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


Comments


2025 年 4 月
 123456
78910111213
14151617181920
21222324252627
282930  

The definition of (real) exponent of bR, 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
    (bm)1/n=(bp)1/q.

    Hence it makes sense to define br=(bm)1/n.
  2. Prove that br+s=brbs if r and s are rational.
  3. If x is real, define B(x) to be the set of all numbers bt, where t is rational and tx. Prove that
    br=supB(r)

    when r is rational. Hence it makes sense to define
    bx=supB(x)

    for every real x.
  4. Prove that bx+y=bxby 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 α=(bm)1/n and β=(bp)1/q, then by the definition of nth roots, we know that
α=(bm)1/nαn=bm,andβ=(bp)1/qβq=bp.


Thus, by the meaning of notation bm (which says b multiples itself by m times), we know that
(bm)p=bmp=(bp)m,

thus
αnp=(αn)p=(bm)p=(bp)m=(βq)m=βqm,

since r=m/n=p/q, we have pn=mq, and
α=β,

by the uniqueness of nth roots of a real number. That is (bm)1/n=(bp)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.,
br+s=bmn+pq=bmq+npnq=(bmq+np)1/nq=(bmqbnp)1/nq=(bmq)1/nq(bnp)1/nq=bm/nbp/q=brbs,


we employ the result (prove in the book) that:
(αβ)1/n=α1/nβ1/n,

for any α,βR and n be a positive integer.

Now, for c., by the definition, we can write B(x) as
B(x)={bt|tQ,tx},


and note that b>1 thus
btbr,

for all t,rQ with tr.

In fact, if bt>br. Set t=m/n, r=p/q, and tr then by the definition of ordered field,
mqnpbmqbnp,


thus by a.
(bt)nq=bmqbnp=(br)nq,

a contradiction to the assumption bt>br.

In conclusion, we know that
sup{bt|tQ,tr}br,


and apparently,
brsupB(r)=sup{bt|tQ,tr},

thus br=supB(r).

For d., firstly, for any r,sQ, if rx, sy, then by the definition of bx,by, we have
brbx,bsby,


thus
br+s=brbsbxby,

since R is a ordered field.

Now, for any tQ, if tx+y, we can always find r,sQ such that rx,sy and r+s=t. In fact, by the definition of x+y, we know that x+y={r+s|rx,sy}, thus the assertion is certainly satisfied. Consequently,
bx+y=sup{bt|tQ,tx+y}bxby.


On the other hand, if bx+y<bxby, then set ε=bxbybx+y>0. For any small enough δ>0, by the definition of bx,by, there exist r0,s0Q, with r0x, s0y, such that
br0>bxδ>0,bs0>byδ>0.

Since R is an ordered field, then
br0bs0>bxbyδ(bx+byδ),

particularly, we can take δ be the positive roots of
δ(bx+byδ)=ε/2,

then
2bx+y2br0bs0>2bxbyε=bxby+bx+y,

that is
bx+y>bxby,

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}

发表回复

您的邮箱地址不会被公开。 必填项已用 * 标注

此站点使用Akismet来减少垃圾评论。了解我们如何处理您的评论数据


Other news

  • 使用Chrome播放本地SWF文件

    两个版本, 一个是选择文件, 一个是直接拖拽。 当然也有合并到一起的办法, 参考这里。直接将下列文件放到和fl…

  • Chrome下载完成后显示病毒扫描失败的解决办法

    很蛋疼的一个提示, 一个pdf下载好后给提示病毒扫描失败。把下载的pdf直接删除了。 解决办法是运行如下的注册…

  • C1驾照学习经验

    历时4个月+15天, 我的C1驾照到手了. 下面分享下经验, 为广大学员解惑。 学驾照, 要趁早 为啥呢, 一…