Completeness

General Mathematics
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Papapetros Vaggelis
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Completeness

#1

Post by Papapetros Vaggelis »

Does the ordered field of the rational functions satisfy the completeness theorem : " All non-empty

sets have a supremum" .
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Tolaso J Kos
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Re: Completeness

#2

Post by Tolaso J Kos »

Papapetros Vaggelis wrote:Does the ordered field of the rational functions satisfy the completeness theorem : " All non-empty

sets have a supremum" .
I give the answer but not the solution so that I don't spoil someone's fun if he/she wants to try.
Answer
The answer to the question would be no. The ordered field of the rational functions does not satisfy the completeness theorem.
Imagination is much more important than knowledge.
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Riemann
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Re: Completeness

#3

Post by Riemann »

Papapetros Vaggelis wrote:Does the ordered field of the rational functions satisfy the completeness theorem : " All non-empty

sets have a supremum" .

No, it does not, because the ordered field of the rational functions does not satisfy the Archimedean property. This is quite known.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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