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The Complex Field

  1. Oct 15, 2009 #1

    jgens

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    Would the following prove that the set of complex numbers do not form and ordered field?

    Clearly [itex]i \neq 0[/itex]. Therefore, if the complex numbers form an ordered field either [itex]i > 0[/itex] or [itex]i < 0[/itex]. Suppose first that [itex]i > 0[/itex], then [itex]i^2 = -1 > 0[/itex], a contradiction. Now suppose that [itex]i < 0[/itex], then [itex]i^2 = -1 > 0[/itex], another contradiction. Thus, the set of complex numbers do not form an ordered field.

    This seems awfully fishy and I wouldn't be surprised to find that it's completely invalid. Feedback and suggestions are welcome. Thanks!
     
  2. jcsd
  3. Oct 15, 2009 #2
    You have i2 = -1 > 0, but -1 < 0.
     
  4. Oct 15, 2009 #3

    jgens

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    Yes, that's why [itex]i > 0[/itex] and [itex]i < 0[/itex] are contradictions as I state in my original post.
     
  5. Oct 15, 2009 #4

    Mark44

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    I understand the logic behind what you're saying, but you could say it in a way that is clearer.

    Suppose that i > 0.
    Multiplying by a positive number preserves the direction of the inequality, so i2 must be positive. This is incorrect, though, because i2 = - 1 < 0. Thus, assuming i > 0 leads to a contradiction.

    And similar for the assumption that i < 0.
     
  6. Oct 15, 2009 #5

    jgens

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    Thanks for the suggestion! It seems like I always need to add details to my proofs in order to make them clearer. Does the method I'm using work?
     
  7. Oct 16, 2009 #6

    HallsofIvy

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    What you have proven is that the usual order on the real numbers, in which -1< 0, cannot be extended to the complex numbers.

    But you can prove more. Can there exist some order? Perhaps one in which 1< 0and -1> 0?

    Suppose such an order existed. The i is not 0 (the additive identity does not depend upon the order). If i> 0 then it follow that (i)(i)= -1> 0. From that i(-1)= -i> i(0)= 0. Now that contradicts i>0. If i< 0, the -i> 0 so (-i)(i)= 1< 0. From that i(1)= i< i(0)= 0 and that contradicts i< 0.
     
  8. Oct 16, 2009 #7
    Ok, that proves that there can be no order where 1<0 and -1>0. How does one show that there can not be any order relation defined on the complex numbers? Or R^n for n>1 for that matter?
     
  9. Oct 16, 2009 #8

    Ben Niehoff

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    What about this one:

    Let [itex]z_1 = r_1 e^{i\theta_1}[/itex] and [itex]z_2 = r_2 e^{i\theta_2}[/itex], where [itex]\theta[/itex] is restricted to be in [itex][0, 2\pi)[/itex]. Then define > as follows:

    1. If [itex]r_1 > r_2[/itex], then [itex]z_1 > z_2[/itex]

    2. If [itex]r_1 = r_2[/itex] and [itex]\theta_1 > \theta_2[/itex], then [itex]z_1 > z_2[/itex]

    3. If [itex]r_1 = r_2[/itex] and [itex]\theta_1 = \theta_2[/itex], then [itex]z_1 = z_2[/itex]

    This looks like an order relation to me...it even has a least element, 0. Now, I know that C is not supposed to have an order relation, so what's wrong with this? Have I implicitly used the axiom of choice or something?
     
  10. Oct 16, 2009 #9
    You can certainly put some order on the complex numbers, Ben, but you can't find one that behaves nicely with their algebraic structure. For a field to be considered an ordered field, it must have an order that is compatible with multiplication and addition. That is, we want

    a < b ==> a + c < b + c
    a < b, c > 0 ==> ac < bc

    for all a, b, c. Your order doesn't satisfy either of these properties, so it doesn't make the complex numbers into an ordered field. Sorry.
     
  11. Oct 16, 2009 #10

    Ben Niehoff

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    Ah, yes, that does change things a bit.
     
  12. Oct 22, 2009 #11
    Order of a field is defined via a subset of that field. Let's denote it P. An ordered field F contains a subset P that satisfies the following properties:

    1) for any a (which is not 0) in F, Either a or -a are in P, but not both.
    2) For any a,b in P, a+b is in P.
    3) For any a,b in P, a*b is in P.

    Then, we say a>b iff (a-b) is in P.

    You can see immediately that the C doesn't contain such a set because
    if i belongs P, it implies that -i=i*i*i belongs to P (3) in contradiction to (1).
    The same if we assume -i belongs to P.
     
  13. Oct 22, 2009 #12

    HallsofIvy

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    As elibj123 said, one of the requirements for an "ordered field", trichotomy, is that one and only one of the following be true of any member of the field, x: (trichotomy)
    a) x= 0
    b) x> 0
    c) x< 0

    You cannot have "-1< 0" and "1> 0" in any ordered field (where "1" is defined as the multiplicative identity and "0" is defined as the additive identity).

    Your reference to [itex]R^n[/itex] doesn't make sense because it is not a field. What arithmetic operations would you wish to define on it to make it a field?
    If you ignore the operations, then the complex numbers can be ordered (any set can be). One such order ("lexicographical") is "If a< c then a+bi< c+ di. If a= c and b< d, then a+bi< c+ di." That's a perfectly good order on C or [itex]R^2[/itex] and can be extended in an obvious way to [itex]R^n[/itex]. As long as you don't look at the operations, there is no problem.
     
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