Mathematical Induction. Part Two
|
|
- Eleanor Griffith
- 5 years ago
- Views:
Transcription
1 Mathematical Induction Part Two
2 The principle of mathematical induction states that if for some property P(n), we have that If it starts P(0) is true and and it keeps going For any n N, we have P(n) P(n + ) Then then it's always true. For any n N, P(n) is true.
3 n(n+ ) Theorem: For any natural number n, i= Proof: By induction. Let P(n) be 2 n n(n+ ) P(n) i= i= 2 For our base case, we need to show P(0) is true, meaning that 0 0(0+ ) i= i= 2 Since the empty sum is defined to be 0, this claim is true. n+ i= n i= n i= For the inductive step, assume that for some n N that P(n) holds, so n(n+ ) i= 2 We need to show that P(n + ) holds, meaning that n+ (n+ )(n+ 2) i= i= 2 To see this, note that n n(n+ ) n(n+ )+ 2(n+ ) (n+ )(n+ 2) i= i+ (n+ )= + n+ = = i= Thus P(n + ) is true, completing the induction.
4 Induction in Practice Typically, a proof by induction will not explicitly state P(n). Rather, the proof will describe P(n) implicitly and leave it to the reader to fill in the details. Provided that there is sufficient detail to determine what P(n) is, that P(0) is true, and that whenever P(n) is true, P(n + ) is true, the proof is usually valid.
5 Theorem: For any natural number n, Proof: By induction on n. For our base case, if n = 0, note that and the theorem is true for 0. For the inductive step, assume that for some n the theorem is true. Then we have that n(n+ ) n(n+ )+ 2(n+ ) (n+ )(n+ 2) i+ (n+ )= + n+ = = n+ i= n i= i= 0 i= i= so the theorem is true for n +, completing the induction. n i= 0(0+ ) =0 2 i= n(n+ ) 2
6 A Variant of Induction
7 n 2 versus 2 n 0 2 = 0 2 = 2 2 = = = = = = = = = 00 < < = > = < < < < < < 2 n is much 2 0 = bigger here. 2 = 2 Does the trend 2 2 = 4 continue? 2 3 = = = = = = = = 024
8 Theorem: For any natural number n 5, n 2 < 2 n. Proof: By induction on n. As a base case, if n = 5, then we have that 5 2 = 25 < 32 = 2 5, so the claim holds. For the inductive step, assume that for some n 5, that n 2 < 2 n. Then we have that (n + ) 2 = n 2 + 2n + Since n 5, we have (n + ) 2 = n 2 + 2n + < n 2 + 2n + n (since < 5 n) = n 2 + 3n < n 2 + n 2 (since 3n < 5n n 2 ) = 2n 2 So (n + ) 2 < 2n 2. Now, by our inductive hypothesis, we know that n 2 < 2 n. This means that (n + ) 2 < 2n 2 (from above) < 2(2 n ) (by the inductive hypothesis) = 2 n + Completing the induction.
9 Theorem: For any natural number n 5, n 2 < 2 n. Proof: By induction on n. As a base case, if n = 5, then we have that 5 2 = 25 < 32 = 2 5, so the claim holds. For the inductive step, assume that for some n 5, that n 2 < 2 n. Then we have that (n + ) 2 = n 2 + 2n + Since n 5, we have Why (n + ) 2 = n 2 + 2n Why is + is this this < n 2 + 2n allowed? allowed? + n (since < 5 n) = n 2 + 3n < n 2 + n 2 (since 3n < 5n n 2 ) = 2n 2 So (n + ) 2 < 2n 2. Now, by our inductive hypothesis, we know that n 2 < 2 n. This means that (n + ) 2 < 2n 2 (from above) < 2(2 n ) (by the inductive hypothesis) = 2 n + Completing the induction.
10 Why is this Legal? Let P(n) be Either n < 5 or n 2 < 2 n. P(0) is trivially true. P() is trivially true, so P(0) P() P(2) is trivially true, so P() P(2) P(3) is trivially true, so P(2) P(3) P(4) is trivially true, so P(3) P(4) We explicitly proved P(5), so P(4) P(5) For any n 5, we explicitly proved that P(n) P(n + ). Thus P(0) and for any n N, P(n) P(n + ), so by induction P(n) is true for all natural numbers n.
11 Induction Starting at k To prove that P(n) is true for all natural numbers greater than or equal to k: Show that P(k) is true. Show that for any n k, that P(n) P(n + ). Conclude P(k) holds for all natural numbers greater than or equal to k. You don't need to justify why it's okay to start from k.
12 An Important Observation
13 One Major Catch In In an an inductive inductive proof, proof, to to prove prove P(5), P(5), we we can can only only assume assume P(4). P(4). We We cannot cannot rely rely on on any any of of our our earlier earlier results! results!
14 Strong Induction
15 The principle of strong induction states that if for some property P(n), we have that Assume Assume that that P(n) P(n) holds holds for for all all natural natural numbers numbers smaller smaller than than n. n. P(0) is true and For any n N with n 0, if P(n') is true for all n' < n, then P(n) is true then For any n N, P(n) is true.
16 Using Strong Induction
17 Induction and Dominoes
18 Strong Induction and Dominoes
19 Weak and Strong Induction Weak induction (regular induction) is good for showing that some property holds by incrementally adding in one new piece. Strong induction is good for showing that some property holds by breaking a large structure down into multiple small pieces.
20 Proof by Strong Induction State that you are attempting to prove something by strong induction. State what your choice of P(n) is. Prove the base case: State what P(0) is, then prove it. Prove the inductive step: State that you assume for all 0 n' < n, that P(n') is true. State what P(n) is. (this is what you're trying to prove) Go prove P(n).
