Solutions for Practice problems on proofs

Size: px
Start display at page:

Download "Solutions for Practice problems on proofs"

Transcription

1 Solutions for Practice problems on proofs Definition: (even) An integer n Z is even if and only if n = 2m for some number m Z. Definition: (odd) An integer n Z is odd if and only if n = 2m + 1 for some number m Z. Definition: (perfect square) A number n N is a perfect square if and only if n = m 2 for some m N. Definition: (power of 5) A number n N is a power of 5 if and only if n = 5 k for some k N. 1. (a) n, m Z If m and n are odd integers, then m + n is even. Proof: We use a direct proof. Let m, n be odd integers, i.e. m = 2k + 1 and n = 2k + 1 for some integers k, k. Then m+n = 2k +1+2k +1 = 2(k +k +1). Since k, k are integers, so is k + k + 1, which means m + n = 2c for some integer c. Therefore m + n is even. (b) n, m Z If the product of n and m is odd, then both m, n are odd. Proof: We prove the contrapositive of the statement, i.e., we prove that if either m or n is an even integer then the product nm is even. Let m be even (The case in which n is even is identical). That is m = 2c for some integer c. Then mn = 2cn. Since c and n are integers, so is cn, which implies mn is even. Therefore, n, m Z If the product of n and m is odd, then both m, n are odd. (c) n, m N If n and m are powers of 5 then n + m is not a power of 5. For the proof of this statement, we take the following lemma for granted Lemma 1: ( k N)5 k = 2c + 1 for some integer c, i.e. every power of 5 is odd. (One can prove this statement by induction. We leave the proof as an exercise.) Remark : In the exam, we will clearly specify what statements you can take for granted in your proofs. A general rule is that you can assume any fact you saw in high school. Of course the exception is that you cannot take for granted the statement you are asked to prove. Proof of (c): We use a direct proof: Let n and m be powers of 5, it follows from the lemma that n and m are odd integers. By part 1

2 (a), this implies n + m is even. Since every power of 5 is odd, n + m cannot be a power of 5. Therefore, n, m N If n and m are powers of 5 then n + m is not a power of 5. (d) For all natural numbers n > 0, If n is a perfect square, then n + 1 is not a perfect square. (Note that there was a typo in the original statement of this problem. The statement claims that the implication is true for every natural number n. This is not true because for n = 0, both n and n + 1 are perfect squares.) We take the following lemma for granted. Lemma 2: a, b N, If a 2 > b 2 then a > b. (Ex: Give a proof by contrapositive of this statement.) Proof of (d): We give a direct proof of this implication. Let n > 0 be a perfect square. That is n = m 2 for some m N. Since n + 1 > n, we have n + 1 > n (by lemma 2). i.e., n + 1 > m. Now, consider (m + 1) 2 = m 2 + 2m + 1. Since n > 0, and n = m 2, m must be greater than 0. Which implies (m 2 +2m+1) > (m 2 +1), i.e. (m 2 +2m+1) > (n+1). Written differently, we have (m+1) 2 > n+1, it follows (by lemma 2) that We have shown that m + 1 > n + 1. m + 1 > n + 1 > m. Since m is an integer, n + 1 is not an integer. This tells us that n + 1 is not a perfect square. Therefore, For all natural numbers n > 0, If n is a perfect square, then n + 1 is not a perfect square. (e) n N If n has a rational square root, then the square root of n is in fact a natural number. (You may assume that if a doesn t divide b then a 2 doesn t divide b 2.) Here by square root, we are referring to the non-negative square root. Proof : We give a proof by contradiction. Assume that statement (e) is not true. That is, assume that there is a natural number n such that n is a rational number but not a natural number. By our assumption n = p/q for some natural numbers p and q > 0. Also, n is not a natural number, which implies q doesn t divide p. Since q doesn t divide p, q 2 doesn t divide p 2. This implies n = p 2 /q 2 is not a natural number. This is a contradiction. Which tells us that our assumption was false. Therefore, n N If n has a rational square root, then the square root of n is in fact a natural number. 2

3 2. (a) Proof: We give a proof by induction. We prove the statement ( n N)P (n) where P (n) is the statement: n 2 = n(n + 1)(2n + 1)/. Base case: We first look at the base case P (0). Note that in P (0), the left hand side is the empty sum 0, and the right hand side is 0(0 + 1)( )/ = 0. So, the base case holds. Inductive step: Now we argue that the inductive step is valid. We prove the implication P (k) P (k + 1) for any k using a direct proof. Let k N and assume that P (k) holds, that is, assume that k 2 = k(k + 1)(2k + 1)/. Consider the sum k 2 + (k + 1) 2. The sum of the first k terms is k(k + 1)(2k + 1)/ by the induction hypothesis. So, we have k 2 + (k + 1) 2 = k(k + 1)(2k + 1)/ + (k + 1) 2. k(k+1)(2k+1)/+(k+1) 2 = (k + 1)(2k 2 + 7k + ) k(k + 1)(2k + 1) + (k + 1)2 = (k + 1)(k + 2)(2k + 3). This shows that k 2 + (k + 1) 2 = (k+1)(k+2)(2k+3), that is P (k + 1) is true. This completes the inductive step. It follows by induction that n N n 2 = n(n + 1)(2n + 1)/. (b) Proof : We give a proof by induction. We prove the statement ( n 5)P (n) where P (n) is the statement: 2 n > n 2 Base case: We first look at the base case P (5). Note that 2 5 = 32 and 5 2 = 25. Therefore, 2 5 > 5 2. So, the base case holds. Inductive step: Now we argue that the inductive step is valid. We prove the implication P (k) P (k + 1) for any k 5 using a direct proof. Let k 5 and assume that P (k) holds, that is, assume that 2 k > k 2. 2 k+1 = 2 2 k > 2 k 2. The final inequality follows from the induction hypothesis. Now, Since k > 4, we have k 2 > 4 k > 2k + k > 2k + 1. Therefore, 2 k 2 > k 2 + 2k + 1, i.e., 2 k 2 > (k + 1) 2. Which implies, 2 k+1 = 2 2 k > 2 k 2 > (k + 1) 2. That is P (k + 1) is true. This completes the induction step. Therefore by induction n > 4 2 n > n 2. = (k + 1)(k(2k + 1) + (k + 1)) 3

