# If $\mathbb{Q}$ is dense in an (Archimedean) ordered field K, is K a complete ordered field?

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.)

• Have you thought about the case $K=\Bbb Q$? – David C. Ullrich Oct 3 '15 at 0:14
• @DavidC.Ullrich: You are right. I have to edit or delete this post... – TCHuang Oct 3 '15 at 0:16
• 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. – André Nicolas Oct 3 '15 at 0:20
• You really might have given the matter a little thought before "fixing" the question by just adding the word "proper"... – David C. Ullrich Oct 3 '15 at 0:24
• @DavidC.Ullrich: I apologize for such possibly low quality question... – TCHuang Oct 3 '15 at 0:29

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.
• Correct. ${}\qquad{}$ – Michael Hardy Oct 3 '15 at 0:26