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

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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)].