4 (c) Proof : We give a proof by induction. We prove the statement ( n 3)P (n) where P (n) is the statement: n 2 7n Base case: We first look at the base case P (3). Note that = 2 > 0. So, the base case holds. Inductive step : Now we argue that the inductive step is valid. We prove the implication P (k) P (k + 1) for any k 3 using a direct proof. Let k 3 and assume that P (k) holds, that is, assume that k 2 7k Consider (k + 1) 2 7(k + 1) (k + 1) 2 7(k + 1) + 12 = k 2 + 2k + 1 7k = k 2 5k + = (k 2 7k + 12) + (2k ). By the induction hypothesis (k 2 7k + 12) 0 and since k 3, the second term (2k ) is also non-negative. Thus, (k + 1) 2 7(k + 1) which is P (k + 1). This completes the inductive step. Therefore, n 3 n 2 7n + 12 is non-negative. (d) Which amounts of money can be formed using just two dollar and five dollar bills? Prove your answers using induction. Solution: 2 dollars can be formed, 1 and 3 dollars cannot be formed, and all amounts greater than 3 can be formed. Let s prove the final statement via induction. Lemma 3: ( n 4) n can be expressed as a combination of 2 and 5 dollar bills. Proof of Lemma 3: The proof is by induction. Let P (n) be the statement n = a 2 + b 5 for some a, b N. We want to prove ( n 4)P (n) Base case: We need to argue that P (4) is true. 4 can be written as Therefore, P (4) is true. Inductive step: In the induction step we prove that P (k) P (k + 1) holds for any k 4 using a direct proof. Assume that for some k 4, P (k) is true. That is assume k can be written as k = a 2 + b 5 for some a, b N. We now show that P (k + 1) is true using the induction hypothesis. We do so by considering 2 cases: Case 1: b = 0. That is k = a Now, k + 1 = a So, k + 1 = (a 2) = (a 2) Therefore k + 1 can be written as a 2 + b 5 where a = (a 2) and b = 1. Since, k 4 and k = a 2, a must greater than or equal to 2, which implies that (a 2) 0. Thus, a, b are both natural numbers. Therefore, P (k + 1) is true. Case 2: b 1. Now, k + 1 = a 2 + b (by the induction hypothesis.) 4

5 So, k + 1 = a 2 + (b 1) = a 2 + (b 1) 5 + = (a+3) 2+(b 1) 5. Therefore k+1 can be written as a 2+b 5 where a = (a + 3) and b = (b 1). Since, b 1 and a N, we have b and a are natural numbers. Which implie P (k + 1). Since in both cases P (k + 1) holds, we have that ( k 4)P (k) P (k + 1). This completes the induction step. Therefore, by induction all amounts greater than 3 can be formed as a combination of 2 and 5 dollar bills. (e) Proof : The proof is by strong induction. Let P (n) be the statement: If n is of the form 4k+1 for some integer k, then the second player has a winning strategy; othewise the first player has a winning strategy. We show that ( n 1) P (n). Our base case is P (1). As we argued above, then the first player has a winning strategy, which is consistent with the fact that 1 can be written as 1 = For the strong induction step, we assume that P (m) holds for all integers m in the range 1 m n, and we want to argue that P (n + 1) holds. Let us call the first player for the game with n + 1 sticks Alice, and the second player Bob. We consider four cases. i. Case n+1 is of the form 4k. If Alice picks 3 sticks, Bob then sees 4(k 1) + 1 sticks, so Alice has a winning strategy, consistent with our conjecture P (n + 1). ii. Case n + 1 is of the form 4k + 1. Since n 1, we know that n Alice can then chose to remove 1, 2, or 3 sticks. If she removes one stick, the remaining number of sticks is n = 4k. By the strong induction hypothesis, the player who plays first at this point has a winning strategy. That player is Bob, so Bob has a winning strategy. Similarly, if Alice removes two sticks, the remaining number is 4(k 1) + 3. Again, Bob has a winning strategy, by the same reasoning. Similarly, if Alice removes 3 sticks, Bob has a winning strategy. So, however Alice moves, Bob has a winning strategy for the subsequent rounds. So, Bob has a winning strategy. This proves our conjecture P (n + 1) in this case. iii. Case n + 1 is of the form n + 1 = 4k + 2. If Alice removes 1 stick, Bob is left with 4k + 1, so Alice has a winning strategy, consistent with our conjecture P (n + 1). iv. Case n + 1 is of the form 4k + 3. If Alice picks 2 sticks, Bob is left with 4k +1 sticks, so Alice has a winning strategy, consistent with P (n + 1). So in any case, P (n + 1) holds, so by strong induction we concluce that P (n) holds for all integers n The key observation is that the square of a natural number can be written as a product of an even number of (not necessarily distinct) primes. The quantity on the left hand side is a product of an odd number of primes, while the quantity on the right hand side is a product of an even number of primes. Since every number has a unique prime factorization, the equality cannot hold. This is a contradiction. 5

