[proofplan] Let $F$ and $F'$ be complete ordered fields. The strategy has three parts. First, every complete ordered field $F$ contains a canonical copy of $\mathbb{Q}$: the prime subfield generated by $1_F$ embeds $\mathbb{Z}$ and then $\mathbb{Q}$ into $F$, and the embedding is order-preserving because any complete ordered field has characteristic $0$ and inherits the strict order from $\mathbb{Q}$. Second, $\mathbb{Q}$ is dense in any complete ordered field (a consequence of the Archimedean property, which itself follows from completeness). Third, any element $x \in F$ is the supremum of its lower rational cuts $\{q \in \mathbb{Q} : \iota_F(q) < x\}$. Defining $\Phi(x) \in F'$ as the supremum of the image cut in $F'$ produces a well-defined order-preserving field isomorphism $\Phi: F \to F'$. [/proofplan]

[step:Embed $\mathbb{Q}$ as an ordered subfield of any complete ordered field] Let $(F, +, \cdot, <)$ be a complete ordered field, with multiplicative identity $1_F$ and additive identity $0_F$. Define $\iota_F: \mathbb{Q} \to F$ as follows. First, the map \begin{align*} \iota_\mathbb{Z}: \mathbb{Z} &\to F, \\ n &\mapsto \begin{cases} \underbrace{1_F + \cdots + 1_F}_{n \text{ times}} & n > 0, \\ 0_F & n = 0, \\ -\iota_\mathbb{Z}(-n) & n < 0 \end{cases} \end{align*} is a unital ring homomorphism. Since $F$ is ordered, $1_F > 0_F$, so $\iota_\mathbb{Z}(n) > 0_F$ for $n > 0$ in $\mathbb{Z}$ (sum of positives is positive in any ordered field). Hence $\iota_\mathbb{Z}$ is injective: if $\iota_\mathbb{Z}(n) = 0_F$ with $n \ne 0$, WLOG $n > 0$, contradicting $\iota_\mathbb{Z}(n) > 0_F$. Therefore $F$ has characteristic $0$, and $\iota_\mathbb{Z}(\mathbb{Z})$ is an integral domain inside $F$. For $p/q \in \mathbb{Q}$ with $p \in \mathbb{Z}$, $q \in \mathbb{Z}_{>0}$, define \begin{align*} \iota_F: \mathbb{Q} &\to F, \\ p/q &\mapsto \iota_\mathbb{Z}(p) \cdot \iota_\mathbb{Z}(q)^{-1}. \end{align*} This is well-defined ($\iota_\mathbb{Z}(q) \ne 0_F$ since $q > 0$) and is the unique extension of $\iota_\mathbb{Z}$ to a field homomorphism $\mathbb{Q} \to F$ (by the universal property of the field of fractions of $\mathbb{Z}$). Field homomorphisms with non-trivial codomain are injective, so $\iota_F$ is injective. Order preservation follows from $\iota_F(p/q) > 0_F \iff p > 0$ (because $\iota_\mathbb{Z}(q) > 0_F$ and the order inverts under negation). Hence $\iota_F: \mathbb{Q} \hookrightarrow F$ is an injective ordered-field homomorphism. The same construction yields $\iota_{F'}: \mathbb{Q} \hookrightarrow F'$. [/step]

[step:Establish the Archimedean property and density of $\mathbb{Q}$ in $F$] We claim $F$ is Archimedean: for every $x, y \in F$ with $x > 0_F$, there exists $n \in \mathbb{N}$ with $\iota_F(n) \cdot x > y$. Suppose, for contradiction, that there exist $x_0, y_0 \in F$ with $x_0 > 0_F$ and $\iota_F(n) \cdot x_0 \le y_0$ for all $n \in \mathbb{N}$. Then the set $A := \{\iota_F(n) \cdot x_0 : n \in \mathbb{N}\} \subset F$ is non-empty (it contains $\iota_F(1) \cdot x_0 = x_0$) and bounded above by $y_0$. By completeness of $F$ (least upper bound property in $F$), $\alpha := \sup A \in F$ exists. Since $x_0 > 0_F$, $\alpha - x_0 < \alpha$, so $\alpha - x_0$ is not an upper bound: there exists $n \in \mathbb{N}$ with $\iota_F(n) \cdot x_0 > \alpha - x_0$, hence $\iota_F(n+1) \cdot x_0 = \iota_F(n) \cdot x_0 + x_0 > \alpha$, contradicting $\alpha = \sup A$. Therefore $F$ is Archimedean. We now claim $\iota_F(\mathbb{Q})$ is dense in $F$: for every $x, y \in F$ with $x < y$, there exists $q \in \mathbb{Q}$ with $x < \iota_F(q) < y$. By the