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I am new to number theory. My question is asking me to prove that $X:=\mathbb{Q}(\sqrt{2})$ is an ordered field that does not follow the Completion Axiom.

I started by showing $X$ was a subring of $\mathbb{R}$, and then showed it was a field (commutative division ring) by calculating the inverse and showing it as a member of $X$. So now all I have left is the order and what seems to be a pretty simple task (hope) of showing it doesn't follow the Completion Axiom.

The order `laws' are like,

O1: given a and b, either $a \leq b$ or $b \leq a$

O2: If $a \leq b$ and $b \leq a$, then $a=b$

O3: If $a \leq b$ and $b \leq c$, then $a \leq c$

O4: If $a \leq b$, then $a+c \leq b+c$

O5: If $a \leq b$ and $0 \leq c$, then $ac \leq bc$.

To compare these I thought about the norm, as in $a^2 + 2\,b^2$. Is this a good way to prove the order of $X$? Any other ideas or suggestions are much appreciated.

I fully understand the Completeness Axiom and why $\mathbb{Q}$ does not follow it, so I am hoping it won't be an issue. :)

Thanks much!

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    For failure of Completeness, one could, in imitation of proof that $\mathbb{Q}$ is not complete, show that $\sqrt{3}$, or maybe $\sqrt[4]{2}$, is not in $\mathbb{Q}(\sqrt{2})$.2012-02-13

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$\mathbb Q(\sqrt 2)$ is a subfield of $\mathbb R$ so you can use the restriction of the usual order of the reals to your field.