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To prove that a field is complex |
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| May31-12, 05:54 PM | #1 |
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To prove that a field is complex
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. |
| May31-12, 06:05 PM | #2 |
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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. |
| May31-12, 06:11 PM | #3 |
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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 |
| May31-12, 06:11 PM | #4 |
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To prove that a field is complex |
| May31-12, 07:58 PM | #5 |
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http://en.wikipedia.org/wiki/Field_(mathematics) 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/ |
| May31-12, 07:59 PM | #6 |
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