21 Application: Binary Numbers
22 Binary Numbers The binary number system is base 2. Every number is represented as s and 0s encoding various powers of two. Examples: 00 2 = = = = 27 Enormously useful in computing; almost all computers do computation on binary numbers. Question: How do we know that every natural number can be written in binary?
23 Justifying Binary Numbers To justify the binary representation, we will prove the following result: Every natural number n can be expressed as the sum of distinct powers of two. This says that there's at least one way to write a number in binary; we'd need a separate proof to show that there's exactly one way to do it. So how do we prove this?
24 One Proof Idea
25 General Idea Repeatedly subtract out the largest power of two less than the number. Can't subtract 2 n twice for any n; otherwise, you could have subtracted 2 n+. Eventually, we reach 0; the number is then the sum of the powers of two that we subtracted. How do we formalize this as a proof?
26 Theorem: Every n N is the sum of distinct powers of two. Proof: By strong induction. Let P(n) be n is the sum of distinct powers of two. We prove that P(n) is true for all n N. As our base case, we prove P(0), that 0 is the sum of distinct powers of 2. Since the empty sum of no powers of 2 is equal to 0, P(0) holds. For the inductive step, assume that for some nonzero n N, that for any n' N where 0 n' < n, that P(n') holds and n' is the sum of distinct powers of two. We prove P(n), that n is the sum of distinct powers of two. Let 2 k be the greatest power of two such that 2 k n. Consider n 2 k. Since 2 k for any natural number k, we know that n 2 k < n. Since 2 k n, we know Notice 0 n 2 k. Thus, by our inductive hypothesis, n 2 Notice the the stronger stronger version version of of k is the sum of distinct powers of two. If S be the set of these powers of two, then n is the the sum the induction of induction these powers hypothesis. hypothesis. of two and 2 k. If we can show that We're We're 2 k now S, now we showing showing will have that that n P(n') P(n') is the is sum is of distinct powers of two (namely, true the elements of S and 2 k ). Then P(n) will hold, completing the induction. true for for all all natural natural numbers numbers in in the We show 2 k S by the range contradiction; range 0 assume n' n' < n. that n. We'll 2We'll k S. Since 2 k S and the sum of the powers use use of two this this in fact S fact is n later later 2 k, this on. on. means that 2 k n 2 k. Thus 2 k + 2 k n, so 2 k + n. This contradicts that 2 k is the largest power of two no greater than n. We have reached a contradiction, so our assumption was wrong and 2 k S, as required.
27 Theorem: Every n N is the sum of distinct powers of two. Proof: By strong induction. Let P(n) be n is the sum of distinct powers of two. We prove that P(n) is true for all n N. As our base case, we prove P(0), that 0 is the sum of distinct powers of 2. Since the empty sum of no powers of 2 is equal to 0, P(0) holds. For the inductive step, assume that for some nonzero n N, that for any n' N where 0 n' < n, that P(n') holds and n' is the sum of distinct powers of two. We prove P(n), that n is the sum of distinct powers of two. Let 2 k be the greatest power of two such that 2 k n. Consider n 2 k. Since 2 k for any natural number k, we know that n 2 k < n. Since 2 k n, we know Here's 0 n 2 k. Thus, by our inductive hypothesis, n 2 k is the sum of Here's the distinct powers the key key step of two. step of If of the S be the proof. the proof. set of these powers of two, then n is the sum If If of we we these can can powers show show of that that two and 2 k. If we can show that 2 k S, we will have that n is the sum of distinct powers of two (namely, the 0 elements of S and 2 k ). Then P(n) will hold, n 2 k completing the induction. k < n We show 2 k S by then then contradiction; we we can can use use assume the the that inductive inductive 2 k S. Since 2 k S and the sum of the powers of two in S is n 2 k, this means that 2 k n 2 k. hypothesis Thus 2 k + 2 k hypothesis to n, so 2 k + to claim n. claim that This contradicts that n 2 that k k is is a 2 k a is the largest power of two no sum sum greater of of distinct than distinct n. We powers powers have reached of of two. two. a contradiction, so our assumption was wrong and 2 k S, as required.
28 Theorem: Every n N is the sum of distinct powers of two. Proof: By strong induction. Let P(n) be n is the sum of distinct powers of two. We prove that P(n) is true for all n N. As our base case, we prove P(0), that 0 is the sum of distinct powers of 2. Since the empty sum of no powers of 2 is equal to 0, P(0) holds. For the inductive step, assume that for some nonzero n N, that for any n' N where 0 n' < n, that P(n') holds and n' is the sum of distinct powers of two. We prove P(n), that n is the sum of distinct powers of two. Let 2 k be the greatest power of two such that 2 k n. Consider n 2 k. Since 2 k for any natural number k, we know that n 2 k < n. Since 2 k n, we know 0 n 2 k. Thus, by our inductive hypothesis, n 2 k is the sum of distinct powers of two. If S be the set of these powers of two, then n is the sum of these powers of two and 2 k. If we can Here Here show is is that where where 2 k strong S, strong we will induction induction have that n kicks kicks is the in. in. sum of We We distinct powers use of two (namely, the elements of S and 2 k ). Then P(n) will hold, completing use the the the fact induction. fact that that any any smaller smaller number number can can be be written We show written as 2 k S by as the contradiction; the sum sum of of distinct assume distinct powers that powers of 2 k S. Since of two two 2 k S and the sum to to of show show the powers that that n of two 2 k k in can can S is be n be 2written k, this means as as the the that sum sum 2 k n 2 k. Thus 2 k + 2 k n, so of 2 k + n. This contradicts that 2 k is the largest power of two no greater of distinct distinct powers than n. powers of We have reached of two. two. a contradiction, so our assumption was wrong and 2 k S, as required.