Archimedean property applied to $1_F > 0_F$ and $1_F/(y - x)$, choose $n \in \mathbb{N}$ with $\iota_F(n)(y - x) > 1_F$. Set $A := \{m \in \mathbb{Z} : \iota_F(m) > \iota_F(n) \cdot x\}$. By Archimedeanness $A$ is non-empty and bounded below; let $m_0 := \min A \in \mathbb{Z}$ (existence of minima of bounded-below subsets of $\mathbb{Z}$ is a property of $\mathbb{Z}$). Then $\iota_F(m_0) > \iota_F(n) \cdot x$ and $\iota_F(m_0 - 1) \le \iota_F(n) \cdot x$, so \begin{align*} \iota_F(n) \cdot x < \iota_F(m_0) = \iota_F(m_0 - 1) + 1_F \le \iota_F(n) \cdot x + 1_F < \iota_F(n) \cdot x + \iota_F(n)(y - x) = \iota_F(n) \cdot y. \end{align*} Dividing by $\iota_F(n) > 0_F$ gives $x < \iota_F(m_0/n) < y$, with $q := m_0/n \in \mathbb{Q}$. The same arguments (with $F$ replaced by $F'$, $\iota_F$ by $\iota_{F'}$) show $F'$ is Archimedean and $\iota_{F'}(\mathbb{Q})$ is dense in $F'$. [/step]

[step:Define the candidate isomorphism $\Phi: F \to F'$ via lower rational cuts]For $x \in F$, define the lower rational cut \begin{align*} L_F(x) := \{q \in \mathbb{Q} : \iota_F(q) < x\} \subseteq \mathbb{Q}. \end{align*} By Step 2 (density), for any $x \in F$ there exists $q \in \mathbb{Q}$ with $\iota_F(q) < x$, so $L_F(x) \ne \varnothing$, and similarly there exists $q' \in \mathbb{Q}$ with $x < \iota_F(q')$, so $L_F(x)$ is bounded above by $q'$ in $\mathbb{Q}$. Moreover $\iota_{F'}(L_F(x)) := \{\iota_{F'}(q) : q \in L_F(x)\}$ is non-empty and bounded above in $F'$ (by $\iota_{F'}(q')$, since $\iota_{F'}$ is order-preserving). By the least upper bound property in $F'$, define \begin{align*} \Phi: F &\to F', \\ x &\mapsto \sup_{F'} \iota_{F'}(L_F(x)). \end{align*}[/step]

[guided]We have established two facts about complete ordered fields. First, $\mathbb{Q}$ embeds canonically into both $F$ and $F'$ via $\iota_F, \iota_{F'}$ (Step 1). Second, $\iota_F(\mathbb{Q})$ is dense in $F$ and $\iota_{F'}(\mathbb{Q})$ is dense in $F'$ (Step 2). We now build the isomorphism $\Phi: F \to F'$ from these ingredients. The strategy is *Dedekind-style*: any element $x \in F$ is determined by which rationals lie below it, because density forces $x = \sup_F \iota_F(L_F(x))$. Indeed, $\iota_F(L_F(x))$ is bounded above by $x$ by definition; if $u < x$ in $F$, density yields $q \in \mathbb{Q}$ with $u < \iota_F(q) < x$, so $q \in L_F(x)$ and $u$ is not an upper bound of $\iota_F(L_F(x))$. Therefore $x$ is the least upper bound. So we **encode** $x$ by its rational cut $L_F(x) \subseteq \mathbb{Q}$, **transport** the cut into $F'$ via $\iota_{F'}$, and **decode** by taking the supremum in $F'$. We verify that each operation is legal. **Rational cut.** For $x \in F$ define \begin{align*} L_F(x) := \{q \in \mathbb{Q} : \iota_F(q) < x\} \subseteq \mathbb{Q}. \end{align*} By density (Step 2), there is $q_- \in \mathbb{Q}$ with $\iota_F(q_-) < x$, so $L_F(x) \ne \varnothing$; and there is $q_+ \in \mathbb{Q}$ with $x < \iota_F(q_+)$, so for every $q \in L_F(x)$ we have $\iota_F(q) < x < \iota_F(q_+)$, hence $q < q_+$ in $\mathbb{Q}$ (injective order-preserving). Thus $L_F(x)$ is non-empty and bounded above in $\mathbb{Q}$. **Transport.** Define \begin{align*} \iota_{F'}(L_F(x)) := \{\iota_{F'}(q) : q \in L_F(x)\} \subseteq F'. \end{align*} This set is non-empty (since $L_F(x)$ is). It is bounded above in $F'$: for every $q \in L_F(x)$, $q < q_+$ in $\mathbb{Q}$, so $\iota_{F'}(q) < \iota_{F'}(q_+)$ by order preservation of $\iota_{F'}$. Hence $\iota_{F'}(q_+)$ is an upper bound. **Decode.