# Math Problem Set

**Problem 1.1.** Find an opponent and play a number of games of 2-

spot and 3-spot sprouts. Find out the best strategy that you can for

2-spot sprouts, and write a short essay describing your finding.

This essay should fit on one side of a sheet of paper.

**Due Friday, January 16, 2009.**

**Problem 1.2.** Prove that every game of sprouts starting with any

finite number of spots (say with N spots) must eventually end, no

matter how the players play. (Of course they must obey the rules.)

If the game starts with N spots, give an upper bound on the number

of moves in any game.

**Due Wednesday, January 21, 2009.**

**Problem 1.3.** Read pages 39 - 51 in our text "An Introduction to

Mathematical Reasoning" and do problems 5.1 to 5.4 on page 51.

**Due Monday, January 26, 2009.**

**Problem 1.4.** Prove by induction that

(1/1x3) + 1/(3 x 5) + 1/(5 x 7) + ... + 1/((2n-1) x (2n + 1)) =

n/(2n + 1).

**Due Monday, January 26, 2009.**

**Problem 1.5.** The Fibonacci Numbers are the numbers in the

sequence 1,1,2,3,5,8,13,21,34,55,144,189,....

We denote these by f_{1} = 1, f_{2} = 1, f_{3} = 3, f_{4} = 5 and

f_{n+1} = f_{n} + f_{n-1}.

Each Fibonacci number is the sum of the previous two Fibonacci

numbers. There are many patterns in the Fibonacci series.

Notice that

Prove by induction that

**Due Friday, January 30,2009.**

**Problem 1.6. **(Returning to Sprouts) Prove that for every natural

numbrer n (n = 1, 2, 3, 4, ...) there is a sprouts game, starting with

n sprouts, that ends in exaactly 3n -1 moves. You can use

mathematical induction in your proof.

**Due Monday, February 2, 2009.**

**Problem 1.7.** In 4-sprouts we use the same rules as in ordinary

sprouts, but we allow 4 lines to touch a spot

as in the diagram below.

A move has the same form as in regular spots, but notice that the

new spot, having two lines going into it, has two freedoms in the

game of 4-spots. Show that some games of 4-spots can go on forever.

Give specific examples. Think about the question of how to

modify the rules of 4-spots so that it will become a game that always

ends in a finite number of moves.

**Due Wednesday, February 4, 2009.**

**Problem 1.8.** The Fibonacci Numbers are the numbers in the

sequence 1,1,2,3,5,8,13,21,34,55,144,189,....

We denote these by f_{1} = 1, f_{2} = 1, f_{3} = 3, f_{4}
= 5 and

f_{n+1} = f_{n} + f_{n-1}. We also take f_{0} =
0 by convention.

Prove the following formula by induction.

For example 8 x 21 = 168 = 169 - 1 = 13^{2} -1.

**Due Friday, February 6,2009.**

**Problem 1.9**. Problem 20. page 56 of our textbook.

**Due, Monday, February 9,2009.**

**Problem 1.9.** Problem 20. page 56 of our textbook.

**Due, Monday, February 9,2009.**

**Problem 1.10. **Prove by induction that

for n = 1,2,3,... .

**Due, Monday, February 9,2009.**

**Problem 1.11.** Suppose that you play a sprouts game with **n**

starting spots and that the game lasts for **3n - 1** moves. How many

Pharisees will there be at the end of the game? Illustrate your result

with a specific example.

**Due, Wednesday, February 11,2009.**

**Problem 1.12. (The Gas-Electricity-Water Problem)**

Three companies, the gas company, the electricity compay and the

water company want to make connections from the gas main (G),

the electrical source (E) and the water main (W) to three houses

(H1, H2 and H3). They wish to lay their lines so that no two lines

meet except at the sources (G,E and W) and at the houses

(H1, H2 and H3). Can you find a solution to this design problem?

If not, then why not?

