Proof of Rules of Arithmetic and Order in an Ordered Field

Since $x,y,z,w$ are arbitrary elements of $r$, each clause above, once proved, applies to any elements of $r$.

Preliminaries. By Ordered Fields §ordered-field, $r$ is a field and an ordered ring. By Fields §field, $r$ is a commutative ring with $0\neq1$; rearrangements below by associativity, commutativity, neutral elements and distributivity are by Commutative Rings §ring. By Ordered Rings §ordered-ring, for all $u,v,s\in r$,

$$\text{if }u\le v\text{ then }u+s\le v+s,\qquad(\mathrm{OA})$$ $$\text{if }0\le u\text{ and }0\le v\text{ then }0\le u\cdot v.\qquad(\mathrm{OM})$$

By Negatives, Differences, Reciprocals and Quotients §negative and Negatives, Differences, Reciprocals and Quotients §reciprocal, $u+(-u)=0$ for every $u\in r$ and $u\cdot u^{-1}=1$ for every $u\neq0$; and by Additive and Multiplicative Inverses Are Unique §negative and Additive and Multiplicative Inverses Are Unique §reciprocal, for all $u,v\in r$,

$$\text{if }u+v=0\text{ then }v=-u,\quad\text{and if }u\neq0\text{ and }u\cdot v=1\text{ then }v=u^{-1}.\qquad(\mathrm{U})$$

The total order $\le$ is a partial order on $r$, so by Partial and Total Orders on a Set and the Associated Strict Relation §partial it is reflexive, antisymmetric and transitive, and by Partial and Total Orders on a Set and the Associated Strict Relation §total any $u,v\in r$ satisfy $u\le v$ or $v\le u$ (totality). As $<$ is the strict relation of $\le$, Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization and Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation give, for all $u,v\in r$,

$$u<v\text{ if and only if }u\le v\text{ and }u\neq v,\qquad(\mathrm{S})$$ $$u\le v\text{ fails if and only if }v<u.\qquad(\mathrm{T})$$

Zero. By Commutative Rings §ring, $0\cdot x=(0+0)\cdot x=0\cdot x+0\cdot x$. Adding $-(0\cdot x)$ to both sides and rearranging,

$$0=0\cdot x+(-(0\cdot x))=(0\cdot x+0\cdot x)+(-(0\cdot x))=0\cdot x+(0\cdot x+(-(0\cdot x)))=0\cdot x+0=0\cdot x.$$

Now let $x\cdot y=0$ with $x\neq0$. By Commutative Rings §ring and the first part,

$$y=y\cdot1=y\cdot(x\cdot x^{-1})=(x\cdot y)\cdot x^{-1}=0\cdot x^{-1}=0.$$

So $x=0$ or $y=0$.

Signs. Since $(-x)+x=x+(-x)=0$, (U) applied to $-x$ gives $x=-(-x)$. By Commutative Rings §ring,

$$(x+y)+((-x)+(-y))=(x+(-x))+(y+(-y))=0+0=0,$$

so $-(x+y)=(-x)+(-y)$ by (U). By distributivity and the clause zero above, $x\cdot y+(-x)\cdot y=(x+(-x))\cdot y=0\cdot y=0$, so $(-x)\cdot y=-(x\cdot y)$ by (U). By this and commutativity, $x\cdot(-y)=(-y)\cdot x=-(y\cdot x)=-(x\cdot y)$; hence, applying it once more and then the first part,

$$(-x)\cdot(-y)=-(x\cdot(-y))=-(-(x\cdot y))=x\cdot y.$$

Reciprocals. Let $x\neq0$ and $y\neq0$. If $x\cdot y=0$, the clause zero above would give $x=0$ or $y=0$; so $x\cdot y\neq0$, and $(x\cdot y)^{-1}$ is defined. By Commutative Rings §ring,

$$(x\cdot y)\cdot(x^{-1}\cdot y^{-1})=(x\cdot x^{-1})\cdot(y\cdot y^{-1})=1\cdot1=1,$$

so $(x\cdot y)^{-1}=x^{-1}\cdot y^{-1}$ by (U). If $x^{-1}=0$, then $1=x\cdot x^{-1}=x\cdot0=0$ by the clause zero above, contrary to $0\neq1$; so $x^{-1}\neq0$, and since $x^{-1}\cdot x=x\cdot x^{-1}=1$, (U) gives $x=(x^{-1})^{-1}$.

Order-sum. Let $x\le y$ and $z\le w$. By (OA), $x+z\le y+z$ and $z+y\le w+y$, that is, $y+z\le y+w$ by commutativity; by transitivity, $x+z\le y+w$.