** Since $F'$ is complete (least upper bound property), $\sup_{F'} \iota_{F'}(L_F(x))$ exists in $F'$. We thus define \begin{align*} \Phi: F &\to F', \\ x &\mapsto \sup_{F'} \iota_{F'}(L_F(x)). \end{align*} What would fail without each ingredient? Without density in $F$, the cut $L_F(x)$ could fail to determine $x$ (two distinct elements could share the same cut). Without completeness of $F'$, the supremum defining $\Phi(x)$ could fail to exist. Without the embedded $\iota_{F'}: \mathbb{Q} \to F'$, there would be no transport map. The construction uses precisely the three core properties: the embedded $\mathbb{Q}$, density (an Archimedean consequence of completeness), and least upper bounds.[/guided]

[step:Verify $\Phi$ extends $\iota_{F'} \circ \iota_F^{-1}$ and is order-preserving] For $r \in \mathbb{Q}$, we show $\Phi(\iota_F(r)) = \iota_{F'}(r)$. We have $L_F(\iota_F(r)) = \{q \in \mathbb{Q} : \iota_F(q) < \iota_F(r)\} = \{q \in \mathbb{Q} : q < r\}$ (using $\iota_F$ order-preserving and injective). Hence \begin{align*} \Phi(\iota_F(r)) = \sup_{F'} \{\iota_{F'}(q) : q < r\}. \end{align*} Since $\iota_{F'}$ is order-preserving, $\iota_{F'}(q) < \iota_{F'}(r)$ for all $q < r$, so $\iota_{F'}(r)$ is an upper bound of $\{\iota_{F'}(q) : q < r\}$. It is the least: if $u < \iota_{F'}(r)$ in $F'$, by density of $\iota_{F'}(\mathbb{Q})$ in $F'$ (Step 2), there exists $q \in \mathbb{Q}$ with $u < \iota_{F'}(q) < \iota_{F'}(r)$, hence $q < r$ (injective order-preserving), so $\iota_{F'}(q) \in \{\iota_{F'}(q'') : q'' < r\}$ exceeds $u$. Therefore $\Phi(\iota_F(r)) = \iota_{F'}(r)$. For order: if $x < y$ in $F$, by density there exist $q_1, q_2 \in \mathbb{Q}$ with $x < \iota_F(q_1) < \iota_F(q_2) < y$. Then $L_F(x) \subsetneq L_F(y)$: indeed $q_1 \notin L_F(x)$ (since $\iota_F(q_1) > x$) but $q_1 \in L_F(y)$ (since $\iota_F(q_1) < y$). Hence $\iota_{F'}(L_F(x)) \subseteq \iota_{F'}(L_F(y))$, and moreover $\iota_{F'}(q_1) \in \iota_{F'}(L_F(y))$ exceeds every element of $\iota_{F'}(L_F(x))$ (since elements of $L_F(x)$ are $< q_1$ in $\mathbb{Q}$). Therefore $\Phi(x) \le \iota_{F'}(q_1) < \iota_{F'}(q_2) \le \Phi(y)$, so $\Phi(x) < \Phi(y)$. In particular, $\Phi$ is injective. [/step]

[step:Verify $\Phi$ is surjective] Let $y \in F'$. Define \begin{align*} L'(y) := \{q \in \mathbb{Q} : \iota_{F'}(q) < y\} \subseteq \mathbb{Q}. \end{align*} By density of $\iota_{F'}(\mathbb{Q})$ in $F'$ (Step 2), $L'(y) \ne \varnothing$ and is bounded above in $\mathbb{Q}$. Hence $\iota_F(L'(y)) \subseteq F$ is non-empty and bounded above (by any $\iota_F(q')$ with $q'$ exceeding all elements of $L'(y)$, using order preservation of $\iota_F$). By completeness of $F$, define $x := \sup_F \iota_F(L'(y)) \in F$. We claim $\Phi(x) = y$, which we establish by showing $L_F(x) = L'(y)$. **$L'(y) \subseteq L_F(x)$.** Fix $q \in L'(y)$. We must show $\iota_F(q) < x$. Since $q \in L'(y)$, by density of $\iota_{F'}(\mathbb{Q})$ in $F'$ applied to $\iota_{F'}(q) < y$, there exists $q' \in \mathbb{Q}$ with $\iota_{F'}(q) < \iota_{F'}(q') < y$. By injective order-preservation of $\iota_{F'}$, $q < q'$ in $\mathbb{Q}$, and $q' \in L'(y)$. Hence $\iota_F(q) < \iota_F(q')$ in $F$, and $\iota_F(q') \in \iota_F(L'(y))$, so $\iota_F(q') \le \sup_F \iota_F(L'(y)) = x$. Therefore $\iota_F(q) < \iota_F(q') \le x$, so $q \in L_F(x)$. **$L_F(x) \subseteq L'(y)$.** Fix $q \in L_F(x)$, so $\iota_F(q) < x = \sup_F \iota_F(L'(y))$. By the supremum characterisation, there exists $q' \in L'(y)$ with $\iota_F(q) < \iota_F(q')$, hence $q < q'$ in $\mathbb{Q}$. Since $q' \in L'(y)$, $\iota_{F'}(q') < y$, and since $q < q'$, $\iota_{F'}(q) < \iota_{F'}(q') < y$, so $q \in L'(y)$. Thus $L_F(x) = L'(y)$. Therefore \begin{align*} \Phi(x) = \sup_{F'} \iota_{F'}(L_F(x)) = \sup_{F'} \iota_{F'}(L'(y)) = y, \end{align*} where the last equality is the same supremum-from-below identity used in Step 4: by density of $\iota_{F'}(\mathbb{Q})$ in $F'$, $y$ is the supremum in $F'$ of $\{\iota_{F'}(q) : q < y \text{ in the cut sense}\} = \iota_{F'}(L'(y))$. [/step]