29 Theorem: Every n N is the sum of distinct powers of two. Proof: By strong induction. Let P(n) be n is the sum of distinct powers of two. We prove that P(n) is true for all n N. As our base case, we prove P(0), that 0 is the sum of distinct powers of 2. Since the empty sum of no powers of 2 is equal to 0, P(0) holds. For the inductive step, assume that for some nonzero n N, that for any n' N where 0 n' < n, that P(n') holds and n' is the sum of distinct powers of two. We prove P(n), that n is the sum of distinct powers of two. Let 2 k be the greatest power of two such that 2 k n. Consider n 2 k. Since 2 k for any natural number k, we know that n 2 k < n. Since 2 k n, we know 0 n 2 k. Thus, by our inductive hypothesis, n 2 k is the sum of distinct powers of two. If S be the set of these powers of two, then n is the sum of the elements of S and 2 k. If we can show that 2 k S, we will have that n is the sum of distinct powers of two (namely, the elements of S and 2 k ). Then P(n) will hold, completing the induction. We show 2 k S by contradiction; assume that 2 k S. Since 2 k S and the sum of the powers of two in S is n 2 k, this means that 2 k n 2 k. Thus 2 k + 2 k n, so 2 k + n. This contradicts that 2 k is the largest power of two no greater than n. We have reached a contradiction, so our assumption was wrong and 2 k S, as required.
30 Theorem: Every n N is the sum of distinct powers of two. Proof: By strong induction. Let P(n) be n is the sum of distinct powers of two. We prove that P(n) is true for all n N. As our base case, we prove P(0), that 0 is the sum of distinct powers of 2. Since the empty sum of no powers of 2 is equal to 0, P(0) holds. For the inductive step, assume that for some nonzero n N, that for any n' N where 0 n' < n, that P(n') holds and n' is the sum of distinct powers of two. We prove P(n), that n is the sum of distinct powers of two. Let 2 k be the greatest power of two such that 2 k n. Consider n 2 k. Since 2 k for any natural number k, we know that n 2 k < n. Since 2 k n, we know 0 n 2 k. Thus, by our inductive hypothesis, n 2 k is the sum of distinct powers of two. If S be the set of these powers of two, then n is the sum of the elements of S and 2 k. If we can show that 2 k S, we will have that n is the sum of distinct powers of two (namely, the elements of S and 2 k ). Then P(n) will hold, completing the induction. We show 2 k S by contradiction; assume that 2 k S. Since 2 k S and the sum of the powers of two in S is n 2 k, this means that 2 k n 2 k. Thus 2 k + 2 k n, so 2 k + n. This contradicts that 2 k is the largest power of two no greater than n. We have reached a contradiction, so our assumption was wrong and 2 k S, as required.
31 Application: Continued Fractions
32 Continued Fractions
33 Continued Fractions A continued fraction is an expression of the form a + a 2 + a a n Formally, a continued fraction is either An integer n, or n + / F, where n is an integer and F is a continued fraction. Continued fractions have numerous applications in number theory and computer science. (They're also really fun to write!)
34 Fun with Continued Fractions Every rational number, including negative rational numbers, has a continued fraction representation. Harder result: every irrational number has an (infinite) continued fraction representation. Even harder result: If we truncate an infinite continued fraction for an irrational number, we can get progressively better approximations of that number.
35 π as a Continued Fraction π=
36 Approximating π π= = And And he he made made the the Sea Sea of of cast cast bronze, bronze, ten ten cubits cubits from from one one brim brim to to the the other; other; it it was was completely completely round. round. [ [ A] A] line line of of thirty thirty cubits cubits measured measured its its circumference. Kings Kings 7:23, 7:23, New New King King James James Translation Translation
37 Approximating π π= = /7 = Greek Greek mathematician Archimedes knew knew of of this this approximation, circa circa BCE BCE
38 Approximating π π= = /7 = /06 = /3 = Chinese Chinese mathematician mathematician 祖沖之祖沖之 (Zu (Zu Chongzhi) Chongzhi) discovered discovered this this approximation approximation in in the the early early fifth fifth century; century; this this was was the the best best approximation approximation of of pi pi for for over over a a thousand thousand years. years.
39 Approximating π π= = /7 = /06 = /3 = /3302 =
40 More Continued Fractions 3 The The Ancient Ancient Greeks Greeks knew knew about about this this connection. connection. 4 They They called called this this procedure procedure anthyphairesis. anthyphairesis. 3 4 =
41 An Interesting Continued Fraction x=
42 An Interesting Continued Fraction x= / 2 / 3 / 2 5 / 3 8 / 5 3 / 8 2 / 3 34 / 2 Each fraction is the ratio of consecutive Fibonacci numbers!
