5.3 Existence of Roots

5.35   Definition (Graph.) Let $f\colon A\to B$ be a function. The graph of $f$ is

$$\{(a,b)\in A\times B\colon b=f(a)\}.$$

5.36   Remark. If $f$ is a function from $\mbox{{\bf R}}$ to $\mbox{{\bf R}}$, then graph $f$ is

$$\{(x,y)\in\mbox{{\bf R}}^2\colon y=f(x)\}.$$

You may find it useful to think of $\mbox{{\bf R}}$ as points on a line, and $\mbox{{\bf R}}^2$ as points in a plane and to represent the graph by a picture. Any such picture is outside the scope of our formal development, but I will draw lots of such pictures informally.

$$\mbox{ graph of } f \mbox{ where } f(x)=x^2 \mbox{ for } x\in (-1,2).$$

5.37   Definition (Sum and product of functions.) Let $F$ be a field, and let $\alpha\in F$. Let $A,B$ be sets and let $f\colon A\to F, g\colon B\to F$ be functions. We define functions $f+g$, $f-g$, $f\cdot g$, $\alpha f$ and $${ {f\over g}}$$ by:

$$\begin{eqnarray*} &\;& f+g\colon A\cap B\to F \qquad (f+g)(a)=f(a)+g(a)\mbox{ fo... ...\over g}\right)(a)={{f(a)}\over {g(a)}} \mbox{ for all }a \in D. \end{eqnarray*}$$

where $D=\{x\in A\cap B\colon g(x)\neq 0\}$.

5.38   Remark. Let $F$ be a field, let $S$ be a set, and let $f\colon S\to F$, $g\colon S\to F$ be functions with the same domain. Then the operations $+,\cdot, -$ are binary operations on the set ${\cal S}$ of all functions from $S$ to $F$. These operations satisfy the same commutative, associative and distributive laws that the corresponding operations on $F$ satisfy; e.g.,

$$f\cdot (g+h)=f\cdot g+f\cdot h \mbox{ for all } f,g,h\in\cal S.$$ (5.39)

Proof of (5.39). For all $x\in S$,

$$\begin{eqnarray*} \left(f\cdot (g+h)\right)(x) &=& f(x)(g+h)(x) \\ &=& f(x)\lef... ...(x)+(f\cdot h)(x) \\ &=&\left( (f\cdot g)+(f\cdot h)\right)(x). \end{eqnarray*}$$

Hence, $f\cdot(g+h)=(f\cdot g)+(f\cdot h)$. (Two functions are equal when they have the same domain, the same codomain, and the same rule.) $\mid\!\mid\!\mid$

5.40   Definition (Increasing and decreasing.) Let $J$ be an interval in $\mbox{{\bf R}}$ and let $f\colon J\to\mbox{{\bf R}}$. We say

$f$ is increasing on $J$ if for all $s,t\in J \left(s\leq t\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(s)\leq f(t)\right)$.

$f$ is strictly increasing on $J$ if for all $s,t\in J\left(s<t\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$} f(s)<f(t)\right)$.

$f$ is decreasing on $J$ if for all $s,t\in J\left(s\leq t\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(s)\geq f(t)\right)$.

$f$ is strictly decreasing on $J$ if for all $s,t\in J\left(s<t\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$} f(s)>f(t)\right)$.

5.41   Remark. Since $s = t \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(s) = f(t)$, we can reformulate the definitions of increasing and decreasing as follows:

$f$ is increasing on $J$ if for all $s,t\in J \left(s< t\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(s)\leq f(t)\right)$.

$f$ is decreasing on $J$ if for all $s,t\in J \left(s< t\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(s)\geq f(t)\right)$.

5.42   Exercise. Is there a function $f\colon\mbox{{\bf R}}\to\mbox{{\bf R}}$ that is both increasing and decreasing? If the answer is yes, give an example. If the answer is no, explain why not.

5.43   Exercise. Give an example of a function $f\colon\mbox{{\bf R}}\to\mbox{{\bf R}}$ such that $f$ is increasing, but not strictly increasing.

