NUMBER THOERY
CHAPTER 1
THE INTEGERS
CONTRIBUTORS:
ROSIE ADIAO
BRIONES LEGOLAS P. ARAGORN
MEL ANGELO L. CORREA
ZHYRLAND Q. DELOSTRICO
ARNELA P. GIGANTE
SHONNE FARQER A. OBAÑA
JENNIFER A. REGENCIA
ARLAN C. SOLOMON JR.
ILLUMINADO YANGAO JR.
ACKNOWLEDGEMENT
This module is made possible through the collaborative effort of every member.
Our collaboration fostered a creative and productive environment, leading to the successful
completion of this module. Please allow us to also dedicate this acknowledgment to
recognize significant contributors:
First and above all, we would like to thank God Almighty who give us knowledge,
wisdom, prosperity, and hope throughout our clerical journey in completing this module
in Number Theory.
We extend our heartfelt gratitude towards Dr. Jose De Leon. His encouragement
and guidance made it possible for us to accomplish this module.
Lastly, to our parents who never cease to love and support us in all our needs, to
our friends, and classmates who helped us in gathering information for this module, our
profound appreciation to all of you.
TABLE OF CONTENTS
Introduction
Chapter 1: The Integers
Real Number System
Sequences
Fibonacci Numbers
The Well Ordering Principle and Mathematical Induction
1.4.1 Adding of Suitable Terms using Mathematical Induction
1.4.2 Divisibility using Mathematical Induction
1.4.3 Inequalities by Comparison using Mathematical Induction
1.5 The Sums and Products
1.6 Divisibility
1.1
1.2
1.3
1.4
INTRODUCTION
CHAPTER
1
THE INTEGERS
1.1 The Real Number System
This section aims to:
1. identify and classify the subsets of real numbers;
2. determine whether or not a set is well-ordered; and
3. enumerate and apply the properties in real numbers in arguments of mathematical statement.
𝑎1
Chapter 1: THE INTEGERS
1.2 Sequences
This section aims to:
1.
2.
Chapter 1: THE INTEGERS
1.3 Fibonacci Numbers
This section aims to:
1.
2.
Chapter 1: THE INTEGERS
1.4 The Well Ordering Principle and Mathematical Induction
This section aims to:
1. Define the well ordering principle and mathematical induction;
2. Enumerate the steps of the mathematical induction; and
3. Prove proposition using the well ordering principle and mathematical induction.
Number theory relies on numerous ideas and methods for proofs. This section introduces
two fundamental principles: the Well-Ordering Principle and Mathematical Induction,
explaining how to utilize them for establishing proofs.
Well-Ordering Principle
Every nonempty set of nonnegative integers has a smallest element.
In other words, for every element of m ∈ S, there exists an integer n ∈ S, such that m ≤ n.
The principle of well-ordering is an existence theorem. It does not tell us which element is
the smallest integer, nor does it tell us how to find the smallest element.
To understand this principle better, let’s look at an example.
Consider the set of positive integers {1, 2, 3, 4, …}. This set is well-ordered because it has
a smallest element, which is 1. Every other element in the set is greater than 1, so 1 is the least
element.
Now let’s consider the set of positive even integers {2, 4, 6, 8, …}. This set is also wellordered because it has a smallest element, which is 2. Every other element in the set is greater
Mathematical Induction
Is a standard procedure for establishing the validity of Mathematical statements which we
shall call as propositions involving a series of positive integers.
than 2, so 2 is the least element.
Theorem 1.4.1 Principle of Mathematical Induction
Let N be a set positive integer such that
i)
1 ∈ N, and
ii)
Whenever the integer k ∈ N, then k + 1 ∈ N.
Then N is the set of positive integers
Proof.
Let N: set of all positive integers; and
Let N1 : set of all positive integers not belong in N.
Assume N1 ≠ Ø. Then by the Well-Ordering Principle, there exist a least element
Let say m ∈ N1 . But by i), 1 ∈ N
This implies that must be greater than 1, m > 1 and 0 < m – 1 < m.
The choice of m being the smallest integer in N1 implies that m – 1 ∉ N, But by ii)
whenever m – 1 ∉ N, then the next integer (m – 1) + 1 = m.
We know that m ∉ N. But this contradicts our claim that m ∉ N1 .
We must claim that N1 = Ø . Thus, N contains the set of all positive integers.
4 Steps Using Mathematical Induction
Step 1. Verification
Verifying the validity of the proposition for a few particular cases of n. We start with the
smallest value of n for which the proposition holds, usually n = 1, unless otherwise stated.
Step 2. Induction Hypothesis
This is the assumption made in carrying out of the proof of the induction. This means that
we have to assume at the statement is true for some positive integers n = k.
Step 3.