Now let $x<y$. By (S), $x\le y$ and $x\neq y$, so $x+z\le y+z$ by (OA). If $x+z=y+z$, then by Commutative Rings §ring

$$x=x+(z+(-z))=(x+z)+(-z)=(y+z)+(-z)=y+(z+(-z))=y,$$

a contradiction; so $x+z\neq y+z$, and $x+z<y+z$ by (S). Conversely, if $x+z<y+z$, then what was just shown, with $-z$ added, gives $(x+z)+(-z)<(y+z)+(-z)$, which is $x<y$ by the same rearrangement.

Finally, recall that $y-x=y+(-x)$ by Negatives, Differences, Reciprocals and Quotients §negative. If $x\le y$, then by (OA), adding $-x$, $0=x+(-x)\le y+(-x)=y-x$. Conversely, if $0\le y-x$, then by (OA), adding $x$, and by Commutative Rings §ring,

$$x=0+x\le(y+(-x))+x=y+((-x)+x)=y+0=y.$$

Order-negative. Let $x\le y$. By (OA), adding $(-x)+(-y)$, and by Commutative Rings §ring,

$$-y=(x+(-x))+(-y)=x+((-x)+(-y))\le y+((-x)+(-y))=(y+(-y))+(-x)=-x.$$

Conversely, if $-y\le-x$, the forward direction gives $-(-x)\le-(-y)$, that is, $x\le y$ by the clause signs above. Since $0+0=0$, (U) gives $-0=0$; so this clause, with $0$ on one side, yields for every $u\in r$

$$u\le0\text{ if and only if }0\le-u,\quad\text{and}\quad0\le u\text{ if and only if }-u\le0.\qquad(\mathrm{N})$$

Order-product. Let $0<z$; by (S), $0\le z$ and $z\neq0$. First, if $x\cdot z=y\cdot z$, then by Commutative Rings §ring

$$x=x\cdot(z\cdot z^{-1})=(x\cdot z)\cdot z^{-1}=(y\cdot z)\cdot z^{-1}=y\cdot(z\cdot z^{-1})=y.\qquad(\ast)$$

Let $x\le y$. By (OA), $0=x+(-x)\le y+(-x)=y-x$, so $0\le(y-x)\cdot z$ by (OM). By (OA), adding $x\cdot z$, and by Commutative Rings §ring and the clause signs above,

$$x\cdot z=0+x\cdot z\le(y+(-x))\cdot z+x\cdot z=y\cdot z+(-(x\cdot z)+x\cdot z)=y\cdot z.$$

If $x<y$, then $x\le y$ and $x\neq y$ by (S), so $x\cdot z\le y\cdot z$ by the above and $x\cdot z\neq y\cdot z$ by $(\ast)$; hence $x\cdot z<y\cdot z$ by (S).

Conversely, let $x\cdot z\le y\cdot z$. If $x\le y$ failed, then $y<x$ by (T), so $y\cdot z<x\cdot z$ by the strict forward direction, and $x\cdot z\le y\cdot z$ would fail by (T), a contradiction. Let $x\cdot z<y\cdot z$. If $x<y$ failed, then $y\le x$, since otherwise $x<y$ by (T); so $y\cdot z\le x\cdot z$ by the weak forward direction, whereas $y\cdot z\le x\cdot z$ fails by (T), a contradiction.

Squares. If $0\le x$, then $0\le x\cdot x$ by (OM). Otherwise $x\le0$ by totality, so $0\le-x$ by (N), and $0\le(-x)\cdot(-x)=x\cdot x$ by (OM) and the clause signs above. In particular $0\le1\cdot1=1$; as $0\neq1$, $0<1$ by (S).

Positive-reciprocal. Let $0<x$. By (S), $x\neq0$, so $x^{-1}$ is defined. If $0<x^{-1}$ failed, then $x^{-1}\le0$ by (T); by the clause order-product above with $x$ as the positive factor, and by commutativity and the clause zero above, $1=x^{-1}\cdot x\le0\cdot x=0$. With $0\le1$ from the clause squares above, antisymmetry gives $0=1$, a contradiction. So $0<x^{-1}$.

