postulated to be zero[but is still not defined], the Wheeler DeWitt equation is: H(psi) = 0
There is a complementary path integral approach for this equation. The brilliant physicist Stephen Hawking, derived the wavefunction of the universe as a path integral, for a complex function of the classical configuration space:
psi(q) = integral exp(-S(g)/hbar) dg
The problem is that "dg" is not well defined either.
"exp" is the base of the natural logarithm "e" raised to a power.
The power in this case, is the quantity -S(g)/hbar, where S(g) is the Einstein Hilbert action. The Einstein Hilbert Action, is defined via the Lagrangian, which is the difference of kinetic and potential energies, and it has a formulation in general relativity:
Lagrangian = R vol
R is the Ricci scalar curvature of the metric g, derived by contracting the Ricci tensor and "vol" is the volume form associated to g.
The Einstein Hilbert action then becomes: S(g) = integral R vol
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