I'm wondering if it's possible to order a field extension $\mathbb{Q}[\sqrt{x}]$ for $0<x\in\mathbb{Q}$ such that it is an ordered field with an ordered subfield isomorphic to $\mathbb{Q}$.
It seemed that the ordering $$\leq_\sqrt{x}=\{(a+b\sqrt{x},c+d\sqrt{x}):a^2+c^2-2ac\leq x(b^2+d^2-2bd)\}$$ was promising, due to the algebraic manipulations $$a+b\sqrt{x}\leq c+d\sqrt{x}\iff a-c\leq\sqrt{x}(d-b)$$ $$\iff (a-c)^2\leq x(b-d)^2\iff a^2+c^2-2ac\leq x(b^2+d^2-2bd).$$ Despite this, if we use the canonical product structure $$(a+b\sqrt{x})(b+d\sqrt{x})=ab+bdx+(ad+bc)\sqrt{x}$$ it is not evident to me via algebra-crunching that orders are well-preserved under multiplication. Specifically, it isn't clear that for $a,b,c,d\in\mathbb{Q}[\sqrt{x}]$ we have that $a<b$ and $c<d$ imply $ac<bd$. Is this the case, and if not is there an ordereing we can explicitly define that makes a field extension like this ordered with $\mathbb{Q}$ as a subfield?
Edit
More specifically, suppose we have (a-priori) the ordered field $\mathbb{Q}=\langle\mathbb{Q},+,-,\times,\div,\leq\rangle$, and we wish to totally order the field extension $\mathbb{Q}[\sqrt{2}]=\langle\mathbb{Q},+_{\sqrt{2}},-_{\sqrt{2}},\times_{\sqrt{2}},\div_{\sqrt{2}},\sqrt{2}\rangle$ with a total ordering $\leq_\sqrt{2}$ expressed in the language of ordered fields such that the subfield $\{a+b\sqrt{2}:b=0\}\subsetneq\mathbb{Q}[\sqrt{2}]$ is isomorphic to $\mathbb{Q}$ as an ordered field under the projection $\pi_0$ of $\mathbb{Q}[\sqrt{2}]$ onto its first factor, $$\pi_0(a+b\sqrt{2})=a.$$ What is an explicit formula for $\leq_\sqrt{2}$? Does the above formula for $\leq_\sqrt{x}$ work in the case that $x=2$?
Note: I'll do the specific case of $\sqrt{2}$.
I think it's best to do it through the set of positive numbers. Remember that in an ordered field $K$, for any $a\in K\setminus\{0\}$ either $a>0$ or $-a>0$. Moreover if $a>0$ and $b>0$ then $a+b>0$ and $ab>0$.
So we have the field $\mathbb Q[\sqrt{2}]$ with elements of the form $a+b\sqrt{2}$ with $a,b\in\mathbb Q$. Now for elements of $\mathbb Q$ we know what numbers are positive (and indeed, it's not hard to show that we don't have a choice there). But what is the sign of $\sqrt{2}$? We have two choices: Either $\sqrt{2}>0$, then $-\sqrt{2}<0$, or $\sqrt{2}<0$, then $-\sqrt{2}>0$. But both choices are actually equivalent, as there's a field isomorphism exchanging $\sqrt{2}$ and $-\sqrt{2}$, and thus the choice just amounts to which of the two solutions of $x^2-2=0$ we call $\sqrt{2}$. Going with the usual convention, we choose $\sqrt{2}$ to be positive.
Now consider a general element $a + b\sqrt{2}$. We can concentrate on the case $b>0$ since for $b<0$ we simply get the opposite sign of $-a-b\sqrt{2}$ which again has a positive value of $b$. And for $b=0$ we just have a rational number, whose sign we already know.
Now we use the fact that with $b>0$ we also have $b^{-1}>0$, and multiply the above number with $b^{-1}$ (remember, multiplying with a positive number does not change the sign). Thus we get the condition $ab^{-1}+\sqrt{2}>0$.
If we write $c=ab^{-1}$, we therefore have to check for which values of $c$ we get $c+\sqrt{2}>0$.
Now if $c\ge 0$, then this is obviously positive. So we need to consider the case $c<0$. To more easily see the sign, let's define $d=-c$, so that $d>0$.
Now we can rewrite the condition as $$0 < {\sqrt{2}-d} = \frac{(\sqrt{2}-d)(\sqrt{2}+d)}{\sqrt{2}+d} = \frac{2-d^2}{\sqrt{2}+d}\quad.$$ The denominator is, again, clearly positive, so we get a positive number iff $d^2<2$.
Substituting everything back and combining the cases, we get for the case $b>0$ the condition: $$a\ge 0 \lor a^2 < 2b^2$$ For $b<0$ we need the opposite condition for $-a$ and $-b$: $$\lnot(-a\ge 0 \lor (-a)^2 < 2(-b)^2))$$ which can be rewritten as $$a>0 \land a^2\ge 2b^2$$ The three cases ($b=0$, $b>0$ and $b<0$) can now of course be combined into a big expression, but I don't see an obvious way to simplify that.
Of course, as usual, you have $u>v$ iff $u-v>0$, which you can insert in the above conditions to make it even more complicated.
Your condition $$a+b\sqrt{2} \le c+d\sqrt{2}\iff a^2+c^2-2ac\leq 2(b^2+d^2-2bd)$$ does not work, as it is symmetric under exchange of the two numbers. A concrete example that fails is $a=b=1, c=d=0$, which gives $$1^2+9^2-2\cdot 1\cdot 0 = 1 \le 2 = 2(1^2+0^2-2\cdot 1 \cdot 0$$ so you'd get $1+\sqrt{2}\le 0$.