The post Solution of Rudin’s Principles of Mathematical Analysis:Chap1. Ex.6 appeared first on 小小泪.

]]>

6. Fix $b>1$.

- 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 $b^r=(b^m)^{1/n}$. - Prove that $b^{r+s}=b^rb^s$ if $r$ and $s$ are rational.
- 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 $x$. - 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,

\]

contradict to the assumption.

本作品采用创作共用版权协议, 要求**署名、非商业用途和保持一致**. 转载本站内容必须也遵循**署名-非商业用途-保持一致**的创作共用协议.

The post Solution of Rudin’s Principles of Mathematical Analysis:Chap1. Ex.6 appeared first on 小小泪.

]]>