In the illustration above the city planners have drawn a graph to

help them design the connections but they have run into a difficulty

with making a water line from W to H1. Everything went fine with

the design up to that point, but then there does not seem to be any

way to conncet from W to H1 without crossing previously created

lines. It will cost the city a great deal to dig tunnels to make lines

cross over one another. So these designers really need to know

whether the job can be done with no crossovers, and if it cannot be

done that way, then they want to know the least number of

crossovers that are needed to do the job.

**Due, Wednesday, February 11,2009.**

**Problem 1.13**. Construct a proof of Euler's formula by induction

on the total number of edges and vertices in the graph G. You

should consider how the graph can be built up from simpler graphs

by adding edges to them. In fact, any connected graph can be built

from a single vertex graph by adding new edges in two ways that I

will now explain, but first we introduce an abbreviation: The

diagram below stands for some vertex in a larger graph.

You can tell when I am using this abbreviation because the edges

that go out of this vertex are not meeting any other vertices in the

picuture. The picture is a shorthand for a possibly larger and more

complete picture. In the abbreviation we show three edges touching

the vertex. In a real situation some edges touch the vertex, but the

number is not necessarily equal to three. Ok?

Now lets use this and illustrate two ways to make a larger graph.

In method number I we add a new edge and a new vertex by

attaching the new edge to an already existing vertex. In method

number II we connect two vertices with a new edge.

**Remark.** We regard the move

as a special case of II.

I claim that any connected graph can be built up by performing

a sequence of operations of these two types. Here is an example.

You can use this claim in your proof, and if you want, you can also

make a proof of the claim. We will discuss why the claim is true in

class.

Now, to prove the Euler Theorem, you can proceed by induction,

showing that V - E + F does not change its value when you perform a

move of type I or type II. You will find that it is very easy to see this

for type I, and that in order to see it for type II you need

to start with a connected graph. If the graph is connected, then a

move of type II will create a new region in the graph. Look at the

example above and see how this works. You can use this fact also in

your proof (that a move of type II will create a new region). You

should then be able to construct an inductive proof of the Euler

formula.

**Here is an example:
We create a triangle graph by adding an edge to a tree.**

Note that adding the edge creates a new region, and V-E + F does

not change from before to after the addition of the new edge.

(c) Discuss your proof of the Euler formula with another student in

the class. Do you both feel that the proof is complete? What might

be missing? In this problem, it worth having the discussion. We will

discuss some of the issues related to the problem in class before the

problem is due.

**Due, Friday, February 20,2009.**

**Supplement to Problem 1.13.**

There is a fact about curves in the plane that you can use in

thinking about regions that are created when graphs are drawn in

the plane.

This fact is called the

**Jordan Curve Theorem:** A closed curve in the plane without any

self-intersections divides the plane into exactly two regions.

Here is an example:

You are not required to prove this result, but you can use it and it is

interesting to see how complex examples can look!

Is the black dot inside or outside this curve?

Of course you can solve this like solving a maze, but look!

An arrow from the dot intesects the curve in an ODD number of

points. I claim that this tells you that the point must be INSIDE.

If the intersection number were EVEN, then the point would be

outside. Can you explain why this works? (I say explain, and of

course I am hoping that your explanation will turn into a

mathematical proof. But lets explore.)

We will discuss in class why and how the Jordan Curve Theorem is

relevant to proving Euler's Formula.

**Remark**. Another approach to the Euler formula uses the concept

of a tree: A graph is said to be a tree if it does not contain any cycles

(a cycle is a sequence of distinct edges such that the each edge

shares its endpoints with the edges before and after it in the

sequence. For example in the graph above, bce is a cycle and abcd is

a cycle. When a plane graph has no cycles then the only region it

can delineate is the rest of the plane other than itself, and so a tree

has **F = 1.
Show that for a connected tree, V - E = 1.**

From this is follows that for connected plane trees

**V - E + F = 2,**

and so we know the Euler formula already for trees.

**The picture above illustrates this result for trees. You can
prove that V - E = 1 for a connected tree by induction on
the number of edges in the tree.**

You can then prove the Euler formula for an arbitrary connected

plane graph by just making that graph by adding edges by our type

II move to a tree. Think about this and try some examples.

