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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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All of the order axioms can just be taken over from the standard order on $\mathbb R$. – Henning Makholm Feb 13 '12 at 2:30
Since it is contained in $\mathbb{R}$, you can use the order inherited from $\mathbb{R}$ by restriction. – Arturo Magidin Feb 13 '12 at 2:32
I see that, because $X \subset \mathbb{R}$, but I thought maybe I needed to show it.... Short of calculating them using elements like $(a'+a''\sqrt{2})<(b'+b''\sqrt{2})$, or stating the fact that they are reals, I didn't know how. If that is the case, that is easy and thank you! – nate Feb 13 '12 at 2:33
If you don't want to explicitly invoke $\mathbb{R}$, then just "hide it". Consider what it takes, in terms of $a',a'',b',b''$, for $a'+a''\sqrt{2}\leq b'+b''\sqrt{2}$ to hold; for example, if $a'-b'\leq 0\leq b''-a''$, then the inequality holds; if $0\leq a'-b'\leq b''-a''$, then the inequality holds; if $a'-b'\leq b''-a''\leq 0$, the inequality holds if and only if $2(b''-a'')^2 \leq (a'-b')^2$. Etc. – Arturo Magidin Feb 13 '12 at 2:40
Okay, I will do that, long as it may be. Thank you! – nate Feb 13 '12 at 2:45
up vote 3 down vote accepted

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

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