2.6.E: Problems on Roots, Powers, and Irrationals (Exercises)

Exercise $\PageIndex{1}$

Let $n \in N$ in $E^{1} ;$ let $p>0$ and $a>0$ be elements of an ordered field $F$.
Prove that
(i) if $p^{n}>a,$ then $(\exists x \in F) p>x>0$ and $x^{n}>a$;
(ii) if $p^{n}<a,$ then $(\exists x \in F) x>p$ and $x^{n}<a$.
[Hint: For (i), put
$$x=p-d, \text { with } 0<d<p.$$
Use the Bernoulli inequality (Problem 5 (ii) in §§5-6) to find $d$ such that
$$x^{n}=(p-d)^{n}>a,$$
i.e.,
$$\left(1-\frac{d}{p}\right)^{n}>\frac{a}{p^{n}}.$$
Solving for $d,$ show that this holds if
$$0<d<\frac{p^{n}-a}{n p^{n-1}}<p . \quad \text { (Why does such a } d \text { exist? } )$$
For $(\mathrm{ii}),$ if $p^{n}<a,$ then
$$\frac{1}{p^{n}}>\frac{1}{a}.$$
Use (i) with $a$ and $p$ replaced by 1$/ a$ and 1$/ p . ]$

Exercise $\PageIndex{2}$

Prove Theorem 1 assuming that
(i) $a>1$;
(ii) $0<a<1$ (the cases $a=0$ and $a=1$ are trivial).
$[\text { Hints: }(\text { i) Let }$
$$A=\left\{x \in F | x \geq 1, x^{n}>a\right\}.$$
Show that $A$ is bounded below (by 1 ) and $A \neq \emptyset$ (e.g., $a+1 \in A-$why?).
By completeness, put $p=$ inf $A .$
Then show that $p^{n}=a$ (i.e., $p$ is the required $\sqrt[n]{a} )$.
Indeed, if $p^{n}>a,$ then Problem 1 would yield an $x \in A$ with
$$x<p=\inf A . \text { (Contradiction!) }$$
Similarly, use Problem 1 to exclude $p^{n}<a$.
To prove uniqueness, use Problem 4(ii) of §§5-6.
Case (ii) reduces to (i) by considering 1$/ a$ instead of $a . ]$

Exercise $\PageIndex{3}$

Prove Note 1.
[Hint: Suppose first that $a$ is not divisible by any square of a prime, i.e.,
$$a=p_{1} p_{2} \cdots p_{m},$$
where the $p_{k}$ are distinct primes. (We assume it known that each $a \in N$ is the product of [possibly repeating] primes.) Then proceed as in the proof of Theorem 2, replacing "even" by "divisible by $p_{k}$."
The general case, $a=p^{2} b,$ reduces to the previous case since $\sqrt{a}=p \sqrt{b} . ]$

Exercise $\PageIndex{4}$

Prove that if $r$ is rational and $q$ is not, then $r \pm q$ is irrational; so also are $r q, q / r,$ and $r / q$ if $r \neq 0$.
[Hint: Assume the opposite and find a contradiction.]

Exercise $\PageIndex{5}$

$\Rightarrow 5 .$ Prove the density of irrationals in a complete field $F :$ If $a<b(a, b \in F)$, there is an irrational $x \in F$ with
$$a<x<b$$
(hence infinitely many such irrationals $x ) .$ See also Chapter 1, §9, Problem $4 .$
[Hint: By Theorem 3 of §10,
$$(\exists r \in R) \quad a \sqrt{2}<r<b \sqrt{2}, r \neq 0 .(\mathrm{Why} ?)$$
Put $x=r / \sqrt{2} ;$ see Problem 4].

Exercise $\PageIndex{6}$

Prove that the rational subfield $R$ of any ordered field is Archimedean.
[Hint: If
$$x=\frac{k}{m} \text { and } y=\frac{p}{q} \quad(k, m, p, q \in N),$$
then $n x>y$ for $n=m p+1 ]$.