**Logic Problems and Other Problems From Part I.**

**Problem 2.1.
Proof by Contradiction.**

Consider the following proof that the square root of 2 is not

a rational number. At those places where we state an exercise, do

the exercise.

**Proof. **Suppose that there is a rational number p/q whose square is

equal to 2. We can assume that p/q is in reduced fractional form

and that it is positive. Thus we assume that p and q are positive

integers and that they do not have a common divisor other than 1.

So (p/q)^{2} = 2.

Whence p^{2}/q^{2}= 2.

Whence p^{2} = 2q^{2}.

Thus p^{2} is even.

This implies that p is even

**[Exercise: Prove that, when p is a natural number, if p ^{2} is
even ,then p is even.]
**Therefore p = 2r for some natural number r.

Hence (2r)

^{2}= 2q

^{2}by substitution into p

^{2}= 2q

^{2}.

This is the same as saying 4r

^{2}= 2q

^{2}.

Dividing this equation by 2, we get

2r

^{2}= q

^{2}. Thus q

^{2}is even.

But this means that q is even, by the exercise.

Therefore both p and q are even.

This means that p and q have the common factor 2.

But this contradicts our assumption that p and q do not have a

common factor.

We conclude that the assumption that there is rational number

whose square is 2, leads to a contradiction. Since we assume that

mathematics is consistent, this is a proof by contradiction that the

square root of 2 is irrational. Q.E.D.

This proof is a very good example of a proof by contradiction. We

wish to prove P and we start by assuming ~P and find that this

assumption leads to an absurdity.

We conclude from this that ~P cannot be true. But if ~P is false, then

P is true!

The result that the square root of 2 is not rational was a scandal for

the ancient Greek mathematicians, and even today there is a

mystery about this. It means that the decimal expansion of the

square root of two does not have a periodic pattern, and to know

the square root of two fully as a decimal you would have to know

infinitely many terms of its expansion.

**[Exercise: Prove that there is no rational number whose
square is equal to 3.]
Remark on the above exercise. You willl need to prove that
3 divides p**

^{2}implies that 3 divides p. In order to do this, note that the natural numbers fall into three disjoint classes:

**A is the class of all numbers that are divisible by 3.
B is the class of all numbers that leave a remainder of 1
upon division by 3.
C is the class of all numbers that leave a remainder of 2
upon division by 3.**

**You can show that p ^{2} is of type A only if p is of type A.
From this it follows that if p^{2} is divisible by 3, then p is
divisible by 3.**

**Due Monday, March 2, 2009.**

**Problem 2.2.
From the book.
p. 9: 1.3, 1.4.
p. 19: 2.1,2.3,2.4,2.5.
p. 53: 1,2.**

**Due Monday, March 2, 2009.**

**Problem 2.3.
p. 53: 3.**

**p. 54: 6.
Due Wednesday, March 4, 2009.**

Remark on Problem 6 of page 54:

Here you are asked to use only the axioms given on pages

18 and 19. This is a very stringent exercise! Here is an example of

how one can proceed.

We want to prove that a x 0 = 0. Here is the proof.

**Proof that a x 0 = 0.
0 = 0 + 0 by (iv)
Hence a x 0 = a x (0 + 0) and
a x (0 + 0) = a x 0 + a x 0 by the distributive law (iii).
Thus a x 0 = a x 0 + a x 0.
Thus
a x 0 + (- (a x 0) ) = (a x 0 + a x 0 ) + (- (a x 0) )
and
(a x 0 + a x 0 ) + (- (a x 0) ) = a x 0 + (a x 0 + (- (a x 0) )
by the associative law (ii).
Thus
a x 0 + (- (a x 0) ) = a x 0 + (a x 0 + (- (a x 0) )
We know that a x 0 + (- (a x 0) ) = 0 by (vi).
Thus
0 = a x 0 + 0.
Since a x 0 + 0 = a x 0 by (iv), we conclude that
0 = a x 0.
QED.**

Note that in applying an axiom like "x + 0 = x for any x"

we can put whatever we like in place of x.

