The Natural Numbers and the Integers inside the Rational Numbers

In the setting of The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion, let $\mathbb{Z}$, its operations, $0_{\mathbb{Z}}$, $1_{\mathbb{Z}}$ and $\iota$ be as in The Integers, and $\mathbb{Q}$, $[x,m]$, its operations and order, $0_{\mathbb{Q}}$, $1_{\mathbb{Q}}$ and $j$ as in The Rational Numbers. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring and The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field, $\mathbb{Z}$ is a commutative ring and $\mathbb{Q}$ an ordered field, with negatives as in Negatives, Differences, Reciprocals and Quotients §negative and, in $\mathbb{Q}$, reciprocals and quotients as in Negatives, Differences, Reciprocals and Quotients §reciprocal. Let $x\in\mathbb{Z}$ and $m,n\in\mathbb{N}$.

¶ The negations of The Integers §operations and The Rational Numbers §operations are the negatives in $\mathbb{Z}$ and in $\mathbb{Q}$.

¶ The map $n\mapsto j(\iota(n))$ from $\mathbb{N}$ to $\mathbb{Q}$ is injective; $j(\iota(1))=1_{\mathbb{Q}}$, $j(\iota(m+n))=j(\iota(m))+j(\iota(n))$ and $j(\iota(mn))=j(\iota(m))\,j(\iota(n))$; $m\le n$ if and only if $j(\iota(m))\le j(\iota(n))$, and likewise for $<$; and $0_{\mathbb{Q}}<j(\iota(n))$.

¶ $[x,m]=j(x)/j(\iota(m))$.