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\centerline{MATH 242 -- Principles of Analysis}
\centerline{Solutions for Problem Set 2 -- due:  Feb. 8}
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\noindent
{\it `A` Section}
\bigskip 
\item{1.}  Let $x \in [1,3]$.  Determine the largest and smallest
values of $|x - 5|$, $|x + 5|$, and $1/|x^2 - 25|$.
\item{} {\it Solution:}  If $x \in [1,3]$, then $x - 5 \in [-4,-2]$, so the largest
and smallest values of $|x - 5|$ are $4$ and $2$ respectively.  Next, $x + 5 \in [6,8]$, 
so the largest and smallest values of $|x + 5|$ are $8$ and $6$ respectively.  Finally,
$|x^2 - 25| = |x - 5||x + 5|$, so the largest and smallest values of 
$1/|x^2 - 25|$ are $1/16$ and $1/24$ respectively.  (Think of the graph $y = 1/|x^2 - 25$
to see this.)
\item{2.} Use the binomial theorem (Theorem 1.4.1) for all parts of this problem.
\itemitem{a.}  Expand using the binomial theorem and simplify as much as possible:
$$(a^2 - 5b^3)^6.$$
\itemitem{} {\it Solution:}  We have 
$$(a^2 - 5b^3)^6 = a^{12} - 30a^{10}b^3 + 375a^8b^6 - 2500a^6b^9 + 9375a^4b^{12} - 18750a^2b^{15} + 15625b^{18}.$$
\itemitem{b.}  What is the coefficient of $x^3$ in the expansion of 
$$\left({x^4 + 7x\over x^2}\right)^3.$$
\itemitem{} {\it Solution:}  Dividing inside the power, we are looking at
$$\left(x^2 + 7 x^{-1}\right)^3$$
The only $x^3$ term will come from squaring the $x^2$ and taking $x^{-1}$ to the first power.
So the coefficient is ${3\choose 1}\cdot 7 = 21$.   
\itemitem{c.}  What is $\sum_{k=0}^n {n\choose k}$?  Explain.
\itemitem{} {\it Solution:}  From the binomial theorem this sum is what we obtain from 
$$(1 + 1)^n = 2^n.$$
\itemitem{d.}  What is $\sum_{k=0}^n (-1)^k {n \choose k}$?  Explain.
 \itemitem{} {\it Solution:}  From the binomial theorem this sum is what we obtain from
$$(1 - 1)^n = 0^n = 0.$$
\item{3.}  For each of the following statements, say whether the statement is true or false.
If it is false, give a counterexample; if it is true, give a short reason.
\itemitem{a.}  A set $A \subset {\bf R}$ is bounded if there exists some $B$ such
that $x\le B$ for all $x \in A$.
\itemitem{} {\it Solution:}  FALSE -- the set is bounded above, but not necessarily below.  An 
example would be the open interval $(-\infty, 0)$ with $B = 0$.  
\itemitem{b.}  If $A, B\subset {\bf R}$ are bounded, then $A \cup B$ is also bounded.
\itemitem{} {\it Solution:}  TRUE -- If $A$ is bounded above by $M_A$ and $B$ is bounded
above by $M_B$, then $A \cup B$ is bounded above by $\max(M_A,M_B)$.  Similarly, 
if $A$ is bounded below by $m_A$ and $B$ is bounded
below by $m_B$, then $A \cup B$ is bounded above by $\min(m_A,m_B)$.
\itemitem{c.}  If $A, B\subset {\bf R}$ are bounded, then $D = \{x - y \mid x \in A, y \in B\}$ 
is also bounded.
\itemitem{} {\it Solution:}  TRUE -- Suppose $A$ is bounded above by $M_A$ and $B$ is bounded
above by $M_B$.  Similarly, suppose $A$ is bounded below by $m_A$ and $B$ is bounded
below by $m_B$.  Then for all $x \in A$ and $y \in B$ we have $m_A \le x \le M_A$ and 
$m_B \le y \le M_B$.  It follows that $m_A - M_B \le x - y \le M_A - m_B$.  Therefore
$D$ is bounded.
\itemitem{d.}  If $A, B\subset {\bf R}_{>0}$ are bounded, then $Q = \{x/y \mid x \in A, y \in B\}$
is also bounded.
\itemitem{} {\it Solution:}  FALSE -- Let $B = (0,1)$ and let $A = \{1\}$.  The set $Q$
is the set $\{1/y \mid y \in (0,1)\}$ which is not bounded above.  
\item{4.}  
\itemitem{a.}  Let $A = [0,4] \cap (1,5)$.  What is $a = {\rm lub}\ A$?  What is $b = {\rm glb}\ A$?
Are $a,b \in A$?
\itemitem{} {\it Solution:}  We have $A = (1,4]$.  Therefore $a = 4$, which is in $A$.  But
$b = 1$ is not in $A$.  
\itemitem{b.}  Let $B = \{x \in {\bf R} \mid 0 < x^2 - 4x + 1 < 4\}$.  What is $a = {\rm lub}\ B$?  What is $b = {\rm glb}\ B$?   Are $a,b \in B$?
\itemitem{} {\it Solution:}  By completing the square or using the quadratic 
formula, we see $B = (2 - \sqrt{7}, 2 - \sqrt{3}) \cup (2 + \sqrt{3},2 + \sqrt{7})$.
Hence $a = 2 + \sqrt{7}$ and $b = 2 - \sqrt{7}$.  Neither is in $B$.
\itemitem{c.}  Let $C = \{x \in {\bf Q} \mid x^2 < 5\}$.   What is $a = {\rm lub}\ C$?  What is $b = {\rm glb}\ C$?   Are $a,b \in C$?
\itemitem{} {\it Solution:}  There are rational numbers arbitrarily close to $\pm\sqrt{5}$, so 
whose squares are arbitrarily close to $5$, but $\pm \sqrt{5}$ are not themselves
rational numbers.  It follows that $a = \sqrt{5}$ and $b = -\sqrt{5}$, but neither is in $C$.  
\bigskip
\noindent
{\it `B` Section}
\bigskip
\item{1.}  Let $x,y$ be any real numbers.  
\itemitem{a.}  Show that $|x| - |y| \le |x - y|$ and deduce that $||x| - |y|| \le |x - y|$.
\itemitem{} {\it Solution:}  From the usual triangle inequality,
$$|x| = |(x-y) + y| \le |x - y| + |y|.$$
Subtracting, we obtain $|x| - |y| \le |x - y|$ as desired.  Similarly, reversing the 
roles of $x,y$, we have $|y| - |x| \le |y - x| = |x - y|$.  Since either $|x| \ge |y|$
or $|y| \ge |x|$ is true, we have either $||x| - |y|| = |x| - |y|$ or $||x| - |y|| = |y| - |x|$.
Since both of those are $\le |x - y|$, it follows that $||x| - |y|| \le |x - y|$
as desired.  
\itemitem{b.}  Show that if $x,y > 0$, then $x < y$ is equivalent to $x^2 < y^2$.
\itemitem{} {\it Solution:}  Since $x > 0$, $x < y$ implies $x^2 < xy$.  Similarly,
$x < y$ implies $xy < y^2$. But then transitivity implies $x^2 < y^2$.  Conversely,
if $x^2 < y^2$, then $0 < y^2 - x^2 = (y - x)(y + x)$.  Since $x,y > 0$, the sum
$x + y > 0$.  This implies that $y - x > 0$ too, or equivalently $x < y$.  
\itemitem{c.}  Show that if $0 < x < y$, then $\sqrt{y} - \sqrt{x} < \sqrt{y - x}$.
\itemitem{} {\it Solution:}  By part b, since $\sqrt{y} - \sqrt{x} > 0$ and 
$\sqrt{y - x} > 0$, it suffices to show that 
$$(\sqrt{y} - \sqrt{x})^2 < (\sqrt{y - x})^2.$$
But the left side here is $y - 2\sqrt{y}\sqrt{x} + x$ and the right side is $y - x$.  
We have 
$$(y - x) - (y - 2\sqrt{y}\sqrt{x} + x) = 2\sqrt{y}\sqrt{x} - 2x = 2\sqrt{x}(\sqrt{y} - \sqrt{x}).$$
This is $>0$ because of the assumption $y > x$ and part b.  Hence the desired inequality follows. 
\item{2.}  Let $a,b$ be any real numbers.  Define $\max(a,b)$ and $\min(a,b)$ 
to be the larger and smaller of the two numbers, respectively.  (That is, $\max(a,b) = a$
if $a \ge b$ and $\max(a,b) = b$ if $b \ge a$.  Similarly for the minimum.)
Show that 
$$\max(a,b) = {a + b\over 2} + {|a - b|\over 2}$$
and 
$$\min(a,b) = {a + b\over 2} - {|a - b|\over 2}.$$
\item{} {\it Solution:}  Geometrically, the average ${a + b\over 2}$ is the midpoint
of the line segment along ${\bf R}$ between $a$ and $b$ and $|a - b|$ is the 
distance between $a$ and $b$ (the length of the line segment).  So starting at
the midpoint and going $1/2$ the distance to the right gives the maximum of the 
endpoints, and going $1/2$ the distance to the left gives the minimum of the endpoints. 
More analytically, we can also prove these by breaking into cases.  Suppose first that 
$a \ge b$ so $a$ is the right endpoint.  Then $|a - b| = a - b$, and 
$${a + b\over 2} + {|a - b|\over 2} = {a + b\over 2} + {a - b\over 2} = a = \max(a,b),$$
while 
$${a + b\over 2} + {|a - b|\over 2} = {a + b\over 2} - {a - b\over 2} = b = \min(a,b).$$
If $b$ is the maximum and $a$ is the minimum, then 
$${a + b\over 2} + {|a - b|\over 2} = {a + b\over 2} + {b - a\over 2} = b = \max(a,b),$$
while
$${a + b\over 2} - {|a - b|\over 2} = {a + b\over 2} - {b - a\over 2} = a = \min(a,b).$$
\item{3.}  Show by mathematical induction that 
$$1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$$
for all $n\ge 1$.  
\item{} {\it Solution:}  With $n = 1$, we have $1 = 1$, so the base case is established.  
Now assume the formula has been proved for $n = k$ and consider the next case $n = k+1$.
We have by the induction hypothesis
$$\eqalign{
(1^3 + 2^3 + \cdots + k^3) + (k+1)^3 &= (1 + \cdots + k)^2 + (k + 1)^3\cr
&= \left({k(k+1)\over 2}\right)^2 + (k+1)^3\cr
&={(k+1)^2(k^2 + 4k + 4)\over 4}\cr
&={(k+1)^2(k+2)^2\over 4}\cr
&=(1 + 2 + \cdots + (k+1))^2\cr}$$
\item{4.}  Show by mathematical induction that $n! \ge 2^n$ for all $n \ge 4$.  
\item{} {\it Solution:}  The base case here is $n = 4$ and $4! = 24 > 16 = 2^4$ is 
true.  For the induction step, assume $k! > 2^k$ and consider $(k+1)!$.  
We see, by the induction hypothesis:
$$(k+1)! = (k+1)k! > (k+1) 2^k > 2\cdot 2^k = 2^{k+1}.$$
(The third step is valid since $k \ge 4$ implies $k + 1 > 2$.) 

\bye
