Proof of Natural Numbers Read in an Ordered Field Agree with the Canonical Embedding of the Rationals

Each result cited is universally quantified over the data in its own statement.

For $m\in\mathbb{N}_{0}$ and a commutative ring $R$, let $m_{R}$ be the image of $m$ in $R$. By Commutative Rings, Fields and Ordered Fields: Standard Notation §numerals, the element $n$ of $F$ is $n_{F}$. The rational number $n$ of The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §identification is $j(\iota_{0}(n))$, which is $n_{\mathbb{Q}}$, the element $n$ of the commutative ring $\mathbb{Q}$ read as in Commutative Rings, Fields and Ordered Fields: Standard Notation §numerals, by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §numerals. So we must show $n_{F}=\kappa_{F}(n_{\mathbb{Q}})$.

By The Canonical Embedding of the Rational Numbers into an Ordered Field §embedding, $\kappa_{F}$ is the map $\varphi$ of The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique: it maps the unit of $\mathbb{Q}$ to the unit of $F$, and $\kappa_{F}(u+v)=\kappa_{F}(u)+\kappa_{F}(v)$ for all $u,v\in\mathbb{Q}$. The image $1_{\mathbb{Q}}$ of $1$ in $\mathbb{Q}$ is the unit of $\mathbb{Q}$ by The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products §constants, so $\kappa_{F}(1_{\mathbb{Q}})$ is the unit of $F$. Moreover $\kappa_{F}$ maps the zero of $\mathbb{Q}$ to the zero of $F$ by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §zero.

Let $A=\{m\in\mathbb{N}_{0}:m_{F}=\kappa_{F}(m_{\mathbb{Q}})\}$. We show $A=\mathbb{N}_{0}$ by induction from $0$, as in The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction, the successor of $m$ being $m+1$ as recorded in The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §recursion.

Base. By The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products §constants, $0_{F}$ is the zero of $F$ and $0_{\mathbb{Q}}$ is the zero of $\mathbb{Q}$, so $\kappa_{F}(0_{\mathbb{Q}})=0_{F}$ by the above. Hence $0\in A$.

Step. Let $m\in A$. By The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products §sum, applied in $F$ and in $\mathbb{Q}$, $(m+1)_{F}=m_{F}+1_{F}$ and $(m+1)_{\mathbb{Q}}=m_{\mathbb{Q}}+1_{\mathbb{Q}}$, and $1_{F}$ is the unit of $F$ by The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products §constants. Therefore, using $m\in A$ and the properties of $\kappa_{F}$ above,

$$(m+1)_{F}=m_{F}+1_{F}=\kappa_{F}(m_{\mathbb{Q}})+\kappa_{F}(1_{\mathbb{Q}})=\kappa_{F}(m_{\mathbb{Q}}+1_{\mathbb{Q}})=\kappa_{F}\big((m+1)_{\mathbb{Q}}\big),$$

so $m+1\in A$.

Hence $A=\mathbb{N}_{0}$; in particular $n\in A$, that is $n_{F}=\kappa_{F}(n_{\mathbb{Q}})$, which is the claim.