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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Exam 1 -- October 7, 2005}
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I.
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A) (10) State the Axiom of Completeness for the real number system.
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\item{\it Solution:}  Every set $A \subset {\bf R}$ that is bounded
above has a least upper bound $s = \sup(A)$ in ${\bf R}$.
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B)  (10) Show that if $s = \sup(A)$ for some $A\subset {\bf R}$
and $\varepsilon > 0$, then there exists some $a \in A$ satisfying
$s - \varepsilon < a \le s$.  
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\item{\it Solution:}  By definition, $s$ is the real smallest number 
that is an upper bound for $A$.  Since $\varepsilon > 0$, 
$s - \varepsilon < 0$ and hence $s - \varepsilon$ is {\it not}
an upper bound for $A$.  This implies that there is some
$a\in A$ with $s - \varepsilon < a$.  $a \le s$ because $s$
is an upper bound for $A$.
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C)  (10) Let $A = \cup_{n=1}^\infty \left(0,1-{1\over 2n}\right)$.
Explain why $A$ is bounded above and determine $\sup(A)$.  
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\item{\it Solution:} 
If $x \in A$, then there is some $n$ such that $x < {1\over 2n}$.
Since ${1\over 2n} < 1$ for all $n\in {\bf N}$, we get $x < 1$ for
all $x\in A$. In fact, $\sup(A) = 1$ since if $y = 1 - \varepsilon < 1$, then
the Archimedean Property implies that there exists $n \in {\bf N}$
such that ${1\over 2n} < \varepsilon$, so 
$1 - {1\over 2n} > 1 - \varepsilon = y$.  Hence there exist $x \in A$
with $x > y$ too, and $y$ cannot be an upper bound for $A$.  
Since 1 is an upper bound for $A$, but no $y < 1$ is, 
$\sup(A) = 1$.
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\noindent
II. 
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A) (5) Find 
$$\lim_{n\to\infty} {2n^2 - 1\over 5n^2 + 2n}$$  
using the Algebraic Limit Theorem.
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\item{\it Solution:}  
$$\lim_{n\to\infty} {2n^2 - 1\over 5n^2 + 2n}
= \lim_{n\to\infty} {2 - {1\over n^2} \over 5 + {2\over n}} = {2\over 5}$$ 
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B) (15) Prove that $x_n = {2n^2 - 1\over 5n^2 + 2n}$ 
converges to your limit from part A {\it using the definition}.
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\item{\it Solution:}  Preliminaries:  We can use some slightly
tricky estimates to reduce the complication of the $N$ that
works for a given $\varepsilon > 0$:
$$\eqalign{\left|{2n^2 - 1\over 5n^2 + 2n} - {2\over 5}\right|
&= {4n+5 \over 25n^2 + 10n}\cr
& < {4n+5 \over 25n^2}\cr
& < {5n + 5\over 25n^2}\cr
& = {n + 1\over 5n^2}\cr
& < {2n\over 5n^2} = {2\over 5n}\cr}$$
Proof:  Let $\varepsilon > 0$, and let $N \in {\bf N}$
satisfy $N > {2\over 5\varepsilon}$.  Then
for all $n \ge N$, 
$$\eqalign{\left|{2n^2 - 1\over 5n^2 + 2n} - {2\over 5}\right|
&= {4n+5 \over 25n^2 + 10n}\cr
& < {4n+5 \over 25n^2}\cr
& < {5n + 5\over 25n^2}\cr
& = {n + 1\over 5n^2}\cr
& < {2n\over 5n^2}\cr
&\le {2\over 5N}\cr
& < \varepsilon\cr}$$
Hence $\lim_{n\to\infty} {2n^2 - 1\over 5n^2 + 2n} = {2\over 5}$.
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III.  Let $(x_n)$ be the sequence defined by $x_1  = 2$ and
$x_{n+1} = {2x_n - 1\over 3}$ for all $n \ge 1$.  
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A) (15) Using mathematical induction, show that 
$x_n \ge -1$ for all $n\ge 1$.  
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\item{\it Solution:}  The base case is $n=1$, and $x_1 = 2 \ge -1$
from the definition of the sequence.  For the induction step,
assume $x_k\ge -1$, and apply the operations on both sides to 
produce $x_{k+1}$ on the left:
$$x_{k+1} = {2x_k - 1\over 3} \ge {2(-1) - 1\over 3} = -1.$$
Hence by induction, $x_n \ge -1$ for all $n \in {\bf N}$.
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B) (15) Using mathematical induction, show that 
$(x_n)$ is strictly decreasing:  $x_{n+1} < x_n$ for all $n\ge 1$.
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\item{\it Solution:}  With $n = 1$, we have $x_2 = {2(2) - 1\over 3} = 1 
< x_1 = 2$.  Now assume that $x_{k+1} < x_k$ and apply the operations
producing $x_k$ from $x_{k+1}$ to both sides:
$$x_{k+2} = {2x_{k+1} - 1\over 3} < {2x_k - 1\over 3} = x_{k+1}.$$
This shows $x_{n+1} < x_n$ for all $n\in {\bf N}$.
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IV.  True-False.  For each true statement give a short proof or 
reason; for each false statement give a counterexample.
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A) (10) If $A$ and $B$ are sets of real numbers 
with the property that $a > b$ for all $a\in A$ and $b \in B$,
then $\inf(A) > \sup(B)$.
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\item{\it Solution:}  This is FALSE.  It can be true that $a > b$ for
all $a \in A$ and $b \in B$, but $\inf(A) = \sup(B)$.  Counterexample
to the statement as given:
$A = \left\{1 + {1\over n} : n \in {\bf N}\right\}$ and 
$B = \left\{1 + {1\over n} : n \in {\bf N}\right\}$.
All elements of $A$ are strictly bigger than 1, while
all elements of $B$ are strictly less than 1.  Hence
$a > b$ for all $a\in A$ and $b \in B$.  But
$\inf(A) = \sup(B) = 1$.
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B)  (10) If $r$ is a nonzero rational number and $t$
is an irrational number, then $rt$ is irrational.
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\item{\it Solution:} This is TRUE.  Arguing by contradiction,
suppose $rt \in {\bf Q}$.  Write $r = a/b, rt = c/d$ with 
$a,b,c,d \in {\bf Z}$.  Then 
$at/b = c/d$, or $t = (bc)/(ad) \in {\bf Q}$.  This is 
a contradiction.  so $t$ must be irrational.
\bye