2.6 Ordered Fields

2.100   Definition (Ordered field axioms.) An ordered field is a pair $(F,F^+)=((F,+,\cdot),F^+)$ where $F$ is a field, and $F^+$ is a subset of $F$ satisfying the conditions

1. For all $a,b\in F^+$, $\;\;a+b\in F^+$.
2. For all $a,b\in F^+$, $\;\;a\cdot b\in F^+$.
3. (Trichotomy) For all $a\in F$, exactly one of the statements
   
   $$a\in F^+,\quad -a\in F^+, \quad a=0$$
   
   
   is true. The set $F^+$ is called the set of positive elements of $F$. A field $F$ is orderable if it has a subset $F^+$ such that 1), 2) and 3) are satisfied.

2.101   Example. The rational numbers $(\mbox{{\bf Q}},\mbox{${\mbox{{\bf Q}}}^{+}$})$ form an ordered field, where $\mbox{${\mbox{{\bf Q}}}^{+}$}$ denotes the familiar set of positive rationals.

2.102   Notation ($F^-$.) Let $(F,F^+)$ be an ordered field. We let

$$F^- = \{x\in F\colon -x\in F^+\}.$$

We call $F^-$ the set of negative elements in $F$. Thus

$$x\in F^-\hspace{1ex}\Longleftrightarrow\hspace{1ex}-x\in F^+,$$

and

$$-x\in F^-\hspace{1ex}\Longleftrightarrow\hspace{1ex}-(-x)\in F^+\hspace{1ex}\Longleftrightarrow\hspace{1ex}x\in F^+.$$

We can restate the Trichotomy axiom as: For all $x\in F$, exactly one of the statements

$$x\in F^+,\qquad x=0,\qquad x\in F^-$$

is true.

2.103   Theorem. Let $(F,F^+)$ be an ordered field. Then for all $x\in F\backslash\{0\}$, $x^2\in F^+$.

Proof: Since $x\neq 0$, we know $x\in F^+$ or $x\in F^-$. Now

$$x\in F^+\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}x\cd... ...^+\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}x^2\in F^+,$$

and

$$x\in F^-\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(-x)... ...grightarrow\hspace{1ex}$}x^2\in F^+.\mbox{ $\mid\!\mid\!\mid$}$$

2.104   Corollary. In any ordered field, $1\in F^+$.

2.105   Example. The field $\mbox{{\bf Z}}_5$ is not orderable.

First Proof: If there were a subset $\mbox{{\bf Z}}_5^+$ of $\mbox{{\bf Z}}_5$ such that $(\mbox{{\bf Z}}_5,\mbox{{\bf Z}}_5^+)$ were an ordered field, we would have $4=2^2\in\mbox{{\bf Z}}_5^+$. But in $\mbox{{\bf Z}}_5$, $4=-1$ so $-1\in\mbox{{\bf Z}}_5^+$ and $1\in\mbox{{\bf Z}}_5^+$, which contradicts trichotomy. $\mid\!\mid\!\mid$

Second Proof: If $(\mbox{{\bf Z}}_5,\mbox{{\bf Z}}_5^+)$ were an ordered field, we would have $1\in\mbox{{\bf Z}}_5^+$, so $1+1=2\in\mbox{{\bf Z}}_5^+$, so $1+2=3\in\mbox{{\bf Z}}_5^+$, so $3+1=4\in\mbox{{\bf Z}}_5^+$ so $4+1=0\in\mbox{{\bf Z}}_5^+$. This contradicts trichotomy. $\mid\!\mid\!\mid$

2.106   Remark. The method used in the second proof above shows that none of the fields $\mbox{{\bf Z}}_n$ are orderable.

2.107   Definition ($<,\leq,>,\geq$) Let $(F,F^+)$ be an ordered field, and let $a,b\in F$. We define

$$\begin{eqnarray*} a<b &\mbox{$\Longleftrightarrow$}& b-a\in F^+.\\ a\leq b &\mb... ....\\ a\geq b &\mbox{$\Longleftrightarrow$}& a>b \mbox{ or } a=b. \end{eqnarray*}$$

2.108   Remark. In any ordered field $(F,F^+)$:

$$\begin{eqnarray*} 0<b &\mbox{$\Longleftrightarrow$}& b\in F^+.\\ b<0 &\mbox{$\L... ... 0-b\in F^+ \hspace{1ex}\Longleftrightarrow\hspace{1ex}b\in F^-. \end{eqnarray*}$$

2.109   Exercise. Let $(F,F^+)$ be an ordered field, and let $a,b\in F$. Show that exactly one of the statements

$$b<a,\qquad b=a, \qquad b>a$$

is true.

2.110   Theorem (Transitivity of $<$.) Let $(F,F^+)$ be an ordered field. Then for all $a,b,c\in F$,

$$\left( (a<b) \mbox{ and } (b<c)\right)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(a<c).$$

Proof: For all $a,b,c\in F$ we have

$$\begin{eqnarray*} (a<b)\mbox{ and } (b<c) &\mbox{$\Longleftrightarrow$}& b-a\in ... ...F^+\\ &\mbox{$\Longrightarrow$}& a<c.\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