Exercise $\PageIndex{7}$

Verify the formulas in $(1)$ for powers with positive rational exponents $r, s .$

Exercise $\PageIndex{8}$

Prove that
(i) $a^{r+s}=a^{r} a^{s}$ and
(ii) $a^{r-s}=a^{r} / a^{s}$ for $r, s \in E^{1}$ and $a \in F(a>0)$.
[Hints: For $(\mathrm{i}),$ if $r, s>0$ and $a>1,$ use Problem 9 in §§8-9 to get
Verify that
$$\begin{aligned} A_{a r} A_{a s} &=\left\{a^{x} a^{y} | x, y \in R, 0<x \leq r, 0<y \leq s\right\} \\ &=\left\{a^{z} | z \in R, 0<z \leq r+s\right\}=A_{a, r+s} \end{aligned}$$
Hence deduce that
$$a^{r} a^{s}=\sup \left(A_{a, r+s}\right)=a^{r+s}$$
by Definition 2.
For $(\mathrm{ii}),$ if $r>s>0$ and $a>1,$ then by $(\mathrm{i})$,
$$a^{r-s} a^{s}=a^{r}$$
so
$$a^{r-s}=\frac{a^{r}}{a^{s}}.$$
For the cases $r<0$ or $s<0,$ or $0<a<1,$ use the above results and Definition 2$(\mathrm{ii})(\mathrm{iii}) . ]$

Exercise $\PageIndex{9}$

From Definition 2 prove that if $r>0\left(r \in E^{1}\right),$ then
$$a>1 \Longleftrightarrow a^{r}>1$$
for $a \in F(a>0)$.

Exercise $\PageIndex{10}$

Prove for $r, s \in E^{1}$ that
(i) $r<s \Leftrightarrow a^{r}<a^{s}$ if $a>1$;
(ii) $r<s \Leftrightarrow a^{r}>a^{s}$ if $0<a<1$.
[Hints: (i) By Problems 8 and 9,
$$a^{s}=a^{r+(s-r)}=a^{r} a^{s-r}>a^{r}$$
since $a^{s-r}>1$ if $a>1$ and $s-r>0$.
(ii) For the case $0<a<1,$ use Definition 2(ii).]

Exercise $\PageIndex{11}$

Prove that
$$(a \cdot b)^{r}=a^{r} b^{r} \text { and }\left(\frac{a}{b}\right)^{r}=\frac{a^{r}}{b^{r}}$$
for $r \in E^{1}$ and positive $a, b \in F$.
[Hint: Proceed as in Problem $8 . ]$

Exercise $\PageIndex{12}$

Given $a, b>0$ in $F$ and $r \in E^{1},$ prove that
(i) $a>b \Leftrightarrow a^{r}>b^{r}$ if $r>0,$ and
(ii) $a>b \Leftrightarrow a^{r}<b^{r}$ if $r<0$.
[Hint:
$$a>b \Longleftrightarrow \frac{a}{b}>1 \Longleftrightarrow\left(\frac{a}{b}\right)^{r}>1$$
if $r>0$ by Problems 9 and 11].

Exercise $\PageIndex{13}$

Prove that
$$\left(a^{r}\right)^{s}=a^{r s}$$
for $r, s \in E^{1}$ and $a \in F(a>0)$.
[Hint: First let $r, s>0$ and $a>1 .$ To show that
$$\left(a^{r}\right)^{s}=a^{r s}=\sup A_{a, r s}=\sup \left\{a^{x y} | x, y \in R, 0<x y \leq r s\right\},$$
use Problem 13 in §§8-9. Thus prove that
(i) $(\forall x, y \in R | 0<x y \leq r s) a^{x y} \leq\left(a^{r}\right)^{s},$ which is easy, and
(ii) $(\forall d>1)(\exists x, y \in R | 0<x y \leq r s)\left(a^{r}\right)^{s}<d a^{x y}$.
Fix any $d>1$ and put $b=a^{r} .$ Then
$$\left(a^{r}\right)^{s}=b^{s}=\sup A_{b s}=\sup \left\{b^{y} | y \in R, 0<y \leq s\right\}.$$
Fix that $y .$ Now
$$a^{r}=\sup A_{a r}=\sup \left\{a^{x} | x \in R, 0<x \leq r\right\};$$
so
$$(\exists x \in R | 0<x \leq r) \quad a^{r}<d^{\frac{1}{2 y}} a^{x} . \quad(\mathrm{Why} ?)$$
Combining all and using the formulas in $(1)$ for rationals $x, y,$ obtain
$$\left(a^{r}\right)^{s}<d^{\frac{1}{2}}\left(a^{r}\right)^{y}<d^{\frac{1}{2}}\left(d^{\frac{1}{2 y}} a^{x}\right)^{y}=d a^{x y},$$
thus proving (ii)].