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

Conventions. By Ordered Fields §ordered-field and Fields §field, $r$ is a set and a field, so $0_{r}\neq1_{r}$, and $r$ is a commutative ring. Computations in $r$ use the laws of Commutative Rings §ring, $w+(-w)=0_{r}$ from Negatives, Differences, Reciprocals and Quotients §negative, $w\cdot w^{-1}=1_{r}$ for $w\neq0_{r}$ from Negatives, Differences, Reciprocals and Quotients §reciprocal, and the rules of Rules of Arithmetic and Order in an Ordered Field cited below. These rules also hold in $\mathbb{Q}$, which is an ordered field by The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field. Classes of natural numbers used for induction are formed by class abstraction, as Sets and Maps: Ordinary Notation §set-builder prescribes, and induction is Arithmetic and Order of the Natural Numbers §induction. We use Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive in $r$ (a total, hence partial, order), which we call (T).

Step 1: the natural numbers. Since $w+1_{r}\in r$ for $w\in r$, Maps and Relations Given by Formulas §binary gives a map $g:\mathbb{N}\times r\to r$ with $g((n,w))=w+1_{r}$. By Recursion on the Natural Numbers Starting at One §recursion, with $a=r$ and $c=1_{r}$, there is a map $\psi:\mathbb{N}\to r$ with $\psi(1)=1_{r}$ and $\psi(n+1)=\psi(n)+1_{r}$ for every $n\in\mathbb{N}$. Let $m,n\in\mathbb{N}$.

(1a) $\psi(m+n)=\psi(m)+\psi(n)$. Fix $m$ and induct on $n$. For $n=1$ this is the recursion. If it holds for $n$, then by Arithmetic and Order of the Natural Numbers §associative, $\psi(m+(n+1))=\psi((m+n)+1)=\psi(m+n)+1_{r}=\psi(m)+(\psi(n)+1_{r})=\psi(m)+\psi(n+1).$

(1b) $\psi(mn)=\psi(m)\psi(n)$. Fix $m$ and induct on $n$. For $n=1$, $\psi(m\cdot1)=\psi(m)=\psi(m)\cdot1_{r}$ by Arithmetic and Order of the Natural Numbers §one. If it holds for $n$, then by Arithmetic and Order of the Natural Numbers §distributive, Arithmetic and Order of the Natural Numbers §one and (1a), $\psi(m(n+1))=\psi(mn+m)=\psi(m)\psi(n)+\psi(m)\cdot1_{r}=\psi(m)(\psi(n)+1_{r})=\psi(m)\psi(n+1).$

(1c) $0_{r}<\psi(n)$. In particular $\psi(n)\neq0_{r}$, so $\psi(n)^{-1}\in r$ exists. For $n=1$, $0_{r}<1_{r}$ by Rules of Arithmetic and Order in an Ordered Field §squares. If $0_{r}<\psi(n)$, then $1_{r}=0_{r}+1_{r}<\psi(n)+1_{r}=\psi(n+1)$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, and $0_{r}<1_{r}<\psi(n+1)$ gives $0_{r}<\psi(n+1)$ by (T).

Step 2: the integers. By The Integers §integers, $\mathbb{Z}=P/{\sim}$ with $P=\mathbb{N}\times\mathbb{N}$, where $\sim$ is the equivalence relation of Construction of the Integers: Pairs of Natural Numbers up to Equal Differences, with Sum, Product, Negation and Order §equivalence and $[a,b]$ is the class of $(a,b)$. By Equivalence Classes Partition the Set: Cover, Disjointness and Representatives; the Quotient Is a Set and the Canonical Projection Is a Surjection §projection, $[a,b]=\pi_{\sim}((a,b))$. By Maps and Relations Given by Formulas §binary there is a map $F:P\to r$ with $F((a,b))=\psi(a)-\psi(b)$. Suppose $(a,b)\sim(c,d)$, that is, $a+d=b+c$. Then $\psi(a)+\psi(d)=\psi(b)+\psi(c)$ by (1a). Adding $(-\psi(b))+(-\psi(d))$ to both sides gives $\psi(a)-\psi(b)=\psi(c)-\psi(d)$. By A Map Constant on Equivalence Classes Factors Uniquely through the Quotient §factorization there is a map $\chi:\mathbb{Z}\to r$ with $\chi\circ\pi_{\sim}=F$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §composition this gives:

(2a) $\chi([a,b])=\psi(a)-\psi(b)$ for all $a,b\in\mathbb{N}$.

