Lesson 0008: Completeness

Let $(F,+,\times,\preceq)$ be an ordered field.
We say that $F$ has the completeness (with respect to suprema) property if for every $X\subseteq F$ a nonempty subset bounded above, $\sup(X)$ exists.
In that case we call $F$ a complete ordered field.
The axiom of the real numbers is the following proposition.
There exists a complete ordered field. We denote the set by $F = \Bbb R$ and its ordering relation by $\le$.
We denote the field of fractions of $\Bbb R$ by the symbol $\Bbb Q$. We assume without proof that this is the same as our intuitive concept of the rational numbers.
If $x\in\Bbb R^{\ge 0}$ and $n\in\Bbb N$ then we define the $n$th root of $x$ as a number $y$ satisfying $$ y^n = x $$ We denote the unique positive solution by $y=\sqrt[n]x$ or by $y = x^{1/n}$. If $r = p/q\in\Bbb Q$ is a rational number, with $p,q\in\Bbb Z$ and $q\ne 0$, then we define $$x^r = (x^{1/q})^p$$
Theorem: Real roots exist
Roots of positives exist uniquely, and so do odd roots of negatives.
Statement and proof. Statement: If $x\in\Bbb R^{\ge 0}$ then for every $n\in\Bbb N$, the $n$th root $\sqrt[n]x$ exists and is unqiue.
Proof: Define the set $L = \{ z\in\Bbb R: z^n < x\}^{\ge 0}$.
$\sup(L)=\alpha$ exists. Clearly $0\in L$ so it is nonempty.
If $1\le x$ then by the integer power comparison theorem, $x^n$ is an upper bound on $L$. If $0\le x < 1$ then 1 is an upper bound on $L$.
Since $L$ is bounded above in every case, then therefore $\sup(L)=\alpha$ exists.
$\alpha^n\le x$. Suppose for contradiction that $x < \alpha^n$, and define $$ S = \left\{\frac{\alpha^n - x}{n\binom n k \alpha^k} : 0\le k\le n\right\} $$ and $$ \varepsilon = \min\{1, \min(S)\}/2 $$ Therefore, for any integer $0\le k\le n$, we have $\varepsilon^k < \varepsilon$ and $\varepsilon < \frac{\alpha^n-x }{n\binom n k \alpha^k}$. Then by the binomial theorem, and the above, $$\begin{aligned} (\alpha-\varepsilon)^n &= \sum_{k=0}^n \binom n k \alpha^k (-\varepsilon)^{n-k} \\ &= \alpha^n + \sum_{k=0}^{n-1}\binom n k \alpha^k(-\varepsilon)^{n-k} \\ &> \alpha^n - \sum_{\substack{0 \le n-k \le n \\ n-k \text{ odd}} } \binom nk \alpha^k \varepsilon^{n-k} \\ &> \alpha^n - \sum_{\substack{0 \le n-k \le n \\ n-k \text{ odd}}} \binom n k \alpha^k \varepsilon \\ &> \alpha^n - \sum_{\substack{0 \le n-k \le n \\ n-k \text{ odd}}} \binom n k \alpha^k \left(\frac{\alpha^n-x}{n\binom n k \alpha^k}\right)\\ &= \alpha^n - \frac 1 n \sum_{\substack{0 \le n-k \le n \\ n-k \text{ odd}}} (\alpha^n - x) \\ &\ge \alpha^n - \frac 1 n \cdot n (\alpha^n-x) \\ &= x \end{aligned}$$ Therefore $\alpha-\varepsilon < \alpha$ and also $\alpha-\varepsilon\in UB_L$. But this contradicts the assumption that $\alpha=\sup(L)$.
