C O M P U L S O R Y
C O M P U L S O R Y
R E A D I N G S
R E A D I N G S 1
1
1 According to the author of the module, the compulsory readings do not infringe known copyright.
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Number Theory-readings, Notas de estudo de Informática
Elementary Number Theory
Tipologia: Notas de estudo
2014
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C O M P U L S O R YC O M P U L S O R Y
R E A D I N G S R E A D I N G S
(^1) According to the author of the module, the compulsory readings do not infringe known copyright.
10. COMPULSORY READINGS
Reading #1:
Complete Reference : Elementary Number Theory, By W. Edwin Clark, University of South Florida, 2003. (File name on CD: Elem_number_theory_Clarke)
Abstract/Rationale : A complete open-source text book in number theory. The complete text is provided as a readable computer file. Specific page references are given in the learning activities to direct the student to activities, readings and exercises.
Reading #2:
Complete Reference : Elementary Number Theory, By William Stein, Harvard University, 2005 (File name on CD: Number_Theory_Stein)
Abstract/Rationale : A complete open-source text book in number theory. The complete text is provided as a readable computer file. Specific page references are given in the learning activities to direct the student to activities, readings and exercises.
Reading #3:
Complete Reference : MIT Open Courseware, Theory of Numbers, Spring 2003, Prof. Martin Olsson (Folder name on CD: MIT_Theory_of_Numbers)
Abstract/Rationale : A collection of lecture notes from Number Theory lectures at MIT in Boston, USA. Each lecture clearly addresses a specific number theory topic to supplement the learning materials.
Reading # 1: MacTutor History of Mathematics (visited 03.11.06)
Complete reference :
http://www-history.mcs.standrews.ac.uk/Indexes/Number_Theory.html
Reading # 3: Wikipedia (visited 03.11.06)
Complete reference : http://en.wikipedia.org/wiki/Number_Theory
Abstract : Wikipedia should be the learners closest source of reference in Number Theory. It is a very powerful resource that all learners must refer to understand abstract mathematics. Moreover, it enables the learner to access various arguments that have puzzled mathematicians over the centuries.
Rationale: It gives definitions, explanations, and examples that learners cannot access in other resources. The fact that wikipedia is frequently updated gives the learner the latest approaches, abstract arguments, illustrations and refers to other soucers to enable the learner acquire other proposed approaches in number theory.
ii
Preface
Number theory is concerned with properties of the integers:
... , − 4 , − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , 4 ,....
The great mathematician Carl Friedrich Gauss called this subject arithmetic and of it he said:
Mathematics is the queen of sciences and arithmetic the queen of mathematics.”
At first blush one might think that of all areas of mathematics certainly arithmetic should be the simplest, but it is a surprisingly deep subject. We assume that students have some familiarity with basic set theory, and calculus. But very little of this nature will be needed. To a great extent the book is self-contained. It requires only a certain amount of mathematical maturity. And, hopefully, the student’s level of mathematical maturity will increase as the course progresses. Before the course is over students will be introduced to the symbolic programming language Maple which is an excellent tool for exploring number theoretic questions. If you wish to see other books on number theory, take a look in the QA 241 area of the stacks in our library. One may also obtain much interesting and current information about number theory from the internet. See particularly the websites listed in the Bibliography. The websites by Chris Caldwell [2] and by Eric Weisstein [11] are especially recommended. To see what is going on at the frontier of the subject, you may take a look at some recent issues of the Journal of Number Theory which you will find in our library.
iii
v
Famous Quotations Related to Number Theory
Two quotations from G. H. Hardy: In the first quotation Hardy is speaking of the famous Indian mathe- matician Ramanujan. This is the source of the often made statement that Ramanujan knew each integer personally.
I remember once going to see him when he was lying ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. “No,” he replied, “it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways. ”
Pure mathematics is on the whole distinctly more useful than ap- plied. For what is useful above all is technique, and mathematical technique is taught mainly through pure mathematics.
Two quotations by Leopold Kronecker
God has made the integers, all the rest is the work of man.
The original quotation in German was Die ganze Zahl schuf der liebe Gott, alles Ubrige ist Menschenwerk.¨ More literally, the translation is “ The whole number, created the dear God, everything else is man’s work.” Note in particular that Zahl is German for number. This is the reason that today we use Z for the set of integers.