1.3 Induction and Other Proof Techniques

1.3 Induction and Other Proof Techniques 4CHAPTER 1. INTRODUCTORY MATERIAL: SETS, FUNCTIONS AND MATHEMATICAL INDU 1.3 Induction and Other Proof Techniques The purpose of this section is to study the proof technique known as mathematical induction.

More information

Climbing an Infinite Ladder

Climbing an Infinite Ladder Section 5.1 Climbing an Infinite Ladder Suppose we have an infinite ladder and the following capabilities: 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder,

More information

Logic and Proof Solutions. Question 1

Logic and Proof Solutions. Question 1 Logic and Proof Solutions Question 1 Which of the following are true, and which are false? For the ones which are true, give a proof. For the ones which are false, give a counterexample. (a) x is an odd

More information

Mathematical Induction

Mathematical Induction Mathematical Induction Victor Adamchik Fall of 2005 Lecture 2 (out of three) Plan 1. Strong Induction 2. Faulty Inductions 3. Induction and the Least Element Principal Strong Induction Fibonacci Numbers

More information

MATHEMATICAL INDUCTION. Mathematical Induction. This is a powerful method to prove properties of positive integers.

MATHEMATICAL INDUCTION. Mathematical Induction. This is a powerful method to prove properties of positive integers. MATHEMATICAL INDUCTION MIGUEL A LERMA (Last updated: February 8, 003) Mathematical Induction This is a powerful method to prove properties of positive integers Principle of Mathematical Induction Let P

More information

Oh Yeah? Well, Prove It.

Oh Yeah? Well, Prove It. Oh Yeah? Well, Prove It. MT 43A - Abstract Algebra Fall 009 A large part of mathematics consists of building up a theoretical framework that allows us to solve problems. This theoretical framework is built

More information

Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof. Harper Langston New York University

Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof. Harper Langston New York University Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof Harper Langston New York University Proof and Counterexample Discovery and proof Even and odd numbers number n from Z is called

More information

Chapter 6 Finite sets and infinite sets. Copyright 2013, 2005, 2001 Pearson Education, Inc. Section 3.1, Slide 1

Chapter 6 Finite sets and infinite sets. Copyright 2013, 2005, 2001 Pearson Education, Inc. Section 3.1, Slide 1 Chapter 6 Finite sets and infinite sets Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 1 Section 6. PROPERTIES OF THE NATURE NUMBERS 013 Pearson Education, Inc.1 Slide Recall that denotes

More information

Proof: A logical argument establishing the truth of the theorem given the truth of the axioms and any previously proven theorems.

Proof: A logical argument establishing the truth of the theorem given the truth of the axioms and any previously proven theorems. Math 232 - Discrete Math 2.1 Direct Proofs and Counterexamples Notes Axiom: Proposition that is assumed to be true. Proof: A logical argument establishing the truth of the theorem given the truth of the

More information

INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS

INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28

More information

Course Notes for Math 320: Fundamentals of Mathematics Chapter 3: Induction.

Course Notes for Math 320: Fundamentals of Mathematics Chapter 3: Induction. Course Notes for Math 320: Fundamentals of Mathematics Chapter 3: Induction. February 21, 2006 1 Proof by Induction Definition 1.1. A subset S of the natural numbers is said to be inductive if n S we have

More information

More Mathematical Induction. October 27, 2016

More Mathematical Induction. October 27, 2016 More Mathematical Induction October 7, 016 In these slides... Review of ordinary induction. Remark about exponential and polynomial growth. Example a second proof that P(A) = A. Strong induction. Least

More information

CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 Solutions CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

More information

The Real Numbers. Here we show one way to explicitly construct the real numbers R. First we need a definition.