(2b) $\chi(x+y)=\chi(x)+\chi(y)$ and $\chi(xy)=\chi(x)\chi(y)$ for all $x,y\in\mathbb{Z}$. By Construction of the Integers: Pairs of Natural Numbers up to Equal Differences, with Sum, Product, Negation and Order §equal, we may write $x=[a,b]$ and $y=[c,d]$ with $a,b,c,d\in\mathbb{N}$. Then Construction of the Integers: Pairs of Natural Numbers up to Equal Differences, with Sum, Product, Negation and Order §operations gives $x+y=[a+c,b+d]$ and $xy=[ac+bd,ad+bc]$. Hence, by (2a), (1a), (1b) and Rules of Arithmetic and Order in an Ordered Field §signs, $\chi(x+y)=(\psi(a)+\psi(c))-(\psi(b)+\psi(d))=(\psi(a)-\psi(b))+(\psi(c)-\psi(d))=\chi(x)+\chi(y),$ $\chi(xy)=(\psi(a)\psi(c)+\psi(b)\psi(d))-(\psi(a)\psi(d)+\psi(b)\psi(c))=(\psi(a)-\psi(b))(\psi(c)-\psi(d))=\chi(x)\chi(y).$ The middle equality of the second line comes from expanding the right side by distributivity, using $(-w)z=-(wz)$ and $(-w)(-z)=wz$ from the same rule.

(2c) $\chi(\iota(n))=\psi(n)$ for $n\in\mathbb{N}$. By The Integers §embedding, $\iota(n)=[n+1,1]$, so $\chi(\iota(n))=\psi(n+1)-\psi(1)=(\psi(n)+1_{r})-1_{r}=\psi(n)$. Since $\iota(1)=1_{\mathbb{Z}}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding, in particular $\chi(1_{\mathbb{Z}})=1_{r}$.

Step 3: existence. By The Rational Numbers §rationals, $\mathbb{Q}=Q/{\approx}$ with $Q=\mathbb{Z}\times\mathbb{N}$, where $\approx$ is the equivalence relation of Construction of the Rationals: Pairs of an Integer and a Natural Number up to Equal Ratios, with Sum, Product, Negation and Order §equivalence. By Equivalence Classes Partition the Set: Cover, Disjointness and Representatives; the Quotient Is a Set and the Canonical Projection Is a Surjection §projection, $[x,m]=\pi_{\approx}((x,m))$. By (1c), $\chi(x)\psi(m)^{-1}\in r$, so Maps and Relations Given by Formulas §binary gives a map $F':Q\to r$ with $F'((x,m))=\chi(x)\psi(m)^{-1}$. Suppose $(x,m)\approx(y,n)$, that is, $x\,\iota(n)=y\,\iota(m)$. Then (2b) and (2c) give $\chi(x)\psi(n)=\chi(x\,\iota(n))=\chi(y\,\iota(m))=\chi(y)\psi(m)$. Multiplying by $\psi(m)^{-1}\psi(n)^{-1}$ gives $\chi(x)\psi(m)^{-1}=\chi(y)\psi(n)^{-1}$. By A Map Constant on Equivalence Classes Factors Uniquely through the Quotient §factorization and Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §composition there is a map $\varphi:\mathbb{Q}\to r$ with:

(3a) $\varphi([x,m])=\chi(x)\psi(m)^{-1}$ for all $x\in\mathbb{Z}$ and $m\in\mathbb{N}$.

By Construction of the Rationals: Pairs of an Integer and a Natural Number up to Equal Ratios, with Sum, Product, Negation and Order §equal, every $u\in\mathbb{Q}$ is some $[x,m]$. Let $u=[x,m]$ and $v=[y,n]$. By Construction of the Rationals: Pairs of an Integer and a Natural Number up to Equal Ratios, with Sum, Product, Negation and Order §operations, $u+v=[x\,\iota(n)+y\,\iota(m),mn]$ and $uv=[xy,mn]$. By (1b) and Rules of Arithmetic and Order in an Ordered Field §reciprocals, $\psi(mn)^{-1}=\psi(m)^{-1}\psi(n)^{-1}$. Hence, by (3a), (2b) and (2c), $\varphi(u+v)=(\chi(x)\psi(n)+\chi(y)\psi(m))\,\psi(m)^{-1}\psi(n)^{-1}=\chi(x)\psi(m)^{-1}+\chi(y)\psi(n)^{-1}=\varphi(u)+\varphi(v),$ $\varphi(uv)=\chi(x)\chi(y)\,\psi(m)^{-1}\psi(n)^{-1}=\varphi(u)\varphi(v).$ By The Rational Numbers §constants, $1=1_{\mathbb{Q}}=[1_{\mathbb{Z}},1]$, so by (2c), $\varphi(1)=\chi(1_{\mathbb{Z}})\psi(1)^{-1}=1_{r}\cdot1_{r}^{-1}=1_{r}$. Thus $\varphi$ has the three properties in The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique. That it is the only such map is shown in Step 6.

