# Solution of Rudin’s Principles of Mathematical Analysis:Chap1. Ex.6

The definition of (real) exponent of $b\in\R$, $b>0$ is give in Rudin’s Principles of Mathematical Analysis (3.ed), Chapter 1 Exercise 6 , which says that:

6. Fix $b>1$.

1. If $m,n,p,q$ are integers, $n>0,q>0$, and $r=m/n=p/q$, prove that
$(b^m)^{1/n}=(b^p)^{1/q}.$
Hence it makes sense to define $b^r=(b^m)^{1/n}$.
2. Prove that $b^{r+s}=b^rb^s$ if $r$ and $s$ are rational.
3. If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t\leq x$. 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 $x$.
4. Prove that $b^{x+y}=b^xb^y$ for all real $x$ and $y$.

It is a litter hard for beginners to solve this problem, so I will try to post my solution here.

For a. suppose $\alpha=(b^m)^{1/n}$ and $\beta=(b^p)^{1/q}$, then by the definition of $n$th roots, we know that
$\alpha=(b^m)^{1/n}\Leftrightarrow \alpha^n=b^m, \quad \text{and}\quad \beta=(b^p)^{1/q}\Leftrightarrow \beta^q=b^p.$
Thus, by the meaning of notation $b^m$ (which says $b$ multiples itself by $m$ times), we know that
$(b^m)^p=b^{mp}=(b^p)^m,$
thus
$\alpha^{np}=(\alpha^n)^p=(b^m)^p=(b^p)^m=(\beta^q)^m=\beta^{qm},$
since $r=m/n=p/q$, we have $pn=mq$, and
$\alpha=\beta,$
by the uniqueness of $n$th roots of a real number. That is $(b^m)^{1/n}=(b^p){1/q}$.

For b. we can assume $r=m/n$ and $s=p/q$ for $n,q>0$ and $m,n,p,q$ are integers. Then by a.,
$b^{r+s}=b^{\frac{m}{n}+\frac{p}{q}}=b^{\frac{mq+np}{nq}}=(b^{mq+np})^{1/nq}=(b^{mq}b^{np})^{1/nq} =(b^{mq})^{1/nq}(b^{np})^{1/nq}=b^{m/n}b^{p/q}=b^rb^s,$
we employ the result (prove in the book) that:
$(\alpha\beta)^{1/n}=\alpha^{1/n}\beta^{1/n},$
for any $\alpha,\beta\in\R$ and $n$　be a positive integer.

Now, for c., by the definition, we can write $B(x)$ as
$B(x)=\set{b^t|t\in Q, t\leq x},$
and note that $b>1$ thus
$b^t\leq b^r,$
for all $t,r\in Q$ with $t\leq r$.

In fact, if $b^t>b^r$. Set $t=m/n$, $r=p/q$, and $t\leq r$ then by the definition of ordered field,
$mq\leq np\Rightarrow b^{mq}\leq b^{np},$
thus by a.
$(b^t)^{nq}=b^{mq}\leq b^{np}=(b^{r})^{nq},$
a contradiction to the assumption $b^t>b^r$.

In conclusion, we know that
$\sup \set{b^t|t\in Q, t\leq r}\leq b^r,$
and apparently,
$b^r\leq\sup B(r)=\sup \set{b^t|t\in Q, t\leq r},$
thus $b^r=\sup B(r)$.

For d., firstly, for any $r,s\in Q$, if $r\leq x$, $s\leq y$, then by the definition of $b^x, b^y$, we have
$b^r\leq b^x,\quad b^s\leq b^y,$
thus
$b^{r+s}=b^rb^s\leq b^xb^y,$
since $\R$ is a ordered field.

Now, for any $t\in Q$, if $t\leq x+y$, we can always find $r,s\in Q$ such that $r\leq x, s\leq y$ and $r+s=t$. In fact, by the definition of $x+y$, we know that $x+y=\set{r+s|r\in x, s\in y}$, thus the assertion is certainly satisfied. Consequently,
$b^{x+y}=\sup\set{b^t|t\in Q, t\leq x+y}\leq b^xb^y.$
On the other hand, if $b^{x+y}< b^xb^y$, then set $\varepsilon=b^xb^y-b^{x+y} >0$. For any small enough $\delta>0$, by the definition of $b^x, b^y$, there exist $r_0,s_0\in Q$, with $r_0\leq x$, $s_0\leq y$, such that
$b^{r_0}>b^x-\delta>0,\quad b^{s_0}>b^y-\delta>0.$
Since $\R$ is an ordered field, then
$b^{r_0}b^{s_0}>b^xb^y-\delta\left(b^x+b^y-\delta\right),$
particularly, we can take $\delta$ be the positive roots of
$\delta\left(b^x+b^y-\delta\right)=\eps/2,$
then
$2b^{x+y}\geq 2b^{r_0}b^{s_0} >2b^xb^y-\eps=b^xb^y+b^{x+y},$
that is
$b^{x+y}>b^xb^y,$

## “Solution of Rudin’s Principles of Mathematical Analysis:Chap1. Ex.6”上的3条回复 Josh说：

Now, for any t∈Q, if t≤x+y, we can always find r,s∈Q such that r≤x,s≤y and r+s=t. I dont think this is true. What if x=sqrt{2} and y=sqrt{2}???

Since by the definition of x+y, that is x+y={r+s|r∈x,s∈y}, we have if t∈x+y, then there exists r∈x, s∈y, such that t=r+s. This implies that there exists r∈Q, and s∈Q, and r≤x, s≤y. Here x,y is the cut (which defined as real number). Josh说：

should be y= -sqrt{2}