The Real Numbers. Here we show one way to explicitly construct the real numbers R. First we need a definition. The Real Numbers Here we show one way to explicitly construct the real numbers R. First we need a definition. Definitions/Notation: A sequence of rational numbers is a funtion f : N Q. Rather than write

More information

MATH 289 PROBLEM SET 4: NUMBER THEORY

MATH 289 PROBLEM SET 4: NUMBER THEORY MATH 289 PROBLEM SET 4: NUMBER THEORY 1. The greatest common divisor If d and n are integers, then we say that d divides n if and only if there exists an integer q such that n = qd. Notice that if d divides

More information

Chapter 1 Basic Number Concepts

Chapter 1 Basic Number Concepts Draft of September 2014 Chapter 1 Basic Number Concepts 1.1. Introduction No problems in this section. 1.2. Factors and Multiples 1. Determine whether the following numbers are divisible by 3, 9, and 11:

More information

CS103X: Discrete Structures Homework Assignment 2: Solutions

CS103X: Discrete Structures Homework Assignment 2: Solutions CS103X: Discrete Structures Homework Assignment 2: Solutions Due February 1, 2008 Exercise 1 (10 Points). Prove or give a counterexample for the following: Use the Fundamental Theorem of Arithmetic to

More information

Inference Rules and Proof Methods

Inference Rules and Proof Methods Inference Rules and Proof Methods Winter 2010 Introduction Rules of Inference and Formal Proofs Proofs in mathematics are valid arguments that establish the truth of mathematical statements. An argument

More information

2.4 Mathematical Induction

2.4 Mathematical Induction 2.4 Mathematical Induction What Is (Weak) Induction? The Principle of Mathematical Induction works like this: What Is (Weak) Induction? The Principle of Mathematical Induction works like this: We want

More information

Introduction. Appendix D Mathematical Induction D1

Introduction. Appendix D Mathematical Induction D1 Appendix D Mathematical Induction D D Mathematical Induction Use mathematical induction to prove a formula. Find a sum of powers of integers. Find a formula for a finite sum. Use finite differences to

More information

8.7 Mathematical Induction

8.7 Mathematical Induction 8.7. MATHEMATICAL INDUCTION 8-135 8.7 Mathematical Induction Objective Prove a statement by mathematical induction Many mathematical facts are established by first observing a pattern, then making a conjecture

More information

Basic Proof Techniques

Basic Proof Techniques Basic Proof Techniques David Ferry dsf43@truman.edu September 13, 010 1 Four Fundamental Proof Techniques When one wishes to prove the statement P Q there are four fundamental approaches. This document

More information

k, then n = p2α 1 1 pα k

k, then n = p2α 1 1 pα k Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

More information

The Limit of a Sequence of Numbers: Infinite Series

The Limit of a Sequence of Numbers: Infinite Series Connexions module: m36135 1 The Limit of a Sequence of Numbers: Infinite Series Lawrence Baggett This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License

More information

MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear: MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

More information

COMP232 - Mathematics for Computer Science

COMP232 - Mathematics for Computer Science COMP3 - Mathematics for Computer Science Tutorial 10 Ali Moallemi moa ali@encs.concordia.ca Iraj Hedayati h iraj@encs.concordia.ca Concordia University, Winter 016 Ali Moallemi, Iraj Hedayati COMP3 - Mathematics

More information

Mathematical Induction

Mathematical Induction Chapter 2 Mathematical Induction 2.1 First Examples Suppose we want to find a simple formula for the sum of the first n odd numbers: 1 + 3 + 5 +... + (2n 1) = n (2k 1). How might we proceed? The most natural

More information

Math 55: Discrete Mathematics

Math 55: Discrete Mathematics Math 55: Discrete Mathematics UC Berkeley, Fall 2011 Homework # 5, due Wednesday, February 22 5.1.4 Let P (n) be the statement that 1 3 + 2 3 + + n 3 = (n(n + 1)/2) 2 for the positive integer n. a) What

More information

Homework until Test #2

Homework until Test #2 MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such

More information

n k=1 k=0 1/k! = e. Example 6.4. The series 1/k 2 converges in R. Indeed, if s n = n then k=1 1/k, then s 2n s n = 1 n + 1 +...

n k=1 k=0 1/k! = e. Example 6.4. The series 1/k 2 converges in R. Indeed, if s n = n then k=1 1/k, then s 2n s n = 1 n + 1 +... 6 Series We call a normed space (X, ) a Banach space provided that every Cauchy sequence (x n ) in X converges. For example, R with the norm = is an example of Banach space. Now let (x n ) be a sequence

More information

Even Number: An integer n is said to be even if it has the form n = 2k for some integer k. That is, n is even if and only if n divisible by 2.

Even Number: An integer n is said to be even if it has the form n = 2k for some integer k. That is, n is even if and only if n divisible by 2. MATH 337 Proofs Dr. Neal, WKU This entire course requires you to write proper mathematical proofs. All proofs should be written elegantly in a formal mathematical style. Complete sentences of explanation

More information

Mathematical induction. Niloufar Shafiei

Mathematical induction. Niloufar Shafiei Mathematical induction Niloufar Shafiei Mathematical induction Mathematical induction is an extremely important proof technique. Mathematical induction can be used to prove results about complexity of

More information

Sample Induction Proofs

Sample Induction Proofs Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand Sample Induction Proofs Below are model solutions to some of the practice problems on the induction worksheets. The solutions given

More information

4. PRINCIPLE OF MATHEMATICAL INDUCTION

4. PRINCIPLE OF MATHEMATICAL INDUCTION 4 PRINCIPLE OF MATHEMATICAL INDUCTION Ex Prove the following by principle of mathematical induction 1 1 + 2 + 3 + + n 2 1 2 + 2 2 + 3 2 + + n 2 3 1 3 + 2 3 + 3 3 + + n 3 + 4 (1) + (1 + 3) + (1 + 3 + 5)