Proof of Induction
This is the basis of induction where the hypothesis is utilized to carry out the desired result.
Step 4. Conclusion
The proposition is true for all positive integral values of n.
ADDING SUITABLE TERMS USING MATHEMATICAL INDUCTION
Example: Prove 1 + 2 + 3 + 4 + 5 + ⋯ + 𝑛 =
𝑛(𝑛+1)
2
by mathematical induction.
Proof.
Step 1.
Verification
Check if the proposition is true for the first few integers.
Let n = 1, then
1=
𝑛(𝑛 + 1)
2
1=
1(1 + 1)
2
1=
1(2)
2
1=1
Let n = 2, then
1+2=
3=
𝑛(𝑛 + 1)
2
2(2 + 1)
2
3=
6
2
3=3
Step 2.
Induction Hypothesis
Assume that the proposition is true for some positive integer n = k.
Step 3.
1+2+3+4+5+⋯+𝑛 =
𝑛(𝑛 + 1)
2
1 + 2 + 3 + 4 + 5 + ⋯+ 𝑘 =
𝑘(𝑘 + 1)
2
Proof of the induction
We shall show that whenever the hypothesis holds, then the proposition also holds for the
next integer n = k + 1.
1+2+3+4+5+⋯+ 𝑛 =
𝑛(𝑛 + 1)
2
Substitute n = k + 1
1 + 2 + 3 + 4 + 5 + ⋯ + 𝑘 + ( 𝑘 + 1) =
We know that 1 + 2 + 3 + 4 + 5 + … + k is the same as
(𝑘 + 1)( 𝑘 + 1 + 1)
2
𝑘(𝑘+1)
2
, replacing it;
𝑘(𝑘 + 1)
(𝑘 + 1)( 𝑘 + 1 + 1)
+ ( 𝑘 + 1) =
2
2
Simplify both sides of an equation
𝑘(𝑘 + 1) + 2𝑘 + 2
(𝑘 + 1)( 𝑘 + 2)
=
2
2
𝑘 2 + 𝑘 + 2𝑘 + 2
(𝑘 + 1)( 𝑘 + 2)
=
2
2
(𝑘 + 1)( 𝑘 + 2)
𝑘 2 + 3𝑘 + 2
=
2
2
Factor the numerator on the left side of the equation
(𝒌 + 𝟏)(𝒌 + 𝟐)
(𝒌 + 𝟏)( 𝒌 + 𝟐)
=
𝟐
𝟐
Step 4.
Conclusion
The proposition holds for n = 1 or k = 1; hence it also holds for k + 1 = 2. Therefore, it
must also hold for k + 1 = 3, and so on.
Example 2: Prove 3 + 7 + 11 + ⋯ + (4𝑛 − 1) = 𝑛(2𝑛 + 1) using mathematical
induction.
Step 1.
Verification
Check if the proposition is true for the first few integers.
Let n = 1, then
3 = 1[2(1) + 1]
3 = 1(3)
3= 3
Let n = 2, then
3 + 7 = 2[2(2) + 1]
10 = 2(5)
10 = 10
Step 2. Induction Hypothesis
Assume that the proposition is true for some positive integer n = k.
3 + 7 + 11 + ⋯ + (4𝑛 − 1) = 𝑛(2𝑛 + 1)
3 + 7 + 11 + ⋯ + (4𝑘 − 1) = 𝑘(2𝑘 + 1)
Step 3. Proof of the induction
We shall show that whenever the hypothesis holds, then the proposition also holds for the
next integer n = k + 1.
3 + 7 + 11 + ⋯ + (4𝑛 − 1) = 𝑛(2𝑛 + 1)
Substituting n = k + 1
3 + 7 + 11 + ⋯ + (4𝑘 − 1) + [4(𝑘 + 1) − 1] = (𝑘 + 1)[2(𝑘 + 1) + 1]
We know that 3 + 7 + 11 + ⋯ + (4𝑘 − 1)𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑘(2𝑘 + 1), replacing it to the
previous equation.
3 + 7 + 11 + ⋯ + (4𝑘 − 1) + [4(𝑘 + 1) − 1] = (𝑘 + 1)[2(𝑘 + 1) + 1]
𝑘(2𝑘 + 1) + (4𝑘 + 4 − 1) = (𝑘 + 1)[2(𝑘 + 1) + 1]
Simplify both sides of the equation
(2𝑘² + 𝑘) + (4𝑘 + 3) = (𝑘 + 1)(2𝑘 + 2 + 1)
2𝑘 2 + 𝑘 + 4𝑘 + 3 = (𝑘 + 1)(2𝑘 + 3)
2𝑘 2 + 5𝑘 + 3 = 2𝑘 2 + 5𝑘 + 3
Step 4.