Step 4: consequences of the three properties. Let $\theta:\mathbb{Q}\to r$ be any map with $\theta(1)=1_{r}$, $\theta(u+v)=\theta(u)+\theta(v)$ and $\theta(uv)=\theta(u)\theta(v)$ for all $u,v\in\mathbb{Q}$. Let $u,v\in\mathbb{Q}$.

(4a) $\theta(0)=0_{r}$. Since $0+0=0$, we have $\theta(0)=\theta(0)+\theta(0)$. Adding $-\theta(0)$ to both sides gives $0_{r}=\theta(0)$.

(4b) $\theta(-u)=-\theta(u)$ and $\theta(u-v)=\theta(u)-\theta(v)$. Since $u+(-u)=0$ by Negatives, Differences, Reciprocals and Quotients §negative, (4a) gives $\theta(u)+\theta(-u)=\theta(0)=0_{r}$. So $\theta(-u)=-\theta(u)$, by the uniqueness in Additive and Multiplicative Inverses Are Unique §negative. Then $\theta(u-v)=\theta(u+(-v))=\theta(u)+(-\theta(v))=\theta(u)-\theta(v)$.

(4c) If $u\neq0$, then $\theta(u)\neq0_{r}$, $\theta(u^{-1})=\theta(u)^{-1}$ and $\theta(v/u)=\theta(v)/\theta(u)$. Since $u\cdot u^{-1}=1$ by Negatives, Differences, Reciprocals and Quotients §reciprocal, $\theta(u)\theta(u^{-1})=\theta(1)=1_{r}$. If $\theta(u)=0_{r}$, this would give $1_{r}=0_{r}\cdot\theta(u^{-1})=0_{r}$ by Rules of Arithmetic and Order in an Ordered Field §zero, which is impossible. So $\theta(u)\neq0_{r}$, and $\theta(u^{-1})=\theta(u)^{-1}$ by the uniqueness in Additive and Multiplicative Inverses Are Unique §reciprocal. Hence $\theta(v/u)=\theta(v\cdot u^{-1})=\theta(v)\theta(u)^{-1}=\theta(v)/\theta(u)$.

Taking $\theta=\varphi$, (4a), (4b) and (4c) prove The Rational Numbers Embed in Exactly One Way into Every Ordered Field §zero, The Rational Numbers Embed in Exactly One Way into Every Ordered Field §negative and The Rational Numbers Embed in Exactly One Way into Every Ordered Field §reciprocal.

Step 5: order. Let $u,v\in\mathbb{Q}$. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field, $\le$ is total on $\mathbb{Q}$ and $<$ is its strict relation. So exactly one of $u<v$, $u=v$ and $v<u$ holds, by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §trichotomy.

(5a) If $0<u$, then $0_{r}<\varphi(u)$. Write $u=[x,m]$. By The Rational Numbers §constants, $0=[0_{\mathbb{Z}},1]$. By Construction of the Rationals: Pairs of an Integer and a Natural Number up to Equal Ratios, with Sum, Product, Negation and Order §operations, $0\le u$ holds if and only if $0_{\mathbb{Z}}\,\iota(m)\le x\,\iota(1)$, that is, if and only if $0_{\mathbb{Z}}\le x$. Here $0_{\mathbb{Z}}\,\iota(m)=0_{\mathbb{Z}}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §negation, and $x\,\iota(1)=x\cdot1_{\mathbb{Z}}=x$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding and The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring. Likewise, Construction of the Rationals: Pairs of an Integer and a Natural Number up to Equal Ratios, with Sum, Product, Negation and Order §equal shows that $u=0$ if and only if $x=0_{\mathbb{Z}}$. Hence $0<u$ gives $0_{\mathbb{Z}}\le x$ and $x\neq0_{\mathbb{Z}}$, that is, $0_{\mathbb{Z}}<x$ by The Integers §operations. So $x=\iota(n)$ for some $n\in\mathbb{N}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §positive, and $\varphi(u)=\psi(n)\psi(m)^{-1}$ by (3a) and (2c). By (1c) and Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal, $0_{r}<\psi(m)^{-1}$. Applying Rules of Arithmetic and Order in an Ordered Field §order-product with $z=\psi(m)^{-1}$ to $0_{r}<\psi(n)$, and using Rules of Arithmetic and Order in an Ordered Field §zero, gives $0_{r}=0_{r}\cdot z<\psi(n)z=\varphi(u)$.

