Re: cantorian algebra
- From: Randy Poe <poespam-trap@xxxxxxxxx>
- Date: Wed, 22 Aug 2007 13:06:05 -0700
On Aug 22, 9:47 am, tommy1729 <tommy1...@xxxxxxxxx> wrote:
lwalke3 wrote:
On Aug 18, 2:15 pm, tommy1729 <tommy1...@xxxxxxxxx>
wrote:
this is how cantor or cantorians solve algebra
I once jokingly came up with similar examples:
1) The Fundamental Theorem of Algebra fails.
For a polynomial equation of degree n has at most n
solutions, but many of the OP's equations have
infinitely many solutions (as given above).
2) The Fundamental Theorem of Arithmetic fails.
For c = 2^aleph_0 = 3^aleph_0 = p^aleph_0 for any
prime number p. Thus c has infinitely many "prime
factorizations" rather than a unique one.
3) Fermat's Last Theorem fails.
For a^n + b^n = c^n has infinitely many solutions,
just by letting n = aleph_0 and a,b,c be any
numbers from 2 to aleph_0. Then the equation
reduces to c + c = c.
Of course, the truth is that the above Diophantine
equations were designed to admit only _finite_
numbers as solutions, not infinite numbers. In
ZFC equations which may lack finite solutions may
have infinite "solutions."
Notice that if we were to switch to the hyperreals,
or any theory involving the Transfer Principle,
than an equation has an infinite solution whenever
it has infinitely many finite solutions, and if no
finite solution exists, than neither can an
infinite solution exist. Thus the Diophantine
equations 2^a = 3^b and a^n + b^n = c^n can have
no solutions other than the trivial ones (so
a,b,c must be >= 1, n must be >= 3), and so have
no solution among the hypernaturals.
And so the OP seems to be arguing against Cantor
because standard cardinalities do not adhere to
the Transfer Principle, and so may appear to be
a "solution" to equations with no finite solutions.
hahaha
very funny
have you ever noticed this:
all brances of math are consistant with eachother
apart cantor set theory !!!
What do you mean by that?
For instance, multiplication of numbers is commutative:
a*b = b*a
But multiplication of matrices is not:
A*B != B*A
What does "consistent with each other" mean when
dealing with different types of objects?
- Randy
.
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