Conclusion
The proposition holds for n = 1 or k = 1; hence it also holds for k + 1 = 2. Therefore, it
must also hold for k + 1 = 3, and so on.
DIVISIBILITY PROBLEM USING MATHEMATICAL INDUCTION
Example 3: Prove 6𝑛 + 4 is divisible by 5 for all positive integers n.
Proof.
Step 1.
Verification
Check if the proposition is true for the first few integers.
Let n = 1, then
6𝑛 + 4
61 + 4 = 10
10 is divisible by 5
Let n = 2, then
6𝑛 + 4
62 + 4
36 + 4 = 40
40 is divisible by 5
Step 2. Induction Hypothesis
Assume that the proposition is true for some positive integer n = k.
6𝑛 + 4
Assume 6𝑘 + 4 is divisible by 5
We can say that 6𝑘 + 4 = 5𝑚 no matter what the value of integer m, it still divisible by 5
6𝑘 = 5𝑚 − 4
Step 3. Proof of the induction
We shall show that whenever the hypothesis holds, then the proposition also holds for the
next integer n = k + 1.
6𝑛 + 4
Substitute k + 1 to n
6𝑘+1 + 4
Assume that 6𝑘+1 + 4 is also divisibile by 5
6𝑘+1 + 4
Using the Law of exponents (product of a power) it can be written as
6𝑘 • 61 + 4
We know that 6𝑘 = 5𝑚 − 4, we can replace it
(5𝑚 − 4)6 + 4
Distributing the 6
30𝑚 − 24 + 4
30𝑚 − 20
We want 6𝑘+1 + 4 = 5𝐿, 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑖𝑛𝑡𝑔𝑒𝑟 𝐿
Using the expression 30𝑚 − 20 we can factor it, giving us
5(6𝑚 − 4)
Since m is an integer, then the expression 6𝑚 − 4 will gives us integer as well.
We can say 6𝑚 − 4 = 𝐿, 𝑡ℎ𝑒𝑛
5(6𝑚 − 4)
5𝐿
We show that 6𝑘+1 + 4 = 5𝐿.
Step 4.
Conclusion
The proposition holds for n = 1 or k = 1; hence it also holds for k + 1 = 2. Therefore, it
must also hold for k + 1 = 3, and so on.
INEQUALITIES BY COMPARISON USING MATHEMATICAL INDUCTION
Example 4: Prove 4𝑛−1 > 𝑛2 𝑓𝑜𝑟 𝑛 ≥ 3 𝑢𝑠𝑖𝑛𝑔 𝑀𝑎𝑡ℎ𝑒𝑚𝑎𝑡𝑖𝑐𝑎𝑙 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
Proof.
Step 1.
Verification
Check if the proposition is true for the first few integers. We cannot say n = 1, because
the restriction says that 3 is the smallest value for n.
Let n = 3, then
4𝑛−1 > 𝑛²
43−1 > 3²
42 > 9
16 > 9
Let n = 4, then
4𝑛−1 > 𝑛2
44−1 > 42
43 > 16
64 > 16
Step 2. Induction Hypothesis
Assume that the proposition is true for some positive integer n = k.
4𝑛−1 > 𝑛2
4𝑘−1 > 𝑘 2
Step 3. Proof of the induction
We shall show that whenever the hypothesis holds, then the proposition also holds for the
next integer n = k + 1.
4𝑛−1 > 𝑛2
Substitute n = k + 1
4𝑘+1−1 > (𝑘 + 1)2
Look 4𝑘+1−1 = 4𝑘−1+1 = 4𝑘−1 • 4
We also multiply 4 on the other side in order to balance the inequality
4𝑘−1 • 4 > 4𝑘 2
4k² can be written as k² + 2k² + k²
If we can show that 4k² is greater than (k + 1) ² then we can prove the inequality
𝑘 2 + 2𝑘 2 + 𝑘 2 > (𝑘 + 1)2
𝑘 2 + 2𝑘 2 + 𝑘 2 > 𝑘 2 + 2𝑘 + 1
Using comparison both sides have k²
The left-hand side has 2k² which is bigger than right-hand side
The left-hand side has k² which is bigger than 1 on the right-hand side because k² can’t
be 1 because 3 is the smallest value.
Therefore, 4k² is bigger than (k + 1) ². We also know that 4𝑘−1+1 > 4𝑘 2
We have proved that 4𝑘+1−1 > (𝑘 + 1)2
Step 4.
Conclusion
The proposition holds for n = 3 or k = 3 hence it also holds for k + 1 = 4. Therefore, it
must also hold for k + 1 = 5, and so on.
Chapter 1: THE INTEGERS
1.5 The Sums and Products
This section aims to:
1.
2.
Chapter 1: THE INTEGERS
1.6 Divisibility
This section aims to:
1.
2.