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To prove that a field is complex

  1. May 31, 2012 #1
    I understand that the complex numbers form a "field" since the complex numbers are closed under addition, subtraction, multiplication, and division. And I understand the complex numbers are not an ordered field since it's not possible to define a relation z1<z2.

    My question is: Are all not ordered fields necessarily complex? Then how would you prove that a field is not ordered, is that something that is observed in a system or imposed? Thanks.
     
    Last edited: May 31, 2012
  2. jcsd
  3. May 31, 2012 #2

    micromass

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    Take F={0,1} with

    [tex]0+0=1+1=0,~1+0=0+1=1[/tex]

    and

    [tex]0*0=1*0=0*1=0,~1*1=1[/tex]

    then F is a field that can not be ordered.
     
  4. May 31, 2012 #3

    No. Any finite field is not orderable (in fact, any field of positive characteristic is not ordered), or any non-real extension of [itex]\,\mathbb{Q}\,[/itex] is not orderable...



    A field can be ordered iff -1 can't be expressed as a sum of squares, or equivalently iff a sum of squares equals zero iff every summand is zero.

    DonAntonio
     
  5. May 31, 2012 #4
    This seems like a very strange way to define + and *. Are you saying that in a field that we can define + and * and way we wish? Or is there some requirements for + and * so that they are consistent with each other?
     
  6. May 31, 2012 #5
    A field is a specific algebraic structure with its own axioms so, no, we can't do anything we wish. What Micromass described is a special (very small) field.

    http://en.wikipedia.org/wiki/Field_(mathematics [Broken])

    I would suggest doing a little reading on Groups and Rings as well, to give Fields some context. Wikipedia might not be the best place for a beginner to start. Try a free textbook like this:

    http://abstract.ups.edu/
     
    Last edited by a moderator: May 6, 2017
  7. May 31, 2012 #6
    It's not that strange at all. It's just like a clock with only two hours: 0 and 1. Take a look at this article: http://en.wikipedia.org/wiki/Modular_arithmetic

    No, there are axioms that + and * must satisfy in order for (F, +, *) to be considered a field. Briefly, (F, +) must be an abelian group, ##(F^\times, *)## must be an abelian group and the distributive law must hold. You can read the axioms in more detail here: http://en.wikipedia.org/wiki/Field_(mathematics).
     
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