$x\le \alpha^n$. Assume for contradiction that $\alpha^n < x$ and define $$ S = \left\{ \frac{x-\alpha^n}{n\binom n k \alpha^k}: 0\le k\le n \right \} $$ and $$ \varepsilon = \min\{1,\min (S)\} $$ Then $$\begin{aligned} (\alpha+\varepsilon)^n &= \sum_{k=0}^n \binom n k \alpha^k\varepsilon^{n-k} \\ & < \alpha^n + \sum_{k=0}^{n-1} \binom n k \alpha^k\varepsilon \\ & < \alpha^n + \sum_{k=0}^{n-1} \binom n k \alpha^k\left(\frac{x-\alpha^n}{n\binom n k \alpha^k}\right) \\ &= \alpha^n + \frac 1 n \sum_{k=0}^{n-1}(x-\alpha) \\ &= x \end{aligned}$$ Therefore $\alpha+\varepsilon\in L$ and $\alpha<\alpha+\varepsilon$. But this contradicts the assumption that $\alpha=\sup(L)$.
Uniqueness. If $a,b\in\Bbb R^+$ are any two positive solutions of the equation $y^n = x$, then $a^n = x = b^n$. From an earlier result, therefore $a=b$.
If either $a$ or $b$ is zero, then $a,x$ and $b$ must all be zero.
$\Box$
Theorem: Real pow frac. indep. ord.
Real numbers to a fractional power are independent of order.
Statement and proof. Statement: If $x\in\Bbb R^{\ge 0}$ and if $p,q\in\Bbb Z$ with $q\ne 0$, then $$ x^{p/q} = (x^p)^{1/q} $$ Proof: By facts already established for integer exponents, $$\begin{aligned} (x^{p/q})^q &= ((x^{1/q})^p)^q \\ &= ((x^{1/q})^q)^p \\ &= x^p \end{aligned}$$ By uniqueness it now follows that $(x^{p/q}) = (x^p)^{1/q}$.
$\Box$
Theorem: Real rational pow. laws
Real numbers with rational exponents follow the power laws.
Statement and proof. Statement: For all $p,q\in\Bbb Q$ and $x,y\in\Bbb R^{\ge 0}$, $$ x^py^p = (xy)^p $$ and $$ (x^p)^q = x^{pq} $$ Proof: We prove this in stages. To begin, we show that if $m,n\in\Bbb N$ then $x^{1/m}y^{1/m} = (xy)^{1/m}$ and $(x^{1/m})^{1/n} = x^{\frac 1{mn}}$.
Afterward, we extend this to rational exponents.
$x^{1/m}y^{1/m} = (xy)^{1/m}$. $$ \begin{aligned} (x^{1/m}y^{1/m})^m &= (x^{1/m})^m (y^{1/m})^m \\ &= x y \end{aligned} $$ By uniqueness, therefore $x^{1/m}y^{1/m} = (xy)^{1/m}$.
$(x^{1/m})^{1/n} = x^{\frac{1}{mn}}$. $$ \begin{aligned} ((x^{1/m})^{1/n})^{mn} &= (((x^{1/m})^{1/n})^n)^m \\ &= (x^{1/m})^{m} \\ &= x \end{aligned}$$ By uniqueness, therefore $(x^{1/m})^{1/n} = x^{\frac 1 {mn}}$.
$x^py^p = (xy)^p$. Let $p\in\Bbb Q$ and $p=\frac a b$ where $a,b\in\Bbb Z$ and $b\ne 0$.
Then $$\begin{aligned} x^py^p &= (x^{1/r})^q (y^{1/r})^q \\ &= (x^{1/r}y^{1/r})^q \\ &= ((xy)^{1/r})^q \\ &= (xy)^{q/r} \\ &= (xy)^p \end{aligned}$$
$(x^p)^q = x^{pq}$. With $p = a/b, q=c/d$ and $a,b,c,d\in \Bbb Z$ with $b\ne 0\ne d$, $$\begin{aligned} (x^p)^q & = (((x^{1/b})^a)^{1/d})^c \\ &= ((x^{1/b})^{a/d})^c \\ &= (((x^{1/b})^{1/d})^a)^c \\ &= (x^{\frac 1 {bd}})^{ac} \\ &= x^{\frac{ac}{bd}} \\ &= x^{pq} \end{aligned}$$
$\Box$