Summations

Question? What is the sum of the first 100 positive integers?

Counting
Question? In how many ways may the first three horses in a
10 horse race finish?
Multiplication Principle: If an event can occur in m ways,
and a second event can occur independently in n ways, then
the two events can occur in mn ways.
Question? Where is the independence in the horse race ex-
ample?
Recursive reasoning warmup
A recursive function is a function that calls itself during its
execution.
Example. Factorials

0! = 1
n! =
n! = n(n − 1)!

1
Example. Fibonacci’s rabbits satisfy

X Begin with one pair of immature rab-

bits,
X In the second month the rabbits are Fibonacci
(c. 1170 – c. 1250)
mature and produce a new pair of
rabbits, and
X Each pair continues to produce a new
pair of rabbits each month forever.

Question? How many pairs of rabbits

does Fibonacci have after n months?

Reference: Leonardo of Pisa, Liber abbaci, 1202.

2
Example. Call a string of symbols a valid arithmetic expres-
sion if

R It uses only the digits 0, 1, 2, ... , 9, and the operators +,

-, /, and *,
R It begins and ends with a digit, and
R It does not have two consecutive operators.

Question? How many such valid arithmetic expressions of

length n are there?
Addition principle: If the things to be counted are sepa-
rated into distinct cases, the total number is the sum of the
numbers in the various cases.
A slight diversion
Question? Can you draw this without lifting your pencil from
the paper? Can you do so covering each line segment once and
return to the starting point?

3
First method of proof
Theorem 2.2.2. (The Principal of Mathematical Induction)
Let P (n) be a statement about the positive integers. If one
can prove that

ä P (1) is true, and that

ä For every positive integer m, whenever P (1), P (2), ..., P (m)
is true, then it follows that P (m + 1) is true,

then P (n) is true for every positive integer n.

Example. Using mathematical induction, prove that

n(n + 1)
1 + 2 + 3 + ··· + n = .
2

Example. Prove that the sum of the cubes of three successive

positive integers is divisible by 9.

Question? In how many ways can n people be seated around

a round table?

4
Definition: A permutation is an ordered arrangement of dis-
tinct objects.

Proposition 3.2.14. The number of permutations of n

things taken r at a time is

n!
P (n, r) = .
(n − r)!

Addition principle: If the things to be counted are sepa-

rated into distinct cases, the total number is the sum of the
numbers in the various cases.

Recall that we added the two cases in the arithmetic expres-

sions.

Question? How many even four digit numbers with distinct

digits can be formed from {0, 1, 2, 3, 4, 5, 6}?

Definition: A combination is an unordered arrangement of

objects.

Proposition 3.2.18. The number of combinations of n

things taken r at a time is
n!
C(n, r) =
r!(n − r)!

5
A probability experiment is an experiment that can be
carried out under repeatable conditions with a set of outcomes
that are equally likely to occur. The sample space of the
experiment is the set of all possible outcomes, and a subset of
the sample space is called an event.

If n(D) denotes the number of elements in a set D, then the

probability of an event E is
n(E)
p(E) =
n(S)
where S is the sample space of the experiment.

Question? An urn contains four red and three blue balls. An

experiment consists of randomly selecting four balls. What is
the probability that two red and two blue balls are selected?

Question? How many full houses are there in poker?

Question? How many straights are there in poker?

Fundamental Theorem of Arithmetic. Every integer

n > 1 is expressible as n = p1p2p3 · · · pr where r is a positive
integer and each pi is a prime.

Question? How many positive factors does 1,361,367 have?

6
Pascal’s Triangle:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1 Blaise Pascal (1623-1662)
Source: Wickimedia Commons
1 7 21 35 35 21 7 1
...

The key identity underlying Pascal’s Triangle:

Proposition. C(n + 1, r) = C(n, r) + C(n, r − 1).
We consider two proofs: One by calculation and a combina-
torial proof. A combinatorial proof is one accomplished
by counting the same set in two different ways to establish an
equality.

7
An earlier source is the Yáng Huī (ca. 1238 - 1298) triangle
below. So in China it is known as the Yáng Huī Triangle.

8
Vandermonde’s Identity. Let m, n, and r be nonnegative
integers with r not exceeding either m or n. Then
r
X
C(m + n, r) = C(m, r − k)C(n, k)
k=0
= C(m, r)C(n, 0) + C(m, r − 1)C(n, 1)
+ · · · + C(m, 0)C(n, r)

To illustrate this identity, the 21 in row seven of the triangle

below is the sum of the products of the entries in rows three
and four having the same color:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
...

C(7, 2) = C(4, 2)·C(3, 0)+C(4, 1)·C(3, 1)+C(4, 0)·C(3, 2)

21 = 6 · 1 + 4 · 3 + 1 · 3

9
The Binomial Theorem. The coefficient of the an−k bk
term in the expansion of (a + b)n is C(n, k), i.e.
n
X
(a + b)n = C(n, k)an−k bk
k=0
= a + C(n, 1)an−1b + C(n, 2)an−2b2 + C(n, 3)an−3b3 + · · · + bn
n

n(n − 1) n−2 2 n(n − 1)(n − 2) n−3 3

= an + nan−1b + a b + a b
1·2 1·2·3
+ · · · + bn

To expand (a + b)n:

• The first term is an

• Each successive term is gotten from the previous one by
– Decreasing the exponent on a by 1,
– Multiplying by the old exponent on a,
– Increasing the exponent on b by 1, and
– Dividing by the new exponent on b.

Example. Expand (2 − x)6.

Proposition. A set with n elements has ? subsets.

10
Question? Suppose that a baseball player has a batting aver-
age of .300 and comes to the plate five times in a game. What
is the probability that he gets three hits? Or that he gets at
least three hits?

Example. Prove that for any integer n ≥ 0 that

11n+2 + 122n+1
is divisible by 133.

