Exercise 1.6

Question

\def\where{\,|\,}
Exercise 6: Fix $b>1$.
    \begin{itemize}
        \item[(a)] If $m$, $n$, $p$, $q$ are integers, $n>0$, $q>0$, and $r=m/n=p/q$, prove that $$(b^m)^{1/n}=(b^p)^{1/q}.$$
        \item[(b)] Prove that $b^{r+s}=b^rb^s$ if $r$ and $s$ are rational.
        \item[(c)] If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t\le x$. Prove that $$b^r=\sup B(r)$$ when $r$ is rational.
Hence it makes sense to define $$b^x=\sup B(x)$$ for every real $x$.
        \item[(d)] Prove that $b^{x+y}=b^xb^y$ for all real $x$ and $y$.
    \end{itemize}

analambanomenos · Accepted Answer

\def\where{\,|\,}
\def\eps{\varepsilon}

(a): Since $np=qm$, $$((b^p)^{1/q})^{np} = ((b^p)^{1/q})^{qm}=b^{mp}.$$ Hence by Theorem 1.21, $$((b^p)^{1/q})^n=b^m.$$ A second application of Theorem 1.21 gives us $$(b^p)^{1/q}=(b^m)^{1/n}.$$

(b): Let $r=m/n$, $s=p/q$.
	\begin{align*}
	    (b^rb^s)^{nq}=((b^m)^{1/n}(b^p)^{1/q})^{nq} &= b^{mq}b^{np} \
	    (b^{r+s})^{nq}=((b^{mq+np})^{1/nq})^{nq} &= b^{mq}b^{np}
	\end{align*}
	Hence by Theorem 1.21, $b^rb^s=b^{r+s}$.

(c): Note that the positive rational powers of $b$ are greater than 1 since the positive integral powers and roots of real numbers greater than 1 are also greater than 1. Hence if $t<r$,
	$$b^t<b^tb^{r-t}=b^{t+r-t}=b^r.$$ Hence, since $b^r\in B(r)$, we have $b^r=\sup B(r)$.

(d): We need to show that $\sup B(x+y)=\sup B(x)\sup B(y)$. If $S$ and $T$ are sets of positive real numbers, it isn't hard to show that $\sup ST = \sup S\sup T$,
where $ST$ is the set of products of elements of $S$ with elements of $T$. Hence we want to show that $\sup B(x+y)=\sup(B(x)B(y))$.

Let $p,q$ be rational numbers such that $p\le x$ and $q\le y$. Then $b^pb^q=b^{p+q}\le\sup B(x+y)$, so $\sup(B(x)B(y))\le\sup B(x+y)$.

To get the reverse inequality, let $t$ be any rational number such that $t<x+y$, and let $\epsilon >0$ such that $t<x+y-\epsilon$. By Theorem 1.20(b), there are rational numbers $p\le x$, and $q\le y$
such that $x-\epsilon/2<p$ and $y-\epsilon/2<q$. Then $t<x+y-\epsilon<p+q$ so that $b^t<b^{p+q}=b^pb^q\le\sup B(x)B(y)$. Hence $\sup B(x+y)\le\sup(B(x)B(y))$.  {\it (thanks to Matt Lundy for
pointing out errors)}

Amit Awasthi · Answer

\begin{proof}[Proof of $(a)$] $m/n = p/q$ implies $mq = np$. So, $b^{mq} = b^{np}$. By Theorem 1.21, $(b^{mq})^{1/nq} = (b^{np})^{1/nq}$, and so $(b^m)^{1/n} = (b^p)^{1/q}$, and $b^r$ is well-defined. \end{proof} \begin{proof}[Proof of $(b)$] Let $r = m/n$, $s = p/q$ where $m,n,p,q \in \mathbb{Z}$, and $n>0$, $q>0$. Then, \begin{align*} (b^{r+s})^{nq} &= (b^{m/n + p/q})^{nq}\ &= b^{mq+np} = b^{mq}b^{np}\ &= (b^{m/n})^{nq}(b^{p/q})^{nq}\ &= (b^{m/n}b^{p/q})^{nq}\ &= (b^rb^s)^{nq}. \end{align*} By Theorem 1.21, \begin{equation*} [(b^{r+s})^{nq}]^{1/nq} = [(b^rb^s)^{nq}]^{1/nq}, \end{equation*} and so $b^{r+s} = b^rb^s$. \end{proof} \begin{proof}[Proof of $(c)$] $b^r = b^tb^{r-t} \ge b^t$ since $b > 1$ and $r-t > 0$, and so $b^r$ is an upper bound of $B(r)$. To prove $b^r$ is the \emph{least} upper bound, assume there is an upper bound $b^s$ of $B(r)$ such that $b^t \le b^s < b^r$. However, by the archimedean property, there is a $t$ such that $s < t < r$, which is a contradiction since it implies $b^s < b^t < b^r$ since $b > 1$ and $b^s$ is an upper bound of $B(r)$. Thus, $b^r = \sup B(r)$. The same proof applies for any real $x$. \end{proof} \begin{proof}[Proof of $(d)$] By $(c)$, $$ b^{x+y} = \sup B(x+y) = \sup \left\{ b^t \right\}. $$ For $t \le x+y, t \in \mathbb{Q}$. Letting $r + s = t$, $r \le x$, $s \le y$, $r,s \in \mathbb{Q}$, we have $ b^{x+y} = \sup \left\{ b^{r+s} \right\} $. Since $b^{r+s} = b^rb^s$ by $(b)$, \begin{align*} b^{x+y} &= \sup \left\{ b^r \right\}\sup \left\{ b^s \right\}\ &= \sup B(x) \sup B(y) = b^xb^y.\qedhere \end{align*} \end{proof}

Exercise 1.6

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Exercise 1.6

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