[step:Verify $\Phi$ preserves addition and multiplication]**Additivity.** We show $\Phi(x + y) = \Phi(x) + \Phi(y)$ for $x, y \in F$. Set $L := L_F(x) + L_F(y) := \{q_1 + q_2 : q_1 \in L_F(x), q_2 \in L_F(y)\} \subseteq \mathbb{Q}$. We claim $L$ and $L_F(x + y)$ are mutually cofinal subsets of $\mathbb{Q}$, hence have the same supremum image under $\iota_{F'}$. First, $L \subseteq L_F(x + y)$: if $q_1 \in L_F(x)$ and $q_2 \in L_F(y)$, then $\iota_F(q_1) < x$ and $\iota_F(q_2) < y$, so $\iota_F(q_1 + q_2) = \iota_F(q_1) + \iota_F(q_2) < x + y$, i.e. $q_1 + q_2 \in L_F(x + y)$. Conversely, every $q \in L_F(x + y)$ is dominated by some element of $L$. Indeed, fix $q \in L_F(x + y)$ and set $\varepsilon := (x + y - \iota_F(q))/2_F \in F$, where $2_F := 1_F + 1_F$. Since $\iota_F(q) < x + y$, $\varepsilon > 0_F$. By density (Step 2), choose $q_1, q_2 \in \mathbb{Q}$ with \begin{align*} x - \varepsilon < \iota_F(q_1) < x, \qquad y - \varepsilon < \iota_F(q_2) < y. \end{align*} Then $q_1 \in L_F(x)$, $q_2 \in L_F(y)$, so $q_1 + q_2 \in L$. Moreover, \begin{align*} \iota_F(q_1 + q_2) = \iota_F(q_1) + \iota_F(q_2) > (x - \varepsilon) + (y - \varepsilon) = x + y - 2_F \varepsilon = \iota_F(q), \end{align*} so $q_1 + q_2 > q$ in $\mathbb{Q}$. Therefore for every $q \in L_F(x + y)$ there exists $q^* \in L$ with $q < q^*$, and conversely $L \subseteq L_F(x + y)$. The two sets have identical least upper bounds in $\mathbb{Q}$ (they have identical sets of upper bounds), so under $\iota_{F'}$ (which is order-preserving), \begin{align*} \sup_{F'} \iota_{F'}(L_F(x + y)) = \sup_{F'} \iota_{F'}(L). \end{align*} Now $\sup_{F'} \iota_{F'}(L) = \sup_{F'} \iota_{F'}(L_F(x)) + \sup_{F'} \iota_{F'}(L_F(y))$ by the standard identity $\sup(A + B) = \sup A + \sup B$ for non-empty bounded-above subsets of an ordered field (which holds because $a + b \le \sup A + \sup B$ for $a \in A$, $b \in B$, and given $\eta > 0_{F'}$ we choose $a > \sup A - \eta/2_{F'}$, $b > \sup B - \eta/2_{F'}$ to get $a + b > \sup A + \sup B - \eta$). Hence \begin{align*} \Phi(x + y) = \sup_{F'} \iota_{F'}(L_F(x + y)) = \sup_{F'} \iota_{F'}(L) = \Phi(x) + \Phi(y). \end{align*} **Multiplicativity.** We show $\Phi(xy) = \Phi(x)\Phi(y)$ for $x, y \in F$. *Reduction to positive arguments.* By Step 4, $\Phi(0_F) = \iota_{F'}(0) = 0_{F'}$, and from additivity, $\Phi(-x) = -\Phi(x)$. The case $x = 0_F$ or $y = 0_F$ is immediate: $\Phi(xy) = \Phi(0_F) = 0_{F'} = \Phi(x)\Phi(y)$. For $x, y$ with $x < 0_F < y$, multiplicativity for $(-x)y$ (positive case, established below) gives $\Phi(-xy) = \Phi(-x)\Phi(y) = -\Phi(x)\Phi(y)$; combined with $\Phi(-xy) = -\Phi(xy)$ this yields $\Phi(xy) = \Phi(x)\Phi(y)$. The case $x, y < 0_F$ follows similarly: $\Phi(xy) = \Phi((-x)(-y)) = \Phi(-x)\Phi(-y) = (-\Phi(x))(-\Phi(y)) = \Phi(x)\Phi(y)$. So we reduce to $x, y > 0_F$. *Positive case.* Fix $x, y > 0_F$. Set \begin{align*} L^+ := \{q_1 q_2 : q_1, q_2 \in \mathbb{Q}_{>0},\ \iota_F(q_1) < x,\ \iota_F(q_2) < y\} \subseteq \mathbb{Q}_{>0}. \end{align*} We claim $L^+$ and $L_F(xy) \cap \mathbb{Q}_{>0}$ are mutually cofinal in $\mathbb{Q}_{>0}$. Forward inclusion: if $q_1, q_2 \in \mathbb{Q}_{>0}$ with $\iota_F(q_1) < x$ and $\iota_F(q_2) < y$, then since $0_F < \iota_F(q_1) < x$ and $0_F < \iota_F(q_2) < y$, \begin{align*} \iota_F(q_1 q_2) = \iota_F(q_1)\iota_F(q_2) < x y, \end{align*} so $q_1 q_2 \in L_F(xy) \cap \mathbb{Q}_{>0}$. Cofinality: we show every $q \in L_F(xy) \cap \mathbb{Q}_{>0}$ is dominated by some element of $L^+$. Fix such $q$, so $0 < \iota_F(q) < xy$. Choose $\varepsilon \in F$ with \begin{align*} 0_F < \varepsilon < \min\!