Proof of The Rational Numbers Form an Archimedean Ordered Field Containing the Integers

Conventions. Within the proof of each clause, the letters $x,y,z,m,n,l,u,v,w$ are reused for arbitrary elements of the indicated sets, independently of the fixed $x,y,m,u$ of the statement except where the clause is applied to them. By The Rational Numbers §rationals, every rational number is $[x,m]$ for some $x\in\mathbb{Z}$ and $m\in\mathbb{N}$. By The Rational Numbers §operations and Construction of the Rationals: Pairs of an Integer and a Natural Number up to Equal Ratios, with Sum, Product, Negation and Order §operations, for $x,y\in\mathbb{Z}$ and $m,n\in\mathbb{N}$,

$$[x,m]+[y,n]=[x\iota(n)+y\iota(m),mn],\quad[x,m][y,n]=[xy,mn],\quad-[x,m]=[-x,m],\quad[x,m]\le[y,n]\iff x\iota(n)\le y\iota(m),\qquad(\mathrm{R})$$

and by Construction of the Rationals: Pairs of an Integer and a Natural Number up to Equal Ratios, with Sum, Product, Negation and Order §equal,

$$[x,m]=[y,n]\iff x\iota(n)=y\iota(m).\qquad(\mathrm{E})$$

Since $\mathbb{Z}$ is a commutative ring by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring, sums and products of integers are rearranged as in any commutative ring, with brackets in iterated products omitted; products of natural numbers are natural numbers by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §operations, and brackets in their iterated products are omitted by Arithmetic and Order of the Natural Numbers §associative. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ordered-ring, $\le$ is a total order on $\mathbb{Z}$ with strict relation $<$, so Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders applies to them.

Preliminaries on $\mathbb{Z}$. Let $z,z'\in\mathbb{Z}$ and $a,b,l\in\mathbb{N}$.

(Z1) $0_{\mathbb{Z}}<\iota(l)$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §positive; $\iota(1)=1_{\mathbb{Z}}$ and $\iota(ab)=\iota(a)\iota(b)$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding; hence $z\iota(1)=z$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring; and $0_{\mathbb{Z}}z=0_{\mathbb{Z}}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §negation.

(Z2) $z\le z'$ if and only if $z\iota(l)\le z'\iota(l)$, by (Z1) and The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §positive-factor.

Two consequences of (E). First, $[0_{\mathbb{Z}},m]=0_{\mathbb{Q}}$ for $m\in\mathbb{N}$: $0_{\mathbb{Q}}=[0_{\mathbb{Z}},1]$ by The Rational Numbers §constants, and $0_{\mathbb{Z}}\iota(1)=0_{\mathbb{Z}}=0_{\mathbb{Z}}\iota(m)$ by (Z1). Second, $[x,1]=[y,1]$ if and only if $x=y$, for $x,y\in\mathbb{Z}$, since $x\iota(1)=x$ and $y\iota(1)=y$ by (Z1).

The clause ordered-field above. $\mathbb{Q}$ is a set by The Rational Numbers §rationals; $+$ and $\cdot$ are binary operations on it by The Rational Numbers §operations and Construction of the Rationals: Pairs of an Integer and a Natural Number up to Equal Ratios, with Sum, Product, Negation and Order §operations; and $0_{\mathbb{Q}},1_{\mathbb{Q}}\in\mathbb{Q}$ by The Rational Numbers §constants and Construction of the Rationals: Pairs of an Integer and a Natural Number up to Equal Ratios, with Sum, Product, Negation and Order §equal, as $0_{\mathbb{Z}},1_{\mathbb{Z}}\in\mathbb{Z}$ (The Integers §constants) and $1\in\mathbb{N}$. Let $u,v,w\in\mathbb{Q}$, written $u=[x,m]$, $v=[y,n]$, $w=[z,l]$ with $x,y,z\in\mathbb{Z}$ and $m,n,l\in\mathbb{N}$. Sums and products below are computed by (R), and $\iota$ of a product is expanded by (Z1).

