2.7 Absolute Value

2.130   Definition (Absolute value.) Let $F$ be an ordered field, and let $x\in F$. Then we define

$$\vert x\vert=\cases{x &if $x>0$,\cr 0 &if $x=0$, \cr -x &if $x<0$.\cr}$$

2.131   Remark. It follows immediately from the definition that

1. $\vert x\vert \geq 0$ for all $x\in F.$
2. $\vert x\vert >0$ for all $x\in F\backslash\{0\}$.
3. $(\vert x\vert=0)\hspace{1ex}\Longleftrightarrow\hspace{1ex}(x=0)$.

2.132   Theorem. Let $F$ be an ordered field. Then for all $x\in F$,

$$-\vert x\vert\leq x\leq \vert x\vert.$$ (2.133)

Proof: If $x=0$ then (2.133) becomes $-0\leq 0\leq 0$, which is true. If $x>0$ then $-\vert x\vert<0<x=\vert x\vert$. If $x<0$ then $-\vert x\vert=-(-x)=x<0\leq \vert x\vert$. Hence (2.133) holds in all cases. $\mid\!\mid\!\mid$

2.134   Exercise. Let $F$ be an ordered field. Prove that $\vert x\vert=\vert-x\vert$ for all $x\in F$ and $\vert x\vert^2=x^2$ for all $x\in F$.

2.135   Exercise (Product formula for absolute value.) A Prove that for all $x,y\in F$,

$$\vert xy\vert=\vert x\vert \vert y\vert.$$

2.136   Theorem. Let $F$ be an ordered field, let $x\in F$, and let $p\in F$ with $p\geq 0$. Then

$$(\vert x\vert\leq p)\hspace{1ex}\Longleftrightarrow\hspace{1ex}(-p\leq x\leq p)$$ (2.137)

and

$$(\vert x\vert>p)\hspace{1ex}\Longleftrightarrow\hspace{1ex}\left((x<-p) \mbox{ or } (x>p)\right).$$ (2.138)

Proof: We first show that

$$(\vert x\vert\leq p)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(-p\leq x\leq p).$$ (2.139)

Case 1.

If $x>0$, then

$$\vert x\vert\leq p\mbox{$\hspace{1ex}\Longrightarrow\hspace{1... ...$\hspace{1ex}\Longrightarrow\hspace{1ex}$}-p\leq 0\leq x\leq p.$$

Case 2.

If $x<0$, then

$$\vert x\vert\leq p\mbox{$\hspace{1ex}\Longrightarrow\hspace{1... ...box{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}-p\leq x<0\leq p.$$

Case 3.

If $x=0$, then $-p\leq x\leq p$ is true, so (2.139) is true. Hence (2.139) is valid in all cases.

Next we show that

$$(-p\leq x\leq p)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(\vert x\vert\leq p).$$ (2.140)

Case 1.

If $x>0$, then

$$-p\leq x\leq p\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$... ...x{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\vert x\vert\leq p.$$

Case 2.

If $x<0$, then

$$-p\leq x\leq p\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$... ...x{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\vert x\vert\leq p.$$

Case 3.

If $x=0$ then $x\leq p$ is true, so (2.140) is true. Hence (2.140) is true in all cases.

We have proved (2.137).

Since $P\mbox{$\Longleftrightarrow$}Q$ is true if and only if $\left((\mbox{ not } P)\mbox{$\Longleftrightarrow$}(\mbox{ not } Q)\right)$ is true,

$$\mbox{ not } (\vert x\vert\leq p)\hspace{1ex}\Longleftrightar... ...1ex}\mbox{ not } \left((-p\leq x)\mbox{ and } (x\leq p)\right);$$

i.e.,

$$\begin{eqnarray*} \vert x\vert>p &\mbox{$\Longleftrightarrow$}& \left(\mbox{ not... ... } x>p, \\ &\mbox{$\Longleftrightarrow$}& x<-p \mbox{ or } x>p. \end{eqnarray*}$$

This is 2.138. $\mid\!\mid\!\mid$

2.141   Remark. I leave it to you to check that (2.137) holds when $\leq$ is replaced by $<$, and (2.138) holds when $>$ and $<$ are replaced by $\geq$ and $\leq$, respectively.

2.142   Theorem (Triangle inequality.)Let $F$ be an ordered field. Then

$$\mbox{ for all } x,y\in F,\qquad \vert x+y\vert\leq \vert x\vert+\vert y\vert.$$ (2.143)

Proof: The obvious way to prove this is by cases. But there are many cases to consider, e.g. ($x<0$ and $y>0$ and $x+y<0$). I will use an ingenious trick to avoid the cases. For all $x,y\in F$, we have

$$-\vert x\vert\leq x\leq \vert x\vert,$$

and

$$-\vert y\vert\leq y\leq \vert y\vert.$$

By adding the inequalities, we get

$$-(\vert x\vert+\vert y\vert)\leq x+y\leq (\vert x\vert+\vert y\vert).$$

By theorem 2.136 it follows that $\vert x+y\vert\leq \vert x\vert+\vert y\vert$. $\mid\!\mid\!\mid$

2.144   Exercise. A Let $F$ be an ordered field. For each statement below, either prove the statement, or explain why it is not true.

a)

for all $x,y\in F$, $\vert x-y\vert\leq \vert x\vert+\vert y\vert$.

b)

for all $x,y\in F$, $\vert x-y\vert\leq \vert x\vert-\vert y\vert$.

