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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Exam 1 -- February 25, 2011}
\bigskip
\noindent
I.  
\item{A)}  State the Axiom of Completeness for the real number system.
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\item{} {\it Solution:}  Every nonempty set of real numbers that is bounded
above has a least upper bound.  (This is an {\it axiom} -- we take it 
as an unproved assertion describing properties of the real number system.)
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\item{B)}  Prove that if $A$ and $B$ are bounded sets of real numbers
with ${\rm lub}(A) < {\rm lub}(B)$, then there exists a single $y \in B$ satisfying
$y > x$ for all $x \in A$.    
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\item{} {\it Solution:}  Since ${\rm lub}(A) < {\rm lub}(B)$, ${\rm lub}(A)$ is
not an upper bound for $B$.  Hence there exists at least one $y_0 \in B$ such that 
${\rm lub}(A) < y_0 \le {\rm lub}(B)$.  (The subscript is one way mathematicians
often indicate that they are referring to one particular element of a set.) 
 But then, by definition,
every $x \in A$ satisfies $x \le {\rm lub}(A)$.  So $x < y_0$ for all $x \in A$.  
\bigskip
\item{C)}  Let $A = \cap_{n=1}^\infty \left({-1\over 2n},1 + {1\over 2n}\right)$.
Explain why $A$ is bounded below and determine ${\rm glb}(A)$.  
\bigskip
\item{} {\it Solution:}  It is certainly true that ${-1\over 2}$ is a lower
bound for $A$.  But in fact, given any $\varepsilon > 0$, there exist
$n\in {\bf N}$ such that ${1\over 2n} < \varepsilon$, and hence ${-1\over 2n} > -\varepsilon$.
We can see that $A = [0,1]$, so ${\rm glb}(A) = 0$.  
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\noindent
II.  
\item{A)}  Prove that every interval $(a,b)$ with $0 < a < b$ contains
a rational number ${m\over n}$.  
\bigskip
\item{} {\it Solution:}  Since ${\bf N}$ is not bounded, there exists 
$n \in {\bf N}$ such that $n > {1\over b - a}$, or equivalently ${1\over n} < b - a$.
Pick any one such $n$ and fix it.
For that fixed $n$, let $M = \{m \in {\bf N} \mid m > na\}$.  This set is not
empty (again since ${\bf N}$ is not bounded).  So by the Well Ordering Property,
it contains a smallest element.  Call that $m_0$.  Now $m_0 > na$, so ${m_0 \over n} > a$.
On the other hand, $m_0 - 1 \ne M$, so $m_0 - 1 \le na$.  This implies
${m_0\over n} \le a + {1\over n} < a + b - a = b$.  Combining these last 
facts shows that the rational number ${m_0\over n} \in (a,b)$. 
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\item{B)}  Use part A to show that every interval $(a,b)$ with $a < b < 0$
contains a rational number ${m\over n}$.   
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\item{} {\it Solution:}  If $a < b < 0$, then $0 < (-b) < (-a)$.  By part A,
there is a rational number ${-m\over n}$ satisfying $(-b) < {-m\over n} < (-a)$.
But then, multiplying these inequalities by $-1$, 
$a < {m\over n} < b$.  Since ${m\over n}$ is also rational, we are done.  
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\noindent
III. Let $f : {\bf N} \to {\bf R}$ be defined by $f(1) = 2$ and
$f(n+1) = {2f(n) - 1\over 3}$ for all $n \ge 1$.  Using mathematical induction, show that 
$f(n) > -1$ for all $n\ge 1$.  
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\item{}  {\it Solution:}  The base case is $n = 1$.  By definition,
$f(1) = 2 > -1$.  Now assume as the induction hypothesis
that $f(k) > -1$, and consider
$f(k+1) = {2f(k) - 1\over 3}$.  Then the properties of order imply
$$\eqalign{
2f(k) &> -2\cr
{\rm hence}\ 2f(k) - 1 &> -3\cr
{\rm hence}\ f(k+1) = {2f(k) - 1\over 3} &> -1,\cr
}$$
which is what we wanted to show.
By the principle of mathematical induction, $f(n) > -1$ for all $n \ge 1$.
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\noindent
IV.  True-False.  For each true statement give a short proof or 
reason; for each false statement give a counterexample.
\bigskip
\item{A)}  If $A$ and $B$ are sets of real numbers 
with the property that $a > b$ for all $a\in A$ and all $b \in B$,
then ${\rm glb}(A) > {\rm lub}(B)$.
\bigskip
\item{}  {\it Solution:}  This is FALSE.  A counterexample -- let 
$B = (-\infty,0)$ and $A = (0,\infty)$ in ${\bf R}$.  Then $a > b$
for all $a\in A$ and all $b \in B$.  But ${\rm glb}(A) = 0 = {\rm lub}(B)$.  
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\item{B)}  (10) For all $n \ge 1$, $\sum_{k=0}^n {n\choose k} {1\over 2^k} = {3^n\over 2^n}$.  
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\item{} {\it Solution:}  This is TRUE.  By the binomial theorem,
$$\sum_{k=0}^n {n\choose k} x^k y^{n-k} = (x + y)^n$$
for all real numbers $x,y$.  We get the desired equality by making $x = {1\over 2}$, $y = 1$.   
\bigskip
\item{C)}  (10) The smallest $c$ such that 
$\{x \in {\bf R} \mid |x + 1| + |x + 10| = c\}$ is not empty is $c = 11$.  
\medskip 
\item{} {\it Solution:}  This is FALSE.  $|x + 1| + |x + 10|$ represents
the sum of the distance from $x$ to $-1$ and the distance from $x$ to $-10$
along the number line.  For any $x$ with $-10 \le x \le -1$, the sum of these
distances will be $9 < 11$.  
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\noindent
{\it Extra Credit}  (10)  {\it Caution: this problem may be ``habit forming.''
Only attempt after finishing the rest of the exam!} 
Can you find a 1-1, onto function $f : {\bf R} \to (-1,1]$?  If
so, describe one by giving a formula or a graph.  If there is no such function, 
prove it.
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\noindent
{\it Solution:}  There is such a function.  We can construct one as follows:
First take any strictly increasing onto function $f: {\bf R} \to (-1,1)$
such as $f(x) = {2\over \pi} \arctan(x)$.  This ``misses'' $1$, of course, 
so we have to modify $f$ to get $1$ in the range.  If we define a new
function $g$ with 
$g(1) = 1$, but $g(x) = f(x)$ for all $x \ne 1$, then we have missed $f(1) = {1\over 2}$
in the range of $f$.  But we can make $g(2) = f(1) = {1\over 2}$, then 
$g(3) = f(2)$, etc. to get a function that is still 1-1 and onto:
$$\cases{g(x) = f(x) = {2\over \pi}\arctan(x) & if $x \notin {\bf N}$\cr
         g(1) = 1\cr
         g(n) = f(n-1) & if $n \ge 2 \in {\bf N}$.\cr}.$$
(This is a mathematical version of the ``Hilbert Hotel" with rooms numbered
by $n\in {\bf N}$ -- there's always room for one more guest if we shift
everyone else down one room!)
\bye