One-to-one correspondences

A function f : A → B is one-to-one if f (x) = f (y)

implies that x = y.
A function f : A → B is onto if for any element b
in B there is an element a in A such that f (a) = b.
A function that is both one-to-one and onto is called
a one-to-one correspondence. Sets A and B with
a one-to-one correspondence between them are said
to be in one-to-one correspondence, or to have the
same number of elements, or to have the same car-
dinality.

Question? How many four element subsets of

{1, 2, 3, ..., 20}
are there?

11
Question? How many integer solutions to
x1 + x2 + x3 + x4 + x5 = 16

with each xi non-negative?

Proposition. The solutions to the two examples

above are the same.

A binary sequence of length n is a sequence of n

0’s and 1’s.

Proposition. There are C(n, r) = C(n, n − r) binary

sequences of length n containing exactly r 1’s.

Proposition. The number of solutions in non-negative

integers to
x1 + x2 + · · · + xn = r
is C(n + r − 1, r) = C(n + r − 1, n − 1).

Question? In how many ways may 75 shares of stock

be distributed among 10 men and 10 women in such
a way that each woman receives at least two shares
and each man at least one share?

12
Now that we have conquered counting solutions in-
volving non-negative integers, how about the case
of positive integers?

Question? How many solutions in positive integers

are there to the equation x1 + x2 + x3 + x4 = 12?

Now what about putting an upper bound on the

positive variables, say 6?

Question? In how many ways may the sum of three

dice add up to 14?

Question? What is the probability of getting a sum

of 11 in a toss of three fair dice?

In general, a generating function is a polynomial

such that the coefficient of xr is the answer to a
counting problem involving r.

13
We have seen one generating function in that we
“generated” C(n, r) as the coefficient of xr in the
expansion of (1 + x)n. For instance,

(1 + x)7 = x7 + 7 x6 + 21 x5 + 35 x4 + 35 x3 + 21 x2 + 7 x + 1

Example. Carry out the multiplication

(1 + x + x2)(1 + x + x3 + x5).

Question? How many ways are there to make change

for 25 cents using pennies, nickels and dimes?

Question? How many terms are there in the expan-

sion of (a + b + c)6?

14
Back to Pascal’s Triangle:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
.
.
.

Recall the identity that governs Pascal’s Triangle:

C(n + 1, r) = C(n, r) + C(n, r − 1).

Example 3.3 no. 11. Compute exact numerical an-

swers for the first two parts and then answer the
third.

1. How many binary sequences of length at most 5

have exactly 3 1’s?
2. How many binary sequences of length 6 have
exactly 4 1’s?
3. Why do the first two questions have the same
answer?
15
Question? (Sec. 3.4 no. 3) What entry in Row 7 of
Pascal’s Triangle is the same as
C(5, 2)C(2, 0) + C(5, 1)C(2, 1) + C(5, 0)C(2, 2)?
Hint: Apply the Binomial Theorem to
(1 + x)2(1 + x)5 = (1 + x)7.

16
Derangements
A derangement of a set of items with a natural order
is a permutation of them such that none of the items
is in its natural position.
Question? How many derangements of a set of n
items are there?

The number Dn of derangements of n objects can

be calculated recursively. But determining the re-
cursive formula is tricky, so here it is in full.
First, a list consisting of only one item cannot be
re-arranged, so D1 = 0. Second, D2 = 1 since the
only way to get two items completely out of order
is to transpose them.
Now consider the number Dn of derangements of
the set of the first n positive integers: {1, 2, 3, . . . , n}.
The number 1 can go to any of n − 1 places, so we
can break the set of derangements into n−1 subsets
depending on the location of 1 and calculate the
number x of derangements in each of these subsets.
So we now know that Dn = (n − 1)x, and we need to
calculate x.
Suppose that 1 goes to position k. There now are
two possibilities, so we will use the Addition Prin-
ciple.

17
The possibilities are that k goes to position 1, or it
doesn’t.
Case 1: k goes to position 1. Then the other n − 2
integers are deranged among themselves and this
can be done in Dn−2 ways.
Case 2: k does not go to position 1. With the posi-
tion of 1 known, each of the other integers has one
restriction on where it can move: Each integer i
other than k cannot go to position i and integer
k cannot go to position 1. This is equivalent to de-
ranging the n − 1 integers other than 1 and can be
done in Dn−1 ways.
Hence, x = Dn−1 + Dn−2 and Dn = (n − 1)(Dn−1 + Dn−2).
Some sources use the notation !n for the number
of derangements. Then here is the formula that we
derived:

 0 if n = 1,
!n = Dn = 1 if n = 2, and
(n − 1)(Dn−1 + Dn−2) if n > 2


Question? When the power goes out in a restaurant

10 diners select their coats from the check room at
random. What is the probability that none of them
select their own coats?

18
The derangements of {1, 2, 3, 4} grouped by the lo-
cation of 1:

2 1 4 3
3 1 4 2
4 1 2 3
3 4 1 2
2 4 1 3
4 3 1 2
4 3 2 1
2 3 4 1
3 4 2 1

Note that for each position of 1 there is D2 = 1 de-

rangement – colored red – where 1 exchanges place
with the integer whose position it occupies and
D3 = 2 derangements where the integer in whose
place 1 is is not in first place.

19
Here are the derangements of {1, 2, 3, 4, 5} where 1
goes to position 2:
There are D3 = 2 where 1 and 2 change places:

2 1 4 5 3
2 1 5 3 4

There are D4 = 9 where 1 and 2 do not change

places but 1 is in second place:

4 1 2 5 3
4 1 5 3 2
4 1 5 2 3
3 1 2 5 4
3 1 4 5 2
3 1 5 2 4
5 1 2 3 4
5 1 4 2 3
5 1 4 3 2

20
There is an interesting relationship between facto-
rials and derangements. Note that the plot below
uses !n for the number of derangements.

Source: Wickimedia Commons

It turns out that n! has a recursion very similar to

Dn:

 1 if n = 1,
n! = 2 if n = 2, and
(n − 1)((n − 1)! + (n − 2)!) if n > 2


21
The Excel spreadsheet below shows that the ratio
of the number of derangements to the number of
1
permutations approaches .
e

22
Closed forms for recursions

A closed form expression for a recursive sequence

is one that allows the evaluation of a term without
referring to prior terms.
The next proposition suggests how to obtain a closed
form for some recursions.