2.145   Exercise (Quotient formula for absolute value.) A Let $F$ be an ordered field. Let $a,b\in F$ with $a\neq 0$. Show that

a)

$${ \left\vert{1\over a}\right\vert={1\over {\vert a\vert}}}$$,

b)

$${ \left\vert{b\over a}\right\vert={{\vert b\vert}\over {\vert a\vert}}}$$.

2.146   Definition (Distance.) Let $F$ be an ordered field, and let $a,b\in F$. We define the distance from $a$ to $b$ to be $\vert b-a\vert$.

2.147   Remark. If $F$ is the ordered field of real or rational numbers, $\vert b-a\vert$ represents the familiar notion of distance between the points $a,b$ on the real line (or the rational line).

2.148   Exercise. Let $F$ be an ordered field. Let $x,a,p\in F$ with $p\geq 0$. Show that

$$(\vert x-a\vert\leq p)\hspace{1ex}\Longleftrightarrow\hspace{1ex}(a-p\leq x\leq a+p).$$ (2.149)

HINT: Use theorem 2.136. Do not reprove theorem 2.136.

2.150   Remark. We can state the result of exercise 2.148 as follows. Let $a\in F$, and let $p\in F^+$. Then the set of points whose distance from $x$ is smaller than $p$, is the set of points between $a-p$ and $a+p$.

2.151   Definition (Intervals and endpoints.) Let $F$ be an ordered field. Let $a,b\in F$ with $a\leq b$. Then we define

$$\begin{eqnarray*} (a,b) &=& \{x\in F\colon a<x<b\} \\ (a,b{]} &=& \{x\in F\colo... ...\infty) &=& \{x\in F\colon x\geq a\} \\ (-\infty,\infty) &=& F. \end{eqnarray*}$$

A set that is equal to a set of any of these nine types is called an interval. Note that $[a,a)=(a,a]=(a,a)=\emptyset$ and $[a,a]=\{a\}$, so the empty set is an interval and so is a set containing just one point. Sets of the first four types have endpoints $a$ and $b$, except that $(a,a)$ has no endpoints. Sets of the second four types have just one endpoint, namely $a$. The interval $(-\infty,\infty)$ has no endpoints.

2.152   Examples. Let $F$ be an ordered field. By exercise 2.148 the set of solutions to $\vert x-3\vert\leq 4$ is

$$\{x\in F\colon 3-4\leq x\leq 3+4\}=\{x\in F\colon -1\leq x\leq 7\}=[-1,7].$$

I can read this result from the figure by counting 4 units to the left and right of 3. This method is just a way of remembering the result of theorem 2.148.

Now suppose I want to find the solutions in $F$ to

$$\vert x-2\vert<\vert x-5\vert$$ (2.153)

Here, thinking of $\vert a-b\vert$ as the distance from $a$ to $b$, I want to find all elements that are nearer to 2 than to 5. From the picture I expect the answer to be $${(-\infty,{7\over 2})}$$. Although this picture method is totally unjustified by anything I've done, it is the method I would use to solve the inequality in practice. If I had to use results we've proved to solve (2.153), I'd say (since $\vert x-2\vert\geq 0$)

$$\begin{eqnarray*} \vert x-2\vert<\vert x-5\vert &\mbox{$\Longleftrightarrow$}& \... ...2} \\ &\mbox{$\Longleftrightarrow$}& x\in (-\infty, {7\over 2}) \end{eqnarray*}$$

which agrees with my answer by picture.

2.154   Exercise. A Let $F$ be an ordered field, let $x,a,p\in F$ with $p>0$. Show that

$$\vert x-a\vert>p\hspace{1ex}\Longleftrightarrow\hspace{1ex}x\in (-\infty,a-p)\cup (a+p,\infty).$$

Interpret the result geometrically on a number line.

2.155   Exercise. Let $F$ be an ordered field. Express each of the following subsets of $F$ as an interval, or a union of intervals. Sketch the sets on a number line.

a)

$A=\{x\in F \colon \vert x-3\vert<2\}$

b)

$B=\{x\in F \colon \vert x+2\vert<3\}$

c)

$C=\{x\in F \colon \vert x-1\vert>1\}$

d)

$D=\{x\in F \colon \vert x+1\vert\geq 1\}$

2.156   Note. Girolamo Cardano (1501-1576), in an attempt to make sense of the square root of a negative number, proposed an alternate law of signs in which the product of two numbers is negative if at least one factor is negative. He concluded that ``plus divided by plus gives plus'', and ``minus divided by plus gives minus'', but ``plus divided by minus gives nothing'' (i.e. zero), since both of the assertions ``plus divided by minus gives plus'' and ``plus divided by minus gives minus'' are contradictory.[40, p 25]

I believe that our axioms for an ordered field are due to Artin and Schreier in 1926 [6, page 259].

Systems satisfying various combinations of algebraic and order axioms were considered by Huntington [28] in 1903.

The notation $\vert x\vert$ for absolute value was introduced by Weierstrass in 1841[15, vol.2, page 123]. It was first introduced for complex numbers rather than real numbers.