\left(x,\ y,\ \frac{xy - \iota_F(q)}{x + y + 1_F}\right). \end{align*} (This minimum is strictly positive: $x > 0_F$, $y > 0_F$, and $xy - \iota_F(q) > 0_F$, with $x + y + 1_F > 0_F$ giving a positive quotient.) By density (Step 2), choose $q_1, q_2 \in \mathbb{Q}$ with \begin{align*} x - \varepsilon < \iota_F(q_1) < x, \qquad y - \varepsilon < \iota_F(q_2) < y. \end{align*} Since $\varepsilon < x$ and $\varepsilon < y$, $\iota_F(q_1) > x - \varepsilon > 0_F$ and $\iota_F(q_2) > y - \varepsilon > 0_F$, so $q_1, q_2 \in \mathbb{Q}_{>0}$. Hence $q_1 q_2 \in L^+$. We bound: \begin{align*} \iota_F(q_1 q_2) = \iota_F(q_1)\iota_F(q_2) > (x - \varepsilon)(y - \varepsilon) = xy - \varepsilon(x + y) + \varepsilon^2 \ge xy - \varepsilon(x + y + 1_F), \end{align*} where the last inequality uses $\varepsilon^2 \ge 0_F$ (so $-\varepsilon^2 \le 0_F$, and we replace $\varepsilon(x+y)$ by the larger quantity $\varepsilon(x+y+1_F)$ on the subtracted side; equivalently, $\varepsilon^2 - \varepsilon \cdot 1_F \le 0_F$ holds when $\varepsilon \le 1_F$, which we may assume by choosing $\varepsilon$ small). By the choice of $\varepsilon$, $\varepsilon(x + y + 1_F) < xy - \iota_F(q)$, so \begin{align*} \iota_F(q_1 q_2) > xy - (xy - \iota_F(q)) = \iota_F(q), \end{align*} giving $q_1 q_2 > q$ in $\mathbb{Q}$. Hence $q$ is dominated by $q_1 q_2 \in L^+$. By cofinality (and the order-preservation of $\iota_{F'}$), \begin{align*} \sup_{F'} \iota_{F'}(L_F(xy) \cap \mathbb{Q}_{>0}) = \sup_{F'} \iota_{F'}(L^+). \end{align*} Since $x > 0_F$, $\Phi(x) > \Phi(0_F) = 0_{F'}$ (Step 4 order preservation), and we have \begin{align*} \Phi(x) = \sup_{F'} \iota_{F'}(L_F(x)) = \sup_{F'} \iota_{F'}(L_F(x) \cap \mathbb{Q}_{>0}), \end{align*} because non-positive rationals in $L_F(x)$ map to non-positive elements of $F'$, while $\iota_{F'}(L_F(x) \cap \mathbb{Q}_{>0})$ already contains positive elements (e.g. $\iota_{F'}(q_1)$ for some positive $q_1 \in L_F(x)$, available by density between $0_F$ and $x$), so they cannot affect the supremum. The same identity holds for $y$. Now $\sup_{F'}\iota_{F'}(L^+) = \sup_{F'}\iota_{F'}(L_F(x) \cap \mathbb{Q}_{>0}) \cdot \sup_{F'}\iota_{F'}(L_F(y) \cap \mathbb{Q}_{>0})$ by the standard identity $\sup(AB) = (\sup A)(\sup B)$ for non-empty subsets $A, B \subseteq F'_{>0}$ that are bounded above (proof: $ab \le \sup A \cdot \sup B$ for $a \in A$, $b \in B$; given $\eta > 0_{F'}$, choose $a > \sup A - \eta'$, $b > \sup B - \eta'$ with $\eta'$ small enough that $ab > \sup A \sup B - \eta$, exploiting positivity). Hence \begin{align*} \sup_{F'}\iota_{F'}(L^+) = \Phi(x)\Phi(y). \end{align*} Finally, the cofinality identity $\sup_{F'}\iota_{F'}(L_F(xy) \cap \mathbb{Q}_{>0}) = \sup_{F'}\iota_{F'}(L^+)$ together with $\Phi(xy) = \sup_{F'}\iota_{F'}(L_F(xy)) = \sup_{F'}\iota_{F'}(L_F(xy) \cap \mathbb{Q}_{>0})$ (same argument: $xy > 0_F$, so positive rationals dominate the supremum) yield \begin{align*} \Phi(xy) = \Phi(x)\Phi(y). \end{align*}[/step]