Proposition. If sn and tn are both solutions of a

recurrence un = aun−1 + bun−2 and c and d are any
constants, then csn + dtn is also a solution.

The expression csn + dtn is for constants c and d is

called a linear combination of sn and tn.

Example. We use the proposition above and the

assumption that Fn = tn to determine the closed
form expression for the Fibonacci recursion
F0 = 1, F1 = 1, Fn = Fn−1 + Fn−2 for n ≥ 2.

23
Fibonacci numbers and Phi
The nautilus, the logarithmic spiral, and additive
squares

Source:Chris 73 / Wikimedia Commons Source:dgleahy.com/dgl/p14.html

By using the closed form expression for Fn, one can

show that the quotient
√
Fn+1 1+ 5
→φ= = 1.61803... as n → ∞
Fn 2

24
Pea tendrils, additive triangles, and the logarithmic
spiral

Source:

easyweb.easynet.co.uk/ iany/patterns/spirals.htm

The steady increase in curvature of the tendril as

it tightens around a support creates a curve which
is roughly equiangular or logarithmic. Such a curve
can be generated mathematically by dividing isosce-
les triangles (bisecting a base angle each time) or
rectangles (the remainder to be similar to the orig-
inal rectangle) of any shape – there is an infinite
family of logarithmic curves of different pitch.

25
The Golden Rectangle in Art and Architecture

Seurat Turner

Parthenon Parthenon

26
To pique your interest, here is a relationship be-
tween Fibonacci numbers and binomial coefficients:

For n ≥ 0, fn = C(n, 0) + C(n − 1, 1) + C(n − 2, 2) + · · ·

n
bX
2c

= C(n − i, i)
i=0

As an illustration, the binomial coefficients sum-

ming to f6 = 13 are red and those summing to
f7 = 21 are blue in the triangle below.
 7  Note that
the sum for f7 stops at C(4, 3) since 2 = 3.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
...

27
Getting ready for fun with Fibonacci
To prepare for things to come regarding Fibonacci
numbers, here is an alternate way to define them:
Let fn count the number of ways for an n by 1
board to be covered by a combination of squares
and dominoes. A square covers one square of the
board, and a domino covers two squares. We start
to count with the empty board which can be covered
in one way, i.e. f0 = 1. The one board can be
covered by a square in one way so that f1 = 1. For
n ≥ 2, fn = fn−1 + fn−2.
The five coverings of a 4-board are illustrated below:

This interpretation of the Fibonacci numbers is iden-

tical to the one involving rabbits except that we
start to count with a zero subscript.
Collectively, squares and dominoes are referred to
as “tiles”. So the middle three examples above each
involve three tiles while the first uses four and the
last only two.
For any n there is obviously a single tiling involv-
ing only squares, and for an odd n any tiling must
involve at least one square.

28
Combinatorial fun with Fibonacci

Proposition. For n ≥ 0,
F2n−1 = C(n, 1)F0 + C(n, 2)F1 + · · · + C(n, k)Fk−1 + · · · + C(n, n)Fn−1
n  
X n
= Fk−1
k
k=1

Call a tiling of a board breakable at square k if the

squares k and k + 1 are not covered by a domino.

2
Proposition. For n ≥ 1, Fn2 + Fn+1 = F2n+2.

Proposition. For n ≥ 0, F0 + F2 + F4 + · · · + F2n = F2n+1

And now here are two identities for which appar-

ently no combinatorial proof is known:

2n  
X 2n
Proposition. F2i−1 = 5nF2n−1.
i=0
i

2n+1
X 
2n + 1 2
Proposition. Fi−1 = 5nF2n.
i=0
i

29
The Towers of Hanoi
In the great temple of Brahma in Benares, India,
on a brass plate under the dome that marks the
center of the world, there are 64 disks of pure gold
that the priests carry one at a time between three
diamond needles according to Brahma’s immutable
law: No disk may be placed on a smaller disk. In
the beginning of the world all 64 disks formed the
Tower of Brahma on one needle. Now, however,
the process of transfer of the tower from one needle
to another is in mid course. When the last disk
is finally in place, once again forming the Tower of
Brahma but on a different needle, then will come
the end of the world and all will turn to dust.

Question? Suppose n disks are to be moved. How

many moves are necessary?

30
A bit of logic
To prove a proposition like “If P , then Q”, abbre-
viated “P ⇒ Q” there are two options: prove the
proposition directly, or prove the logically equiva-
lent contrapositive of the proposition ¬Q ⇒ ¬P .
In the second method, called proof by contradiction,
one assumes that Q is false and shows that that
assumption leads to a contradiction of P .

Proposition. The square root of 2 is irrational.

The Pigeonhole Principle: If k + 1 or more pigeons

are placed into k pigeonholes, then there must be
one pigeonhole containing two or more pigeons.

Finally, we will need a bit of number theory:

Proposition. If two numbers have the same remain-
der when divided by c, then their difference is di-
visible by c.

Problem. A group of 30 students wrote a dictation.

John Bull made 13 errors, and each of the rest made
fewer than 13 errors. Prove that at least 3 students
made the same number of errors.

31
Problem. Prove that, given any 12 distinct natural
numbers, we can choose two of them such that their
difference is divisible by 11.

Problem. Given a set S of 10 distinct numbers be-

tween 1 and 100, inclusive, there exist two distinct,
disjoint subsets A and B of S whose elements sum
to the same number.

32
Modular arithmetic

Modular arithmetic was devel-

oped by Carl Friedrich Gauss
and published in his Disquisi-
tiones Arithmeticae in 1801 when
he was 21. It offers a conve-
nient approach to questions of
divisibility.
Carl Friedrich
Gauss (1777 – 1855)
Definition: Let m be a fixed integer. For integers
a and b we say that a is congruent to b modulo m
and write

a ≡ b (mod m)
whenever m|(a − b). If m 6 |(a − b), we write a 6≡ b
(mod m).

