TITLE: Ordering on $R[\sqrt{n}]$ for an ordered ring $R$
QUESTION [11 upvotes]: I'm interested in showing that if $R$ is an ordered ring (with ordering $\leq$), and $n \geq 0$, then $R[\sqrt{n}]$ is also an ordered ring.
In the reals, $0 \leq a_1 + a_n\sqrt n$ iff either (1) $0 \leq a_1$ and $na_n^2 \leq a_1^2$, or (2) $0 \leq a_n$ and $a_1^2 \leq na_n^2$.  So it seems natural to define the set $P$ of nonnegative elements of $R[\sqrt{n}]$ by the rule

$a_1 + a_n\sqrt{n} \in P$ iff (1) or (2) above hold

However, I'm having trouble proving that $x \in P$ and $y \in P$ implies $x + y \in P$.  It seems like it should be elementary, but I'm finding myself running around in circles.
I've been trying to split the proof into cases, depending on whether $x$ and $y$ satisfy rules (1) or (2).  The cases where they both satisfy (1) or both satisfy (2) are easy, but the casework is getting overwhelming when $x$ satisfies (1) and $y$ satisfies (2).
I wonder if anyone can suggest something I'm missing, or maybe point me to a more general theorem that would help (although I'd like to avoid getting too esoteric since my eventual goal is to formulate the proof in coq).
Thanks!
Update: I’ve made progress but am still stuck.  Unless I’m mistaken, the only case I need is to show that $0 \leq (x_1 + x_n\sqrt{n}) + (y_1 + y_n\sqrt{n})$ when the following conditions all hold:

$0 \leq x_1$
$0 \leq -x_n$
$nx_n^2 \leq x_1^2$
$0 \leq -y_1$
$0 \leq y_n$
$y_1^2 \leq ny_n^2$
$0 \leq x_1 + y_1$
$0 \leq -x_n - y_n$

Combining these assumptions gives $0 \leq y_1^2 \leq ny_n^2 \leq n x_n^2 \leq x_1^2$.
In this case I need to show that $n(x_n + y_n)^2 \leq (x_1 + y_1)^2$.
I’ve tried playing with fractions, derivatives, absolute values, and just can’t solve this case.  Any help would be appreciated.

REPLY [4 votes]: There are scores of details to check, but the following seems to handle the most difficult cases.
Below $a,b,c,d$ will always denote arbitrary positive elements of $R$, so for example $-a<0$.

Assuming that $R$ is an integral domain we can extend its ordering to its field of fractions $F$. Next I define that a fraction $a/b$, both $a,b$ positive elements of $R$, is $>\sqrt n$ if and only if $a^2>n b^2$ in $R$. Obviously the negative fractions are all $<\sqrt n$. It is trivial to show that equal fractions behave compatibly.
At this point we have, in a sense, defined where $\sqrt{n}$ lies on the number line of $F$. We still need to check that this is compatible with the ordering of $F$. To that end we prove the result.
Lemma 1. If $c/d>\sqrt n$ and $a/b>c/d$ then $a/b>\sqrt n$.
Proof. We have $c^2/d^2>n$ and $a^2/b^2>c^2/d^2$ so $a^2/b^2>n$ and the claim follows.
Next we define the set of positive elements of $R[\sqrt n]$ by declaring that

$a+b\sqrt n$ is always positive, $-a-b\sqrt n$ is always negative,
and then in the mixed cases
$$
\begin{aligned}
a-b\sqrt n>0&\Longleftrightarrow a/b>\sqrt n,\\
-a+b\sqrt n>0&\Longleftrightarrow a/b<\sqrt n.
\end{aligned}
$$
Handle the cases $a=0$ and $b=0$ in the obvious way. I think this is equivalent to the OP's definition of positive elements of $R[\sqrt n]$, but I choose to work in $F$, so rephrase it this way.

The goal is to show that the set of positive numbers thus defined is closed under addition. I will only handle the mixed case numbers, for I think those are the most delicate ones.

Remembering that the mediant $(a+c)/(b+d)$ lies between $a/b$ and $c/d$ will play a key role. We also need to look at what happens with $(a-c)/(b-d)$ when $b>d$.
The model I kept in mind is to compare the following rationals with $\sqrt3$:
$$
\frac21>\frac74>\sqrt 3>\frac{17}{10}>\frac53.
$$

Here the mediant $(7+2)/(1+4)=9/5$ will automatically be $>\sqrt3$ as the mediant is $>7/4$. Similarly the mediant $(17+5)/(10+3)$ will automatically be $<\sqrt 3$ as it is below $17/10$.
With $a/b=7/4$ and $c/d=5/3$, so $b>d$, we see that $(a-c)/(b-d)=2/1$ is actually above $a/b$. This is also because $7/4$ is the mediant of $2/1$ and $5/3$.
Similarly with $a/b=17/10$ and $c/d=7/4$ (again $b>d$) the fraction
$(a-c)/(b-d)=10/6=5/3$ is below $a/b$. This is because $17/10$ is below $7/4$, and also the mediant of $7/4$ and $10/6$.

These inequalities generalize easily in the indicated way and give us the claim. Behold.

Lemma 2. If $x$ and $y$ are positive elements of $R[\sqrt n]$, so is $x+y$.
My proof only covers the mixed cases for I think the rest are easier. I split it into cases according to what types of mixed cases we have.
Case I. $x=a-b\sqrt n, y=c-d\sqrt n$.
In this case we have $a/b>\sqrt n$ and $c/d>\sqrt n$. The mediant
$$
\frac{a+c}{b+d}=\alpha \frac ab+\beta \frac cd
$$
with $\alpha=b/(b+d)$, $\beta=d/(b+d)$ both positive. Furthermore $\alpha+\beta=1$. These imply that $(a+b)/(c+d)$ is between $a/b$ and $c/d$. By Lemma 1 we can conclude that
$(a+b)/(c+d)>\sqrt n$. Consequently $x+y=(a+b)-(c+d)\sqrt n$ is positive.
Case II. $x=a-b\sqrt n$, $y=-c+d\sqrt n$.
This time we know that $a/b>\sqrt n$ and $c/d<\sqrt n$.
Here $x+y=(a-c)-(b-d)\sqrt n$, so we need to compare $(a-c)/(b-d)$ with $\sqrt n$. We need to further split this into two cases according to which denominator, $b$ or $d$, is larger (the case $b=d$ is trivial).
Assume first that $b>d$. Our goal is to show that $(a-c)/(b-d)>\sqrt n$. We can reuse the result about the mediant from Case I simply by observing that $a/b$ is the mediant of $(a-c)/(b-d)$ and $c/d$. Therefore $a/b$ lies in between the other two fractions. As $a/b>c/d$ we can conclude that $(a-c)/(b-d)>a/b$.
Again, by Lemma 1, we have $(a-c)/(b-d)>\sqrt n$ proving the claim.
On the other hand when $d>b$ we have $x+y=-(c-a)+(d-b)\sqrt n$, and thus we need to show that $(c-a)/(d-b)<\sqrt n$. This follows similarly by observing that $c/d$ is the mediant of $a/b$ and $(c-a)/(d-b)$. This time the conclusion is that $(c-a)/(d-b)<c/d$.
As $c/d<\sqrt n$ the obvious counterpart of Lemma 1 implies that also $(c-a)/(d-b)<\sqrt n$.