43 The Golden Ratio ϕ= = ϕ
44 The Golden Ratio
45 The Golden Spiral
46 How do we prove all rational numbers have continued fractions?
47 Constructing a Continued Fraction = = = 3+ 2
48 Constructing a Continued Fraction =
49 Constructing a Continued Fraction = > 7 > 2 > 2
50 The Division Algorithm For any integers a and b, with b > 0, there exists unique integers q and r such that and a = qb + r 0 r < b q is the quotient and r is the remainder. Given a = and b = 4: = Given a = -37 and b = 42: -37 =
51 Theorem: Every rational has a continued fraction. Proof: By strong induction. Let P(d) be any rational with denominator d has a continued fraction. We prove that P(d) is true for all positive natural numbers. Since all rationals can be written with a positive denominator, this proves that all rationals have continued fractions. For our base case, we prove P(), that any rational with denominator has a continued fraction. Consider any rational with denominator ; let it be n /. Since n is a continued fraction and n = n /, P() holds. For our inductive step, assume that for some d N with d >, that for any d' N where d' < d, that P(d') is true, so any rational with denominator d' has a continued fraction. We prove P(d) by showing that any rational with denominator d has a continued fraction. Take any rational with denominator d; let it be n / d. Using the division algorithm, write n = qd + r, where 0 r < d. We consider two cases: Case : r = 0. Then n = qd, so n / d = q. Then q is a continued fraction for n / d. n Case 2: r 0. Given that n = qd + r, we have d =q+ r d =q+ d /r. Since r < d, by our inductive hypothesis there is some continued fraction for d / r; call it F. Then q + / F is a continued fraction for n / d. In either case, we find a continued fraction for n / d, so P(d) holds, completing the induction.
52 For more on continued fractions:
53 Next Time Graphs and Relations Representing structured data. Categorizing how objects are connected.
Mathematical Induction. Part Two
Mathematical Induction Part Two Announcements Problem Set due Friday, January 8 at the start of class. Problem Set checkpoints graded, will be returned at end of lecture. Afterwards, will be available
More informationMathematical Induction
Mathematical Induction MAT30 Discrete Mathematics Fall 018 MAT30 (Discrete Math) Mathematical Induction Fall 018 1 / 19 Outline 1 Mathematical Induction Strong Mathematical Induction MAT30 (Discrete Math)
More informationMathematical Induction Assignments
1 Mathematical Induction Assignments Prove the Following using Principle of Mathematical induction 1) Prove that for any positive integer number n, n 3 + 2 n is divisible by 3 2) Prove that 1 3 + 2 3 +
More information1 Continued Fractions
Continued Fractions To start off the course, we consider a generalization of the Euclidean Algorithm which has ancient historical roots and yet still has relevance and applications today.. Continued Fraction
More informationTHE DIVISION THEOREM IN Z AND R[T ]
THE DIVISION THEOREM IN Z AND R[T ] KEITH CONRAD 1. Introduction In both Z and R[T ], we can carry out a process of division with remainder. Theorem 1.1. For any integers a and b, with b nonzero, there
More informationMathematical Induction
Mathematical Induction MAT231 Transition to Higher Mathematics Fall 2014 MAT231 (Transition to Higher Math) Mathematical Induction Fall 2014 1 / 21 Outline 1 Mathematical Induction 2 Strong Mathematical
More informationMathematics 228(Q1), Assignment 2 Solutions
Mathematics 228(Q1), Assignment 2 Solutions Exercise 1.(10 marks) A natural number n > 1 is said to be square free if d N with d 2 n implies d = 1. Show that n is square free if and only if n = p 1 p k
More informationHence, the sequence of triangular numbers is given by., the. n th square number, is the sum of the first. S n
Appendix A: The Principle of Mathematical Induction We now present an important deductive method widely used in mathematics: the principle of mathematical induction. First, we provide some historical context
More informationStandard forms for writing numbers
Standard forms for writing numbers In order to relate the abstract mathematical descriptions of familiar number systems to the everyday descriptions of numbers by decimal expansions and similar means,
More informationInfinite Continued Fractions
Infinite Continued Fractions 8-5-200 The value of an infinite continued fraction [a 0 ; a, a 2, ] is lim c k, where c k is the k-th convergent k If [a 0 ; a, a 2, ] is an infinite continued fraction with
More informationDiscrete Mathematics. Spring 2017
Discrete Mathematics Spring 2017 Previous Lecture Principle of Mathematical Induction Mathematical Induction: rule of inference Mathematical Induction: Conjecturing and Proving Climbing an Infinite Ladder
More informationInduction 1 = 1(1+1) = 2(2+1) = 3(3+1) 2
Induction 0-8-08 Induction is used to prove a sequence of statements P(), P(), P(3),... There may be finitely many statements, but often there are infinitely many. For example, consider the statement ++3+
More informationWhat is proof? Lesson 1
What is proof? Lesson The topic for this Math Explorer Club is mathematical proof. In this post we will go over what was covered in the first session. The word proof is a normal English word that you might
More informationPUTNAM TRAINING NUMBER THEORY. Exercises 1. Show that the sum of two consecutive primes is never twice a prime.
PUTNAM TRAINING NUMBER THEORY (Last updated: December 11, 2017) Remark. This is a list of exercises on Number Theory. Miguel A. Lerma Exercises 1. Show that the sum of two consecutive primes is never twice
More informationDiscrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 3
EECS 70 Discrete Mathematics and Probability Theory Spring 014 Anant Sahai Note 3 Induction Induction is an extremely powerful tool in mathematics. It is a way of proving propositions that hold for all
More informationSome Review Problems for Exam 1: Solutions
Math 3355 Fall 2018 Some Review Problems for Exam 1: Solutions Here is my quick review of proof techniques. I will focus exclusively on propositions of the form p q, or more properly, x P (x) Q(x) or x
More informationINTEGERS. In this section we aim to show the following: Goal. Every natural number can be written uniquely as a product of primes.