Example. Clocks and modular arithmetic.

33
 Proposition. Let m be a fixed integer. Let a, b, and
c be integers. Then

(i) a ≡ a (mod m).

(ii) If a ≡ b (mod m), then b ≡ a (mod m).
(iii) If a ≡ b (mod m) and b ≡ c (mod m), then a ≡ c
(mod m).

The properties enumerated in the Proposition are

called, reflexive, symmetric, and transitive, respec-
tively.

Proposition. If a ≡ a0, (mod m) and b ≡ b0 (mod m),

then

(i) a + b ≡ a0 + b0 (mod m)
(ii) a − b ≡ a0 − b0 (mod m)
(iii) a · b ≡ a0 · b0 (mod m)

Question? So addition, subtraction, and multipli-

cation behave nicely! How about division? The
following example illustrates the difficulty with di-
vision.

34
Example. Note that 28 ≡ 10 (mod 6). Also, 2 divides
both 28 and 10 so we might expect to divide by 2
and have 14 be equivalent to 5 modulo 6.
But 14 6≡ 5 (mod 6).
What happened? Why can’t we“cancel” the 2 in
this case?

The answer requires a couple of definitions. . .

The greatest common divisor of two integers a and

b is the largest integer d such that d|a and d|b. The
greatest common divisor of two integers is denoted
by gcd(a, b).

Integers a and b are said to be relatively prime if

gcd(a, b) = 1.

This allows an answer to when we can “cancel” in

modular arithmetic. . .

Proposition. If ac ≡ bc (mod m) and gcd(c, m) = 1,

then a ≡ b (mod m).

So we should not have expected 14 to be equivalent

to 5 modulo 6 since gcd(2, 6) = 2.

Example. 20 · 2 ≡ 9 · 2 (mod 11) and therefore 20 ≡ 9

(mod 11) since gcd(11, 2) = 1.

35
So we can cancel a constant in a modular equiva-
lence only when the constant and the modulus are
relatively prime.

Proposition. a ≡ b (mod m) if and only if a and b

have the same remainder when divided by m.
Proposition. An integer n is divisible by 9 if and
only if the sum of its digits is divisible by 9.

Example. Determine all solutions to 8x + 12y = b

where x and y are positive integers and 75 < b < 80.

Example. Determine the remainder when 237 is di-

vided by 7.

36
Introduction to Graph Theory
A graph is a set of vertices and a set of edges such
that each edge is associated with an (unordered)
pair of vertices.

A graph is simple if it contains at most one edge

between any pair of vertices and no loops, i.e., no
edges that start and end on the same vertex.

A path is a sequence of vertices such that consecu-

tive vertices are adjacent via an edge and no edge
is used twice.

A graph is connected if there is a path joining each

pair of distinct vertices.

A path is Eulerian if it uses every edge in the graph

exactly once.

A circuit is a path that returns to its starting point.

The degree of a vertex is the number of edges inci-

dent on the vertex.

37
The Konigsberg Bridge Problem
The river Pregel flows through Konigsberg, and in
Euler’s day there were seven bridges connecting the
North and South shores and the two islands in the
river. A popular puzzle in the town was whether or
not it was possible to walk in such a way as to cross
each bridge exactly once.
The Bridges of Konigsberg
Leonhard
Euler (1707 – 1783)

Euler’s Theorem. A graph has an Euler path if and

only if . . .

38
A graph is planar if it can be drawn in the plane
without its edges crossing.

A graph is complete if every pair of distinct vertices

is connected by an edge. The complete graph with
n vertices is denoted by Kn.

Example. K4 is planar while K5 is not.

Utilities Puzzle
Connect all of the circles with all of the squares
without arcs crossing.

Unsolved problem. The question of the the mini-

mum number of edge crossings required to draw Kn
is unsolved for most cases where n ≥ 13.
A graph is bipartite if it includes two disjoint sets
of vertices such that all edges connect a vertex in
one of the sets with a vertex in the other.
39
Kn,m denotes the bipartite graph that includes all
possible edges joining a set of n vertices with a set
of m vertices.

A graph H is a subgraph of a graph G if every

vertex and every edge of H is also a vertex or an
edge of G.

Kuratowski’s Theorem. A graph is planar if and

only if it does not contain K5 or K3,3 as a subgraph.

Two graphs are isomorphic if (ignoring their vertex

labels) one can be re-drawn to look like the other.

Thus, unlike a one-to-one function that establishes

an isomorphism for sets, the one-to-one correspon-
dence between the vertices of isomorphic graphs
must be accompanied by a one-to-one correspon-
dence between their edges in such a way that the
incidence relationships are preserved.

40
Question? So who cares if a graph is planar, or at
least, close to planar?

In addition to the number of edges, e, and the num-

ber of vertices, v, for a planar graph there is also
the number f of faces.

When a planar graph is drawn without edge inter-

sections, the plane is divided into contiguous re-
gions called faces. There is also an unbounded face
consisting of the region lying entirely outside the
graph.

Theorem. (Euler’s Formula) In a connected, planar

graph
e − v + 2 = f,
i.e., the number of edges minus the number of ver-
tices plus two equals the number of faces.
Proof. Use induction on the number of edges . . .

Theorem 6.1.7. In any graph G, the sum of the

degrees of the vertices is twice the number of edges.

Corollary 6.1.8. In any graph G, the number of

nodes with odd degree is even.

41
Question? Can there exist a hydrocarbon with five
carbon atoms and three hydrogen atoms?

Two hydrocarbons

H
•
..
...
..
...
...
...
...
...
...
...
....
METHANE
H• .....................................................
•
...
.....................................................
•H CH4
...
...
...
C
...
...
...
...
...
...
...
...

• H H
H •
...
...
• ...
...
... ...
... ...
... ...
... ...
... ...
... ...

ETHYLENE ...
...
... .....................................
............
... .............. ........
.......
...
...
...
...

