Proof of The Natural Numbers and the Integers inside the Rational Numbers

By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring, $\mathbb{Z}$, with $+$, $\cdot$, $0_{\mathbb{Z}}$ and $1_{\mathbb{Z}}$, is a commutative ring. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field, $\mathbb{Q}$, with $+$, $\cdot$, $0_{\mathbb{Q}}$, $1_{\mathbb{Q}}$ and $\le$, is an ordered field; so by Ordered Fields §ordered-field it is a field, and by Fields §field a commutative ring. For $u,v$ both in $\mathbb{Z}$ or both in $\mathbb{Q}$, $u<v$ means $u\le v$ and $u\neq v$, by The Integers §operations and The Rational Numbers §operations.

Negation. Let $x\in\mathbb{Z}$, and let $-x$ be its negation as in The Integers §operations. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring, $x+(-x)=0_{\mathbb{Z}}$. By Additive and Multiplicative Inverses Are Unique §negative, applied to the commutative ring $\mathbb{Z}$, there is exactly one $w\in\mathbb{Z}$ with $x+w=0_{\mathbb{Z}}$, and by Negatives, Differences, Reciprocals and Quotients §negative the negative of $x$ in $\mathbb{Z}$ is this $w$. Since the negation $-x$ is such an element, it equals the negative of $x$. Likewise, for $u\in\mathbb{Q}$ with negation $-u$ as in The Rational Numbers §operations, $u+(-u)=0_{\mathbb{Q}}$ by The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field, so by Additive and Multiplicative Inverses Are Unique §negative and Negatives, Differences, Reciprocals and Quotients §negative, applied to the commutative ring $\mathbb{Q}$, $-u$ is the negative of $u$ in $\mathbb{Q}$. This proves the clause negation above.

Naturals. Let $m,n\in\mathbb{N}$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §composition, applied to the maps $\iota:\mathbb{N}\to\mathbb{Z}$ of The Integers §embedding and $j:\mathbb{Z}\to\mathbb{Q}$ of The Rational Numbers §embedding, $n\mapsto j(\iota(n))$ is the map $j\circ\iota:\mathbb{N}\to\mathbb{Q}$. If $j(\iota(m))=j(\iota(n))$, then $\iota(m)=\iota(n)$ because $j$ is injective by The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §embedding, and then $m=n$ because $\iota$ is injective by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding; so $j\circ\iota$ is injective by Injective, Surjective and Bijective Functions between Classes §injective.

Here $1$, $m+n$ and $mn$ are natural numbers by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §sets and The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §operations. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding and The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §embedding,

$$j(\iota(1))=j(1_{\mathbb{Z}})=1_{\mathbb{Q}},\qquad j(\iota(m+n))=j(\iota(m)+\iota(n))=j(\iota(m))+j(\iota(n)),\qquad j(\iota(mn))=j(\iota(m)\,\iota(n))=j(\iota(m))\,j(\iota(n)).$$

For the order, let $y,z\in\mathbb{Z}$. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §embedding, $y\le z$ if and only if $j(y)\le j(z)$; and $y\neq z$ if and only if $j(y)\neq j(z)$, since $j$ is injective and, conversely, $y=z$ gives $j(y)=j(z)$. Hence $y<z$ if and only if $j(y)<j(z)$. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding, $m<n$ if and only if $\iota(m)<\iota(n)$, so taking $y=\iota(m)$ and $z=\iota(n)$, $m<n$ if and only if $j(\iota(m))<j(\iota(n))$.

For $\le$, put $p=j(\iota(m))$ and $q=j(\iota(n))$. By the injectivity of $j\circ\iota$ shown above, and since $m=n$ gives $p=q$, we have $m=n$ if and only if $p=q$. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field, $\le$ is a total order on $\mathbb{Q}$, hence reflexive by Partial and Total Orders on a Set and the Associated Strict Relation §total and Partial and Total Orders on a Set and the Associated Strict Relation §partial. So $p\le q$ if and only if $p<q$ or $p=q$: if $p\le q$ and $p\neq q$, then $p<q$; conversely, $p<q$ gives $p\le q$, and $p=q$ gives $p\le q$ by reflexivity. By Arithmetic and Order of the Natural Numbers §partial-order, $m\le n$ if and only if $m<n$ or $m=n$. Combining these with the strict form and the equivalence of $m=n$ and $p=q$, $m\le n$ if and only if $p\le q$, that is, $j(\iota(m))\le j(\iota(n))$.

Finally, $0_{\mathbb{Z}}<\iota(n)$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §positive, so $j(0_{\mathbb{Z}})<j(\iota(n))$, and $j(0_{\mathbb{Z}})=0_{\mathbb{Q}}$ by The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §embedding; thus $0_{\mathbb{Q}}<j(\iota(n))$. This proves the clause naturals above.

Fractions. Let $x\in\mathbb{Z}$ and $m\in\mathbb{N}$, and put $c=j(\iota(m))$. By the clause naturals above, $0_{\mathbb{Q}}<c$, so $c\neq0_{\mathbb{Q}}$. As $\mathbb{Q}$ is a field, Negatives, Differences, Reciprocals and Quotients §reciprocal gives the reciprocal $c^{-1}\in\mathbb{Q}$ with $c\cdot c^{-1}=1_{\mathbb{Q}}$ and the quotient $j(x)/c=j(x)\cdot c^{-1}$. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §fraction, $[x,m]\cdot c=j(x)$. Using $u\cdot1_{\mathbb{Q}}=u$ and the associativity of $\cdot$ from Commutative Rings §ring,

$$[x,m]=[x,m]\cdot1_{\mathbb{Q}}=[x,m]\cdot(c\cdot c^{-1})=([x,m]\cdot c)\cdot c^{-1}=j(x)\cdot c^{-1}=j(x)/c=j(x)/j(\iota(m)).$$

This proves the clause fractions above.