5.44   Exercise. Let $f\colon\mbox{{\bf R}}\to\mbox{{\bf R}}$ and $g\colon\mbox{{\bf R}}\to\mbox{{\bf R}}$ be increasing functions. Either prove that $f+g$ is increasing or give an example to show that $f+g$ is not necessarily increasing

5.45   Exercise. Let $f\colon\mbox{{\bf R}}\to\mbox{{\bf R}}$ and $g\colon\mbox{{\bf R}}\to\mbox{{\bf R}}$ be increasing functions. Either prove that $f\cdot g$ is increasing or give an example to show that $f\cdot g$ is not necessarily increasing.

5.46   Theorem. Let $m\in\mbox{{\bf Z}}_{\geq 1}$, let $a\in\mbox{{\bf R}}$, $a\geq 1$. Then $a^m\geq a$.

The proof is by induction, and is omitted.

5.47   Theorem. Let $m\in\mbox{{\bf Z}}_{\geq 1}$. Let $f_m(x)=x^m$ for all $x\in [0,\infty)$ in $\mbox{{\bf R}}$. Then $f_m$ is strictly increasing on $[0,\infty)$.

Proof: The proof follows from induction on $m$ or by factoring $x^m-y^m$, and is omitted.

5.48   Exercise. A Let $J$ be an interval in $\mbox{{\bf R}}$ and let $f\colon J\to\mbox{{\bf R}}$ be a strictly increasing function on $J$. Show that for each $a\in\mbox{{\bf R}}$ the equation $f(x)=a$ has at most one solution $x$ in $J$.

5.49   Theorem. Let $p\in\mbox{{\bf Z}}_{\geq 1}$ and let $a\in[0,\infty)$ in $\mbox{{\bf R}}$. Then there is a unique $c\in[0,\infty)$ in $\mbox{{\bf R}}$ such that

$$c^p=a.$$

Proof: First I will construct a binary search sequence $\{[a_n,b_n]\}$ in $\mbox{{\bf R}}$ such that

$$a_n^p\leq a\leq b_n^p \mbox{ for all } n\in\mbox{{\bf N}}.$$

By completeness of $\mbox{{\bf R}}$, I'll have $\{[a_n,b_n]\}\to c$ for some $c\in\mbox{{\bf R}}$. I'll show $c^p=a$, and the proof will be complete.

Let $[a_0,b_0]=[0,(1+a)]$. Then

$$a_0^p=0\leq a < (1+a)\leq (1+a)^p=b_0^p.$$

For $n\in\mbox{{\bf N}}$, define

$$[a_{n+1},b_{n+1}]=\cases{ \left[a_n,{{a_n+b_n}\over 2}\right]... ...2}, b_n\right] &if $\left( {{a_n+b_n}\over 2}\right)^p <a$.\cr}$$

The proof that $\{[a_n,b_n]\}$ is a binary search sequence and that $a_n^p\leq a\leq b_n^p$ for all $n\in\mbox{{\bf N}}$ is the same as the proof given in example 5.16 for $a=p=2$, and will not be repeated here. By completeness $\{[a_n,b_n]\}\to c$ for some $c\in\mbox{{\bf R}}$. Since $0\leq a_n\leq c\leq b_n$, we have $a_n^p\leq c^p\leq b_n^p$. It follows that

$$\vert a-c^p\vert\leq b_n^p-a_n^p \mbox{ for all }n\in\mbox{{\bf N}}.$$

By the formula for factoring $b^p - a^p$ (cf. (3.78)), we have

$$\begin{eqnarray*} \vert a - c^p\vert & \leq & (b_n - a_n)\sum_{j=0}^{p-1}b_n^j a... ...ver 2^n}\cdot p b_0^{p-1} \leq {(b_0 - a_0)p b_0^{p-1} \over n} \end{eqnarray*}$$

for all $n\in\mbox{{\bf Z}}_{\geq 1}$. By Archimedean property 3 (cf corollary 5.28), it follows that $a - c^p = 0$, i.e $c^p=a$.

Let $f_p(x)=x^p$. Since $f_p$ is strictly increasing on $\mbox{{\bf R}}$, it follows from exercise 5.48A that $x^p=a$ has at most one solution in $\mbox{{\bf R}}^+$ and this completes the proof of the theorem. $\mid\!\mid\!\mid$

5.50   Notation ($a^{1\over p}$.) If $p\in\mbox{{\bf Z}}_{\geq 1}$ and $a\in[0,\infty)$, then the unique number $c$ in $[0,\infty)$ such that $c^p=a$ is denoted by $${a^{{1\over p}}}$$, and is called the $p$th root of $a$. An alternative notation for $a^{1\over 2}$ is $\sqrt{a}$.