• ...
•C
..
C
C 2 H4 ... .........
...
...
...
........
.............
....................................
..
.......
........
...
...
...
...
... ...
... ...
... ...
... ...
... ...
... ...
... ...
... ...

•
H
•
H

What is the degree of each hydrogen atom?

What is the degree of each carbon atom?

Question? How many edges are there in Kn? How

about Kn,m?

Proposition. C(n + m, 2) = C(n, 2) + C(m, 2) + nm.

42
Question? (The Party Problem.) Six guests are
invited to a party, and each pair of guests are ei-
ther friends or strangers. Is it true that among the
guests there are always three who are all friends or
three who are all strangers?

Example. Provide an example of a graph for a party

with five guests without a set of three who are all
friends or three who are all strangers.

Ramsey’s Theorem. For any pair of positive inte-

gers (r, s), there exists a least positive integer R(r, s)
such that for any complete graph on R(r, s) vertices,
whose edges are coloured red or blue, there exists
either a complete subgraph on r vertices which is
entirely blue, or a complete subgraph on s vertices
which is entirely red.

Unsolved problems. The Ramsey number R(r, s) is

known for small values of r and s, and there are
estimates for others, for example 40 ≤ R(10, 3) ≤ 42,
or 205 ≤ R(7, 7) ≤ 540. And 43 ≤ R(5, 5) ≤ 49. But the
search for exact values is full of open problems.

43
Just how hard are these problems? Here is a quote
from Paul Erdös:
“Imagine an alien force, vastly more powerful than
us landing on Earth and demanding the value of
R(5, 5) or they will destroy our planet. In that case,
we should marshal all our computers and all our
mathematicians and attempt to find the value. But
suppose, instead, that they asked for R(6, 6), we
should attempt to destroy the aliens.”

44
So here is a problem likely not so hard, just “...
difficult and is not likely to be solvable by an ama-
teur.”
Unsolved problem. (The 3n + 1 Problem)
In this problem, a sequence is generated starting
with an initial natural number n. The rule for gen-
erating the next natural number in the sequence
is; if n is even, the next natural number in the
sequence is n/2, or if n is odd, the next natural
number in the sequence is 3n + 1. For example, if
the initial value of n is 17, the sequence generated
is {17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, . . .}. If the
natural number 4 is encountered in the sequence,
the sequence starts repeating ({4, 2, 1} is generated
over and over). There are three possibilities;

1. The natural number 4 is encountered in the se-

quence,
2. The sequence starts repeating, but for a differ-
ent cycle than 4, 2, 1, or
3. The sequence doesn’t repeat.

The 3n + 1 conjecture states that only the first pos-

sibility can occur.
Reference: http://home.graysoncable.com/dkcox/ which is copyrighted by Dar-

rell Cox.

45
Instant Insanity
Four blocks have their faces colored red, white, green
and blue. The problem is to stack the blocks so that
all four colors show on each side of the stack.

Question? How many ways are there to stack the

blocks?

Question? The solution is determined by . . .

Trees
A tree is a graph with a unique path joining any
pair of vertices.

Intuitively, a tree is a graph with just enough edges

to be connected.

Proposition. A tree with n vertices has ??? edges.

Proposition. A tree is planar.

One use of trees is guidance in making decisions. In

the following example, we will see a use of a “deci-
sion tree” in the solution to a counting problem:

46
Counting “words”
Question? Using the 26 letters of the alphabet
and considering the letters AEIOUY to be vow-
els and the others consonants, how many five letter
“words”, i.e., five letter lists, are there subject to
the following constraints?

H No vowels are adjacent,

H There are never three adjacent consonants, and
H Adjacent consonants are always different.

47
In a graph, a component is a maximal connected
subgraph. Two vertices are in the same connected
component if and only if there exists a path between
them
Two useful facts:
Lemma. Removing an edge that belongs to a circuit
of a connected graph does not disconnect the graph.

Lemma. (Maximally circuit free) Adding an edge

to a tree creates a unique circuit.

Proposition. (Characterization Theorem) The fol-

lowing are equivalent for a graph T having n ver-
tices, no loops, and no parallel edges:

(a.) T is a tree.
(b.) T is connected and contains no cycles.
(c.) T is connected and has n − 1 edges.
(d.) T contains no cycles and has n − 1 edges.

48
A weighted graph is a graph with lengths assigned
to the edges, e.g., distances, times, costs, etc.

A spanning tree for a graph G is a subgraph which

is a tree and includes all vertices of G.

If the edges of G are weighted, a minimal spanning

tree is a spanning tree of minimal total edge weight.

Proposition. Any edge of the least weight belongs

to a minimal spanning tree.
Proof. Suppose not; i.e., suppose that edge ij has
the minimal weight but belongs to no minimal span-
ning tree. Let T be a minimal spanning tree. Then
adding edge ij to T produces a graph containing
a cycle involving ij. Removing any edge from the
cycle other than ij produces a spanning tree with
total weight no larger than that of T , and hence a
minimal spanning tree. Contradiction. 

49
Prim’s algorithm
We assume that G is a weighted, connected graph
without loops or parallel edges and having n ver-
tices. Let Tk be the tree formed by the algorithm
at step k.

INIT: Choose the shortest edge from G and the

vertices v0 and v1 on which it is incident to
form T0.
ADD EDGE: Select the edge of minimal weight which
is incident on a vertex in Tk and a vertex not
in Tk . Add this edge and vertex to Tk to form
Tk+1.
END?: Repeat ADD EDGE until all vertices of G
are in the tree.

50
Theorem. Prim’s algorithm is correct.
Proof. We must show that the algorithm actually
produces a minimal spanning tree. Letting Ti be the
tree formed by the algorithm at step i, we proceed
by induction.

Induction hypothesis: For each i, Ti is contained in some

minimal spanning tree.

i = 1: This is clear from the proposition above since

the shortest edge belongs to some minimal spanning
tree.

Now assume Ti−1 ⊂ T, T a minimal spanning tree.

