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Old Jun29-09, 10:56 AM                  #1
Starwatcher16

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Fundamental Theorum of Algebra an i.

Does the Fundamental Theorum of Algebra prove that imaginery numbers have to exist for our number system to be complete?
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Old Jun29-09, 11:35 AM                  #2
g_edgar

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Re: Fundamental Theorum of Algebra an i.

Something other than real numbers have to exist for the equation LaTeX Code: x^2+1=0 to be solvable. But I wouldn't say it the way you did.
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Old Jun29-09, 11:53 AM                  #3
statdad

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Re: Fundamental Theorum of Algebra an i.

Originally Posted by Starwatcher16 View Post
Does the Fundamental Theorum of Algebra prove that imaginery numbers have to exist for our number system to be complete?
No. The Fundamental Theorem of Algebra implicitly assumes the existence of complex numbers: it states that every polynomial of degree LaTeX Code:  n  with complex coefficients has at least one zero. (You sometimes see this written to say that if you count the zeros' multiplicities then the number of zeros equals the degree of the polynomial).
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Old Jun29-09, 03:03 PM                  #4
Geomancer

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Re: Fundamental Theorum of Algebra an i.

As usual the answer depends on what question you're asking. You can create an algebraically complete field containing the integers that is a proper subset of the complex numbers (indeed, a countable set, whereas the complex numbers are uncountable). Note though that your definition of a good number system probably includes the rational numbers. The minute you try to topologize these, you get the reals as a completion (assuming you want the topology to behave in the usual way). Introducing a root to x^2+1 give the complex numbers. Then the FTA tells you that you don't need anything else to be algebraically complete. Indeed, extending the complex numbers is rather hard if you want everything to still behave sanely.
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