The Rational Numbers Form an Archimedean Ordered Field Containing the Integers

In the setting of The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion, let $\mathbb{Z}$, its operations and order, $0_{\mathbb{Z}}$, $1_{\mathbb{Z}}$ and $\iota$ be as in The Integers §integers, The Integers §operations, The Integers §constants and The Integers §embedding, and let $\mathbb{Q}$, $[x,m]$, its operations and order, $0_{\mathbb{Q}}$, $1_{\mathbb{Q}}$ and $j$ be as in The Rational Numbers §rationals, The Rational Numbers §operations, The Rational Numbers §constants and The Rational Numbers §embedding. Let $x,y\in\mathbb{Z}$, $m\in\mathbb{N}$ and $u\in\mathbb{Q}$.

¶ $\le$ is a total order on $\mathbb{Q}$, with $<$ as its strict relation, and $\mathbb{Q}$, with $+$, $\cdot$, $0_{\mathbb{Q}}$, $1_{\mathbb{Q}}$ and $\le$, is an ordered field; moreover $u+(-u)=0_{\mathbb{Q}}$.

¶ $j$ is injective, $j(0_{\mathbb{Z}})=0_{\mathbb{Q}}$, $j(1_{\mathbb{Z}})=1_{\mathbb{Q}}$, $j(x+y)=j(x)+j(y)$, $j(xy)=j(x)\,j(y)$, $j(-x)=-j(x)$, and $x\le y$ if and only if $j(x)\le j(y)$.

¶ $[x,m]\cdot j(\iota(m))=j(x)$.

¶ There is $k\in\mathbb{N}$ with $u<j(\iota(k))$.