Let pq be the edge and q the node to be added to
form Ti.
If Ti 6⊂ T, we must show that Ti is contained in some
minimal spanning tree. Consider the graph H =
T ∪ {pq}. H is not a tree since it contains the extra
edge pq. Further, since pq joins a node not in Ti−1
with one of Ti−1, H contains a cycle that involves
another edge rs joining a node of Ti−1 with a node
not in Ti−1.
Removing rs from H produces a spanning tree T 0
containing Ti. By Prim’s algorithm, the weight of
rs is at least as great as that of pq. Hence, the
weight of T 0 is no greater than that of T, so T 0 is
a minimal spanning tree containing Ti.

51
Theorem. The number of edges examined in execut-
ing Prim’s algorithm is at worst a cubic polynomial
of the number of edges.

The proof will require the following result from the

first assignment:

n(n + 1)(2n + 1)
12 + 22 + · · · + n2 =
6

An application
A symmetric function of several variables is one in
which an exchange of variables does not alter the
value of the function.
Interesting property. A minimal spanning tree min-
imizes an increasing symmetric function of the edge
lengths.
Question? A message is to be passed to all mem-
bers of an underground organization. Each member
knows certain other members and has a means of
contacting them. Associated with each contact i
to j is a probability pij that the message will be
compromised.
Assuming that these probabilities are independent,
how should a message be distributed to minimize
the overall chance of compromise?

52
The Dual of a Graph and The Four Color Theorem
The dual G0 of a planar graph G is the graph ob-
tained by placing a vertex of G0 in each face of G
and connecting the vertices of G0 if the correspond-
ing faces of G share an edge.

Source: http://en.wikipedia.org/wiki/File:Duals graphs.svg

Viewing G as a map with each face a country, col-

oring the map so that countries sharing a border
have different colors can be reduced to coloring the
vertices of the dual graph so that the vertices on
opposite ends of each edge have different colors.
For many years, it was known that any planar graph
could be colored with five colors, and that there
were graphs that could not be colored with three
colors. No one could find a graph that required five
colors or a proof that all maps could be colored with
four colors.
The problem was resolved by Appel and Haken in
1976 who reduced the problem to showing that 1955

53
cases needed to be colored and used a computer for
a case-by-case analysis of those cases.
Theorem. (Four Color Theorem) Any planar graph
can be colored by four colors.

Maps on surfaces that are not flat can require more

colors. For instance, by joining the single arrows
together and the double arrows together in the fig-
ure below, one obtains a torus with seven mutually
touching regions; therefore seven colors are neces-
sary.

Source: http://en.wikipedia.org/wiki/Four color theorem

www.meru.org/Posters/hextorus.html www.parabola.unsw.edu.au/vol35 no2/node2.html

54
A bit of perfection
An integer n is perfect if it is the sum of its positive
divisors other than itself.

Examples. 6, 28, and 496 are perfect.

Proposition. (Euclid) If N = 2k−1(2k − 1) and 2k − 1

is prime, then N is perfect.
Marin Mersenne
2k−1(2k −1) being perfect can be
(1588 – 1648)
shown to be equivalent to 2k −1
being prime. Such primes are
called Mersenne primes and
much work has been devoted
to finding them. It can also be
shown that if 2k − 1 is prime,
then k is prime.
http://commons.wikimedia.org

On January 25, 2013, the 48th Mersenne prime

257,885,161 − 1 was discovered by Curtis Cooper and
has 17,425,170 digits. The associated perfect num-
ber has 34,850,339 digits.

You can be a part of the search for Mersenne primes:

http://www.mersenne.org/

55
Three unsolved problems:
Unsolved problem. Are there infinitely many Mersenne
primes?

Unsolved problem. Is every Mersenne number 2p −1

square free?
Source: http://primes.utm.edu/mersenne/index.html

Unsolved problem. Are there any odd perfect num-

bers?

Many properties that an odd perfect number must

satisfy have been determined, among them its largest
prime factor is greater than 108, and it has at least
101 prime factors and at least 9 distinct prime fac-
tors.

56
A little about prime numbers and their distribution
Theorem. (Euclid) There are infinitely many primes.
Theorem. There are arbitrarily large gaps between
successive primes.
Proof. Can be done in one line!

H The primes 29 and 31 are called twin primes.

Another pair is 59 and 61. Two others, each
having 4,932 digits, are 697, 053, 813 · 216,352 + 1 and
697, 053, 813 · 216,352 − 1. How many such pairs are
there?
H (Dirichlet, 1837) Any arithmetic progression whose
terms do not all share a common factor contains
an infinite number of primes.
H The Goldbach Conjecture. Every even integer
greater than 2 can be written as the sum of two
primes.
H For every positive integer m there is an even
number 2n such that there are more than m
pairs of consecutive primes with difference 2n.

“God may not play dice with the universe,

but something strange is going on with the
prime numbers.”

Paul Erdös (1913-1996)

57
Attitudes toward infinity at the time of Cantor
Expressions like this basic limit
1
lim =0
n→∞ n

were understood by saying that if n grew arbitrarily

then n1 would get arbitrarily close to zero.
But the notion of an infinite set was not accepted
or understood. Descartes said “The infinite is rec-
ognizable, but not comprehensible.”
Gauss wrote that “In mathematics infinite magni-
tude may never be used as something final; infinity
is only a façon de parler, meaning a limit to which cer-
tain ratios may approach as closely as desired when
others are permitted to increase indefinitely.”
Henri Poincare called Cantor’s work a “grave dis-
ease” affecting mathematics. Cantor was called a
“scientific charlatan” and a “corruptor of youth”.
Cantor realized that he was breaking sharply with
his predecessors and said in 1883: “I place myself
in a certain opposition to widespread views on the
mathematical infinite and to oftdefended opinions
of the essence of number.”
But Cantor had his supporters. David Hilbert de-
fended him saying “No one shall expel us from the
Paradise that Cantor has created.”
58
Infinite sets
We say that two sets have the same size, or the same
cardinality, if the elements of the two sets can be
placed into one-to-one correspondence.
A set is finite if it can be placed in one-to-one cor-
respondence with the set of the first n integers for
some n; otherwise, a set is infinite.
Cantor developed a special notation to express the
cardinal numbers, i.e. the sizes, of various sets:

j ℵ0 denotes the cardinality of the integers. Sets

having this cardinality are said to be countable.
j ℵ1 denotes the cardinality of the next larger in-
finite set. Sets having this or larger cardinal
number are said to be uncountable.
j c denotes the cardinality of the set of all real
numbers.

