\magnification=\magstep1
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Exam 1 -- February 22, 2013}
\bigskip
\noindent
I.
\item{A)} (10)  Let
$$A = \bigcup_{n=1}^\infty \left({1\over n} ,2 - {1\over n}\right)$$
Explain why $A$ is bounded and determine upper and lower bounds for $A$.
\item{} {\it Solution:}  $A$ is bounded below since ${1\over n} > 0$ for all $n$.
Therefore all elements of $A$ are strictly positive, or $x > 0$ for all $x \in A$.  On the other hand $2 - {1\over n} < 2$
for all $n$, so all elements of $A$ satisfy $x < 2$.
\item{B)} (10)  {\it Define:}  The real number $a$ is a {\it least upper bound} of $A \subset {\bf R}$, and
state the Least Upper Bound Axiom for ${\bf R}$.
\item{} {\it Solution:}  $a$ is a least upper bound of $A$ if (1) $a \ge x$ for all $x \in A$, and
(2)  If $b \ge x$ for all $x \in A$, then $b \ge a$.  The LUB Axiom states that every nonempty
set of real numbers that is bounded above has a least upper bound in ${\bf R}$.   (The existence
is not guaranteed just by defining a term by giving the properties that a lub should
satisfy.)
\item{C)} (10)  Let $A$ be a bounded subset of ${\bf R}$ and let $B = \{4\cdot x - 3\mid x \in A\}$.
What can be said about ${\rm lub}(B)$?  Prove your assertion.
\item{} {\it Solution:}  If $a = {\rm lub}(A)$, then we claim ${\rm lub}(B) = 4a - 3$.   To prove
this, note that $a$ exists in ${\bf R}$ by the LUB Axiom.  Since $a \ge x$ for all $x \in A$,
we have $4a \ge 4x$ and $4a -3 \ge 4x - 3$ by properties of the order relation.  This
shows that $4a - 3$ is an upper bound for $B$.  Then, if $b$ is any upper bound for $B$,
we have $b \ge 4x - 3$ for all $x \in A$, so ${b + 3\over 4} \ge x$ for all $x\in A$.
This implies ${b + 3\over 4} \ge a$ by definition of an lub.  Hence $b \ge 4a - 3$.
This shows that $4a - 3 = 4{\rm lub}(A) - 3 = {\rm lub}(B)$.
\bigskip
\noindent
II.  (20) Let $x_n$ be the sequence defined by the rules $x_1 = 1$ and $x_{n+1} = {1\over 3} x_n + 1$
for all $n \ge 1$.  Show by mathematical induction that
$$x_n = {1 - {1\over 3^n} \over 1 - {1\over 3}}\ {\rm for\ all\ } n\ge 1.$$
\noindent
{\it Solution:}  The base case is $n = 1$.  We have $x_1 = 1 = {1 - {1\over 3} \over 1 - {1\over 3}}$,
so the formula is true in this case.  Now assume that
$$x_k =  {1 - {1\over 3^k} \over 1 - {1\over 3}}.$$
By definition,
$$\eqalign{
x_{k+1} &= {1\over 3}x_k + 1 \quad {\rm (by\ definition\ of\ the\ sequence)}\cr
&= {1\over 3} \cdot  {1 - {1\over 3^k} \over 1 - {1\over 3}} + 1\quad{\rm (by\ the\ induction\ hypothesis)} \cr
&= {{1\over 3} - {1\over 3^{k+1}} \over 1 - {1\over 3}} + 1\quad \cr
&= {{1\over 3} - {1\over 3^{k+1}} + 1 - {1\over 3} \over 1 - {1\over 3}}\quad {\rm (common\ denominator)}\cr
&= {1 - {1 \over 3^{k+1}}\over 1 - {1\over 3}} \quad {\rm (by\ algebra)}\cr
}$$
This shows that $x_n$ is given by the formula above for all $n\ge 1$.
\bigskip
\noindent
III.   Let $x_n = {5^n \over 7^n + 3}$ for all natural numbers $n \ge 1$.
\item{A)}  (10) Determine $\lim_{n\to\infty} x_n$ intuitively.
