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!