Cantor’s Theorem(s):
Theorem 1. The set of integers has the same size
as the set of even integers.
This result caused Cantor to define a set as infinite if
could be placed in one-to-one correspondence with
one of its proper subsets.

59
Theorem 2. The set of rational numbers has the
same size as the set of natural numbers.
Proof. Establish a one-to-one correspondence be-
tween the representations of the rational numbers
and the natural numbers as below:

n Rational representations
1 1 1 1 1 1
1 1 2 3 4 5 6 ···
2 2 2 2 2
3 1 2 3 4 5
3 3 3 3
6 1 2 3 4
10 4 4 4 ...
1 2 3
5 5
15 1 2
6
21 1
... ...

Question? Do you recognize the numbers in the n

column?

Question? Can you associate them with a geometric

shape?

Theorem 3. The set of all possible computer pro-

grams has the same size as the set of natural num-
bers.
60
Theorem 4. There are more real numbers in the
interval (0, 1) than there are natural numbers.
Proof. We argue by contradiction. Suppose not.
Then one could subscript all the reals in the interval
(0, 1) with natural numbers and list them together
with their decimal expansions. For example,
a0 = 0. 1 2 1 3 4 5 4 . . . Georg Cantor
a1 = 0. 0 7 7 2 2 1 6 . . . (1845 – 1918)
a2 = 0. 0 0 0 1 1 1 5 . . .
a3 = 0. 9 9 9 7 1 1 3 . . .
a4 = 0. 3 3 3 3 3 3 9 . . .
a5 = 0. 2 5 0 0 0 0 8 . . .
a6 = 0. 1 1 1 2 2 2 4 . . .
...

Now construct a number x accord-

ing to the following rules:
If digit n of an is even, make digit
n of x 1. If digit n of an is odd, Source:

make digit n of x 2. http://i12bent.tumblr.com/post/362218072

For example, for the above start we cantor-german-

would have x = 0.2212211... Now x is mathematician-and-

a real number in the interval (0, 1) philosopher

which differs in at least one decimal

place from every number on the list,
and is therefore not accounted for
in the one-to-one correspondence, a
contradiction. 

61
Another method of proof

The Well Ordering Principle is another of Cantor’s

contributions.

The Well Ordering Principle. Every non-empty set

of non-negative integers has a least element.

Fundamental Theorem of Arithmetic. Every integer

n > 1 is expressible as n = p1p2p3 · · · pr where r is a
positive integer and each pi is a prime.
Proof. We use contradiction and well-ordering. Assume that not all integers
have prime factorizations, and let n be the smallest positive integer without a
prime factorization.

n cannot be prime because then it would be its own factorization. So n is

composite.

Thus n = ab with a and b both strictly between 1 and n. Since n is the

smallest integer without a prime factorization and a and b are smaller, then

they each have prime factorizations a = p1 p2 p3 · · · pr and b = q1 q2 q3 · · · qs . But

this means that n = ab = p1 p2 p3 · · · pr q1 q2 q3 · · · qs gives a prime factorization of n,

which is a contradiction.

Lehman’s Lemma. There are no positive integer

solutions to the following equation
8a4 + 4b4 + 2c4 = d4.

62
Peano’s Axioms for Arithmetic
The following axioms were developed by Guiseppe
Peano (1858-1932) as the culmination of his effort to
formalize arithmetic. They are included here only
because they are referenced in the next major the-
orem.

1. 1 is a natural number.
2. For every natural number x there exists another
natural number x0 called the successor of x.
3. 1 6= x0 for every natural number x (x0 being the
successor of x).
4. If x0 = y 0 then x = y.
5. If Q is a property such that:
(a) If 1 has the property Q, and
(b) if x has property Q then x0 has property Q,
then property Q holds for all natural numbers.

63
 Kurt Gödel
(1906 – 1978)
Gödel’s Incompleteness Theorem.
In any consistent formal mathemati-
cal system that is expressive enough
to state Peano’s Axioms for arith-
metic, there is a statement that is
true but not provable.

Source:

https://en.wikipedia.org/wiki/Kurt Gödel

Two undecideable statements:

The Continuum Hypothesis states that there is no
subset of the real line, classically called the contin-
uum, that has more elements than the integers but
fewer than the whole line.
The Axiom of Choice states that given any collec-
tion of non-empty sets, however large, one element
can be picked from each of the sets.

64
Problems potpourri

Proposition.
1 · C(n, 1) + 2 · C(n, 2) + 3 · C(n, 3) + · · · + n · C(n, n) = n2n−1.

Problem. A winning configuration in the game of

Mini-tetris is a complete tiling of a 2 × n board by
2 × 2 squares and 2 × 1 dominos. A domino can be
placed so that it lies on a side 2 units long or stands
on a 1 unit end.

1. Determine a recursion to count the number Tn

of winning configurations of a 2×n board, n ≥ 1.
2n+1 + (−1)n
2. Show that Tn = for n ≥ 1.
3

Proposition. If n + 1 integers are selected from the

set {1, 2, 3, . . . , 2n}, then one of the integers divides
another.

Proposition. Every positive integer can be written

as a sum of distinct terms in the Fibonacci sequence.

Proposition. All persons have the same eye color.

Proposition. If all vertices of a simple graph have

degree less than or equal to n, then the graph can be
65
colored with n + 1 colors so that no pair of adjacent
vertices have the same color.

Proposition. Some two vertices in a graph with n

vertices must have the same degree.