[guided]We prove preservation of both operations. The structure is the same in each case: encode by rational approximations from below, transport to $F'$, decode by supremum, and exploit that approximations from below combine well under $+$ and $\cdot$ (provided we control signs for multiplication). **Step A: Additivity, $\Phi(x + y) = \Phi(x) + \Phi(y)$.** The candidate is to compare $L_F(x + y)$ with the *sumset* \begin{align*} L := L_F(x) + L_F(y) = \{q_1 + q_2 : q_1 \in L_F(x),\ q_2 \in L_F(y)\}. \end{align*} Why this set? Because if we take $q_1 < x/\iota_F$ and $q_2 < y/\iota_F$ (informally), their sum $q_1 + q_2 < x + y$ should give a typical element of $L_F(x+y)$. Conversely, an arbitrary $q < x + y$ should be approachable as $q_1 + q_2$. *Forward inclusion $L \subseteq L_F(x+y)$.* If $\iota_F(q_1) < x$ and $\iota_F(q_2) < y$, adding (the ordered field axiom: order is preserved under addition) gives $\iota_F(q_1) + \iota_F(q_2) < x + y$. Since $\iota_F$ is a ring homomorphism, $\iota_F(q_1 + q_2) = \iota_F(q_1) + \iota_F(q_2)$, so $q_1 + q_2 \in L_F(x+y)$. *Cofinality of $L$ over $L_F(x+y)$.* The reverse inclusion $L_F(x+y) \subseteq L$ may fail: a "near-extremal" $q$ very close to $x+y$ may have no decomposition $q_1 + q_2$ inside $L_F(x) \times L_F(y)$. But we do not need equality of sets — only that $L$ is **cofinal** in $L_F(x+y)$, meaning every element of $L_F(x+y)$ is dominated by some element of $L$. Why does this suffice? Because cofinality is precisely the condition under which two non-empty bounded-above subsets of an ordered field have the same supremum. To prove cofinality, fix $q \in L_F(x+y)$. The "slack" between $\iota_F(q)$ and $x+y$ is the positive quantity $x + y - \iota_F(q) > 0_F$. We split this slack between two halves: set \begin{align*} \varepsilon := \frac{x + y - \iota_F(q)}{2_F}, \end{align*} where $2_F = 1_F + 1_F > 0_F$. By density, pick $q_1, q_2 \in \mathbb{Q}$ approximating $x$ and $y$ from below by less than $\varepsilon$: \begin{align*} x - \varepsilon < \iota_F(q_1) < x, \qquad y - \varepsilon < \iota_F(q_2) < y. \end{align*} Then $q_1 \in L_F(x)$ and $q_2 \in L_F(y)$, so $q_1 + q_2 \in L$. Adding the two lower bounds: \begin{align*} \iota_F(q_1 + q_2) = \iota_F(q_1) + \iota_F(q_2) > (x - \varepsilon) + (y - \varepsilon) = x + y - 2_F\varepsilon = \iota_F(q). \end{align*} So $q_1 + q_2 > q$ in $\mathbb{Q}$, i.e. $q$ is dominated by an element of $L$. *Suprema agree.* Since $L \subseteq L_F(x+y)$ and every element of $L_F(x+y)$ is dominated by an element of $L$, the two non-empty bounded-above subsets of $\mathbb{Q}$ have identical sets of upper bounds, hence identical least upper bounds in $\mathbb{Q}$. Applying $\iota_{F'}$ (order-preserving) and taking suprema in $F'$ (which exist by completeness), \begin{align*} \sup_{F'} \iota_{F'}(L_F(x+y)) = \sup_{F'} \iota_{F'}(L). \end{align*} *Sumset supremum identity.* In any ordered field, for non-empty bounded-above subsets $A, B$, \begin{align*} \sup(A + B) = \sup A + \sup B, \end{align*} where $A + B := \{a + b : a \in A, b \in B\}$. The "$\le$" direction is clear: $a + b \le \sup A + \sup B$. For "$\ge$": fix $\eta > 0_{F'}$; by characterisation of supremum, choose $a \in A$ with $a > \sup A - \eta/2_{F'}$ and $b \in B$ with $b > \sup B - \eta/2_{F'}$; then $a + b > \sup A + \sup B - \eta$, and $a + b \in A + B$, so $\sup(A + B) > \sup A + \sup B - \eta$ for all $\eta > 0_{F'}$, giving $\sup(A + B) \ge \sup A + \sup B$. Applying this identity to $A := \iota_{F'}(L_F(x))$ and $B := \iota_{F'}(L_F(y))$ — and noting $A + B = \iota_{F'}(L)$ because $\iota_{F'}$ is additive — we conclude \begin{align*} \Phi(x+y) = \sup_{F'} \iota_{F'}(L) = \sup_{F'} A + \sup_{F'} B = \Phi(x) + \Phi(y). \end{align*} **Step B: Multiplicativity, $\Phi(xy) = \Phi(x)\Phi(y)$.