The Rational Numbers as an Ordered Field

Well-definedness. We must first check the operations do not depend on the choice of representative. Suppose $(a,b)\sim(a',b')$ and $(c,d)\sim(c',d')$, i.e. $ab'=a'b$ and $cd'=c'd$. For addition, we must show $(ad+bc,bd) \sim (a'd'+b'c', b'd')$, i.e.

Expand the left side: $adb'd' + bcb'd' = (ab')dd' + (cd')bb'$. Using $ab'=a'b$ and $cd'=c'd$, this equals $a'b\,dd' + c'd\,bb' = a'd'\,bd + b'c'\,bd$, which is the right side. A parallel (shorter) computation checks multiplication: $(ac)(b'd') = (ab')(cd') = (a'b)(c'd) = (a'c')(bd)$. So $+$ and $\cdot$ are well-defined on equivalence classes.

Field axioms. Additive identity is $\tfrac{0}{1}$; additive inverse of $\tfrac ab$ is $\tfrac{-a}{b}$, since $\tfrac{a}{b}+\tfrac{-a}{b} = \tfrac{ab-ab}{b^2}=\tfrac{0}{b^2}\sim\tfrac01$. Commutativity and associativity of $+$ and $\cdot$ follow by direct computation from the corresponding properties of $+,\cdot$ on $\mathbb{Z}$; for instance

Multiplicative identity is $\tfrac11$. If $\tfrac ab \neq \tfrac 01$ then $a\neq 0$, and $\tfrac ba$ is a valid representative (since $a \ne 0$) with $\tfrac ab \cdot \tfrac ba = \tfrac{ab}{ab} \sim \tfrac 11$, giving multiplicative inverses. Distributivity is a direct check:

(cancelling the common factor $b$ from numerator and denominator, valid since $b\neq0$ in $\mathbb{Z}$ and equivalence of fractions permits this cancellation). The two sides agree, so distributivity holds. Also $0\ne 1$ since $0\cdot 1 \ne 1\cdot 1$ in $\mathbb{Z}$ (as $0\ne 1$ there). This verifies all field axioms.

Order axioms. First check well-definedness of $<$: represent every class with positive denominator (possible since if $b<0$, replace $(a,b)$ by $(-a,-b)$). If $\tfrac ab \sim \tfrac{a'}{b'}$ with $b,b'>0$ then $ab'=a'b$; one checks the truth value of $ad<bc$ (for a fixed $\tfrac cd$, $d>0$) is unchanged when $(a,b)$ is replaced by $(a',b')$, using that multiplying both sides of an inequality in $\mathbb{Z}$ by a positive integer preserves it — this is exactly (O4) for the ordered ring $\mathbb{Z}$, which we take as previously established. (O1) Trichotomy. For $\tfrac ab,\tfrac cd$ with $b,d>0$, exactly one of $ad<bc$, $ad=bc$, $bc<ad$ holds by trichotomy in $\mathbb{Z}$, and these correspond respectively to $\tfrac ab<\tfrac cd$, $\tfrac ab=\tfrac cd$, $\tfrac cd<\tfrac ab$. (O2) Transitivity. Suppose $\tfrac ab<\tfrac cd<\tfrac ef$ with $b,d,f>0$. Then $ad<bc$ and $cf<de$. Multiply the first inequality by $f>0$ and the second by $b>0$ (valid by (O4) in $\mathbb Z$): $adf < bcf$ and $bcf < bde$. By transitivity in $\mathbb Z$, $adf<bde$, and cancelling the common positive factor $d$ (dividing an inequality by a positive integer preserves it) gives $af<be$, i.e. $\tfrac ab<\tfrac ef$. (O3) Compatibility with addition. Suppose $\tfrac ab<\tfrac cd$, $b,d>0$, so $ad<bc$. For any $\tfrac ef$ ($f>0$),

We must show $(af+be)(df) < (bf)(cf+de)$, i.e. $adf^2+bdef < bcf^2+bdef$, i.e. $adf^2<bcf^2$. This follows from $ad<bc$ by multiplying both sides by $f^2>0$. (O4) Compatibility with multiplication. Suppose $\tfrac ab<\tfrac cd$ ($ad<bc$, $b,d>0$) and $0<\tfrac ef$ (so $e>0$ since $f>0$ and $0\cdot f<e\cdot 1$ reduces to $0<e$). Then