Problem. Prove the power rule of differentiation:

d n
(x ) .
dx

Proposition. For n ≥ 0, prove that

X X n − in − j 
= f2n+1.
i≥0 j≥0
j i

       
n n−2 n−2 n−2
Proposition. = +2 + .
k k k−1 k−2

Proposition. For n ≥ 2, 6 divides 2n3 − 3n2 + n.

Proposition. A positive integer n is divisible by 11

if and only if the alternating sum of its digits, e.g.
for n = 8, 237, −8 + 2 − 3 + 7, is divisible by 11.

Problem. If 17! = 355, 687, ad8, 096, 000 what are a

and b.

Problem. Compute 521 (mod 31).

66
Problem. Show that if a country has only three cent
and five cent stamps then it can arrange postage for
any amount over eight cents.

1 1 1 1 1
Proposition. + 2 + 3 + · · · + k = 1 − k.
2 2 2 2 2

Problem. Show that if five points are located in the

interior √of a 1 × 1 square then two of them must be
within 22 units of each other.

Student submitted problem. If opposite corners are

removed from a standard checker board can the re-
maining board be tiled by dominos that cover two
adjacent squares?

Proposition. If a sequence of positive integers is

defined by a0 = 1, a1 = 2, a2 = 3 and an = an−1 + an−2 +
an−3, show that for all n ≥ 2, an ≤ 2n.

Poker again. Answer each of the following:

v How many poker hands have exactly three cards

higher than 8?
v How many poker hands contain cards from ex-
actly two ranks?

67
Remarks on Calculus and Discrete Mathematics
Disclaimers. Given the time constraints, the defini-
tions, etc., are a bit casual. Being a discrete course,
limits are, of course, avoided.
The derivative of a function f is the rate of change
of the function, or geometrically, the slope of the
tangent line to the graph of the function. If the
derivative of a function f exists for every point
in an interval it defines a function on the interval
whose value at x is denoted by f 0(x).
The process of obtaining a derivative is called dif-
ferentiation.
A substantial part of early calculus involves cal-
culating the derivatives of functions built up from
combinations of other functions. In other words,
differentiation is recursive.
One such rule is the product rule:

d d d
(u(x)v(x)) = u(x) (v(x)) + (u(x)) v(x)
dx dx dx
There are of course rules for the differentiation of
specific functions:

d n
v For n other than 0, (x ) = nxn−1
dx
d
v (sin(x)) = cos(x)
dx
68
 d
v (cos(x)) = − sin(x)
dx

Another part of basic calculus is the calculation of

areas. This historically is the first step in calculus
going back to Archimedes. It is based on Riemann
sums, which involves summing an increasingly large
number of rectangles.
Some discrete math connections:

k The power rule can be proven by mathematical

induction.
k Differentiation is a recursive operation.
k The typical examples of Riemann sums rely on
formulas that most calculus students do not re-
ally understand, but you realize that they are
just the results of routine mathematical induc-
tion exercises and that they have their origins
in Pascal’s Triangle:
n(n + 1)
u 1 + 2 + ... + n =
2
n(n + 1)(2n + 1)
u 12 + 22 + ... + n2 =
6
 2
n(n + 1)
u 13 + 23 + ... + n3 =
2
k The harmonic series diverges: 1 + 12 + 13 + 14 + · · · +
1 n
2n + · · · ≥ 1 + 2 for all n.

69
Calculus Points to Ponder

X Why is the continuity hypothesis appropriate on

a closed interval whereas the differentiability hy-
pothesis applies on an open interval?
X How is the continuity hypothesis used in proving
the Fundamental Theorem of calculus?
X The Intermediate Value Theorem and the exis-
tence of absolute maxima and minima are two
important consequences of continuity to under-
stand.
X Pay close attention to how the inverse of vari-
ous functions and implicit differentiation can be
used to derive differentiation formulae for addi-
tional functions, e.g., the natural log and the ex-
ponential function, and trig functions and their
inverses.
X Keep things straight - don’t assume that the
converse of a proposition is automatically true.
It helps to have counterexamples handy, e.g.,
u a continuous function that is not differen-
tiable.
u a point where a function has a zero derivative
but no local maximum or minimum.
u an important continuous function with no el-
ementary antiderivative.

70
 u a divergent infinite series whose nth term ap-
proaches 0.
X Appreciate the beauty of the connections with
science, e.g.,
u Fermat’s Principle implies Snell’s Law,
u The percentage increase in the flux through
an artery is four times the increase in the
radius.
X Pay attention to how the Mean Value Theorem
can be applied in the development of the meth-
ods of locating maxima and minima. You can
ignore it in many courses, but there are MANY
applications of the Mean Value Theorem later
in analysis, so learn it now!
X Don’t ignore Riemann Sums and the definition
of the definite integral just because the Funda-
mental Theorem of Calculus seems so much eas-
ier. It may happen that the functions that turn
out to be important to you don’t show up in con-
venient closed form with a nice neat antideriva-
tive.
X Watch for the analogy between the solution of
the closed form of the Fibonacci numbers and
the solution of an initial value problem involving
a second order linear differential equation with
constant coefficients.

71
Selected references:

Benjamin, Art, and Jennifer Quinn, “Proofs That Really Count:

The Art of Combinatorial Proof”, Mathematical Association
of America, 2003.

Bona, Miklos, “A Walk Through Combinatorics”, World Sci-

entific, 2008.

Chartrand, Gary, “Introductory Graph Theory”, Dover Books

on Mathematics, 1984.

Erickson, Martin, “Pearls of Discrete Mathematics”, CRC

Press, 2010.

Gilbert, William J., and Scott A. Vanstone, “An Introduction

to Mathematical Thinking: Algebra and Number Systems”,
Pearson Education, Inc, 2005.

Kline, Morris, “Mathematics: The Loss of Certainty”, Oxford

U. Press, 1980.

Prim, R. G.,“Shortest Connection Networks and Some Gen-

eralizations”, Bell Systems Technical Journal, 36, 1957, pp.
1389 – 1401.

Walker, Russell, “Introduction to Mathematical Programming”,

Pearson Learning Solutions, 2013.

72