More information

Solutions to Homework 6 Mathematics 503 Foundations of Mathematics Spring 2014

Solutions to Homework 6 Mathematics 503 Foundations of Mathematics Spring 2014 Solutions to Homework 6 Mathematics 503 Foundations of Mathematics Spring 2014 3.4: 1. If m is any integer, then m(m + 1) = m 2 + m is the product of m and its successor. That it to say, m 2 + m is the

More information

Section 6-2 Mathematical Induction

Section 6-2 Mathematical Induction 6- Mathematical Induction 457 In calculus, it can be shown that e x k0 x k k! x x x3!! 3!... xn n! where the larger n is, the better the approximation. Problems 6 and 6 refer to this series. Note that

More information

MAT2400 Analysis I. A brief introduction to proofs, sets, and functions

MAT2400 Analysis I. A brief introduction to proofs, sets, and functions MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take

More information

Mathematical Induction

Mathematical Induction Mathematical S 0 S 1 S 2 S 3 S 4 S 5 S 6 S 7 S 8 S 9 S 10 Like dominoes! Mathematical S 0 S 1 S 2 S 3 S4 S 5 S 6 S 7 S 8 S 9 S 10 Like dominoes! S 4 Mathematical S 0 S 1 S 2 S 3 S5 S 6 S 7 S 8 S 9 S 10

More information

SECTION 10-2 Mathematical Induction

SECTION 10-2 Mathematical Induction 73 0 Sequences and Series 6. Approximate e 0. using the first five terms of the series. Compare this approximation with your calculator evaluation of e 0.. 6. Approximate e 0.5 using the first five terms

More information

Appendix F: Mathematical Induction

Appendix F: Mathematical Induction Appendix F: Mathematical Induction Introduction In this appendix, you will study a form of mathematical proof called mathematical induction. To see the logical need for mathematical induction, take another

More information

Mathematical induction & Recursion

Mathematical induction & Recursion CS 441 Discrete Mathematics for CS Lecture 15 Mathematical induction & Recursion Milos Hauskrecht milos@cs.pitt.edu 5329 Sennott Square Proofs Basic proof methods: Direct, Indirect, Contradiction, By Cases,

More information

SCORE SETS IN ORIENTED GRAPHS

SCORE SETS IN ORIENTED GRAPHS Applicable Analysis and Discrete Mathematics, 2 (2008), 107 113. Available electronically at http://pefmath.etf.bg.ac.yu SCORE SETS IN ORIENTED GRAPHS S. Pirzada, T. A. Naikoo The score of a vertex v in

More information

Theorem 2. If x Q and y R \ Q, then. (a) x + y R \ Q, and. (b) xy Q.

Theorem 2. If x Q and y R \ Q, then. (a) x + y R \ Q, and. (b) xy Q. Math 305 Fall 011 The Density of Q in R The following two theorems tell us what happens when we add and multiply by rational numbers. For the first one, we see that if we add or multiply two rational numbers

More information

3. Mathematical Induction

3. Mathematical Induction 3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)

More information

31 is a prime number is a mathematical statement (which happens to be true).

31 is a prime number is a mathematical statement (which happens to be true). Chapter 1 Mathematical Logic In its most basic form, Mathematics is the practice of assigning truth to welldefined statements. In this course, we will develop the skills to use known true statements to

More information

WRITING PROOFS. Christopher Heil Georgia Institute of Technology

WRITING PROOFS. Christopher Heil Georgia Institute of Technology WRITING PROOFS Christopher Heil Georgia Institute of Technology A theorem is just a statement of fact A proof of the theorem is a logical explanation of why the theorem is true Many theorems have this

More information

APPLICATIONS OF THE ORDER FUNCTION

APPLICATIONS OF THE ORDER FUNCTION APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and

More information

Mathematical Induction

Mathematical Induction Mathematical Induction Tan Conghui Department of Systems Engineering and Engineering Management, The Chinese University of Hong Kong September 15, 2016 Tan Conghui (ENGG 2440B) Tutorial 1 September 15,

More information

Finite and Infinite Sets

Finite and Infinite Sets Chapter 9 Finite and Infinite Sets 9. Finite Sets Preview Activity (Equivalent Sets, Part ). Let A and B be sets and let f be a function from A to B..f W A! B/. Carefully complete each of the following

More information

CS 2336 Discrete Mathematics

CS 2336 Discrete Mathematics CS 2336 Discrete Mathematics Lecture 5 Proofs: Mathematical Induction 1 Outline What is a Mathematical Induction? Strong Induction Common Mistakes 2 Introduction What is the formula of the sum of the first

More information

MATHEMATICAL INDUCTION

MATHEMATICAL INDUCTION MATHEMATICAL INDUCTION MATH 5A SECTION HANDOUT BY GERARDO CON DIAZ Imagine a bunch of dominoes on a table. They are set up in a straight line, and you are about to push the first piece to set off the chain

More information

Mathematical Induction

Mathematical Induction Mathematical Induction Victor Adamchik Fall of 2005 Lecture 1 (out of three) Plan 1. The Principle of Mathematical Induction 2. Induction Examples The Principle of Mathematical Induction Suppose we have

More information

CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12 CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

More information

Mathematical Induction. Lecture 10-11

Mathematical Induction. Lecture 10-11 Mathematical Induction Lecture 10-11 Menu Mathematical Induction Strong Induction Recursive Definitions Structural Induction Climbing an Infinite Ladder Suppose we have an infinite ladder: 1. We can reach

More information

The Prime Numbers. Definition. A prime number is a positive integer with exactly two positive divisors.