Number theorists are like lotus-eaters – having once tasted of this food they can never give it up.
A quotation by contemporary number theorist William Stein:
A computer is to a number theorist, like a telescope is to an astronomer. It would be a shame to teach an astronomy class without touching a telescope; likewise, it would be a shame to teach this class without telling you how to look at the integers through the lens of a computer.
vi PREFACE
- R e a d i n g ( s )R e a d i n g ( s ) # 1#
- 1 Basic Axioms for Z Preface iii
- 2 Proof by Induction
- 3 Elementary Divisibility Properties
- 4 The Floor and Ceiling of a Real Number
- 5 The Division Algorithm
- 6 Greatest Common Divisor
- 7 The Euclidean Algorithm
- 8 Bezout’s Lemma
- 9 Blankinship’s Method
- 10 Prime Numbers
- 11 Unique Factorization
- 12 Fermat Primes and Mersenne Primes
- 13 The Functions σ and τ
- 14 Perfect Numbers and Mersenne Primes
- 15 Congruences viii CONTENTS
- 16 Divisibility Tests for 2 , 3 , 5 , 9 ,
- 17 Divisibility Tests for 7 and
- 18 More Properties of Congruences
- 19 Residue Classes
- 20 Zm and Complete Residue Systems
- 21 Addition and Multiplication in Zm
- 22 The Groups Um
- 23 Two Theorems of Euler and Fermat
- 24 Probabilistic Primality Tests
- 25 The Base b Representation of n
- 26 Computation of aN mod m
- 27 The RSA Scheme
- A Rings and Groups
Chapter 1
Basic Axioms for Z
Since number theory is concerned with properties of the integers, we begin by setting up some notation and reviewing some basic properties of the integers that will be needed later:
N = { 1 , 2 , 3 , · · · } (the natural numbers or positive integers) Z = {· · · , − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , · · · } (the integers)
Q =
{ (^) n m
| n, m ∈ Z and m 6 = 0
(the rational numbers)
R = the real numbers
Note that N ⊂ Z ⊂ Q ⊂ R. I assume a knowledge of the basic rules of high school algebra which apply to R and therefore to N, Z and Q. By this I mean things like ab = ba and ab + ac = a(b + c). I will not list all of these properties here. However, below I list some particularly important properties of Z that will be needed. I call them axioms since we will not prove them in this course.
Some Basic Axioms for Z
- If a, b ∈ Z, then a + b, a − b and ab ∈ Z. (Z is closed under addition, subtraction and multiplication.)
- If a ∈ Z then there is no x ∈ Z such that a < x < a + 1.
- If a, b ∈ Z and ab = 1, then either a = b = 1 or a = b = −1.
- Laws of Exponents: For n, m in N and a, b in R we have
Chapter 2
Proof by Induction
In this section, I list a number of statements that can be proved by use of The Principle of Mathematical Induction. I will refer to this principle as PMI or, simply, induction. A sample proof is given below. The rest will be given in class hopefully by students.
A sample proof using induction: I will give two versions of this proof. In the first proof I explain in detail how one uses the PMI. The second proof is less pedagogical and is the type of proof I expect students to construct. I call the statement I want to prove a proposition. It might also be called a theorem, lemma or corollary depending on the situation.
Proposition 2.1. If n ≥ 5 then 2 n^ > 5 n.
Proof #1. Here we use The Principle of Mathematical Induction. Note that PMI has two parts which we denote by PMI (a) and PMI (b). We let P (n) be the statement 2n^ > 5 n. For n 0 we take 5. We could write simply:
P (n) = 2n^ > 5 n and n 0 = 5.
Note that P (n) represents a statement, usually an inequality or an equation but sometimes a more complicated assertion. Now if n = 4 then P (n) be- comes the statement 2^4 > 5 · 4 which is false! But if n = 5, P (n) is the statement 2^5 > 5 · 5 or 32 > 25 which is true and we have established PMI (a).
4 CHAPTER 2. PROOF BY INDUCTION
Now to prove PMI (b) we begin by assuming that
P (n) is true for 5 ≤ n ≤ k.
That is, we assume
(2.1) 2 n^ > 5 n for 5 ≤ n ≤ k.
