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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Discussion 2 -- $\sup(S)$ continued}
\centerline{September 29, 2004}
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\noindent
{\it Background}
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Last time, we introduced the {\it supremum} or {\it least upper 
bound} of a subset $S \subseteq {\bf R}$.  Recall, we say that
$s = \sup(S)$ if 
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\item{$\bullet$} $s$ is an upper bound of $S$ -- $x \le s$
is true for all $x \in S$, and
\item{$\bullet$} if $s' < s$, then there exists $x \in S$
such that $x > s'$.
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\noindent
The {\it completeness axiom}, the crucial property of 
the real numbers, is:

\proclaim Completeness Axiom.  If $S$ is nonempty and
bounded above
in ${\bf R}$, then $\sup(S)$ exists in ${\bf R}$.\par

\noindent
That is, {\it every nonempty set that is bounded above 
in ${\bf R}$ has a least upper bound in ${\bf R}$}.
Today, we want to study further properties of the supremum.
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\noindent
{\it Discussion Questions}
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\item{A)}  The second defining property of the supremum $s = \sup(S)$
above says: if $s' < s$, then there exists $x \in S$
such that $x > s'$.  Show that this statement is equivalent 
to saying:  If $t$ is any upper bound for $S$, then $t \ge s$.  
(Note and Hint:  In fact, many analysis books {\it define} the
supremum using this alternate form, because it clearly shows
that $\sup(S)$ is the ``least upper bound'' for $S$.
The relation between these statements comes from
one of our basic logical equivalences!)
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\item{B)}  Let $S$ be a nonempty bounded subset of ${\bf R}$.  Show
that $s = \sup(S)$ is {\it unique}.  Method: Assume that two 
real numbers $s,s'$ both satisfy the definition, then deduce 
that $s$ must equal $s'$.
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\noindent
\item{C)}  Prove that if $x < y$ are any real numbers, then 
there exist {\it infinitely many} rational numbers $r$
with $x < r < y$.  Hint:  Last time we showed that there is one
such rational number.  Try to extend that proof.
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\noindent
\item{D)}  Let $a/b$ be a rational number written as a 
fraction in lowest terms with $0 < a/b < 1$.
\medskip
\itemitem{1)}  Show using the Archimedean Property that
there exists an integer $n$ such that
$${1\over n+1} \le {a\over b} \le {1 \over n}.$$
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\itemitem{2)} If $n$ is chosen as in part 1, show that
$a/b - 1/(n+1)$ is a fraction that when written in lowest terms 
has numerator is less than $a$.
\itemitem{3)}  Use the Principle of Strong Induction from
Problem Set 3 to show that every rational number $a/b$ as
above can be written as a sum:
$${a\over b} = {1\over n_1} + \cdots + {1\over n_k}$$
for some distinct natural numbers $n_i$.
(For instance $49/90 = 1/3 + 1/9 + 1/10$.)
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\item{} Such fractions are often called ``Egyptian fractions''
because the ancient Egyptians did all of their rational number 
arithmetic using them (for some reason they didn't like fractions 
with a numerator different from 1).  They wrote all fractions
this way, then computed sums and products via tabulations
that have survived in ancient papyrus manuscripts -- the 
``Rhind papyrus'' is probably the most famous example.  It is
one of the oldest surviving ancient mathematical documents.
There is an interesting discussion of this online at
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\centerline{\tt 
http://www-gap.dcs.st-and.ac.uk/\~{}history/HistTopics/Egyptian\_papyri.html}
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\noindent
{\it Assignment}
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Group writeups due by 5:00pm on Monday, October 4.
\bye  