Now let also $x<y$. Then $0<y$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive, so $y^{-1}$ is defined and $0<y^{-1}$ by what was just shown. By the clause order-product above with $y^{-1}$ as the positive factor, $0<x^{-1}$ gives $0=0\cdot y^{-1}<x^{-1}\cdot y^{-1}$. By the same clause with $x^{-1}\cdot y^{-1}$ as the positive factor, $x<y$ gives, rearranging by Commutative Rings §ring,

$$y^{-1}=(x\cdot x^{-1})\cdot y^{-1}=x\cdot(x^{-1}\cdot y^{-1})<y\cdot(x^{-1}\cdot y^{-1})=(y\cdot y^{-1})\cdot x^{-1}=x^{-1}.$$

Midpoint. By the clause squares above, $0<1$; by the clause order-sum above, $1=0+1<1+1$; so $0<1+1$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive. Let $t=1+1$. Then $t\neq0$ by (S), and $0<t^{-1}$ by the clause positive-reciprocal above. Let $x<y$. By the clause order-sum above, adding $x$ and adding $y$ respectively, $x+x<y+x=x+y$ and $x+y<y+y$. By distributivity, $u+u=u\cdot1+u\cdot1=u\cdot t$ for every $u\in r$, so $x\cdot t<x+y<y\cdot t$. By the clause order-product above with $t^{-1}$ as the positive factor, and since $(u\cdot t)\cdot t^{-1}=u\cdot(t\cdot t^{-1})=u$ for every $u\in r$,

$$x<(x+y)\cdot t^{-1}<y,$$

and $(x+y)\cdot t^{-1}=(x+y)/(1+1)$ by Negatives, Differences, Reciprocals and Quotients §reciprocal.

Absolute value. By Absolute Value in an Ordered Field §absolute-value, $|u|=u$ if $0\le u$ and $|u|=-u$ otherwise. Moreover, if $u\le0$ then $|u|=-u$: either $0\le u$ fails, or $0\le u$ and $u\le0$, so $u=0$ by antisymmetry and $|u|=u=0=-0=-u$. Call this (A).

Sign. If $0\le x$, then $0\le x=|x|$. Otherwise $x\le0$ by totality, and $0\le-x=|x|$ by (N).

Zero. If $x=0$, then $0\le x$ by reflexivity, so $|x|=x=0$. Conversely, let $|x|=0$. If $0\le x$, then $x=|x|=0$; otherwise $-x=0$, and $x=-(-x)=-0=0$ by the clause signs above.

Negation. If $0\le x$ and $0\le-x$, then $x\le0$ by (N), so $x=0$ by antisymmetry, and $|-x|=-x=-0=0=x=|x|$. If $0\le x$ and $0\le-x$ fails, then $|-x|=-(-x)=x=|x|$ by the clause signs above. If $0\le x$ fails, then $x\le0$ by totality, so $0\le-x$ by (N), and $|-x|=-x=|x|$.

Bounds. We show $-|x|\le x\le|x|$. If $0\le x$, then $x\le x=|x|$ by reflexivity, and $-|x|=-x\le0\le x$ by (N) and transitivity. Otherwise $x\le0$ by totality and $0\le-x=|x|$ by (N), so $x\le|x|$ by transitivity, and $-|x|=-(-x)=x$ by the clause signs above, so $-|x|\le x$ by reflexivity. In both cases, $-|x|\le x\le|x|$.

Characterization. Let $|x|\le y$. Then $x\le|x|\le y$, and by the clause order-negative above $-y\le-|x|\le x$; so $-y\le x$ and $x\le y$ by transitivity. Conversely, let $-y\le x$ and $x\le y$. If $0\le x$, then $|x|=x\le y$. Otherwise $|x|=-x$, and $-x\le-(-y)=y$ by the clause order-negative above and the clause signs above.

Product. By totality, $0\le x$ or $x\le0$, and $0\le y$ or $y\le0$. If $0\le x$ and $0\le y$, then $0\le x\cdot y$ by (OM), so $|x\cdot y|=x\cdot y=|x|\cdot|y|$. If $0\le x$ and $y\le0$, then $0\le-y$ by (N), so $0\le x\cdot(-y)=-(x\cdot y)$ by (OM), the clause signs above and commutativity; hence $x\cdot y\le0$ by (N), and by (A) $|x\cdot y|=-(x\cdot y)=x\cdot(-y)=|x|\cdot|y|$. If $x\le0$ and $0\le y$, the previous case with $x$ and $y$ interchanged and commutativity give $|x\cdot y|=|y\cdot x|=|y|\cdot|x|=|x|\cdot|y|$. If $x\le0$ and $y\le0$, then $0\le-x$ and $0\le-y$ by (N), so $0\le(-x)\cdot(-y)=x\cdot y$ by (OM) and the clause signs above, and by (A) $|x\cdot y|=x\cdot y=(-x)\cdot(-y)=|x|\cdot|y|$. This proves the clause absolute-value above.

Triangle. By the clause absolute-value above, $-|x|\le x\le|x|$ and $-|y|\le y\le|y|$, so by the clause order-sum above and the clause signs above, $x+y\le|x|+|y|$ and $-(|x|+|y|)=(-|x|)+(-|y|)\le x+y$. By the clause absolute-value above, with $x+y$ and $|x|+|y|$ in place of $x$ and $y$, $|x+y|\le|x|+|y|$. This proves the clause triangle above.