2.111   Exercise (Addition of inequalities.) Let $(F,F^+)$ be an ordered field, and let $a,b,c,d\in F$. Show that

$$\left((a<b) \mbox{ and } (c<d)\right) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(a+c)<(b+d)$$

and

$$a<b \hspace{1ex}\Longleftrightarrow\hspace{1ex}a+c<b+c$$

2.112   Exercise. Let $(F,F^+)$ be an ordered field, and let $a,b\in F$. Show that

$$a<b\hspace{1ex}\Longleftrightarrow\hspace{1ex}-b<-a.$$

2.113   Notation. Let $(F,F^+)$ be an ordered field, and let $a,b,c,d\in F$. We use notation like

$$a\leq b<c=d$$ (2.114)

to mean $(a\leq b)$ and $(b<c)$ and $(c=d)$, and similarly we write

$$a>b=c\geq d$$ (2.115)

to mean $a>b$ and $b=c$ and $c\geq d$. By transitivity of $=$ and of $<$, you can conclude $a<c$ from (2.114), and you can conclude $b\geq d$ and $a>c$ from (2.115). A chain of inequalities involving both $<$ and $>$ shows bad style, so you should not write

$$a<b\geq c.$$

2.116   Exercise (Laws of signs.) Let $(F,F^+)$ be an ordered field, and let $a,b\in F$. Show that

1. $(a\in F^+$ and $b\in F^-) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}ab\in F^-$
2. $(a\in F^-$ and $b\in F^+) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}ab\in F^-$
3. $(a\in F^-$ and $b\in F^-) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}ab\in F^+$

These laws together with the axiom

$$a\in F^+ \mbox{ and } b\in F^+\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}ab\in F^+$$

are called the laws of signs.

2.117   Notation. Let $F$ be an ordered field, and let $a,b$ be non-zero elements of $F$. We say $a$ and $b$ have the same sign if either ($a,b$ are both in $F^+$) or ($a,b$ are both in $F^-$). Otherwise we say $a$ and $b$ have opposite signs.

2.118   Corollary (of the law of signs.)Let $(F,F^+)$ be an ordered field and let $a,b\in F\backslash\{0\}$. Then

$$\begin{eqnarray*} a\cdot b\in F^+ &\mbox{$\Longleftrightarrow$}& a \mbox{ and } ... ...leftrightarrow$}& a \mbox{ and } b \mbox{ have opposite signs. } \end{eqnarray*}$$

2.119   Notation. I will now start to use the convention that `` let $F$ be an ordered field" means ``let $\left(F, F^+\right)$ be an ordered field''; i.e., the set of positive elements of $F$ is assumed to be called $F^+$.

2.120   Exercise. Let $F$ be an ordered field and let $a,b,c\in F$. Prove that

$$\begin{eqnarray*} \left((a<b)\mbox{ and } (c<0)\right) &\mbox{$\Longrightarrow$}... ...a<b))\mbox{ and } (c>0)\right) &\mbox{$\Longrightarrow$}& ac<bc. \end{eqnarray*}$$

2.121   Theorem (Multiplication of inequalities.)Let $F$ be an ordered field and let $a,b,c,d$ be elements of $F$. Then

$$\left((0<a<b)\mbox{ and } (0<c<d)\right)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}0<ac<bd.$$

Proof: By the previous exercise we have

$$\begin{eqnarray*} (0<a<b)\mbox{ and } (0<c<d)&\mbox{$\Longrightarrow$}&\left((ca... ...ght)\\ &\mbox{$\Longrightarrow$}&((ac<ad)\mbox{ and } (ad<bd)). \end{eqnarray*}$$

Hence, by transitivity of $<$,

$$(0<a<b)\mbox{ and } (0<c<d)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}ac<bd.\mbox{ $\mid\!\mid\!\mid$}$$

2.122   Exercise. Let $F$ be an ordered field, and let $a\in F\backslash\{0\}$. Show that $a$ and $a^{-1}$ have the same sign.

2.123   Exercise. A Let $F$ be an ordered field, and let $a,b\in F\backslash\{0\}$. Under what conditions (if any) can you say that

$$a<b\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}b^{-1}<a^{-1}?$$

Under what conditions (if any) can you say that

$$a<b\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}a^{-1}<b^{-1}?$$

2.124   Definition (Square root.) Let $F$ be a field, and let $x\in F$. A square root for $x$ is any element $y$ of $F$ such that $y^2=x$.

2.125   Examples. In $\mbox{{\bf Z}}_5$, the square roots of $-1$ are $2$ and $3$.

In an ordered field $F$, no element in $F^-$ has a square root.

In $\mbox{{\bf Q}}$, there is no square root of $2$. (See theorem 3.45 for a proof.)

2.126   Theorem. Let $F$ be an ordered field and let $x\in F^+$. If $x$ has a square root, then it has exactly two square roots, one in $F^+$ and one in $F^-$, so if $x$ has a square root, it has a unique positive square root.

Proof: Suppose $x$ has a square root $y$. Then $y\neq 0$, since $x\in F^+$. If $z$ is any square root of $x$, then $z^2=x=y^2$, so, as we saw in theorem 2.95, $z=y$ or $z=-y$. By trichotomy, one of $y, -y$ is in $F^+$, and the other is in $F^-$. $\mid\!\mid\!\mid$

2.127   Theorem. Let $F$ be an ordered field and let $x,y$ be elements of $F$ with $x\geq 0$ and $y\geq 0$. Then

$$x<y\hspace{1ex}\Longleftrightarrow\hspace{1ex}x^2<y^2.$$ (2.128)

Proof: Let $x,y$ be elements of $F^+\cup\{0\}$. Then $x+y > 0$, unless $x=y=0$, so $x^2 < y^2 \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}x+y > 0$. Hence

$$\begin{eqnarray*} x^2<y^2 &\mbox{$\Longleftrightarrow$}& y^2-x^2>0 \\ &\mbox{$\... ...\ &\mbox{$\Longleftrightarrow$}& x<y.\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

2.129   Remark. The implication (2.128) is also true when $<$ is replaced by $\leq$ in both positions. I'll leave this to you to check.