Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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
1  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.