(5b) Proof of The Rational Numbers Embed in Exactly One Way into Every Ordered Field §order. Let $u<v$. By Rules of Arithmetic and Order in an Ordered Field §order-sum in $\mathbb{Q}$, $0=u+(-u)<v+(-u)=v-u$. Then by (5a) and (4b), $0_{r}<\varphi(v-u)=\varphi(v)-\varphi(u)$. By Rules of Arithmetic and Order in an Ordered Field §order-sum in $r$, $\varphi(u)=0_{r}+\varphi(u)<(\varphi(v)-\varphi(u))+\varphi(u)=\varphi(v)$. Conversely, let $\varphi(u)<\varphi(v)$. If $u=v$, then $\varphi(u)=\varphi(v)$. If $v<u$, then $\varphi(v)<\varphi(u)$ by the first part, so $\varphi(u)=\varphi(v)$ by antisymmetry. Both contradict $\varphi(u)<\varphi(v)$, so $u<v$.

(5c) Proof of The Rational Numbers Embed in Exactly One Way into Every Ordered Field §homomorphism. Let $u\le v$. Then $u<v$ or $u=v$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict, so (5b) gives $\varphi(u)<\varphi(v)$ or $\varphi(u)=\varphi(v)$, and in either case $\varphi(u)\le\varphi(v)$. Since $\mathbb{Q}$ is an ordered field by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §rationals, Step 3 and this monotonicity show that $\varphi$ is a homomorphism of ordered fields from $\mathbb{Q}$ to $r$, as defined in Homomorphisms and Isomorphisms of Ordered Fields §homomorphism.

(5d) Proof of The Rational Numbers Embed in Exactly One Way into Every Ordered Field §injective. Let $u\neq v$. Then $u<v$ or $v<u$, so by (5b), $\varphi(u)<\varphi(v)$ or $\varphi(v)<\varphi(u)$. Either way $\varphi(u)\neq\varphi(v)$, as Injective, Surjective and Bijective Functions between Classes §injective requires.

(5e) Proof of The Rational Numbers Embed in Exactly One Way into Every Ordered Field §absolute, using Absolute Value in an Ordered Field §absolute-value in $\mathbb{Q}$ and in $r$. If $0\le u$, then $|u|=u$, and $0_{r}=\varphi(0)\le\varphi(u)$ by (4a) and (5c). So $|\varphi(u)|=\varphi(u)=\varphi(|u|)$. Otherwise, $u<0$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, and $|u|=-u$. By (5b) and (4a), $\varphi(u)<\varphi(0)=0_{r}$, so $0_{r}\le\varphi(u)$ fails, since it would give $\varphi(u)=0_{r}$ by antisymmetry. Hence $|\varphi(u)|=-\varphi(u)=\varphi(-u)=\varphi(|u|)$ by (4b).

Step 6: uniqueness. Let $\theta:\mathbb{Q}\to r$ be a map with the three properties of The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique. Step 4 applies to $\theta$.

(6a) $\theta(j(\iota(n)))=\psi(n)$ for all $n\in\mathbb{N}$, by induction on $n$. By The Natural Numbers and the Integers inside the Rational Numbers §naturals, $j(\iota(1))=1_{\mathbb{Q}}$, so $\theta(j(\iota(1)))=1_{r}=\psi(1)$. Again by The Natural Numbers and the Integers inside the Rational Numbers §naturals, $j(\iota(n+1))=j(\iota(n))+j(\iota(1))=j(\iota(n))+1_{\mathbb{Q}}$. So if the claim holds for $n$, then $\theta(j(\iota(n+1)))=\psi(n)+1_{r}=\psi(n+1)$.

(6b) $\theta(j(x))=\chi(x)$ for all $x\in\mathbb{Z}$. Write $x=[a,b]$. Then $x=\iota(a)-\iota(b)=\iota(a)+(-\iota(b))$, by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §difference and The Integers §operations. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §embedding, $j(x)=j(\iota(a))+(-j(\iota(b)))$. Here the negation of $\mathbb{Q}$ is the negative, by The Natural Numbers and the Integers inside the Rational Numbers §negation. Hence by (6a), (4b) and (2a), $\theta(j(x))=\psi(a)-\psi(b)=\chi(x)$.

(6c) $\theta=\varphi$. Let $u\in\mathbb{Q}$ and write $u=[x,m]$. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §fraction, $u\cdot j(\iota(m))=j(x)$, so (6a) and (6b) give $\theta(u)\psi(m)=\chi(x)$. Multiplying by $\psi(m)^{-1}$ and using (3a) gives $\theta(u)=\chi(x)\psi(m)^{-1}=\varphi(u)$. Both maps have domain $\mathbb{Q}$, so $\theta=\varphi$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §equality. Together with Step 3, this proves The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique.