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\begin{document}

\pset{6}{April 1, 2003}{Start of class on Tuesday, April 8}

%\newcommand{\solution}[1]{}                          % hide solutions
\newcommand{\solution}[1]{\textbf{Solution.} #1}    % show solution

\newcommand{\beqn}{\begin{eqnarray*}}
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\newcommand{\ex}[1]{\text{Ex} \left[ #1 \right] }
\newcommand{\pr}[1]{\text{Pr} \left\{ #1 \right\} }

\begin{problems}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem The following equation holds for all $n \in \mathbb{N}$.

\beqn
1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \cdots + n (n+1) (n+2)
	& = &	\frac{n(n+1)(n+2)(n+3)}{4}
\eeqn

\begin{problemparts}

\problempart Prove this fact using induction.

\solution{The proof is by induction.  Let $P(n)$ be the proposition
that:

\beqn
1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \cdots + n (n+1) (n+2)
	& = &	\frac{n(n+1)(n+2)(n+3)}{4}
\eeqn

First, note that $P(0)$ holds, because both sides of this equation are
equal to zero.

Now we must show that for all $n \geq 0$, $P(n)$ implies $P(n+1)$.
Suppose that $P(n)$ is true, where $n \geq 0$.  Then we can reason
as follows:

\beqn
\lefteqn{
1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \cdots + n (n+1) (n+2) +
		(n+1) (n+2) (n+3)} \hspace{1in} \\
	& = &	\frac{n(n+1)(n+2)(n+3)}{4} + (n+1) (n+2) (n+3) \\
	& = &	\frac{(n+1)(n+2)(n+3)(n+4)}{4}
\eeqn

This shows that $P(n)$ implies $P(n+1)$.  Therefore, the claim holds
by induction.
}

\problempart Prove this fact using well-ordering.

\solution{The proof uses well-ordering.  Suppose that the claim is
false.  Let $S$ be the set of all $n \in \mathbb{N}$ such that the
equation does not hold.  By our supposition, $S$ is nonempty, and by
the well-ordering priniciple, $S$ contains a smallest element.  This
smallest element cannot be 0; in the case $n = 0$, both sides of the
equation are equal to zero and are therefore equal to each other.
Therefore, the smallest element of $S$ must be some integer $s > 0$.
Since $s - 1$ is in $\mathbb{N}$ and is not in $S$, the equation must
hold for $n = s - 1$.  Thus we have:

\beqn
\lefteqn{
1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \cdots + (s-1) s (s+1) +
		s (s+1) (s+2)} \hspace{1in} \\
	& = &	\frac{(s-1)s(s+1)(s+2)}{4} + s (s+1) (s+2) \\
	& = &	\frac{s(s+1)(s+2)(s+3)}{4}
\eeqn

\noindent The first step uses the fact that the equation holds for $n
= s - 1$, and the second step uses only algebra.  This shows that the
equation holds for $n = s$, contradicting the definition of $s$.
Therefore, our original supposition was wrong, and the claim is true.
}

\end{problemparts}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem Is the proof of the following claim valid?  If not, identify
the first erroneous statement and explain why it is wrong.

{\bf Claim } {\em Every nonegative rational number is equal to $a^2 /
b^2$ for some pair of nonegative integers $a$ and $b$.}

\begin{proof}
The proof uses well-ordering.  Suppose that the claim is false; that
is, there is at least one nonnegative rational number that cannot be
written as $a^2 / b^2$ for some pair of nonnegative integers $a$ and
$b$.  Let $q$ be the smallest such number.  Then, since $q' = q / 4$
is a nonegative rational number smaller than $q$, there exist
nonnegative integers $a$ and $b$ such that $q' = a^2 / b^2$.  But then
$q$ is also equal to a ratio of squares, namely $(2 a)^2 / b^2$.  This
contradicts the definition of $q$.  Therefore, our original
supposition must be false, and so the claim is true.
\end{proof}

\solution{The proof is not valid.  The first erroneous phrase is ``Let
$q$ be the smallest such number.''  There is no assurance that such a
$q$ exists.  The well-ordering principle guarantees that every subset
of $\mathbb{N}$ has a smallest element, but it does not guarantee that
every subset of the positive rational numbers has a smallest element.

In fact, the set of counterexamples to the claim has no smallest
element; every term in the following, strictly-decreasing sequence is
a counterexample:

\[
\frac{1}{2}, \quad \frac{1}{2^3}, \quad \frac{1}{2^5}, \quad
\frac{1}{2^7}, \quad \ldots
\]

\noindent In other words, we are claiming that $1 / 2^{2k+1}$ is not a
ratio of perfect squares for any $k \in \mathbb{N}$.  Suppose this
were not the case; that is, there existed $a$ and $b$ such that:

\beqn
\frac{1}{2^{2k+1}} & = & \frac{a^2}{b^2}
\eeqn

\noindent Then it follows that:

\beqn
b^2 & = & a^2 \cdot 2^{2k+1}
\eeqn

\noindent But this is impossible; 2 divides the left side an even
number of times, but divides the right side an odd number of times.}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem Try drawing the figures below without lifting your pencil or
running over the same segment twice.

\bigskip
\centerline{\resizebox{!}{1in}{\includegraphics{pset6-euler.eps}}}
\bigskip

You should be able to draw the first, but not the second.  Note that
these figures can be interpreted as graphs.  The little circles
represent vertices and the lines between them indicate edges.  Call a
graph that can be drawn without lifting your pencil or retracing a
line {\em drawable}.

Now consider the following two conditions on a graph:

\begin{enumerate}

\item The graph is connected; that is, there is a path between every
pair of vertices.

\item The number of vertices with odd degree is either zero or two.

\end{enumerate}

\begin{problemparts}

\problempart Prove that every drawable graph satisfies both of these
conditions.

\solution{
To solve this problem, we need a precise idea of what a path is. A path is
a sequence $(a_1, \cdots, a_n)$ of vertices, such that $a_i \ne a_{i+1}$. 
If the sequence contains $n$
vertices, then we will say that the path has a length of $n-1$. Therefore,
we will conventionally consider that the empty path has a length of $-1$. A
path is closed if $a_1=a_n$, it is open otherwise.
A graph is drawable if there is a path $(a_1, \cdots, a_n)$ such that each
node is $a_i$ for some $i$, and for each vertex, there is exactly one $i$
such that $\{a_i, a_{i+1}\}$ is that vertex (note that $\{u,v\}=\{v,u\}$).

Property 1 is easy to show. Indeed, if a graph is drawable, then there
exists a path $p$ that goes through every vertex and edge of the graph without
going through an edge twice. For any pair $(a, b)$ of vertices in the graph, those
vertices must therefore appear at least once each in $p$. Consider the subpath of
$p$ that lies between some instance of $a$ and some instance of $b$. That
subpath goes from $a$ to $b$, or $b$ to $a$. The existence of such a subpath for any pair of
vertices in the graph implies that the graph is connected.

The empty graph trivially satisfies Property 2. For other drawable graphs,
the path needed to draw the graph has length at least 0. We shall show
property 2 by induction on that length.
Let $P(n)$ be the predicate ``a drawable graph
that can be drawn in $n$ edges is of one of two types: either it only has
even degree vertices, and it is drawn by a closed path; either it has exactly
two odd degree vertices, and can be drawn by an open path''.

\begin{itemize}

\item Base case: The only graphs that can be drawn with a path of length of 0 is
the single-node graph. Its single node has an even number of edges, and
the path $(v)$ used to draw it starts and ends on the same vertex. So $P(0)$ holds.

\item Inductive step: Assume that $P(n)$ holds for $n>0$. Consider any drawable graph 
$G$ that can be drawn by a path $p$ of length $n+1$. Let $s$ and $e$ be the
first and last vertices in $p$. Let $q$ be the path
that results from removing the last virtex from $p$, it has a length of $n$.
Let $H$ be the graph that is drawn by $q$. By definition, $H$ is drawable, therefore
it satisfise the conditions laid out in $P(n)$.

\begin{itemize}

%#### FIGURE

\item Case 1: If $H$ only has even-degree vertices, then let $s$ be the last
vertex of $q$ (the first and last elements of $q$ are the same by the induction
hypothesis). In $H$, $s$ is an even degree edge, and if $e$ is present it is also an even
degree edge. Because they are consecutive in $p$, $s$ and $e$ are distinct,
so $p$ is an open path.
Moreover, by adding the edge $\{s, e\}$ to $H$ to make $G$, $s$ becomes an
odd-degree vertex, and whether it was present in $H$ or not, $e$ becomes an
odd-degree vertex. All the other vertices in $H$ are unchanged by the
operation and remain even-degree edges. Therefore, $G$ is a graph with
exactly two odd degree vertices, and the odd degree vertices are $s$ and $e$ which
are the first and last vertices of the open path $p$ which defines $G$.

%#### FIGURE

\item Case 2: If $H$ has exactly two odd-degree vertices, then let $a$ be the
last vertex in $q$. By the induction hypothesis $s$ and $a$ are odd degree
vertices in $H$. When we add $\{a, e\}$ to $H$ to make it into $G$, $a$
becomes an even degree vertex since $a$ and $e$, being consecutive in $p$, 
must be distinct.

\begin{itemize}

\item Case 2.1: If $s$ and $e$ are distinct vertices, then when going from $H$ to
$G$, $s$ remains an odd degree vertex, while $e$ becomes an odd degree
vertex (it was even degree or absent in $H$ being distinct from $s$ and
$a$). All unmentioned vertices preserve their parity from $H$ to $G$. Therefore,
there are exactly 2 odd degree vertices $s$ and $e$ in $G$, and the path
that defines $G$ is open since $s$ and $e$ are distinct.

\item Case 2.2: If $s=e$ then when going from $H$ to $G$, $s$ becomes an even
degree vertex. All unmentioned vertices preserve their parity from $H$ to
$G$. Therefore $G$ only has even degree vertices, and is defined by a
closed path.

\end{itemize}

\end{itemize}

Thus in each case we have seen that the requirements for P(n+1) are met, so
P(n+1) holds.

\end{itemize}

We may conclude by induction on $n$ that $P(n)$ holds for all $n$. Thus for any
drawable graph Property 2 holds, either because the graph is empty, or
because of $P(n)$ where $n$ is the number of edges in the graph.
}

\problempart Prove that every graph satisfying both of these
conditions is drawable.

\solution{

Will be provided later.

%%We begin by a simple lemma: if $G$ is a connected graph, then removing an
%%edge from $G$ produces a graph that contains at most two connected
%%components.
%%
%%Indeed, every element $e$ of $G$ has a
%%path $r$ to $a$. Consider the set $R$ of prefixes of $r$ that contain $a$ or $b$. 
%%$R \ne \emptyset$ because $r \in R$. By well ordering there is a shortest
%%path $r'$ in $R$. The vertices $a$ and $b$ cannot appear in $r'$ before the last
%%vertex or else there would be a shorter path than $r'$ in $R$. appear at the last position and
%%only at the last position in $r'$ or else, 
%
%%#####
%
%Let $P(n)$ be the following predicate ``a graph with $n$ edges is drawable if
%properties 1 and 2 are satisfied''. Let us prove $P(n)$ by induction on n.
%
%\begin{itemize}
%
%\item Base case: A connected graph with no edges may not have more than one
%node, as there are no edges to support a path between two distinct nodes.
%It must therefore be the empty graph or a single node graph, both of which
%are drawable. So $P(0)$ holds.
%
%\item Inductive step: Suppose that $P(n)$ holds for all $k<=n$. Let $G$ be any 
%$n+1$ edge graph that
%satisfies properties 1 and 2.
%
%We remove an edge $\{a,b\}$ from $G$ to produce a graph $H$. If $G$ has two
%odd degree vertices, then we arrange for $a$ to be one of those vertices,
%otherwise, we choose the edge arbitrairly. The degree of $a$ and $b$ goes
%down by one in this operation. Thus, if there were no odd degree edges,
%there are now two, whereas if there were two odd degree edges there can be
%either none or two (because of the choice of $a$, and depending on whether
%$b$ is an odd degree node or not). Thus $H$ satisfies Property 2.
%Furthermore, we note that in either case
%where $H$ has two odd degree edges, $b$ must be one of them. 
%By Lemma 1, $H$ has either one or two connected components.
%
%\begin{tabular}[|c|c|c|]
%\hline
%Odd degree edges in $G$	& degree of $a$ & degree of $b$	& $H$ \\
%\hline
%0			& even	 	& even          & Connected \\
%\hline
%2			& odd		& even          & Connected \\
%\hline
%2			& odd		& even          & One odd part, one even part \\
%\hline
%2			& odd           & odd           & Connected \\
%\hline
%
%
%\end{tabular}
%
%\begin{itemize}
%
%\item Case 1: $H$ is connected. In that case $H$ is drawable by the
%induction hypothesis. If $H$ has two odd degree edges then $b$, being odd, is either
%the first or the last vertex in a path that draws $H$ (result proven in part a). 
%Otherwise, $H$ has 
%no odd degree nodes, so a path that draws $H$ is in fact a loop (result
%proven in part a), so by
%breaking the loop at $b$, $b$ can be made to be an extremity of a path that
%draws $H$. Edge $\{a, b\}$ can be added at an extremity where $b$ appears
%to make a path that draws $G$, so $G$ is drawable. 
%
%\item Case 2: $H$ has two connected components. Each component has an even
%number of odd degree nodes by Lemma 2. Since there are at most 2 odd degree
%nodes all together, both connected components satisfy Property 2.
%Therefore, both components are drawable by the induction hypothesis. 
%
%%#### FIGURE
%
%\item Case 1: Suppose that $G$ only has even degree vertices. We construct
%a graph $H$ by removing some arbitrairy edge $\{a,b\}$ from $G$ ($G$ has
%$n+1>0$ edges so we can always do this). $H$ is a $n$ node graph. All of
%its edges have even degree inherited from $G$, except $a$ and $b$ which
%have odd degree because one of their edges has been removed, so $H$
%satisfies Property 2. 
%##### $H$ is connected, so it satisfies Property 1. Therefore, by the
%induction hypothesis, $H$ is drawable. And by the predicate we proved in
%(a), there is a path $p$ that draws $H$ that has $a$ and $b$ as its
%extremities (because $a$ and $b$ have odd degree). The path $p$ covers all
%the edges of $G$ except $\{a, b\}$, we simply extend by adding that edge at
%the end of the path to get a path that draws $G$. So $G$ is drawable.
%
%\item Case 2: Suppose that $G$ has two odd degree vertices $a$ and $b$. We construct
%$H$ by removing some edge $\{a, c\}$ from $G$, and removing $a$ if it no
%longer has any edges remaining.
%
%\begin{itemize}
%
%\item Case 2.1 Suppose $c \ne b$
%
%\item Case 2.2 Suppose $c=b$. In that case
%
%\end{itemize}
%
%
%\end{itemize}
%
%\end{itemize}

}

\end{problemparts}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem How many numbers must be selected from the set $\{ 1, 3, 5,
7, 9, 11, 13, 15 \}$ to guarantee that some pair adds up to 16?
Justify your answer using the pigeonhole principle; in particular,
what are the pigeons and what are the holes?

\solution{
We shall show using pigeonhole principle that we need to pick five elements
to be sure that two of them add up to 16. 

We have 4 holes: A,
B, C and D. The pigeons are the elements of $S=\{1, 3, 5, 7, 9, 11, 13,
15\}$. Given a subset of S, we place 1 and 15 in A, 3 and 13 in B, 5 and 11
in C, and 7 and 9 in D. By the pigeonhole principle, if we pick a subset of
S with at least 5 elements, then two (distinct) elements will end up in the same hole.
Because of the way the pigeons have been assigned to holes, those two elements 
must add up to 16. Therefore, if we pick 5 elements, we are sure that two
of them add up to 15.

Noting that $\{1, 3, 5, 7\}$ is a 4 element subset of S that does not have
a pair of elements that add up to 16, we can conclude that 5 is the
smallest number of elements we need to pick to be sure to find a pair that
adds up to 16.
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem There is a group of people who were all born in 1983.  Below
is a list of properties that the group may or may not possess.  For
each property, either give the minimum number of people that must be
in the group to ensure that the property holds, or else indicate that
the property need not hold even if the group is arbitrarily large.  No
proofs are required.

\begin{problemparts}

\problempart At least 2 people were born on the same day.

\solution{This property holds for every group of 366 or more people. 
With only 365 people, the people could all be born on different days. But by
the pigeonhole principle, where the 365 days of the year are holes, and the people 
are pigeons, the property must hold for 366 people or more.

Note that there are two parts in this proof. First we see that 365 is
not enough, then that 366 is sufficient. If we only applied pigeonhole, we
would know that 366 is sufficient, but we wouldn't sure that 365 isn't
sufficient.
}

\problempart At least 2 people were born on January 1.

\solution{This property need not hold, even for arbitrarily large
groups of people. Indeed, they may all be born on April $1^{st}$, for example.}

\problempart At least 3 people were born on the same day of the week.

\solution{
This property holds for every group of 15 or more people. With 14 people, each day
of the week can correspond to two birthdays, but with 15, we can apply
the general pigeonhole principle with days of the week as holes and people as pigeons.
}

\problempart At least 4 people were born in the same month.

\solution{
This property holds for every group of 37 or more people. With 36 people,
three can be born on each month, but with 37, we can apply the general
pigeonhole principle with months as holes and people as pigeons.}

\problempart At least 2 people were born exactly one week apart.

\solution{This property need not hold.  For example, everyone might be
born on the same day of the year.
}

\problempart At least 2 people were born on the same day or else 2
people were born exactly one week apart.

\solution{184 people. Indeed, 183 people can be accomodated without
satisfying the property by having them born on alternate days of the year
(Jan 1, Jan 3,~\dots, Dec 29, Dec 31). Let us see pairs of days, one of which is in
1983, and which are one week apart, as holes. There are $365+7=372$ holes: one from 
each day in 1983 to the day one week before, and one from each of the seven first
days of 1984 to the day in 1983 that was one week before. 
Each person counts as two pigeons, mobilizing the two holes corresponding to the
person's birthday. For 373 pigeons, we are sure that some pair will be
picked twice. For that we need $\lfloor 373/2 \rfloor=184$ people.
}

\problempart At least 9 people were born on a weekend or at least 11
people were born during the week.

\solution{This property holds for every group of at least $(9-1) +
(11-1) + 1 = 19$ or more people. For 18 people, 9 can be born on the
weekend and 11 on a weekday. But if there are 19 people, either 9 or more 
are born on the weekend, or at most 8 are born on a week end, which
implies that $19-8=11$ or more are born during the week. Therefore, the
property must hold for 19 people.}

\problempart At least $n$ people were born on the $n$-th day of the
year for some $n$, where days are numbered 1, 2, \ldots, 365.

