{"id":2353,"date":"2012-08-10T20:40:37","date_gmt":"2012-08-10T12:40:37","guid":{"rendered":"http:\/\/vanabel.sinaapp.com\/?p=2353"},"modified":"2012-08-10T20:40:37","modified_gmt":"2012-08-10T12:40:37","slug":"solution-of-rudins-principles-of-mathematical-analysischap1-ex-6","status":"publish","type":"post","link":"https:\/\/lttt.vanabel.cn\/?p=2353","title":{"rendered":"Solution of Rudin’s Principles of Mathematical Analysis:Chap1. Ex.6"},"content":{"rendered":"

The definition of (real) exponent of $b\\in\\R$, $b>0$ is give in Rudin’s Principles of Mathematical Analysis (3.ed), Chapter 1\u00a0<\/em>Exercise 6 <\/em>, which says that:<\/p>\n

6. Fix $b>1$.<\/p>\n

    \n
  1. If $m,n,p,q$ are integers, $n>0,q>0$, and $r=m\/n=p\/q$, prove that
    \n\\[
    \n(b^m)^{1\/n}=(b^p)^{1\/q}.
    \n\\]
    \nHence it makes sense to define $b^r=(b^m)^{1\/n}$.<\/li>\n
  2. Prove that $b^{r+s}=b^rb^s$ if $r$ and $s$ are rational.<\/li>\n
  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
    \n\\[
    \nb^r=\\sup B(r)
    \n\\]
    \nwhen $r$ is rational. Hence it makes sense to define
    \n\\[
    \nb^x=\\sup B(x)
    \n\\]
    \nfor every real $x$.<\/li>\n
  4. Prove that $b^{x+y}=b^xb^y$ for all real $x$ and $y$.<\/li>\n<\/ol>\n


    \nIt is a litter hard for beginners to solve this problem, so I will try to post my solution here.
    \n
    \nFor a. suppose $\\alpha=(b^m)^{1\/n}$ and $\\beta=(b^p)^{1\/q}$, then by the definition of $n$th roots, we know that
    \n\\[
    \n\\alpha=(b^m)^{1\/n}\\Leftrightarrow \\alpha^n=b^m, \\quad \\text{and}\\quad
    \n\\beta=(b^p)^{1\/q}\\Leftrightarrow \\beta^q=b^p.
    \n\\]
    \nThus, by the meaning of notation $b^m$ (which says $b$ multiples itself by $m$ times), we know that
    \n\\[
    \n(b^m)^p=b^{mp}=(b^p)^m,
    \n\\]
    \nthus
    \n\\[
    \n\\alpha^{np}=(\\alpha^n)^p=(b^m)^p=(b^p)^m=(\\beta^q)^m=\\beta^{qm},
    \n\\]
    \nsince $r=m\/n=p\/q$, we have $pn=mq$, and
    \n\\[
    \n\\alpha=\\beta,
    \n\\]
    \nby the uniqueness of $n$th roots of a real number. That is $(b^m)^{1\/n}=(b^p){1\/q}$.<\/p>\n

    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.,
    \n\\[
    \nb^{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}
    \n=(b^{mq})^{1\/nq}(b^{np})^{1\/nq}=b^{m\/n}b^{p\/q}=b^rb^s,
    \n\\]
    \nwe employ the result (prove in the book) that:
    \n\\[
    \n(\\alpha\\beta)^{1\/n}=\\alpha^{1\/n}\\beta^{1\/n},
    \n\\]
    \nfor any $\\alpha,\\beta\\in\\R$ and $n$\u3000be a positive integer.<\/p>\n

    Now, for c., by the definition, we can write $B(x)$ as
    \n\\[
    \nB(x)=\\set{b^t|t\\in Q, t\\leq x},
    \n\\]
    \nand note that $b>1$ thus
    \n\\[
    \nb^t\\leq b^r,
    \n\\]
    \nfor all $t,r\\in Q$ with $t\\leq r$.<\/p>\n

    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,
    \n\\[
    \nmq\\leq np\\Rightarrow b^{mq}\\leq b^{np},
    \n\\]
    \nthus by a.
    \n\\[
    \n(b^t)^{nq}=b^{mq}\\leq b^{np}=(b^{r})^{nq},
    \n\\]
    \na contradiction to the assumption $b^t>b^r$.<\/p>\n

    In conclusion, we know that
    \n\\[
    \n\\sup \\set{b^t|t\\in Q, t\\leq r}\\leq b^r,
    \n\\]
    \nand apparently,
    \n\\[
    \nb^r\\leq\\sup B(r)=\\sup \\set{b^t|t\\in Q, t\\leq r},
    \n\\]
    \nthus $b^r=\\sup B(r)$.<\/p>\n

    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
    \n\\[
    \nb^r\\leq b^x,\\quad b^s\\leq b^y,
    \n\\]
    \nthus
    \n\\[
    \nb^{r+s}=b^rb^s\\leq b^xb^y,
    \n\\]
    \nsince $\\R$ is a ordered field. <\/p>\n

    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,
    \n\\[
    \nb^{x+y}=\\sup\\set{b^t|t\\in Q, t\\leq x+y}\\leq b^xb^y.
    \n\\]
    \nOn 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
    \n\\[
    \nb^{r_0}>b^x-\\delta>0,\\quad
    \nb^{s_0}>b^y-\\delta>0.
    \n\\]
    \nSince $\\R$ is an ordered field, then
    \n\\[
    \nb^{r_0}b^{s_0}>b^xb^y-\\delta\\left(b^x+b^y-\\delta\\right),
    \n\\]
    \nparticularly, we can take $\\delta$ be the positive roots of
    \n\\[
    \n\\delta\\left(b^x+b^y-\\delta\\right)=\\eps\/2,
    \n\\]
    \nthen
    \n\\[
    \n2b^{x+y}\\geq 2b^{r_0}b^{s_0} >2b^xb^y-\\eps=b^xb^y+b^{x+y},
    \n\\]
    \nthat is
    \n\\[
    \nb^{x+y}>b^xb^y,
    \n\\]
    \ncontradict to the assumption.<\/p>\n","protected":false},"excerpt":{"rendered":"

    The definition of (real) exponent of $b\\ …<\/p>\n

    Solution of Rudin’s Principles of Mathematical Analysis:Chap1. Ex.6<\/span> \u9605\u8bfb\u66f4\u591a »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12],"tags":[377,604,612,627,800],"class_list":["post-2353","post","type-post","status-publish","format-standard","hentry","category-mathnotes","tag-exponent","tag-real-analysis","tag-rudin","tag-solution","tag-lecture-notes"],"_links":{"self":[{"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts\/2353","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2353"}],"version-history":[{"count":0,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts\/2353\/revisions"}],"wp:attachment":[{"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2353"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2353"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2353"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}