A field that is an ordered field in two distinct ways

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SUMMARY

The field F, defined as the set of all numbers of the form r + s√2 where r, s are rational numbers, can be ordered in two distinct ways. The first ordering uses the subset P, which consists of positive members of the real numbers ℝ. The second proposed ordering involves the subset B, defined by the condition r + s√2 ε B if and only if r - s√2 ε P. The discussion reveals confusion regarding whether B can be empty, but it concludes that the existence of a second ordering is valid regardless of the cardinality of B.

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Homework Statement



Find a field that is an ordered field in two distinct ways.

Homework Equations



The set F of all numbers of form r + s√2 , where r,s ε Q and the operations of addition and multiplication are those of the real number system ℝ of which F is a subset, is an ordered field in that there is a subset P of F such that addition and multiplication are closed in P, and x ε F →
exactly one of the following:

x ε P; x=0; -x ε P

This means that P is the set of all members of F that are positive members of ℝ, so P is the positive reals in F.


The Attempt at a Solution



A second way in which F can be ordered is by way of the subset B, such that

r + s√2 ε B iff r - s√2 ε P .

This I find troublesome, as (I am under the impression) that this implies that B = ∅.
My reasoning for this is that numbers in B will end in the digits that r - s√2 will end in (which would be, e.g., of form 1 - √2), and these are different than the numbers that r + s√2 span (for instance, 1 + √2).

Is B=∅? Are there faults in my reasoning?
 
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The deluge of responses has me chagrined!

It is irrelevant that B would be the null set, as the above demonstration is just supposed to exemplify that it CAN be ordered in another way, irrespective of its number of members. The rationals and reals can but be ordered in one way, so in particular this field is a special case.
 

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