The Prime Numbers. Definition. A prime number is a positive integer with exactly two positive divisors. The Prime Numbers Before starting our study of primes, we record the following important lemma. Recall that integers a, b are said to be relatively prime if gcd(a, b) = 1. Lemma (Euclid s Lemma). If gcd(a,

More information

Chapter 2 Limits Functions and Sequences sequence sequence Example

Chapter 2 Limits Functions and Sequences sequence sequence Example Chapter Limits In the net few chapters we shall investigate several concepts from calculus, all of which are based on the notion of a limit. In the normal sequence of mathematics courses that students

More information

Assignment 3 Solutions Problem1: (Section 1.2 Exercise 8) Consider the following assertions.

Assignment 3 Solutions Problem1: (Section 1.2 Exercise 8) Consider the following assertions. Assignment 3 Solutions Problem1: (Section 1.2 Eercise 8) Consider the following assertions. A: There eists a real number y such that y > for every real number B: For every real number, there eists a real

More information

Triangle deletion. Ernie Croot. February 3, 2010

Triangle deletion. Ernie Croot. February 3, 2010 Triangle deletion Ernie Croot February 3, 2010 1 Introduction The purpose of this note is to give an intuitive outline of the triangle deletion theorem of Ruzsa and Szemerédi, which says that if G = (V,

More information

Section 1. Statements and Truth Tables. Definition 1.1: A mathematical statement is a declarative sentence that is true or false, but not both.

Section 1. Statements and Truth Tables. Definition 1.1: A mathematical statement is a declarative sentence that is true or false, but not both. M3210 Supplemental Notes: Basic Logic Concepts In this course we will examine statements about mathematical concepts and relationships between these concepts (definitions, theorems). We will also consider

More information

New Zealand Mathematical Olympiad Committee. Induction

New Zealand Mathematical Olympiad Committee. Induction New Zealand Mathematical Olympiad Committee Induction Chris Tuffley 1 Proof by dominoes Mathematical induction is a useful tool for proving statements like P is true for all natural numbers n, for example

More information

Math 140a - HW 1 Solutions

Math 140a - HW 1 Solutions Math 140a - HW 1 Solutions Problem 1 (WR Ch 1 #1). If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Given that r is rational, we can write r = a b for some

More information

Homework 6, Fri CS 2050, Intro Discrete Math for Computer Science Due Mon , Note: This homework has 2 pages and 5 problems.

Homework 6, Fri CS 2050, Intro Discrete Math for Computer Science Due Mon , Note: This homework has 2 pages and 5 problems. Homework 6, Fri 0-8- CS 050, Intro Discrete Math for Computer Science Due Mon -7-, Note: This homework has pages and 5 problems. Problem : 0 Points What is wrong with this proof? Theorem: For every positive

More information

3.3. INFERENCE 105. Table 3.5: Another look at implication. p q p q T T T T F F F T T F F T

3.3. INFERENCE 105. Table 3.5: Another look at implication. p q p q T T T T F F F T T F F T 3.3. INFERENCE 105 3.3 Inference Direct Inference (Modus Ponens) and Proofs We concluded our last section with a proof that the sum of two even numbers is even. That proof contained several crucial ingredients.

More information

Discrete Mathematics, Chapter 5: Induction and Recursion

Discrete Mathematics, Chapter 5: Induction and Recursion Discrete Mathematics, Chapter 5: Induction and Recursion Richard Mayr University of Edinburgh, UK Richard Mayr (University of Edinburgh, UK) Discrete Mathematics. Chapter 5 1 / 20 Outline 1 Well-founded

More information

Discrete Mathematics: Solutions to Homework (12%) For each of the following sets, determine whether {2} is an element of that set.

Discrete Mathematics: Solutions to Homework (12%) For each of the following sets, determine whether {2} is an element of that set. Discrete Mathematics: Solutions to Homework 2 1. (12%) For each of the following sets, determine whether {2} is an element of that set. (a) {x R x is an integer greater than 1} (b) {x R x is the square

More information

MATH10040 Chapter 2: Prime and relatively prime numbers

MATH10040 Chapter 2: Prime and relatively prime numbers MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positive

More information

Geometry Chapter 2: Geometric Reasoning Lesson 1: Using Inductive Reasoning to Make Conjectures Inductive Reasoning:

Geometry Chapter 2: Geometric Reasoning Lesson 1: Using Inductive Reasoning to Make Conjectures Inductive Reasoning: Geometry Chapter 2: Geometric Reasoning Lesson 1: Using Inductive Reasoning to Make Conjectures Inductive Reasoning: Conjecture: Advantages: can draw conclusions from limited information helps us to organize

More information

LESSON 1 PRIME NUMBERS AND FACTORISATION

LESSON 1 PRIME NUMBERS AND FACTORISATION LESSON 1 PRIME NUMBERS AND FACTORISATION 1.1 FACTORS: The natural numbers are the numbers 1,, 3, 4,. The integers are the naturals numbers together with 0 and the negative integers. That is the integers

More information

This section demonstrates some different techniques of proving some general statements.