Commutative ring. $(u+v)+w=[(x\iota(n)+y\iota(m))\iota(l)+z\iota(mn),mnl]$ and $u+(v+w)=[x\iota(nl)+(y\iota(l)+z\iota(n))\iota(m),mnl]$; both first components equal $x\iota(n)\iota(l)+y\iota(m)\iota(l)+z\iota(m)\iota(n)$, so $(u+v)+w=u+(v+w)$. By Arithmetic and Order of the Natural Numbers §commutative, $u+v=[x\iota(n)+y\iota(m),mn]=[y\iota(m)+x\iota(n),nm]=v+u$ and $uv=[xy,mn]=[yx,nm]=vu$. By The Rational Numbers §constants, (Z1) and Arithmetic and Order of the Natural Numbers §one, $u+0_{\mathbb{Q}}=[x\iota(1)+0_{\mathbb{Z}}\iota(m),m\cdot1]=[x,m]=u$ and $u\cdot1_{\mathbb{Q}}=[x1_{\mathbb{Z}},m\cdot1]=u$. Next, $(uv)w=[xyz,mnl]=u(vw)$. For distributivity, $u(v+w)=[x(y\iota(l)+z\iota(n)),mnl]$ and $uv+uw=[xy,mn]+[xz,ml]=[xy\iota(ml)+xz\iota(mn),mnml]$; by (E) these are equal if and only if

$$x(y\iota(l)+z\iota(n))\,\iota(mnml)=(xy\iota(ml)+xz\iota(mn))\,\iota(mnl),$$

and both sides equal $xy\iota(m)\iota(m)\iota(n)\iota(l)\iota(l)+xz\iota(m)\iota(m)\iota(n)\iota(n)\iota(l)$. Finally, $-u=[-x,m]\in\mathbb{Q}$, and $u+(-u)=[x\iota(m)+(-x)\iota(m),mm]$; since $(-x)\iota(m)=-(x\iota(m))$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §negation and $x\iota(m)+(-(x\iota(m)))=0_{\mathbb{Z}}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring, $u+(-u)=[0_{\mathbb{Z}},mm]=0_{\mathbb{Q}}$ by the first consequence of (E). Hence $\mathbb{Q}$, with $+$, $\cdot$, $0_{\mathbb{Q}}$ and $1_{\mathbb{Q}}$, is a commutative ring by Commutative Rings §ring, and $u+(-u)=0_{\mathbb{Q}}$.

Field. $1_{\mathbb{Z}}=\iota(1)$ by (Z1), so $1_{\mathbb{Z}}\neq0_{\mathbb{Z}}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §trichotomy; hence $0_{\mathbb{Q}}=[0_{\mathbb{Z}},1]\neq[1_{\mathbb{Z}},1]=1_{\mathbb{Q}}$ by the second consequence of (E). Let $u\neq0_{\mathbb{Q}}$. Then $x\neq0_{\mathbb{Z}}$ by the first consequence of (E), so by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §trichotomy there is $k\in\mathbb{N}$ with $x=\iota(k)$ or $x=-\iota(k)$. In the first case put $v'=[\iota(m),k]$, so $uv'=[\iota(k)\iota(m),mk]$. In the second put $v'=[-\iota(m),k]$; by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §negation and The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring, $(-\iota(k))(-\iota(m))=-(\iota(k)(-\iota(m)))=-((-\iota(m))\iota(k))=-(-(\iota(m)\iota(k)))=\iota(k)\iota(m)$, so again $uv'=[\iota(k)\iota(m),mk]$. As $\iota(k)\iota(m)\iota(1)=\iota(mk)=1_{\mathbb{Z}}\iota(mk)$ by (Z1), (E) and The Rational Numbers §constants give $uv'=[1_{\mathbb{Z}},1]=1_{\mathbb{Q}}$. Hence $\mathbb{Q}$ is a field by Fields §field.

