2.6.E: Problems on Roots, Powers, and Irrationals (Exercises)
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\(/ p . ]\)
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 \(a . ]\)
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 \(\sqrt{a}=p \sqrt{b} . ]\)
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.]
\(\Rightarrow 5 .\) 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].
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 \(n=m p+1 ]\).
Verify the formulas in \((1)\) for powers with positive rational exponents \(r, s .\)
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\((\mathrm{ii})(\mathrm{iii}) . ]\)
From Definition 2 prove that if \(r>0\left(r \in E^{1}\right),\) then
\[
a>1 \Longleftrightarrow a^{r}>1
\]
for \(a \in F(a>0)\).
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).]
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 \(8 . ]\)
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].
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)].