We must show $ae\cdot df < bf\cdot ce$, i.e. $(ad)(ef) < (bc)(ef)$. Since $ad<bc$ and $ef = e\cdot f>0$ (product of two positive integers), multiplying the inequality $ad<bc$ by the positive integer $ef$ preserves it, which is exactly this claim. All axioms hold, so $\mathbb Q$ is an ordered field. $\blacksquare$

(a) By (O3), adding $-x-y$ to both sides of $x<y$ gives $x+(-x-y) < y+(-x-y)$, i.e. $-y<-x$. The converse follows the same way applied to $-y<-x$. (b) Since $z<0$, (a) gives $0<-z$. By (O4) applied to $x<y$ and $0<-z$: $x(-z)<y(-z)$, i.e. $-xz<-yz$. Applying (a) again gives $yz<xz$. (c) Suppose $0<x$ but $x^{-1}\le 0$. If $x^{-1}=0$ then $1=x\cdot x^{-1}=0$, contradicting $1\ne0$ (field axiom). So $x^{-1}<0$. By (b) (multiplying $0<x$ by the negative number $x^{-1}$, reversing direction): $x\cdot x^{-1} < 0 \cdot x^{-1}$, i.e. $1<0$. But from (d) below, $1=1^2>0$, contradicting trichotomy (O1). Hence $0<x^{-1}$. Now suppose $0<x<y$. Both $x^{-1},y^{-1}>0$ by the above. Multiply $x<y$ by $x^{-1}y^{-1}>0$ (a product of positives is positive, shown in (d)'s proof pattern below applied to $x^{-1}$ and $y^{-1}$, or directly: apply (O4) twice), using (O4): $x\cdot(x^{-1}y^{-1}) < y\cdot (x^{-1}y^{-1})$, i.e. $y^{-1} < x^{-1}$. (d) If $x>0$: apply (O4) to $0<x$ and $0<x$, giving $0\cdot x < x\cdot x$, i.e. $0<x^2$. If $x<0$: by (a), $0<-x$, so by the previous case $0<(-x)^2 = x^2$ (since $(-x)(-x)=x\cdot x$ by standard field algebra: $(-x)(-x) = -(x\cdot(-x)) = -(-(x\cdot x)) = x^2$). Either way $x^2>0$. (e) From $x<y$, (O3) gives $x+z<y+z$. From $z<w$, (O3) gives $y+z<y+w$. By transitivity (O2), $x+z<y+w$. $\blacksquare$

Define $\varphi(n) = \underbrace{1_F+\cdots+1_F}_{n}$ for positive integers $n$, $\varphi(0)=0_F$, $\varphi(-n)=-\varphi(n)$; this is the unique ring homomorphism $\mathbb Z\to F$ sending $1\mapsto1_F$ (forced by induction using the recursive definition of integer multiples, since any homomorphism must satisfy $\varphi(n+1)=\varphi(n)+1_F$). By Lemma (d) and induction, $\varphi(n) > 0_F$ for every positive integer $n$ (a finite sum of positive elements is positive, by repeated use of (O3)), so $\varphi$ is injective on $\mathbb{Z}$: if $m\ne n$ then WLOG $m<n$, so $n-m$ is a positive integer, and $\varphi(n)-\varphi(m) = \varphi(n-m) > 0_F$, giving $\varphi(m)\ne\varphi(n)$. Extend to $\mathbb Q$ by $\varphi(a/b) := \varphi(a)\varphi(b)^{-1}$, which is well-defined (if $ab'=a'b$ then applying $\varphi$, using that it is a ring homomorphism on $\mathbb Z$, gives $\varphi(a)\varphi(b')=\varphi(a')\varphi(b)$, hence $\varphi(a)\varphi(b)^{-1}=\varphi(a')\varphi(b')^{-1}$ after rearranging, valid since $\varphi(b),\varphi(b')\ne0$). One checks directly from the definitions of $+,\cdot$ on $\mathbb Q$ that $\varphi$ preserves both operations, using that it does so on $\mathbb{Z}$. For order: if $\tfrac ab<\tfrac cd$ with $b,d>0$ then $ad<bc$ in $\mathbb Z$, so $\varphi(ad)<\varphi(bc)$ (order-preservation on $\mathbb Z$, from injectivity plus the fact $\varphi$ sends positive integers to positive elements and is additive), i.e. $\varphi(a)\varphi(d)<\varphi(b)\varphi(c)$. Since $\varphi(b),\varphi(d)>0$, dividing (Lemma (c) type manipulation, multiplying by $\varphi(b)^{-1}\varphi(d)^{-1}>0$) gives $\varphi(a)\varphi(b)^{-1} < \varphi(c)\varphi(d)^{-1}$, i.e. $\varphi(a/b)<\varphi(c/d)$. This also shows $\varphi$ is injective on all of $\mathbb Q$ by trichotomy. Uniqueness follows because any field homomorphism fixing $1$ is forced on all of $\mathbb Z$ by induction and forced on $\mathbb Q$ by $\varphi(a/b)\varphi(b)=\varphi(a)$. $\blacksquare$