Elephants + 0 = Elephants?

Well, not quite anything we like. It has to be about numbers for

these axioms. So we can say

27 + 0 = 27

and we can say

(xy + 2z) + 0 = (xy + 2z).

In justifying our steps in the proof, we have not explicitly

indicated such substitutions.

**Problem 2.4. p. 9 - 1.3, 1.4, 1.5**

**p. 29, 3.1, 3.2, 3.7
Due Monday, March 10, 2009.**

**Problem 2.5. This problem introduces an interpretation of
logic in terms of switching circuits. This application of
logic to switching circuits is the discovery of
Claude Shannon**

The Journal of Symbolic Logic, Vol. 18, No. 4 (Dec., 1953), p. 347

**and forms the basis for the design of computers to this
day. Here is an abstract switch:**

**A signal can go from left to right through the switch when
it is closed, and no signal can go through when the switch
is open. We choose to designate a switch by a label such as
A above and we let A = T correspond to the closed switch
position (T for "transmit" if you like!) and we let A = F
when the switch is in the open position (F = ~ T = not
transmit).**

**The position of a switch can control a device such as a
lamp.**

**Two basic ways to put switches together are Series and
Parallel Connection:**

**As you can see, the only way for a signal to get from left
to right in the series connection of A and B is if both A and
B are T (closed). Thus the series connection corresponds to
A and B which we write in logic notation as A ^ B.
Similarly, the only way for a signal to get from left to right
in a parallel connection is if one of A or B is closed. Hence
this corresponds to A or B, which we write as A v B.
Thus our two basic logical operations are mirrored in the
behaviour of networks that carry signals.**

**What about negation? An example will show you how we
handle this. What we do is, we allow multiple appearances
of a given label or its negation. We will write either
~A or A' for the negation of A. Here we will use A' ok?**

**Then in the mutiple appearances, all A's will be either
closed or open and if you have A' and A is closed, then A'
will be open. If A is open then A' will be closed.
Look at this example:**

**Each of these circuits has an A and an A'. In the series
connection, this means that one switch is always open and
so the value of the whole circuit is the same as a simple
open circuit. On the other hand in the parallel connection
of A with A' either one line will transmit, or the other line
will transmit. So the circuit as a whole behaves like a
single switch that is closed. Thus we see that the series
connecction corresponds to the identity A ^ A' = F, while
the parallel connection corresponds to the identity
A v A' = T.**

**Terminology: Since a label A in one of our circuits can
appear in many places we will still refer to A as a 'switch'
even though it may be composed of a number of
elementary switches. In such a multiple switch, if you make
one, there has to be a mechanical way to make sure that all
the different parts of the switch work together. Also these
switching patterns may, in practice, happen in elecronic
circuitry that syncronizes actions at separate locations.
For our purposes, we shall imagine simple mechanical
switching devices.**

**Problem: Illustrate the two identities (distribution laws in
logic) with the correspoinding switching circuits.**

**(i) a ^ (b v c) = (a ^ b) v (a ^ c)
(ii) a v (b ^ c) = (a v b) ^ (a v c).**

**In each case, draw the corresponding circuits and explain
why they have the same signal transmission behaviour.
Due Wednesday, March 11, 2009.**

**Problem 2. 6. Design a switching circuit with three
switch labels a,b,c such that each of a, b and c individually
control transmission of the signal. That is if the circuit is
open, then changing the state of any one of a,b,c will close
the circuit and if the circuit is closed, then changing the
state of any one of a,b,c will open the circuit.
(We discussed how to do this with in analogous case of two
labels and how it is related to (a ^ b) v (a' ^ b') in class.)**

**Due Monday, March 16, 2009.
Problem 2.7. p. 72, 6.1, 6.4, 6.5, 6.7.
Due Monday, March 16, 2009.
Problem 2.8. p. 86, 7.6, 7.7.
Due Wednesday, March 18, 2009.**