** Multiplication is more delicate than addition because order is preserved by multiplication only when factors have controlled signs ($a < b$ and $c > 0_F$ imply $ac < bc$, but the analogous statement fails for $c < 0_F$). The natural fix is to **reduce to the case where both factors are positive**, then handle signs separately. *Why reduce to positives?* Suppose $x, y > 0_F$ and we want a rational lower bound for $xy$. If $0 < q_1 < x$ and $0 < q_2 < y$ (positive rationals), then $0 < q_1 q_2 < xy$ — multiplication of positives is order-preserving in both arguments. Without positivity, e.g. $q_1 < 0 < x$ and $q_2 < 0 < y$, the product $q_1 q_2 > 0$ might exceed $xy$ entirely. So we restrict our cuts to *positive* rationals and assume $x, y > 0_F$. *Sign reduction.* Step 4 gives $\Phi(0_F) = \iota_{F'}(0) = 0_{F'}$. From additivity (Step A), $\Phi(-x) + \Phi(x) = \Phi(0_F) = 0_{F'}$, so $\Phi(-x) = -\Phi(x)$. The four sign cases are: - $x = 0_F$ or $y = 0_F$: both sides are $0_{F'}$. - $x, y > 0_F$: this is the positive case proved below. - $x < 0_F < y$ (or symmetrically): $-x > 0_F$, $y > 0_F$, so $\Phi((-x)y) = \Phi(-x)\Phi(y) = (-\Phi(x))\Phi(y) = -\Phi(x)\Phi(y)$ by the positive case; meanwhile $(-x)y = -(xy)$, so $\Phi((-x)y) = -\Phi(xy)$, giving $\Phi(xy) = \Phi(x)\Phi(y)$. - $x, y < 0_F$: $\Phi(xy) = \Phi((-x)(-y)) = \Phi(-x)\Phi(-y) = (-\Phi(x))(-\Phi(y)) = \Phi(x)\Phi(y)$ by the positive case applied to $-x, -y > 0_F$. So it suffices to prove the positive case. *Positive case set-up.* Fix $x, y > 0_F$. Define the *positive rational product set* \begin{align*} L^+ := \{q_1 q_2 : q_1, q_2 \in \mathbb{Q}_{>0},\ \iota_F(q_1) < x,\ \iota_F(q_2) < y\}. \end{align*} We will compare $L^+$ to $L_F(xy) \cap \mathbb{Q}_{>0}$ (the *positive* part of the cut of $xy$). Note $xy > 0_F$ since both factors are positive, so by density there are positive rationals in $L_F(xy)$, hence $L_F(xy) \cap \mathbb{Q}_{>0} \ne \varnothing$. *Forward inclusion $L^+ \subseteq L_F(xy) \cap \mathbb{Q}_{>0}$.* If $q_1, q_2 \in \mathbb{Q}_{>0}$ with $\iota_F(q_1) < x$ and $\iota_F(q_2) < y$, then $\iota_F(q_1) > 0_F$ (since $q_1 > 0$ and $\iota_F$ is order-preserving) and $\iota_F(q_2) > 0_F$. Multiplying the positive lower bounds (positivity is essential — multiplication of positives is order-preserving in both arguments): \begin{align*} \iota_F(q_1 q_2) = \iota_F(q_1)\iota_F(q_2) < xy, \end{align*} where the inequality is strict because, say, $\iota_F(q_1)\iota_F(q_2) < x \cdot \iota_F(q_2) < xy$ using $\iota_F(q_2) > 0_F$ first then $x > 0_F$. So $q_1 q_2 \in L_F(xy) \cap \mathbb{Q}_{>0}$. *Cofinality of $L^+$ over $L_F(xy) \cap \mathbb{Q}_{>0}$.* Fix $q \in L_F(xy) \cap \mathbb{Q}_{>0}$. We must produce $q^* \in L^+$ with $q^* > q$. We mimic the additive argument but use the bound $q_1 q_2 > (x - \varepsilon)(y - \varepsilon)$. Expanding, \begin{align*} (x - \varepsilon)(y - \varepsilon) = xy - \varepsilon(x + y) + \varepsilon^2. \end{align*} We want this to exceed $\iota_F(q)$. Discarding the (non-negative) $\varepsilon^2$ term, it suffices to pick $\varepsilon > 0_F$ small enough that \begin{align*} \varepsilon(x + y) < xy - \iota_F(q). \end{align*} We also need positivity of the approximants: $\iota_F(q_1) > x - \varepsilon > 0_F$ requires $\varepsilon < x$, and similarly $\varepsilon < y$. So we choose \begin{align*} \varepsilon := \frac{1}{2_F}\min\!