$\mathbb{Q}(\sqrt{2})$ carries exactly two distinct field orders, exchanged by the conjugation $\sqrt{2}\mapsto -\sqrt{2}$

$u\in \mathbb{R}$ satisfies $u>0$ and $u^2=2$, and $u\mathrm{\notin }\mathbb{Q}$.

In every ordered field the positivity of a rational is forced: $n\cdot 1>0$ for $n\ge 1$, so for positive integers $p,q$ the element $(p\cdot 1)(q\cdot 1{)}^{-1}$ is positive, and hence a rational is in the positive cone exactly when it is positive in the usual sense.

$F$ is a subfield of $\mathbb{R}$: it contains $0$ and $1$, is closed under subtraction, and $(a+bu)(c+du)=(ac+2bd)+(ad+bc)u$ gives closure under multiplication; for $a+bu\mathrm{\ne }0$ one has $a^2-2b^2\mathrm{\ne }0$, since $b\mathrm{\ne }0$ would otherwise give $(a\mathrm{/}b{)}^2=2$ against [L2] while $b=0$ forces $a\mathrm{\ne }0$, and then $(a+bu{)}^{-1}=(a-bu)(a^2-2b^2{)}^{-1}\in F$.

The representation is unique: $a+bu=a^{\mathrm{\prime }}+b^{\mathrm{\prime }}u$ with $b\mathrm{\ne }{b}^{\mathrm{\prime }}$ would give $u=(a-a^{\mathrm{\prime }})(b^{\mathrm{\prime }}-b{)}^{-1}\in \mathbb{Q}$, against step 1.1; so $b=b^{\mathrm{\prime }}$ and then $a=a^{\mathrm{\prime }}$.

$\sigma$ is therefore a well-defined map $F\to F$, and it is a field automorphism: it is additive by inspection, $\sigma (1)=1$, and $\sigma (a+bu)\sigma (c+du)=(ac+2bd)-(ad+bc)u=\sigma ((a+bu)(c+du))$; moreover $\sigma \circ \sigma$ is the identity, so $\sigma$ is a bijection.

Let $Q$ be any positive cone on $F$. Since $u\mathrm{\ne }0$, exactly one of $u\in Q$, $-u\in Q$ holds.

$Q$ is determined by that choice. Suppose $u\in Q$ (the other case is the same with $u$ replaced by $-u$, which also squares to $2$). Let $x=a+bu\mathrm{\ne }0$. If $b=0$ then $x$ is a nonzero rational and step 1.2 decides it. If $b\mathrm{\ne }0$ then $x=b(u+c)$ with $c:=a\mathrm{/}b\in \mathbb{Q}$, and by [L4] the membership of $x$ is decided by those of $b$ and of $u+c$; for $c\ge 0$ one has $u+c\in Q$, while for $c<0$, writing $e:=-c>0$, the identity $(u-e)(u+e)=2-e^2$ with $u+e\in Q$ and $2-e^2\mathrm{\ne }0$ shows that $u-e\in Q$ exactly when $2-e^2>0$, a condition on a rational decided by step 1.2. So $Q$ is uniquely determined, and there are at most two positive cones on $F$.

Both occur. $P_1$ is a positive cone on $F$, being the restriction to the subfield $F$ of the positive cone of $\mathbb{R}$; and $P_2={\sigma }^{-1}(P_1)$ is one because $\sigma$ is a field automorphism, so trichotomy and closure transfer along it. They are distinct: $u\in P_1$ by step 1.1, whereas $\sigma (u)=-u\mathrm{\notin }{P}_1$, so $u\mathrm{\notin }{P}_2$.

Hence $F$ carries exactly two positive cones, $P_1$ and $P_2$, and since $\sigma$ is an involution, $P_2=\sigma (P_1)$ and $P_1=\sigma (P_2)$: the conjugation exchanges the two orders.