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Let K be an ordered field. Then K contains the smallest ordered filed $\mathbb{Q}$.

  1. If $\mathbb{Q}$ is dense and proper in K, is K a complete ordered field?
  2. If $\mathbb{Q}$ is dense and proper in K, and K has the Archimedean property, is K a complete ordered field?

(i.e. satisfying any one of the equivalent definitions of a complete ordered field.) (If you know the answer you may give hints first.)

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    $\begingroup$ Have you thought about the case $K=\Bbb Q$? $\endgroup$ – David C. Ullrich Oct 3 '15 at 0:14
  • $\begingroup$ @DavidC.Ullrich: You are right. I have to edit or delete this post... $\endgroup$ – TCHuang Oct 3 '15 at 0:16
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    $\begingroup$ There are many such ordered fields, such as the reals of the form $a+b\sqrt{3}$, where $a$ and $b$ are rational, or the real algebraics. $\endgroup$ – André Nicolas Oct 3 '15 at 0:20
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    $\begingroup$ You really might have given the matter a little thought before "fixing" the question by just adding the word "proper"... $\endgroup$ – David C. Ullrich Oct 3 '15 at 0:24
  • $\begingroup$ @DavidC.Ullrich: I apologize for such possibly low quality question... $\endgroup$ – TCHuang Oct 3 '15 at 0:29
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No. For example, consider the smallest field that includes every rational number and also $\sqrt 2$. That is an ordered field that is not complete. You can show it is not complete by using the fact that it is countable, as in this page.

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  • $\begingroup$ Thank you! :) So if an ordered filed is complete then it contains a dense sub ordered filed. But the converse is not true. $\endgroup$ – TCHuang Oct 3 '15 at 0:25
  • $\begingroup$ Correct. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 3 '15 at 0:26

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