The assumption (2.1) is called the induction hypothesis. We want to use it to prove that P (n) holds when n = k + 1. So here’s what we do. By (2.1) letting n = k we have
2 k^ > 5 k.
Multiply both sides by two and we get
(2.2) 2 k+1^ > 10 k.
Note that we are trying to prove 2k+1^ > 5(k + 1). Now 5(k + 1) = 5k + 5 so if we can show 10k ≥ 5 k + 5 we can use (2.2) to complete the proof.
Now 10k = 5k + 5k and k ≥ 5 by (2.1) so k ≥ 1 and hence 5k ≥ 5. Therefore
10 k = 5k + 5k ≥ 5 k + 5 = 5(k + 1).
Thus
2 k+1^ > 10 k ≥ 5(k + 1)
so
(2.3) 2 k+1^ > 5(k + 1).
that is, P (n) holds when n = k + 1. So assuming the induction hypothesis (2.1) we have proved (2.3). Thus we have established PMI (b). We have established that parts (a) and (b) of PMI hold for this particular P (n) and n 0. So the PMI tells us that P (n) holds for n ≥ 5. That is, 2n^ > 5 n holds for n ≥ 5.
I now give a more streamlined proof.
Proposition 2.2. If n ≥ 5 then 2 n^ > 5 n.
6 CHAPTER 2. PROOF BY INDUCTION
- Use the induction hypothesis and anything else that is known to be true to prove that P (n) holds when n = k + 1.
- Conclude that since the conditions of the PMI have been met then P (n) holds for n ≥ n 0.
- Write QED or or // or something to indicate that you have com- pleted your proof.
Exercise 2.1. Prove that 2n^ > 6 n for n ≥ 5.
Exercise 2.2. Prove that 1 + 2 + · · · + n =
n(n + 1) 2
for n ≥ 1.
Exercise 2.3. Prove that if 0 < a < b then 0 < an^ < bn^ for all n ∈ N.
Exercise 2.4. Prove that n! < nn^ for n ≥ 2.
Exercise 2.5. Prove that if a and r are real numbers and r 6 = 1, then for n ≥ 1
a + ar + ar^2 + · · · + arn^ =
a (rn+1^ − 1) r − 1
This can be written as follows
a(rn+1^ − 1) = (r − 1)(a + ar + ar^2 + · · · + arn).
And important special case of which is
(rn+1^ − 1) = (r − 1)(1 + r + r^2 + · · · + rn).
Exercise 2.6. Prove that 1 + 2 + 2^2 + · · · + 2n^ = 2n+1^ − 1 for n ≥ 1.
Exercise 2.7. Prove that 111︸ ︷︷ · · · 1 ︸
n 1 ’s
10 n^ − 1 9
for n ≥ 1.
Exercise 2.8. Prove that 1^2 + 2^2 + 3^2 + · · · + n^2 =
n(n + 1)(2n + 1) 6
if n ≥ 1.
Exercise 2.9. Prove that if n ≥ 12 then n can be written as a sum of 4’s and 5’s. For example, 23 = 5 + 5 + 5 + 4 + 4 = 3 · 5 + 2 · 4. [Hint. In this case it will help to do the cases n = 12, 13 , 14 , and 15 separately. Then use induction to handle n ≥ 16 .]
Exercise 2.10. (a) For n ≥ 1, the triangular number tn is the number of dots in a triangular array that has n rows with i dots in the i-th row. Find a formula for tn, n ≥ 1. (b) Suppose that for each n ≥ 1. Let sn be the number of dots in a square array that has n rows with n dots in each row. Find a formula for sn. The numbers sn are usually called squares.
Exercise 2.11. Find the first 10 triangular numbers and the first 10 squares. Which of the triangular numbers in your list are also squares? Can you find the next triangular number which is a square?
Exercise 2.12. Some propositions that can be proved by induction can also be proved without induction. Prove Exercises 2.2 and 2.5 without induction. [Hints: For 2.2 write s = 1+2+· · ·+(n−1)+n. Directly under this equation write s = n+(n−1)+· · ·+2+1. Add these equations to obtain 2 s = n(n+1). Solve for s. For Exercise 2.5 write p = a+ar +ar^2 +· · ·+arn. Then multiply both sides of this equation by r to get a new equation with rp as the left hand side. Subtract these two equation to obtain pr − p = arn+1^ − a. Now solve for p.]