\item{} {\it Solution:}  We have
$${5^n \over 7^n + 3} = {(5/7)^n \over 1 + 3 (1/7)^n}$$
Since $0 < 5/7 < 1$ and $0 < 1/7 < 1$, $(5/7)^n \to 0$ and $(1/7)^n \to 0$.
We expect the limit should be $0$.
\item{B)}  (20)  Use the $\varepsilon$, $n_0$ definition of convergence
to prove that $\{x_n\}$ converges to the number you identified in part A.
\item{} {\it Solution:}  Given $\varepsilon > 0$, let $n_0$ be any natural
number satisfying $n_0 > {\ln(\varepsilon)\over \ln(5/7)}$.  (Note: If $\varepsilon < 1$,
then both the top and the bottom of the quotient are negative so the lower bound
on $n_0$ is positive, and the smaller $\varepsilon$ is the larger $n_0$ will be.)
The for all $n \ge n_0$ we will have
$$\eqalign{
\left| {5^n \over 7^n + 3} - 0\right| &= {5^n\over 7^n + 3}\cr
&< \left({5\over 7}\right)^n\cr
&<    \left({5\over 7}\right)^{n_0} \quad ({\rm since\ } 5/7 < 1\ {\rm and\ } n \ge n_0)\cr
&<  \left({5\over 7}\right)^{{\ln(\varepsilon)\over \ln(5/7)}} \cr
&= e^{ \ln(5/7)\cdot \ln(\varepsilon)/\ln(5/7)}\cr
&= e^{\ln(\varepsilon)}\cr
&=\varepsilon.\cr}$$
This shows that $x_n \to 0$.
\bigskip
\noindent
IV.  True-False.  For each true statement, give a short proof or reason.  For each false
statement give an explicit counterexample.
\item{A)} (10)  Let $A$ be a bounded set of real numbers and let $a = {\rm lub}(A)$.
Then for each $\varepsilon > 0$, there exists $x\in A$ such that $a - \varepsilon < x < a$.
\item{} {\it Solution:}  This is FALSE.  For example let $A = [0,1] \cup \{2\}$.
Then $a = {\rm lub}(A) = 2$.  But if $\varepsilon < 1$, then there are no
elements of $A$ in the open interval $(2 - \varepsilon, 2)$.  (The statement would
be true if the inequalities were $a - \varepsilon < x \le a$, but it is false if we do not
allow $x = a$. )
\item{B)} (10)  If $r$ is a nonzero rational number and $s$ is a nonzero irrational
number, then $r/s$ is irrational.
\item{} {\it Solution:}  This is TRUE.  Since $r \ne 0$ and $s \ne 0$, $r/s = t \ne 0$.
Suppose $t$ is rational.  Then we can solve to get $s = r/t \in {\bf Q}$, which
contradicts the assumption $s \notin {\bf Q}$.  Hence $t$ must be irrational.
\bigskip
\noindent
{\it Extra Credit} (10)  Is it possible to produce a sequence $x_n$ whose
terms include all the rational numbers $p/q$ with $p, q\in {\bf Z}$ and
$p,q > 0$?   If so, give an indication how to construct such a sequence.
If not, give a reason why there cannot exist such a sequence.
\bigskip
\noindent
{\it Solution:}  There is such a sequence, and we can construct one as
follows.  List all the ratios of positive integers in a two-dimensional table
with denominators constant across the rows and numerators constant
along the columns:
$$\matrix{
1/1 &  & 2/1 & & 3/1 & & 4/1 & & \cdots \cr
1/2 & & 2/2 & & 3/2 & & 4/2 & & \cdots \cr
1/3 & & 2/3 & & 3/3 & & 4/3 & & \cdots \cr
1/4 & & 2/4 & & 3/4 & & 4/4 & & \cdots \cr
\vdots & & \vdots & & \vdots & & \vdots & & \ddots \cr}$$
We can then construct a sequence by starting at the upper left and listing
the elements in this ``zig-zag order'' (down the first diagonal, up the second, and
and alternating that way forever):
$$x_1 = 1/1, x_2 = 2/1, x_3 = 1/2, x_4 = 1/3, x_5 = 2/2, x_6 = 3/1, x_7 = 4/1, x_8=3/2,\ldots$$
Every positive rational will appear somewhere in this sequence (in fact infinitely many
times each, do you see why?)
\bye  