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I am trying to find a field (I'll settle for other stuff - some type of ring) that is ordered and $a>0,\,b>0$ implies $a+b > 0$ yet $a,b > 0$ doesn't imply $ab > 0$.

Much of the problem seems to be working to satisfy the distributive law.

Cheers.

edit: and $a>0$ or $a=0$ or $a<0$ holds for all $a$ and only one relation does in-fact hold.

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The statement mentioned in the edit is usually called the "Law of Trichotomy". // @all: While retagging this post, I just found that we do not have an "ordered-field" tag (or other similar tags), so I have resorted to a compromise. I welcome any suggestions on the tagging. –  Srivatsan Jan 29 '12 at 6:33
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A terminological point, more for the benefit of future readers than the asker: many textbooks define an "ordered field" to be a field endowed with a linear order satisfying additional properties that ensure the product of positive elements is positive. To repeat the essential point, when this definition is used, an "ordered field" is more simply a field endowed with a linear order (and the answer to the asker's question is then "no"). This type of definition is understood when people say that the field of complex numbers cannot be endowed with an order making it an "ordered field." –  leslie townes Jan 29 '12 at 7:20
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2 Answers 2

up vote 4 down vote accepted

Take $\mathbb{C}$ with lexicographic order; that is, $a + bi \le c + di$ if $a \le c$ or $a = c$ and $b \le d$.

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Another example: Reverse the usual order on $\mathbb R$ (or any other ordered field).

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