INTEGERS PETER MAYR (MATH 2001, CU BOULDER) In this section we aim to show the following: Goal. Every natural number can be written uniquely as a product of primes. 1. Divisibility Definition. Let a, b
More informationCSC236H Lecture 2. Ilir Dema. September 19, 2018
CSC236H Lecture 2 Ilir Dema September 19, 2018 Simple Induction Useful to prove statements depending on natural numbers Define a predicate P(n) Prove the base case P(b) Prove that for all n b, P(n) P(n
More informationCS1800: Mathematical Induction. Professor Kevin Gold
CS1800: Mathematical Induction Professor Kevin Gold Induction: Used to Prove Patterns Just Keep Going For an algorithm, we may want to prove that it just keeps working, no matter how big the input size
More informationGrade 8 Chapter 7: Rational and Irrational Numbers
Grade 8 Chapter 7: Rational and Irrational Numbers In this chapter we first review the real line model for numbers, as discussed in Chapter 2 of seventh grade, by recalling how the integers and then the
More informationProof Techniques (Review of Math 271)
Chapter 2 Proof Techniques (Review of Math 271) 2.1 Overview This chapter reviews proof techniques that were probably introduced in Math 271 and that may also have been used in a different way in Phil
More informationGrade 7/8 Math Circles Winter March 20/21/22 Types of Numbers
Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Grade 7/8 Math Circles Winter 2018 - March 20/21/22 Types of Numbers Introduction Today, we take our number
More informationFall 2017 Test II review problems
Fall 2017 Test II review problems Dr. Holmes October 18, 2017 This is a quite miscellaneous grab bag of relevant problems from old tests. Some are certainly repeated. 1. Give the complete addition and
More informationDiscrete Mathematics and Probability Theory Fall 2013 Vazirani Note 1
CS 70 Discrete Mathematics and Probability Theory Fall 013 Vazirani Note 1 Induction Induction is a basic, powerful and widely used proof technique. It is one of the most common techniques for analyzing
More informationLecture 2: Continued fractions, rational approximations
Lecture 2: Continued fractions, rational approximations Algorithmic Number Theory (Fall 204) Rutgers University Swastik Kopparty Scribe: Cole Franks Continued Fractions We begin by calculating the continued
More informationSEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION
CHAPTER 5 SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Copyright Cengage Learning. All rights reserved. SECTION 5.4 Strong Mathematical Induction and the Well-Ordering Principle for the Integers Copyright
More informationCHAPTER 4: EXPLORING Z
CHAPTER 4: EXPLORING Z MATH 378, CSUSM. SPRING 2009. AITKEN 1. Introduction In this chapter we continue the study of the ring Z. We begin with absolute values. The absolute value function Z N is the identity
More informationUNIT 4 NOTES: PROPERTIES & EXPRESSIONS
UNIT 4 NOTES: PROPERTIES & EXPRESSIONS Vocabulary Mathematics: (from Greek mathema, knowledge, study, learning ) Is the study of quantity, structure, space, and change. Algebra: Is the branch of mathematics
More informationWhat can you prove by induction?
MEI CONFERENCE 013 What can you prove by induction? Martyn Parker M.J.Parker@keele.ac.uk Contents Contents iii 1 Splitting Coins.................................................. 1 Convex Polygons................................................
More informationTHE DIVISION THEOREM IN Z AND F [T ]
THE DIVISION THEOREM IN Z AND F [T ] KEITH CONRAD 1. Introduction In the integers we can carry out a process of division with remainder, as follows. Theorem 1.1. For any integers a and b, with b 0 there
More informationClimbing an Infinite Ladder
Section 5.1 Section Summary Mathematical Induction Examples of Proof by Mathematical Induction Mistaken Proofs by Mathematical Induction Guidelines for Proofs by Mathematical Induction Climbing an Infinite
More informationx 2 + 6x 18 x + 2 Name: Class: Date: 1. Find the coordinates of the local extreme of the function y = x 2 4 x.
1. Find the coordinates of the local extreme of the function y = x 2 4 x. 2. How many local maxima and minima does the polynomial y = 8 x 2 + 7 x + 7 have? 3. How many local maxima and minima does the
More information1. Introduction to commutative rings and fields
1. Introduction to commutative rings and fields Very informally speaking, a commutative ring is a set in which we can add, subtract and multiply elements so that the usual laws hold. A field is a commutative
More information1.2 The Well-Ordering Principle
36 Chapter 1. The Integers Exercises 1.1 1. Prove the following theorem: Theorem. Let m and a be integers. If m a and a m, thenm = ±a. 2. Prove the following theorem: Theorem. For all integers a, b and
More informationA NEW SET THEORY FOR ANALYSIS
Article A NEW SET THEORY FOR ANALYSIS Juan Pablo Ramírez 0000-0002-4912-2952 Abstract: We present the real number system as a generalization of the natural numbers. First, we prove the co-finite topology,
More informationNotes on arithmetic. 1. Representation in base B
Notes on arithmetic The Babylonians that is to say, the people that inhabited what is now southern Iraq for reasons not entirely clear to us, ued base 60 in scientific calculation. This offers us an excuse
More informationBasic Principles of Algebra
Basic Principles of Algebra Algebra is the part of mathematics dealing with discovering unknown numbers in an equation. It involves the use of different types of numbers: natural (1, 2, 100, 763 etc.),
More informationHomework 4, 5, 6 Solutions. > 0, and so a n 0 = n + 1 n = ( n+1 n)( n+1+ n) 1 if n is odd 1/n if n is even diverges.
