The Rational Numbers Embed in Exactly One Way into Every Ordered Field

In the setting of The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals, let $r$, with $+$, $\cdot$, $0_{r}$, $1_{r}$ and $\le$, be an ordered field, with negatives, differences, reciprocals and quotients as in Negatives, Differences, Reciprocals and Quotients §negative and Negatives, Differences, Reciprocals and Quotients §reciprocal, absolute values as in Absolute Value in an Ordered Field §absolute-value, and $<$ the strict relation of $\le$. The operations and orders of $\mathbb{Q}$ and $r$ are written alike, as The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §overloading allows; below, those applied to $u$ and $v$ are the ones of $\mathbb{Q}$, and those applied to values of $\varphi$ are the ones of $r$, even if $r$ and $\mathbb{Q}$ have the same elements. Let $u,v\in\mathbb{Q}$.

¶ There is exactly one map $\varphi:\mathbb{Q}\to r$ such that $\varphi(1)=1_{r}$ and, for all $u,v\in\mathbb{Q}$, $\varphi(u+v)=\varphi(u)+\varphi(v)$ and $\varphi(u\cdot v)=\varphi(u)\cdot\varphi(v)$.

¶ This map $\varphi$ is a homomorphism of ordered fields from $\mathbb{Q}$ to $r$.

¶ $\varphi$ is injective.

¶ $u<v$ if and only if $\varphi(u)<\varphi(v)$.

¶ $\varphi(0)=0_{r}$.

¶ $\varphi(-u)=-\varphi(u)$ and $\varphi(u-v)=\varphi(u)-\varphi(v)$.

¶ If $u\neq0$, then $\varphi(u)\neq0_{r}$, $\varphi(u^{-1})=\varphi(u)^{-1}$ and $\varphi(v/u)=\varphi(v)/\varphi(u)$.

¶ $\varphi(|u|)=|\varphi(u)|$.