Re: fourrier-budan

tommy1729 wrote:
tommy1729 wrote:
who knows alot about fourrier-budan methods.

i wish to solve a multivariable equation:

(continue) f(x;..)=0

over an uncountable ( transcendentally extended )
ring larger than
the reals but smaller than the entire complex
field. Good. Now construct a field that

a. Properly contains the field of real numbers. b. Is a proper
subset of the field of complex x numbers.

Betcha can't do it. Hint: a field is a vector space over any subfield. The dimension of R as a vector field over R is 1, while
the dimension of C as a vector field over R is 2. Oh, and the
dimension of a vector space is either infinite or an actual
integer.

I suppose you might postulate an integer that is greater than 1 yet
less than 2. On the other hand, you might note that any transcendental extension of R needs to be an infinite extension, pretty much ruling out your condition about the field being smaller
than the "entire complex field".

i need to find at least 1 solution or
prove/disprove its existance.
i prefer a method similar to fourrier-budan.
although other
approaches are appreciated too.

(the question is for a general method , so dont ask
me for the
function f. , since i want a general 'tool' to
solve these kind of
equations)

so algebra experts , this one is for you !!! :-)

tommy1729
Yeah, they'll get right on it.

Dale

wanna bet i can give a ring smaller than the complex and bigger than
the reals ???

I wouldn't take that side of the bet. I'd take the side that
claims you don't know what you're talking about.

Here. I'll spell it out for you:

Take any non-real complex number, and adjoin it to R. If the
resulting set is closed under addition and under multiplication,
then you get the full set of complex numbers:

Let z = a + bi be a non-real complex number (with
a and b real numbers). Let K = R[z], the smallest
ring containing R and z.

Then b is nonzero, and z - a = bi is in K.
Since b is nonzero, (1/b) is in R, and since
K is closed under multiplication by elements of
R, (1/b)*bi = i is also in K.

Thus, K contains i. Since K is closed under
addition and under multiplication by elements
of R, you can produce any complex number

u + vi

as an element of K.



with a dimension between 1 and 2 ???

it easier than you might think :-)

Since it's so easy, let's see you do it.


ever heard of fractional dimensions ? (guess not)

i think you might be considering this too "cantorian"

or to " cartesian " since these 2 dont know about fractional
dimension.

one you see your mistake , you can use these 2 to blame...

tommy1729

Umm, I've known about non-integral dimension for
several decades.

So, I don't see that you've constructed a field of
fractional dimension as a vector space over the reals.
I don't even see that you've constructed a ring that
contains the reals and is of fractional dimension. You
surely must note that a ring of this sort will still be
a vector space over R (because it must be closed under
addition, as well as under multiplication by elements of
R), and the dimension of a vector space is either an
integer, or it's infinite. Being a subset of the field
of complex numbers forces the dimension to be no greater
than 2, so it must be finite.

Of course, I've already shown you that any ring that is
contained in the set of complex numbers C, and that properly
contains R, the set of real numbers, must be all of C.

Dale
.

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