**Problem 2.9. Generalize Problem 2. 6. to an arbitrary
number N of switch labels. That is, you would like to be
able to control a single lamp with any one of N switches.
You want to find a design that will work in principle for
(say) a building with N floors, so that you can control the
entrance light from any of the floors. The switch on any
given floor will turn the entrance light on or off. That
switch on a given floor will have two positions just as our
labels have two states T or F. This is a hard problem.
The notes on Logic Circuits should make it accessible to
you. See these notes on the website.
Due Monday, March 30, 2009.**

**Problem 2.10. p. 99, 8.1, 8.2, 8.3
Due Friday, March 20, 2009.
Problem 2.11. p. 113, 9.1,9.2,9.3,9.4,9.5,9.6.**

**Due Wednesday, April 1, 2009.
Problem 2.12. p. 115, 2., 3., 6., 8.
Due Friday, April 3, 2009.**

**Problem 2.13.
(a) page 132 #10.3.
(b) Prove that for all sets x,y,z: If {x.y} = {x,z}, then y = z.
(c) Prove that for all sets a,b,c,d:
If {{a},{a,b}} = {{c},{c,d}}, then a = c and b = d.
Due Wednesday, April 8, 2009.**

**Cantor's Theorem**

In class we covered Cantor's proof that P(X) > X for any set X,

where P(X) is the set of subsets of X.

Here is the gist of the proof:

Suppse F:X ----> P(X) is any mapping of these sets.

Let C = {x in X| x is not in F(x)}.

Then C cannot equal F(z) for any z.

To see this, consider a set F(z).

If z is in F(z) then z is not in C and so C and F(z) are different.

If z is not in F(z) then z is in C and so C and F(z) are different.

Therefore C is not of the form F(z) and so F is not surjective.

We have shown that no mapping F:X ----> P(X) is surjective. Since it

is easy to construct an injection from X to P(X) (e.g. x is sent to {x})

this proves that X < P(X).

**Problem 2.14.
(a) Let N _{n} = {1,2,3,...,n} so that N_{n} has cardinality n.
We write |X| for the cardinality of a set X. And so write
|N_{n}| = n. Now, let P(X) denote the set of subseets of a given
set X. Prove by induction on n that P(N_{n}) has cardinality
2^{n}. That is, |P(N_{n})| = 2^{n}.**

**(b) Let X = {a, b, c}. Make an explicit form for P(X).
That is, make an explicit list of all the subsets of X.
Choose a function F: X ----> P(X) (make this quite explicit).
Then apply Cantor's process to F and form the
set C = {x in X | x is not in F(x)}. You should find that
C is not of the form F(z) for any z in X. Discuss.**

**(c) Consider the collection Aleph of all possible sets
whose
members are themselves sets . Sets in Aleph can be infinite
or finite. In a sense, Aleph is "the collection of all sets".
Aleph begins with the empty set { }. Then you get
{ { } }. After that, come things like
{ { }, {{ }}, { { }, {{ }} }, {{{ }}} }.**

**And there are lots of infinite sets of sets. For
example
{{}, {{}}, {{{}}}, {{{{}}}}, ...}. And you can take sets of these.
It grows and grows and whenever you have a element X of
Aleph, you can look inside and see that all the members of
X are themselves sets, and hence members of Aleph. So any
member of Aleph is itself a subset of Aleph.**

**Any subset of Aleph is a collection of sets and so is
also a
member of Aleph. Thus it seems that we have proved that
Aleph = P(Aleph) in contradiction to Cantor's Theorem that
says that the power set of a set is always bigger than the
set. Think about this. This is not exactly a puzzle. It is
really a philosophical and mathematical problem about the
use of the word "all". We will discuss this further in class.**

**Due Friday, April 10, 2009.**

**Problem 2.15.
Let X be a finite set and Y another finite set.
Let X^Y denote the intersection of X and Y and let
XvY denote the union of X and Y. Let |X| denote the
cardinality of X. Prove, in your own words, the formula
|XvY| = |X| + |Y| - |X^Y|.**

**Due Monday, April 13,2009.**