2..2(a) lim a n = 0. Homework 4, 5, 6 Solutions Proof. Let ɛ > 0. Then for n n = 2+ 2ɛ we have 2n 3 4+ ɛ 3 > ɛ > 0, so 0 < 2n 3 < ɛ, and thus a n 0 = 2n 3 < ɛ. 2..2(g) lim ( n + n) = 0. Proof. Let ɛ >
More informationThe Inductive Proof Template
CS103 Handout 24 Winter 2016 February 5, 2016 Guide to Inductive Proofs Induction gives a new way to prove results about natural numbers and discrete structures like games, puzzles, and graphs. All of
More informationCarmen s Core Concepts (Math 135)
Carmen s Core Concepts (Math 135) Carmen Bruni University of Waterloo Week 4 1 Principle of Mathematical Induction 2 Example 3 Base Case 4 Inductive Hypothesis 5 Inductive Step When Induction Isn t Enough
More informationCSC 344 Algorithms and Complexity. Proof by Mathematical Induction
CSC 344 Algorithms and Complexity Lecture #1 Review of Mathematical Induction Proof by Mathematical Induction Many results in mathematics are claimed true for every positive integer. Any of these results
More informationWriting proofs for MATH 61CM, 61DM Week 1: basic logic, proof by contradiction, proof by induction
Writing proofs for MATH 61CM, 61DM Week 1: basic logic, proof by contradiction, proof by induction written by Sarah Peluse, revised by Evangelie Zachos and Lisa Sauermann September 27, 2016 1 Introduction
More informationSolutions Quiz 9 Nov. 8, Prove: If a, b, m are integers such that 2a + 3b 12m + 1, then a 3m + 1 or b 2m + 1.
Solutions Quiz 9 Nov. 8, 2010 1. Prove: If a, b, m are integers such that 2a + 3b 12m + 1, then a 3m + 1 or b 2m + 1. Answer. We prove the contrapositive. Suppose a, b, m are integers such that a < 3m
More informationMAT115A-21 COMPLETE LECTURE NOTES
MAT115A-21 COMPLETE LECTURE NOTES NATHANIEL GALLUP 1. Introduction Number theory begins as the study of the natural numbers the integers N = {1, 2, 3,...}, Z = { 3, 2, 1, 0, 1, 2, 3,...}, and sometimes
More informationDiscrete Mathematics. Spring 2017
Discrete Mathematics Spring 2017 Previous Lecture Principle of Mathematical Induction Mathematical Induction: Rule of Inference Mathematical Induction: Conjecturing and Proving Mathematical Induction:
More informationDiscrete Mathematics for CS Spring 2008 David Wagner Note 4
CS 70 Discrete Mathematics for CS Spring 008 David Wagner Note 4 Induction Induction is an extremely powerful tool in mathematics. It is a way of proving propositions that hold for all natural numbers,
More informationAssignment #2 COMP 3200 Spring 2012 Prof. Stucki
Assignment #2 COMP 3200 Spring 2012 Prof. Stucki 1) Construct deterministic finite automata accepting each of the following languages. In (a)-(c) the alphabet is = {0,1}. In (d)-(e) the alphabet is ASCII
More informationChapter 2. Mathematical Reasoning. 2.1 Mathematical Models
Contents Mathematical Reasoning 3.1 Mathematical Models........................... 3. Mathematical Proof............................ 4..1 Structure of Proofs........................ 4.. Direct Method..........................
More informationDR.RUPNATHJI( DR.RUPAK NATH )
Contents 1 Sets 1 2 The Real Numbers 9 3 Sequences 29 4 Series 59 5 Functions 81 6 Power Series 105 7 The elementary functions 111 Chapter 1 Sets It is very convenient to introduce some notation and terminology
More informationSection 4.2: Mathematical Induction 1
Section 4.: Mathematical Induction 1 Over the next couple of sections, we shall consider a method of proof called mathematical induction. Induction is fairly complicated, but a very useful proof technique,
More information3 Finite continued fractions
MTH628 Number Theory Notes 3 Spring 209 3 Finite continued fractions 3. Introduction Let us return to the calculation of gcd(225, 57) from the preceding chapter. 225 = 57 + 68 57 = 68 2 + 2 68 = 2 3 +
More informationOutline Goals and Assumptions Real Numbers Rational and Irrational. L11: Numbers. Alice E. Fischer
L11: Numbers Alice E. Fischer CSCI 1166 Discrete Mathematics for Computing March 5-8, 2018 1 Goals and Assumptions 2 Real Numbers 3 Rational and Irrational Assumptions We rely the following assumptions:
More informationCHAPTER 1 NUMBER SYSTEMS. 1.1 Introduction
N UMBER S YSTEMS NUMBER SYSTEMS CHAPTER. Introduction In your earlier classes, you have learnt about the number line and how to represent various types of numbers on it (see Fig..). Fig.. : The number
More informationProof: If (a, a, b) is a Pythagorean triple, 2a 2 = b 2 b / a = 2, which is impossible.