$S$ is nonempty and bounded above. $1\in S$ since $1^2=1<2$. And $2$ is an upper bound: if $q>2$ then $q^2>4>2$ (using Lemma (d)-style reasoning: $q>2>0 \Rightarrow q\cdot q > 2\cdot q$ by (O4), and $2\cdot q > 2\cdot 2=4$ similarly, so $q^2>4$), so any $q\in S$ satisfies $q\le 2$.

No rational is the least upper bound. Suppose for contradiction $p\in\mathbb Q$ is a least upper bound of $S$. By Example 2 / the classical argument, $p^2\ne 2$, so by trichotomy either $p^2<2$ or $p^2>2$. Case $p^2<2$. We construct a rational $p+h$, with $h>0$ rational and small, such that $(p+h)^2<2$; this shows $p+h\in S$ and $p+h>p$, contradicting that $p$ is an upper bound of $S$. Since $p\in S$ we may assume $0<p<2$ (as shown above every element of $S$ is $\le 2$, and if $p\le 0$ it cannot be an upper bound of $S\ni 1>0$, so $p>0$). Let

a positive rational since $2-p^2>0$ and $2p+1>0$. Then $0<h\le 1$ and $h\le \tfrac{2-p^2}{2p+1}$, so $h(2p+1) \le 2-p^2$. Compute

using $h\le 1$ so $h^2\le h$, hence $h(2p+h)\le h(2p+1)$ (as $2p+h \le 2p+1$, and $h>0$, so multiplying preserves the inequality by (O4)). Continuing,

We need strict inequality: since $h\le \tfrac{2-p^2}{2p+1}$ was not required to be equality, choose instead $h' = \tfrac12 h>0$; repeating the estimate with $h'$ in place of $h$ (still $h'\le 1$, $h'\le \tfrac{2-p^2}{2p+1}$ since $h'<h$) gives $(p+h')^2 \le p^2 + h'(2p+1) \le p^2 + \tfrac12(2-p^2) < p^2+(2-p^2)=2$ (strict since $\tfrac12(2-p^2) < 2-p^2$ as $2-p^2>0$). So $(p+h')^2<2$, $p+h'>0$, hence $p+h'\in S$ with $p+h'>p$: contradiction. Case $p^2>2$. We construct $0<h<p$ rational with $(p-h)^2>2$ and $p-h$ still an upper bound for $S$, contradicting minimality of $p$ as least upper bound. Let

(positive since $p>0$ and $p^2-2>0$). Then $0<h<p$ (as $h\le p/2 <p$), and $h\le \tfrac{p^2-2}{2p}$ gives $2ph \le p^2-2$. Compute

using $h^2>0$ (Lemma (d), since $h\ne0$) for the strict step. So $(p-h)^2>2$ and $p-h>0$. Now if any $q\in S$ satisfied $q > p-h$, then since also $q\le p$ (as $p$ is an upper bound and $q\in\mathbb{Q}$, $q>0$) — actually we argue directly: for $q\in S$, $q^2<2<(p-h)^2$, and since $q,p-h>0$, this forces $q<p-h$ (if instead $q\ge p-h>0$ then squaring, licensed by (O4) applied twice, would give $q^2 \ge (p-h)^2$, contradiction). So $p-h$ is an upper bound for $S$ strictly smaller than $p$: contradicting that $p$ is the least upper bound. Both cases are impossible, so no rational least upper bound for $S$ exists. $\blacksquare$