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- Thread starter mitcho
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Any help would be appreciated

Thanks

- #2

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Any help would be appreciated

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Think in terms of a set of greater cardinality having all sets of lesser cardinality as proper subsets.

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disregardthat

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Then you continue to construct the integers Z = N x N / ~ where ~ is the equivalence relation such that (a,b) ~ (c,d) if a+c = d+b, where n = [(n,0)], and -n = [(0,n)] for positive n. We induce an ordering on Z by [(a,b)] < [(c,d)] if a+d > c+b. Note that this gives the natural order on Z we are used to.

Then we construct Q = Z x (N-{0}) / ~, where ~ is defined by (a,b) ~ (c,d) if ad = bc, and [(a,b)] < [(c,d)] if ad < bc, where < is the order of Z.

Finally, we construct R by looking at the cauchy-sequences of Q^N (i.e. sequences or rational numbers that are cauchy). A sequence (q_n) of rational numbers is cauchy if for every rational number e, there is a natural number N such that |q_n-q_m| < e for all n,m >= N. Let this set of cauchy-sequences be C.

We define R = C /~ where (q_n) = (p_n) if the sequence (q_n -p_n) converge to 0. The order of R induced is defined as [(q_n)] < [(p_n)] if there is a rational number e such that there exists a natural number N such that p_n-q_n >= e for all n >= N. It will require a proof of that this in fact is a well-defined ordering, but when you do that it will be the natural ordering of R we are used to.

From this definition of R we can prove all the known axioms of R we need, most importantly the least-upper-bound property.

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