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\pagestyle{fancy} \lhead{{\sf Homework 3 $|$ Due May 25 (Friday) }}
\rhead{\thepage} \cfoot{{\sf MthSc 453 $|$ Real Analysis I $|$ Summer
    I 2012 $|$ M.~Macauley}}


\begin{document}

$\;$

\noindent\emph{Read}: Rudin, Chapter 1.  Do not wory about
understanding every detail, but get a good sense of the big ideas. \\


\begin{enumerate}

\item In no more than four sentences, summarize the main concepts and
  ideas of Chapter 1 in Rudin. Pretend that you are writing for a
  friend who is an English major (and naturally, took math throughout
  high school, but maybe not in college). They should be able to
  understand it \emph{and} appreciate the clarity and style in which
  it is written. \\



\item Suppose that we were to define a cut \emph{without} the
  requirement that it contain no largest element. That is, we have a
  set $S$ that contains all cuts $\alpha$, but now we additionally
  have for each $q\in\Q$, the set $c(q)=\{r\in\Q:r^2\leq q\}$ as an
  element of $S$. For this problem, we can refer to $c(q)$ as a
  ``closed cut,'' and elements in $S$ as \emph{generalized cuts}. Keep
  the same definitions of order and addition.
  \begin{enumerate}
  \item Show that the resulting ordered set has the least-upper-bound
    property.
  \item Show that $S$ is closed under addition (that is, verify that
    addition satisfies axiom (A1) from Rudin; see Definition 1.12).
  \item Verify that addition satisfies axioms (A2) and (A3) --
    symmetry and associativity.
  \item Verify that addition satisfies axiom (A4), i.e., that $S$ has
    an additive identity. What is the additive identity? [\emph{Hint}:
      It is \emph{different} than the zero element of the set of cuts!]
  \item Show that axiom (A5) fails. That is, not all elements in $S$
    have an addidtive inverse. \\
  \end{enumerate}



\item Fix $b>1$. This problem builds off of Theorem 1.21 in Rudin,
  which defined (and proved the existence of) for each $n\in\N$, the
  (positive) $n^{\rm th}$ root of $b$, as
  \[
  b^{1/n}:=\sup\{x\in\Q:x^n<b\}\,.
  \]
  You may use the result of the Corollary to this theorem (see page 11
  of Rudin): If $a$ and $b$ are positive real numbers and $n\in\N$,
  then $(ab)^{1/n}=a^{1/n}b^{1/n}$.
  \begin{enumerate}
    \item If $m,n,p,q\in\Z$, and $n,q>0$, and $r=m/n=p/q$, prove that
      \[
      (b^m)^{1/n}=(b^p)^{1/q}\,.
      \]
      Hence it makes sense to define $b^r=(b^m)^{1/n}$, i.e., this
      definition is \emph{well-defined}.
    \item Prove that $b^{r+s}=b^rb^s$ if $r$ and $s$ are rational.
    \item 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 $x$.
    \item Prove that $b^{x+y}=b^xb^y$ for all real $x$ and $y$. \\
  \end{enumerate}



\item Prove that no order can be defined in the complex field that
  turns it into an ordered field. [\emph{Hint}: $-1$ is a square.] \\



\item Suppose $z=a+bi$, $w=c+di$. Define $z<w$ if $a<c$, and also if
  $a=c$ but $b<d$. Prove that this turns the set of all complex
  numbers into an ordered set (This type of order relation is called a
  \emph{dictionary order}, or \emph{lexicographic order}, for obvious
  reasons.) Does this ordered set have the least-upper-bound property? \\



\item If $z_1,\dots,z_n$ are complex, prove that
  $|z_1+z_2+\cdots+z_n|\leq|z_1|+|z_2|+\cdots+|z_n|$. [\emph{Hint}: Use
  induction.] \\



\item Under what conditions does equality hold in the Cauchy-Schwarz
  inequality? Justify your answer.


\end{enumerate}

\end{document}