\left(x,\ y,\ \frac{xy - \iota_F(q)}{x + y}\right) > 0_F. \end{align*} The minimum is strictly positive: $x > 0_F$, $y > 0_F$, and $xy - \iota_F(q) > 0_F$ (since $q \in L_F(xy)$) divided by $x + y > 0_F$. By density, pick $q_1, q_2 \in \mathbb{Q}$ with \begin{align*} x - \varepsilon < \iota_F(q_1) < x, \qquad y - \varepsilon < \iota_F(q_2) < y. \end{align*} Since $\varepsilon < x$, $\iota_F(q_1) > x - \varepsilon > 0_F$, so $q_1 > 0$ in $\mathbb{Q}$, i.e. $q_1 \in \mathbb{Q}_{>0}$. Similarly $q_2 \in \mathbb{Q}_{>0}$. Hence $q_1 q_2 \in L^+$. Multiplying the two positive lower bounds (legitimate because both bounds are positive): \begin{align*} \iota_F(q_1 q_2) = \iota_F(q_1)\iota_F(q_2) > (x - \varepsilon)(y - \varepsilon) = xy - \varepsilon(x+y) + \varepsilon^2 \ge xy - \varepsilon(x+y). \end{align*} By the choice of $\varepsilon$, $2_F\varepsilon \le (xy - \iota_F(q))/(x+y)$, so $\varepsilon(x+y) \le (xy - \iota_F(q))/2_F < xy - \iota_F(q)$. Therefore \begin{align*} \iota_F(q_1 q_2) > xy - (xy - \iota_F(q)) = \iota_F(q), \end{align*} giving $q_1 q_2 > q$. This is the required dominator. *Suprema agree.* By cofinality (and order-preservation of $\iota_{F'}$), \begin{align*} \sup_{F'} \iota_{F'}(L_F(xy) \cap \mathbb{Q}_{>0}) = \sup_{F'} \iota_{F'}(L^+). \end{align*} Moreover, since $xy > 0_F$, $\Phi(xy) > 0_{F'}$, and the negative or zero rationals in $L_F(xy)$ map to elements of $\iota_{F'}(L_F(xy))$ that are dominated by any positive image $\iota_{F'}(q)$ with $q \in L_F(xy) \cap \mathbb{Q}_{>0}$ (such $q$ exists by density between $0_F$ and $xy$). Hence they do not affect the supremum: \begin{align*} \Phi(xy) = \sup_{F'} \iota_{F'}(L_F(xy)) = \sup_{F'} \iota_{F'}(L_F(xy) \cap \mathbb{Q}_{>0}). \end{align*} The same identity (with $x$ in place of $xy$) gives $\Phi(x) = \sup_{F'}\iota_{F'}(L_F(x) \cap \mathbb{Q}_{>0})$, and similarly for $y$. *Product supremum identity.* In any ordered field, for non-empty subsets $A, B \subseteq F'_{>0}$ that are bounded above, \begin{align*} \sup(A B) = \sup A \cdot \sup B, \end{align*} where $AB := \{ab : a \in A, b \in B\}$. The "$\le$" direction: $ab \le \sup A \cdot \sup B$ since both factors are positive and bounded by their respective suprema. For "$\ge$": fix $\eta > 0_{F'}$, set $\eta' := \min(1, \eta/(\sup A + \sup B + 1_{F'}))/2_{F'}$, choose $a > \sup A - \eta'$ and $b > \sup B - \eta'$; then \begin{align*} ab > (\sup A - \eta')(\sup B - \eta') = \sup A \sup B - \eta'(\sup A + \sup B) + \eta'^2 > \sup A \sup B - \eta, \end{align*} by the choice of $\eta'$. Since $ab \in AB$, $\sup(AB) > \sup A \sup B - \eta$ for every $\eta > 0_{F'}$, giving the desired bound. Applying this to $A := \iota_{F'}(L_F(x) \cap \mathbb{Q}_{>0})$ and $B := \iota_{F'}(L_F(y) \cap \mathbb{Q}_{>0})$, both subsets of $F'_{>0}$ — and noting $AB = \iota_{F'}(L^+)$ because $\iota_{F'}$ is multiplicative — we conclude \begin{align*} \Phi(xy) = \sup_{F'} \iota_{F'}(L^+) = \sup_{F'} A \cdot \sup_{F'} B = \Phi(x)\Phi(y). \end{align*} This proves multiplicativity in the positive case, hence by sign reduction in all cases.[/guided]

[step:Conclude that $\Phi$ is an order-preserving field isomorphism] By Step 4, $\Phi$ is order-preserving (hence injective) and extends $\iota_{F'} \circ \iota_F^{-1}$ on $\iota_F(\mathbb{Q})$. By Step 5, $\Phi$ is surjective. By Step 6, $\Phi$ preserves $+$ and $\cdot$; since order preservation has been verified and $\Phi(0_F) = 0_{F'}$, $\Phi(1_F) = \iota_{F'}(1) = 1_{F'}$, $\Phi$ is a unital ordered field isomorphism $F \to F'$. In particular, taking $F'$ to be $\mathbb{R}$ (the Cauchy completion of $\mathbb{Q}$, which is a complete ordered field by the [Completeness of $\mathbb{R}$ via Cauchy Sequences](/theorems/3226) and the [Completeness Axiom](/theorems/3227)), every complete ordered field $F$ is isomorphic to $\mathbb{R}$ as an ordered field. This completes the proof. [/step]