This section demonstrates some different techniques of proving some general statements. Section 4. Number Theory 4.. Introduction This section demonstrates some different techniques of proving some general statements. Examples: Prove that the sum of any two odd numbers is even. Firstly you

More information

Proofs Crash Course. Winter 2011

Proofs Crash Course. Winter 2011 Proofs Crash Course Winter 2011 Today s Topics O Why are Proofs so Hard? O Proof by Deduction O Proof by Contrapositive O Proof by Contradiction O Proof by Induction Why are Proofs so Hard? If it is a

More information

Discrete Math in Computer Science Homework 7 Solutions (Max Points: 80)

Discrete Math in Computer Science Homework 7 Solutions (Max Points: 80) Discrete Math in Computer Science Homework 7 Solutions (Max Points: 80) CS 30, Winter 2016 by Prasad Jayanti 1. (10 points) Here is the famous Monty Hall Puzzle. Suppose you are on a game show, and you

More information

6.2 Permutations continued

6.2 Permutations continued 6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of

More information

Elementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.

Elementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook. Elementary Number Theory and Methods of Proof CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.edu/~cse215 1 Number theory Properties: 2 Properties of integers (whole

More information

MA2C03 Mathematics School of Mathematics, Trinity College Hilary Term 2016 Lecture 59 (April 1, 2016) David R. Wilkins

MA2C03 Mathematics School of Mathematics, Trinity College Hilary Term 2016 Lecture 59 (April 1, 2016) David R. Wilkins MA2C03 Mathematics School of Mathematics, Trinity College Hilary Term 2016 Lecture 59 (April 1, 2016) David R. Wilkins The RSA encryption scheme works as follows. In order to establish the necessary public

More information

arxiv:1112.0829v1 [math.pr] 5 Dec 2011

arxiv:1112.0829v1 [math.pr] 5 Dec 2011 How Not to Win a Million Dollars: A Counterexample to a Conjecture of L. Breiman Thomas P. Hayes arxiv:1112.0829v1 [math.pr] 5 Dec 2011 Abstract Consider a gambling game in which we are allowed to repeatedly

More information

Reading 7 : Program Correctness

Reading 7 : Program Correctness CS/Math 240: Introduction to Discrete Mathematics Fall 2015 Instructors: Beck Hasti, Gautam Prakriya Reading 7 : Program Correctness 7.1 Program Correctness Showing that a program is correct means that

More information

PART I. THE REAL NUMBERS

PART I. THE REAL NUMBERS PART I. THE REAL NUMBERS This material assumes that you are already familiar with the real number system and the representation of the real numbers as points on the real line. I.1. THE NATURAL NUMBERS

More information

AN ANALYSIS OF A WAR-LIKE CARD GAME. Introduction

AN ANALYSIS OF A WAR-LIKE CARD GAME. Introduction AN ANALYSIS OF A WAR-LIKE CARD GAME BORIS ALEXEEV AND JACOB TSIMERMAN Abstract. In his book Mathematical Mind-Benders, Peter Winkler poses the following open problem, originally due to the first author:

More information

ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

More information

Finite Sets. Theorem 5.1. Two non-empty finite sets have the same cardinality if and only if they are equivalent.

Finite Sets. Theorem 5.1. Two non-empty finite sets have the same cardinality if and only if they are equivalent. MATH 337 Cardinality Dr. Neal, WKU We now shall prove that the rational numbers are a countable set while R is uncountable. This result shows that there are two different magnitudes of infinity. But we

More information

In a triangle with a right angle, there are 2 legs and the hypotenuse of a triangle.

In a triangle with a right angle, there are 2 legs and the hypotenuse of a triangle. PROBLEM STATEMENT In a triangle with a right angle, there are legs and the hypotenuse of a triangle. The hypotenuse of a triangle is the side of a right triangle that is opposite the 90 angle. The legs

More information

CARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE

CARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE CARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE With the notion of bijection at hand, it is easy to formalize the idea that two finite sets have the same number of elements: we just need to verify

More information

Chapter 2 Sequences and Series

Chapter 2 Sequences and Series Chapter 2 Sequences and Series 2. Sequences A sequence is a function from the positive integers (possibly including 0) to the reals. A typical example is a n = /n defined for all integers n. The notation

More information

p 2 1 (mod 6) Adding 2 to both sides gives p (mod 6)

p 2 1 (mod 6) Adding 2 to both sides gives p (mod 6) .9. Problems P10 Try small prime numbers first. p p + 6 3 11 5 7 7 51 11 13 Among the primes in this table, only the prime 3 has the property that (p + ) is also a prime. We try to prove that no other

More information

Logic, Sets, and Proofs

Logic, Sets, and Proofs Logic, Sets, and Proofs David A. Cox and Catherine C. McGeoch Amherst College 1 Logic Logical Statements. A logical statement is a mathematical statement that is either true or false. Here we denote logical

More information

HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!

HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)! Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following

More information

3.1. Sequences and Their Limits Definition (3.1.1). A sequence of real numbers (or a sequence in R) is a function from N into R.

