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


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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 nth 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 nth 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.

《 “Solution of Rudin’s Principles of Mathematical Analysis:Chap1. Ex.6” 》 有 3 条评论

  1. Josh 的头像

    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}???

    1. 艾子 的头像

      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).

  2. Josh 的头像

    should be y= -sqrt{2}

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