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math771

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Now, it seems fairly clear that the additive inverse of A will be -A={-a'|a' belongs to Q\A}. Furthermore, it's relatively easy to show that A + (-A) is a subset of 0*. For a belonging to cut A and a' belonging to cut Q\A, a < a'. Thus, a - a' < 0. Because A + (-A)={a - a'|a belongs to A and a' belongs to Q\A}, A + (-A) is a subset of 0*.

Now, I need show that 0* is a subset of A + (-A). To do that, I was thinking that it would suffice to show that for any x<0, there exist a belonging to A and -a' belonging to -A such that x [itex]\leq[/itex] a - a' < 0. Because A + (-A) is itself a cut (something we proved earlier) x [itex]\leq[/itex] a - a' < 0 would imply that all members of 0* belong to A + (-A).

I'll give my start at the proof. x [itex]\leq[/itex] a - a' < 0 is equivalent to -x [itex]\geq[/itex] a' - a > 0. Suppose, then, that 0 < -x < a' - a for all a and -a' beloning to A and -A respectively. At this point, perhaps we could use something like the Archimedian property but for rational numbers so that a' - a < k(-x) and (a' - a)/k < -x for some natural number k. Though, it seems we would have to show that -a'/k and a/k belong to -A and A respectively (correct me if I'm wrong), which would be odd if k were a natural number (and difficult (for me) to prove even if k were not natural). One question I would like to pose, then, is whether the Archimedean property requires that k be a natural number. In most proofs I've seen, it's stated that k is natural, but I fail to see the necessity of this requirement.

Even more troubling however is the fact that I've been led to believe that the Archimedean property is a consequence of what I'm trying to prove here. But I don't see how this could be so. The proof of the Archimedean property that I am familiar with relies on the connectedness of real numbers, but, if I'm not mistaken, it is possible to prove that the real numbers (taken to be Dedekind cuts) are connected without ever talking of the additive identities of cuts.

Any advice would be much appreciated. Thanks!