Total order. By The Rational Numbers §operations, $\le$ is a relation on $\mathbb{Q}$; the order of $\mathbb{Z}$ is reflexive, antisymmetric and transitive by Partial and Total Orders on a Set and the Associated Strict Relation §partial. Reflexivity: $x\iota(m)\le x\iota(m)$, so $u\le u$ by (R). Antisymmetry: if $u\le v$ and $v\le u$, then $x\iota(n)\le y\iota(m)$ and $y\iota(m)\le x\iota(n)$, so $x\iota(n)=y\iota(m)$, and $u=v$ by (E). Transitivity: if $u\le v$ and $v\le w$, then $x\iota(n)\le y\iota(m)$ and $y\iota(l)\le z\iota(n)$; by (Z2), with $\iota(l)$ and with $\iota(m)$, $x\iota(l)\iota(n)\le y\iota(m)\iota(l)$ and $y\iota(m)\iota(l)\le z\iota(m)\iota(n)$, so $x\iota(l)\iota(n)\le z\iota(m)\iota(n)$, hence $x\iota(l)\le z\iota(m)$ by (Z2) with $\iota(n)$, that is, $u\le w$. Totality: $x\iota(n)\le y\iota(m)$ or $y\iota(m)\le x\iota(n)$ by Partial and Total Orders on a Set and the Associated Strict Relation §total for $\mathbb{Z}$, that is, $u\le v$ or $v\le u$. By Reflexive, Symmetric, Antisymmetric and Transitive Relations on a Set §reflexive, Reflexive, Symmetric, Antisymmetric and Transitive Relations on a Set §antisymmetric, Reflexive, Symmetric, Antisymmetric and Transitive Relations on a Set §transitive, Partial and Total Orders on a Set and the Associated Strict Relation §partial and Partial and Total Orders on a Set and the Associated Strict Relation §total, $\le$ is a total order on $\mathbb{Q}$.

Strict relation. By Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization, $(u,v)$ lies in the strict relation of $\le$ on $\mathbb{Q}$ if and only if $u\le v$ and $u\neq v$, which is what $u<v$ means by The Rational Numbers §operations; so $<$ is that strict relation.

Ordered ring and ordered field. Suppose $u\le v$, that is, $x\iota(n)\le y\iota(m)$. By (Z2) with $\iota(ll)=\iota(l)\iota(l)$, $x\iota(n)\iota(l)\iota(l)\le y\iota(m)\iota(l)\iota(l)$; adding $c=z\iota(m)\iota(n)\iota(l)$ to both sides, by Ordered Rings §ordered-ring for $\mathbb{Z}$ (The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ordered-ring), gives

$$(x\iota(l)+z\iota(m))\,\iota(nl)\le(y\iota(l)+z\iota(n))\,\iota(ml).$$

As $u+w=[x\iota(l)+z\iota(m),ml]$ and $v+w=[y\iota(l)+z\iota(n),nl]$, this says $u+w\le v+w$. Next, by The Rational Numbers §constants, (R) and (Z1), $0_{\mathbb{Q}}\le[x,m]$ if and only if $0_{\mathbb{Z}}\iota(m)\le x\iota(1)$, that is, if and only if $0_{\mathbb{Z}}\le x$. So if $0_{\mathbb{Q}}\le u$ and $0_{\mathbb{Q}}\le v$, then $0_{\mathbb{Z}}\le x$ and $0_{\mathbb{Z}}\le y$, hence $0_{\mathbb{Z}}\le xy$ by Ordered Rings §ordered-ring for $\mathbb{Z}$, and $0_{\mathbb{Q}}\le[xy,mn]=uv$. Hence $\mathbb{Q}$, with $+$, $\cdot$, $0_{\mathbb{Q}}$, $1_{\mathbb{Q}}$ and $\le$, is an ordered ring, and, being a field with a total order, an ordered field by Ordered Fields §ordered-field.