3.1. Sequences and Their Limits Definition (3.1.1). A sequence of real numbers (or a sequence in R) is a function from N into R. CHAPTER 3 Sequences and Series 3.. Sequences and Their Limits Definition (3..). A sequence of real numbers (or a sequence in R) is a function from N into R. Notation. () The values of X : N! R are denoted

More information

ELEMENTARY PROBLEMS AND SOLUTIONS. Edited by A. P. HiLLMAN University of New Mexico, Albuquerque, NM 87131

ELEMENTARY PROBLEMS AND SOLUTIONS. Edited by A. P. HiLLMAN University of New Mexico, Albuquerque, NM 87131 ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A. P. HiLLMAN University of New Mexico, Albuquerque, NM 87131 Send all communications regarding ELEMENTARY PROBLEMS AND SOLUTIONS to PROFESSOR A. P. HILLMAN,

More information

Mathematical Induction

Mathematical Induction Mathematical Induction (Handout March 8, 01) The Principle of Mathematical Induction provides a means to prove infinitely many statements all at once The principle is logical rather than strictly mathematical,

More information

MATH 22. THE FUNDAMENTAL THEOREM of ARITHMETIC. Lecture R: 10/30/2003

MATH 22. THE FUNDAMENTAL THEOREM of ARITHMETIC. Lecture R: 10/30/2003 MATH 22 Lecture R: 10/30/2003 THE FUNDAMENTAL THEOREM of ARITHMETIC You must remember this, A kiss is still a kiss, A sigh is just a sigh; The fundamental things apply, As time goes by. Herman Hupfeld

More information

CSE373: Data Structures and Algorithms Lecture 2: Proof by Induction. Linda Shapiro Winter 2015

CSE373: Data Structures and Algorithms Lecture 2: Proof by Induction. Linda Shapiro Winter 2015 CSE373: Data Structures and Algorithms Lecture 2: Proof by Induction Linda Shapiro Winter 2015 Background on Induction Type of mathematical proof Typically used to establish a given statement for all natural

More information

Chapter 7 - Roots, Radicals, and Complex Numbers

Chapter 7 - Roots, Radicals, and Complex Numbers Math 233 - Spring 2009 Chapter 7 - Roots, Radicals, and Complex Numbers 7.1 Roots and Radicals 7.1.1 Notation and Terminology In the expression x the is called the radical sign. The expression under the

More information

Case #1 Case #2 Case #3. n = 3q n = 3q + 1 n = 3q + 2

Case #1 Case #2 Case #3. n = 3q n = 3q + 1 n = 3q + 2 Problems 1, 2, and 3 are worth 15 points each (5 points per subproblem). Problems 4 and 5 are worth 30 points each (10 points per subproblem), for a total of 105 points possible. 1. The following are from

More information

The Foundations: Logic and Proofs. Chapter 1, Part III: Proofs

The Foundations: Logic and Proofs. Chapter 1, Part III: Proofs The Foundations: Logic and Proofs Chapter 1, Part III: Proofs Rules of Inference Section 1.6 Section Summary Valid Arguments Inference Rules for Propositional Logic Using Rules of Inference to Build Arguments

More information

San Jose Math Circle October 17, 2009 ARITHMETIC AND GEOMETRIC PROGRESSIONS

San Jose Math Circle October 17, 2009 ARITHMETIC AND GEOMETRIC PROGRESSIONS San Jose Math Circle October 17, 2009 ARITHMETIC AND GEOMETRIC PROGRESSIONS DEFINITION. An arithmetic progression is a (finite or infinite) sequence of numbers with the property that the difference between

More information

Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate.

Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate. Advice for homework: Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate. Even if a problem is a simple exercise that doesn t

More information

SYSTEMS OF PYTHAGOREAN TRIPLES. Acknowledgements. I would like to thank Professor Laura Schueller for advising and guiding me

SYSTEMS OF PYTHAGOREAN TRIPLES. Acknowledgements. I would like to thank Professor Laura Schueller for advising and guiding me SYSTEMS OF PYTHAGOREAN TRIPLES CHRISTOPHER TOBIN-CAMPBELL Abstract. This paper explores systems of Pythagorean triples. It describes the generating formulas for primitive Pythagorean triples, determines

More information

WOLLONGONG COLLEGE AUSTRALIA. Diploma in Information Technology

WOLLONGONG COLLEGE AUSTRALIA. Diploma in Information Technology First Name: Family Name: Student Number: Class/Tutorial: WOLLONGONG COLLEGE AUSTRALIA A College of the University of Wollongong Diploma in Information Technology Mid-Session Test Summer Session 008-00

More information

conditional statement conclusion Vocabulary Flash Cards Chapter 2 (p. 66) Chapter 2 (p. 69) Chapter 2 (p. 66) Chapter 2 (p. 76)

conditional statement conclusion Vocabulary Flash Cards Chapter 2 (p. 66) Chapter 2 (p. 69) Chapter 2 (p. 66) Chapter 2 (p. 76) biconditional statement conclusion Chapter 2 (p. 69) conditional statement conjecture Chapter 2 (p. 76) contrapositive converse Chapter 2 (p. 67) Chapter 2 (p. 67) counterexample deductive reasoning Chapter

More information

PRINCIPLE OF MATHEMATICAL INDUCTION

PRINCIPLE 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 information