5.51   Exercise. A Let $a\in[0,\infty)$, let $q,r\in\mbox{{\bf Z}}_{\geq 1}$, and let $p,s\in\mbox{{\bf Z}}$.

a)

Show that $${ \left(a^{1\over q}\right)^p=\left(a^p\right)^{1\over q}}$$.

b)

Show that if $${ {p\over q}={s\over r}}$$, then $${ \left(a^{1\over q}\right)^p=\left(a^{1\over r}\right)^s}$$.

5.52   Definition ($a^r$.) If $a\in\mbox{{\bf R}}^+$ and $r\in\mbox{{\bf Q}}$ we define $${a^r=\left(a^{1\over q}\right)^p}$$ where $q\in\mbox{{\bf Z}}_{\geq 1}$, $p\in\mbox{{\bf Z}}$ and $${r={p\over q}}$$. The previous exercise shows that this definition does not depend on what representation we use for writing $r$.

5.53   Theorem (Laws of exponents.), For all $a,b\in[0,\infty)$ and all $r,s\in\mbox{{\bf Q}}$

a)

$(ab)^r=a^rb^r.$

b)

$a^ra^s=a^{r+s}.$

c)

$(a^r)^s=a^{(rs)}.$

Proof: [of part b)] Let $${r={p\over q},\; s={u\over v}}$$ where $u,v$ are integers and $q,v$ are positive integers. Then (by laws of exponents for integer exponents),

$$\begin{eqnarray*} (a^ra^s)^{q\cdot v}&=&\left( a^{p\over q}\cdot a^{u\over v}\ri... ...ght){^q} =(a^p)^v\cdot(a^u)^q=a^{(pv)}a^{(uq)} \\ &=&a^{pv+uq}. \end{eqnarray*}$$

Also,

$$\begin{eqnarray*} (a^{r+s})^{q\cdot v}&=&\left(a^{\left( {p\over q}+{u\over v}\r... ...v+uq)}\right){^{ {1\over {qv}}}}\right){^{qv}} \\ &=&a^{pv+uq}. \end{eqnarray*}$$

Hence, $(a^r a^s)^{q\cdot v}=(a^{r+s})^{q\cdot v}$, and hence $a^r a^s=a^{r+s}$ by uniqueness of $q\cdot v$ roots.

5.54   Exercise. A Prove parts a) and c) of theorem 5.53.

5.55   Entertainment. Show that of the two real numbers

$$\sqrt{\textstyle{9\over 2} + \sqrt{8}} + \sqrt{\textstyle{9\o... ...9\over 2} + \sqrt{8}} - \sqrt{\textstyle{9\over 2} - \sqrt{8}},$$

one is in $\mbox{{\bf Q}}$, and the other is not in $\mbox{{\bf Q}}$.

5.56   Note. The Archimedean property was stated by Archimedes in the following form:

$\cdots$ the following lemma is assumed: that the excess by which the greater of (two) unequal areas exceeds the less can, by being added to itself, be made to exceed any given finite area. The earlier geometers have also used this lemma.[2, p 234]

Euclid indicated that his arguments needed the Archimedean property by using the following definition:

Magnitudes are said to have a ratio to one another which are capable, when multiplied, of exceeding one another.[19, vol 2, p114]

Here ``multiplied'' means ``added to itself some number of times'', i.e. ``multiplied by some positive integer''.

Rational exponents were introduced by Newton in 1676.

Since algebraists write $a^2,a^3,a^4,$ etc., for $aa,aaa,aaaa,$ etc., so I write $a^{1\over 2},a^{3\over 2}, a^{5\over 3},$ for $\raisebox{.6ex}{$\sqrt{}$}a,\raisebox{.6ex}{$\sqrt{}$}a^3, \raisebox{.6ex}{$\sqrt{}$}ca^5;$ and I write $a^{-1}, a^{-2}, a^{-3},$ etc. for ${1\over a},{1\over aa}, {1\over aaa},$ etc.[14, vol 1, p355]

Here $\raisebox{.6ex}{$\sqrt{}$}ca$ denotes the cube root of $a$.

Buck's Advanced Calculus[12, appendix 2] gives eight different characterizations of the completeness axiom and discusses the relations between them.

The term completeness is a twentieth century term. Older books speak about the continuity of the real numbers to describe what we call completeness.