The clause embedding above. By The Rational Numbers §embedding, $j:\mathbb{Z}\to\mathbb{Q}$ is a map with $j(z)=[z,1]$. If $j(z)=j(z')$, then $z=z'$ by the second consequence of (E); so $j$ is injective by Injective, Surjective and Bijective Functions between Classes §injective. By The Rational Numbers §constants, $j(0_{\mathbb{Z}})=0_{\mathbb{Q}}$ and $j(1_{\mathbb{Z}})=1_{\mathbb{Q}}$. By (R), (Z1) and $1\cdot1=1$ (Arithmetic and Order of the Natural Numbers §one),

$$j(x)+j(y)=[x\iota(1)+y\iota(1),1\cdot1]=[x+y,1]=j(x+y),\qquad j(x)j(y)=[xy,1\cdot1]=j(xy),$$

$-j(x)=[-x,1]=j(-x)$, and $j(x)\le j(y)$ if and only if $x\iota(1)\le y\iota(1)$, that is, if and only if $x\le y$.

The clause fraction above. By The Rational Numbers §embedding, (R) and $m\cdot1=m$ (Arithmetic and Order of the Natural Numbers §one), $[x,m]\,j(\iota(m))=[x,m][\iota(m),1]=[x\iota(m),m]$. Since $x\iota(m)\iota(1)=x\iota(m)$ by (Z1), (E) gives $[x\iota(m),m]=[x,1]=j(x)$.

The clause archimedean above. Write $u=[x,m]$ with $x\in\mathbb{Z}$ and $m\in\mathbb{N}$, and let $k\in\mathbb{N}$. Then $j(\iota(k))=[\iota(k),1]$ by The Rational Numbers §embedding, and as $x\iota(1)=x$ and $\iota(k)\iota(m)=\iota(km)$ by (Z1), (R) and (E) give: $u\le j(\iota(k))$ if and only if $x\le\iota(km)$, and $u=j(\iota(k))$ if and only if $x=\iota(km)$. So, by the meaning of $<$ on $\mathbb{Q}$ and on $\mathbb{Z}$ (The Rational Numbers §operations and The Integers §operations), $u<j(\iota(k))$ if and only if $x<\iota(km)$. By Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §trichotomy, $0_{\mathbb{Z}}<x$, $x=0_{\mathbb{Z}}$ or $x<0_{\mathbb{Z}}$.

If $x=0_{\mathbb{Z}}$ or $x<0_{\mathbb{Z}}$, take $k=1$. By (Z1), $0_{\mathbb{Z}}<\iota(m)$; so $x<\iota(m)$ directly if $x=0_{\mathbb{Z}}$, and by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive from $x<0_{\mathbb{Z}}<\iota(m)$ if $x<0_{\mathbb{Z}}$. As $\iota(m)=\iota(1\cdot m)$ by Arithmetic and Order of the Natural Numbers §one, $u<j(\iota(1))$.

If $0_{\mathbb{Z}}<x$, then $x=\iota(n')$ for some $n'\in\mathbb{N}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §positive; take $k=n'+1\in\mathbb{N}$ (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 Arithmetic and Order of the Natural Numbers §least, $1\le m$, so $1\cdot(n'+1)\le m(n'+1)$ by Arithmetic and Order of the Natural Numbers §order, that is, $n'+1\le(n'+1)m$ by Arithmetic and Order of the Natural Numbers §one and Arithmetic and Order of the Natural Numbers §commutative. Since $n'<n'+1$ by Arithmetic and Order of the Natural Numbers §successor, Arithmetic and Order of the Natural Numbers §partial-order gives $n'<(n'+1)m$. Hence $x=\iota(n')<\iota((n'+1)m)$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding, so $u<j(\iota(n'+1))$.

In either case there is $k\in\mathbb{N}$ with $u<j(\iota(k))$.