CS103 Handout 07 Fall 2013 October 2, 2013 Guide to Proofs Thanks to Michael Kim for writing some of the proofs used in this handout. What makes a proof a good proof? It's hard to answer this question
More informationCSE 1400 Applied Discrete Mathematics Proofs
CSE 1400 Applied Discrete Mathematics Proofs Department of Computer Sciences College of Engineering Florida Tech Fall 2011 Axioms 1 Logical Axioms 2 Models 2 Number Theory 3 Graph Theory 4 Set Theory 4
More informationPart I, Number Systems. CS131 Mathematics for Computer Scientists II Note 1 INTEGERS
CS131 Part I, Number Systems CS131 Mathematics for Computer Scientists II Note 1 INTEGERS The set of all integers will be denoted by Z. So Z = {..., 2, 1, 0, 1, 2,...}. The decimal number system uses the
More informationCS 360, Winter Morphology of Proof: An introduction to rigorous proof techniques
CS 30, Winter 2011 Morphology of Proof: An introduction to rigorous proof techniques 1 Methodology of Proof An example Deep down, all theorems are of the form If A then B, though they may be expressed
More informationa = qb + r where 0 r < b. Proof. We first prove this result under the additional assumption that b > 0 is a natural number. Let
2. Induction and the division algorithm The main method to prove results about the natural numbers is to use induction. We recall some of the details and at the same time present the material in a different
More informationMathematical Induction
Mathematical Induction Representation of integers Mathematical Induction Reading (Epp s textbook) 5.1 5.3 1 Representations of Integers Let b be a positive integer greater than 1. Then if n is a positive
More information5) ) y 20 y 10 =
Name Class Date 7.N.4 Develop the laws of exponents for multiplication and division Directions: Rewrite as a base with an exponent. 1) 3 6 3-4 = 2) x 7 x 17 = 3) 10-8 10 3 = 5) 12-3 = -3 12 6) y 20 y 10
More informationAt the start of the term, we saw the following formula for computing the sum of the first n integers:
Chapter 11 Induction This chapter covers mathematical induction. 11.1 Introduction to induction At the start of the term, we saw the following formula for computing the sum of the first n integers: Claim
More information1 Basic Combinatorics
1 Basic Combinatorics 1.1 Sets and sequences Sets. A set is an unordered collection of distinct objects. The objects are called elements of the set. We use braces to denote a set, for example, the set
More informationChapter 1 A Survey of Divisibility 14
Chapter 1 A Survey of Divisibility 14 SECTION C Euclidean Algorithm By the end of this section you will be able to use properties of the greatest common divisor (gcd) obtain the gcd using the Euclidean
More information1 Adeles over Q. 1.1 Absolute values
1 Adeles over Q 1.1 Absolute values Definition 1.1.1 (Absolute value) An absolute value on a field F is a nonnegative real valued function on F which satisfies the conditions: (i) x = 0 if and only if
More informationThe Dynamics of Continued Fractions
The Dynamics of Continued Fractions Evan O Dorney May 3, 20 The Story I was first introduced to the Intel Science Talent Search in ninth grade. I knew I would have no trouble entering this contest, as
More information0.Axioms for the Integers 1
0.Axioms for the Integers 1 Number theory is the study of the arithmetical properties of the integers. You have been doing arithmetic with integers since you were a young child, but these mathematical
More informationThe area of a geometric figure is a measure of how big a region is enclosed inside the figure.
59 CH 7 GEOMETRY Introduction G eo: Greek for earth, and metros: Greek for measure. These roots are the origin of the word geometry, which literally means earth measurement. The study of geometry has gone
More informationWORKSHEET ON NUMBERS, MATH 215 FALL. We start our study of numbers with the integers: N = {1, 2, 3,...}
WORKSHEET ON NUMBERS, MATH 215 FALL 18(WHYTE) We start our study of numbers with the integers: Z = {..., 2, 1, 0, 1, 2, 3,... } and their subset of natural numbers: N = {1, 2, 3,...} For now we will not
More informationMATH10040: Numbers and Functions Homework 1: Solutions
MATH10040: Numbers and Functions Homework 1: Solutions 1. Prove that a Z and if 3 divides into a then 3 divides a. Solution: The statement to be proved is equivalent to the statement: For any a N, if 3
More informationAssignment: Summer Assignment Part 1 of 8 Real Numbers and Their Properties. Student: Date:
Student: Date: Assignment: Summer Assignment Part of 8 Real Numbers and Their Properties. Identify to which number groups (natural numbers, whole numbers, integers, rational numbers, real numbers, and
More informationPRINCIPLE OF MATHEMATICAL INDUCTION
Chapter 4 PRINCIPLE OF MATHEMATICAL INDUCTION 4.1 Overview Mathematical induction is one of the techniques which can be used to prove variety of mathematical statements which are formulated in terms of
More informationMATH 131A: REAL ANALYSIS
MATH 131A: REAL ANALYSIS NICKOLAS ANDERSEN The textbook for the course is Ross, Elementary Analysis [2], but in these notes I have also borrowed from Tao, Analysis I [3], and Abbott, Understanding Analysis
More informationMathematical Fundamentals
Mathematical Fundamentals Sets Factorials, Logarithms Recursion Summations, Recurrences Proof Techniques: By Contradiction, Induction Estimation Techniques Data Structures 1 Mathematical Fundamentals Sets
More informationReading and Writing. Mathematical Proofs. Slides by Arthur van Goetham
Reading and Writing Mathematical Proofs Slides by Arthur van Goetham What is a proof? Why explanations are not proofs What is a proof? A method for establishing truth What establishes truth depends on
More informationSequences of Real Numbers
Chapter 8 Sequences of Real Numbers In this chapter, we assume the existence of the ordered field of real numbers, though we do not yet discuss or use the completeness of the real numbers. In the next
More information1. Introduction to commutative rings and fields
1. Introduction to commutative rings and fields Very informally speaking, a commutative ring is a set in which we can add, subtract and multiply elements so that the usual laws hold. A field is a commutative
More informationSEVENTH EDITION and EXPANDED SEVENTH EDITION
SEVENTH EDITION and EXPANDED SEVENTH EDITION Slide 5-1 Chapter 5 Number Theory and the Real Number System 5.1 Number Theory Number Theory The study of numbers and their properties. The numbers we use to
More informationCSE 311: Foundations of Computing. Lecture 14: Induction
CSE 311: Foundations of Computing Lecture 14: Induction Mathematical Induction Method for proving statements about all natural numbers A new logical inference rule! It only applies over the natural numbers
More informationNumbers. 2.1 Integers. P(n) = n(n 4 5n 2 + 4) = n(n 2 1)(n 2 4) = (n 2)(n 1)n(n + 1)(n + 2); 120 =
2 Numbers 2.1 Integers You remember the definition of a prime number. On p. 7, we defined a prime number and formulated the Fundamental Theorem of Arithmetic. Numerous beautiful results can be presented
More informationBeautiful Mathematics
Beautiful Mathematics 1. Principle of Mathematical Induction The set of natural numbers is the set of positive integers {1, 2, 3,... } and is denoted by N. The Principle of Mathematical Induction is a
More information1 Proof by Contradiction
In these notes, which will serve as a record of what we covered in weeks 3-4, we discuss various indirect methods of proof. While direct proof is often preferred it is sometimes either not possible or
More informationEvery subset of {1, 2,...,n 1} can be extended to a subset of {1, 2, 3,...,n} by either adding or not adding the element n.