\solution{
This property holds for at least $1+\sum_{i=0}^{364}i=364*365/2+1=66431$
people. Indeed, if $a_i$ is the number of people born on the $i^{th}$ day, the
total number of people is $n=\sum_{i=1}^{365}a_i$. If $a_i=i-1$ then the
property doesn't hold, and $n=\sum_{i=1}^{365}(i-1)=66430$, so the property
doesn't hold for that many people. For the property not to hold, we must
have for each $i$, $a_i<=i-1$, therefore $n<=\sum_{i=1}^{365}(i-1)=66430$.
By the contrapositive, for 66431 people or more, the property must hold.  
}

\end{problemparts}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem Let $D$ be the set of all points $(x, y)$ in the real plane
such that $|x| + |y| \leq 1$.  Prove that among every set of five
points in $D$, there exist two points whose distance from one another
is at most 1.

\solution{
We solve this problem using the pigeonhole principle. A little
geometry shows that D is in fact a square of edge size $\sqrt{2}$, which can 
be tiled by four smaller squares of edge size $\sqrt{2}/2$.
The diagonal of each smaller square is therefore $1$, so each one is inscribed in a
circle of diameter $1$. The whole construction is shown here:

\centerline{\resizebox{!}{1.5in}{\includegraphics{pset6-prob6}}}

We shall take the four small squares to be our holes, and the five points
we wish to place in $D$ to be our pigeons. Since the four small squares
cover $D$, each pigeon must fall into a hole (it might actually fall on
more than one because adjacent squares share an edge). By the pigeonhole 
principle, two points must be in the same square. They are therefore in the 
same circle of diameter $1$, and thus must be within $1$ of each other.
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem There is a meeting of $n$ people.  Each pair of people at the
meeting either shakes hands or does not shake hands.  Prove that there
must exist two people at the meeting who shook the same number of hands.

\solution{
We solve this problem using the pigeonhole principle. Let the pigeons be
the $n$ people at the meeting, and the holes be number of times times that a
person has shaken hands. 

First we consider the case where some person $a$ hasn't shaken any hands. In that
case the most hands any other person can shake is $n-2$ (people don't shake
hands with themselves or with $a$). So the number of hands a person can
have shaken is between $0$ and $n-2$. Thus there are $n-1$ possible values
for the number of handshakes a person has exchanged. By the pigeonhole
principle, at least two of the $n$ people must have shaken the same number
of hands.

We now consider the complementary case where everybody has exchanged at least one
handshake. Now the number of handshakes for a person is between $1$ and
$n-1$. Thus there are still $n-1$ possible values for the number of
handshakes, and the pigeonhole principle allows to conclude that two people
have shaken hands the same number of times. 

We can conclude that in any case, two people must have shaken the same
number of hands.
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem Suppose that during each day of an 13-week term, an MIT
student feels overwhelmed at least once.  (Unfortunately, this is
going to be one of those times.)  However, suppose that the student
never feels overwhelmed more than 12 times in any one of the 13 weeks.
Prove that there exists a sequence of consecutive days during which
the student feels overwhelmed exactly 25 times.

\solution{Let $P_i$ be the total number of times the student feels
overwhelmed in the first $i$ days.  Since the student feels
overwhelmed at least once every day, the sequence $P_0, P_2, \ldots,
P_{91}$ is strictly increasing.  Since the student is overwhelmed at
most 12 times per week, the final term, $P_{91}$, is at most $13 \cdot
12 = 156$.  Now the sequence $P_0 + 25, P_1 + 25, \ldots, P_{91} + 25$
is also strictly increasing, and the largest term, $P_{91} + 25$, is
at most $156 + 25 = 181$.  Therefore, the 184 values $P_0, \ldots,
P_{91}$, $P_0 + 25, \ldots, P_{91} + 25$ all lie between 0 and 181.
By the pigeonhole principle, at least two of these values must be the
same.  Since no two of the numbers $P_1, \ldots, P_{91}$ are equal,
and no two of the numbers $P_1 + 25, \ldots, P_{91} + 25$ are equal,
there must be an $i$ and a $j$ such that $P_i = P_j + 25$.  This
implies that on days $j+1, j+2, \ldots, i$, the student is overwhelmed
exactly 25 times.}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{problems}
\end{document}