11 Recurrences A recurrence equation or recurrence counts things using recursion. 11.1 Recurrence Equations We start with an example. Example 11.1. Find a recurrence for S(n), the number of subsets of
More information12 Sequences and Recurrences
12 Sequences and Recurrences A sequence is just what you think it is. It is often given by a formula known as a recurrence equation. 12.1 Arithmetic and Geometric Progressions An arithmetic progression
More information3.1 Induction: An informal introduction
Chapter 3 Induction and Recursion 3.1 Induction: An informal introduction This section is intended as a somewhat informal introduction to The Principle of Mathematical Induction (PMI): a theorem that establishes
More informationWhat Fun! It's Practice with Scientific Notation!
What Fun! It's Practice with Scientific Notation! Review of Scientific Notation Scientific notation provides a place to hold the zeroes that come after a whole number or before a fraction. The number 100,000,000
More informationFall 2017 November 10, Written Homework 5
CS1800 Discrete Structures Profs. Aslam, Gold, & Pavlu Fall 2017 November 10, 2017 Assigned: Mon Nov 13 2017 Due: Wed Nov 29 2017 Instructions: Written Homework 5 The assignment has to be uploaded to blackboard
More informationMATH 2112/CSCI 2112, Discrete Structures I Winter 2007 Toby Kenney Homework Sheet 5 Hints & Model Solutions
MATH 11/CSCI 11, Discrete Structures I Winter 007 Toby Kenney Homework Sheet 5 Hints & Model Solutions Sheet 4 5 Define the repeat of a positive integer as the number obtained by writing it twice in a
More informationHomework #2 Solutions Due: September 5, for all n N n 3 = n2 (n + 1) 2 4
Do the following exercises from the text: Chapter (Section 3):, 1, 17(a)-(b), 3 Prove that 1 3 + 3 + + n 3 n (n + 1) for all n N Proof The proof is by induction on n For n N, let S(n) be the statement
More informationA number that can be written as, where p and q are integers and q Number.
RATIONAL NUMBERS 1.1 Definition of Rational Numbers: What are rational numbers? A number that can be written as, where p and q are integers and q Number. 0, is known as Rational Example:, 12, -18 etc.
More informationProofs Not Based On POMI
s Not Based On POMI James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University February 1, 018 Outline Non POMI Based s Some Contradiction s Triangle
More informationCS1800: Strong Induction. Professor Kevin Gold
CS1800: Strong Induction Professor Kevin Gold Mini-Primer/Refresher on Unrelated Topic: Limits This is meant to be a problem about reasoning about quantifiers, with a little practice of other skills, too
More informationMa/CS 6a Class 2: Congruences
Ma/CS 6a Class 2: Congruences 1 + 1 5 (mod 3) By Adam Sheffer Reminder: Public Key Cryptography Idea. Use a public key which is used for encryption and a private key used for decryption. Alice encrypts
More informationStudy Guide for Math 095
Study Guide for Math 095 David G. Radcliffe November 7, 1994 1 The Real Number System Writing a fraction in lowest terms. 1. Find the largest number that will divide into both the numerator and the denominator.
More informationInduction and recursion. Chapter 5
Induction and recursion Chapter 5 Chapter Summary Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms Mathematical Induction Section 5.1
More informationCYCLICITY OF (Z/(p))
CYCLICITY OF (Z/(p)) KEITH CONRAD 1. Introduction For each prime p, the group (Z/(p)) is cyclic. We will give seven proofs of this fundamental result. A common feature of the proofs that (Z/(p)) is cyclic
More informationChapter 3 Basic Number Theory
Chapter 3 Basic Number Theory What is Number Theory? Well... What is Number Theory? Well... Number Theory The study of the natural numbers (Z + ), especially the relationship between different sorts of
More informationChapter Summary. Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms
1 Chapter Summary Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms 2 Section 5.1 3 Section Summary Mathematical Induction Examples of
More informationDiscrete Mathematics and Probability Theory Fall 2014 Anant Sahai Note 7
EECS 70 Discrete Mathematics and Probability Theory Fall 2014 Anant Sahai Note 7 Polynomials Polynomials constitute a rich class of functions which are both easy to describe and widely applicable in topics
More information1 Numbers. exponential functions, such as x 7! a x ; where a; x 2 R; trigonometric functions, such as x 7! sin x; where x 2 R; ffiffi x ; where x 0:
Numbers In this book we study the properties of real functions defined on intervals of the real line (possibly the whole real line) and whose image also lies on the real line. In other words, they map
More informationClimbing an Infinite Ladder
Section 5.1 Section Summary Mathematical Induction Examples of Proof by Mathematical Induction Mistaken Proofs by Mathematical Induction Guidelines for Proofs by Mathematical Induction Climbing an Infinite
More information