Sample Induction Proofs

Size: px
Start display at page:

Transcription

1 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 illustrate all of the main types of induction situations that you may encounter and that you should be able to handle. Use these solutions as models for your writing up your own solutions in exams and homework. For additional examples, see the following examples and exercises in the Rosen text: Section 4., Examples 0, Exercises 3, 5, 7, 3, 5, 9,, 3, 5, 45. Section 4.3, Example 6, Exercises 3, 5.. Prove by induction that, for all n Z +, n ( )i i = ( ) n n(n + )/. () n ( ) i i = ( )n n(n + ). Base case: When n =, the left side of () is ( ) =, and the right side is ( )(+)/ =, so both sides are equal and () is true for n =. Induction step: Let k Z + be given and suppose () is true for n = k. Then ( ) i i = ( ) i i + ( ) (k + ) = ( )k k(k + ) = ( )k (k + ) + ( ) (k + ) (by induction hypothesis) (k (k + )) = ( )k (k + ) ( k ) = ( ) (k + ). Thus, () holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of induction, () is true for all n Z +.. Find and prove by induction a formula for n i(i+), where n Z +. () n i(i + ) = n n +. Base case: When n =, the left side of () is /( ) = /, and the right side is /, so both sides are equal and () is true for n =.

2 Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand Induction step: Let k Z + be given and suppose () is true for n = k. Then i(i + ) = i(i + ) + (k + )(k + ) = k k + + (by induction hypothesis) (k + )(k + ) k(k + ) + = (by algebra) (k + )(k + ) (k + ) = (k + )(k + ) = k + k +. (by algebra) Thus, () holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of induction, () is true for all n Z Find and prove by induction a formula for n n Z +. (i ) (i.e., the sum of the first n odd numbers), where () n (i ) = n. Base case: When n =, the left side of () is, and the right side is =, so both sides are equal and () is true for n =. Induction step: Let k Z + be given and suppose () is true for n = k. Then (i ) = (i ) + ((k + ) ) = k + (k + ) (by induction hypothesis) = k + k + = (k + ). Thus, () holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of induction, () is true for all n. 4. Find and prove by induction a formula for n i= ( i ), where n Z + and n. Proof: We will prove by induction that, for all integers n, () n ( i ) i= = n + n. Base case: When n =, the left side of () is / = 3/4, and the right side is (+)/4 = 3/4, so both sides are equal and () is true for n =.

3 Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand Induction step: Let k be given and suppose () is true for n = k. Then i= ( i ) = k i= = k + k ( i ) ( ( (k + ) = k + k (k + ) (k + ) = k + k k(k + ) = k + (k + ). ) (k + ) ) (by induction hypothesis) Thus, () holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of induction, () is true for all n Z + with n. 5. Prove that n! > n for n 4. Proof: We will prove by induction that n! > n holds for all n 4. Base case: Our base case here is the first n-value for which is claimed, i.e., n = 4. For n = 4, the left and right sides of are 4 and 6, respectively, so is true in this case. Induction step: Let k 4 be given and suppose is true for n = k. Then (k + )! = k!(k + ) > k (k + ) (by induction hypothesis) k (since k 4 and so k + )) =. Thus, holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of induction, it follows that is true for all n Prove that for any real number x > and any positive integer x, ( + x) n + nx. Proof: Let x be a real number in the range given, namely x >. We will prove by induction that for any positive integer n, ( + x) n + nx. holds for any n Z +. Base case: For n =, the left and right sides of are both + x, so holds. Induction step: Let k Z + be given and suppose is true for n = k. We have ( + x) = ( + x) k ( + x) ( + kx)( + x) (by ind. hyp. and since x > and thus ( + x) > 0) = + (k + )x + kx (by algebra) + (k + )x (since kx 0). Hence holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of induction, it follows that holds for all n Z +. 3

4 Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand 7. Prove that n f i = f nf n+ for all n Z +. Proof: We seek to show that, for all n Z +, n fi = f n f n+. Base case: When n =, the left side of is f =, and the right side is f f = =, so both sides are equal and is true for n =. Induction step: Let k Z + be given and suppose is true for n = k. Then fi = fi + f = f k f + f (by ind. hyp. with n = k) = f (f k + f ) (by algebra) = f f k+ (by recurrence for f n ). Thus, holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of induction, it follows that is true for all n Z Prove that f n (3/) n for all n Z +. Proof: We will show that for all n Z +, f n (3/) n Base cases: When n =, the left side of is f =, and the right side is (3/) = /3, so holds for n =. When n =, the left side of is f =, and the right side is (3/) 0 =, so both sides are equal and is true for n =. Thus, holds for n = and n =. Induction step: Let k be given and suppose is true for all n =,,..., k. Then f = f k + f k (by recurrence for f n ) (3/) k + (3/) k 3 (by induction hypothesis with n = k and n = k ) = (3/) k ( (3/) + (3/) ) (by algebra) ( = (3/) k ) 9 = (3/) k 0 9 > (3/)k. Thus, holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of strong induction, it follows that is true for all n Z +. Remarks: Number of base cases: Since the induction step involves the cases n = k and n = k, we can carry out this step only for values k (for k =, k would be 0 and out of range). This in turn forces us to include the cases n = and n = in the base step. Such multiple bases cases are typical in proofs involving recurrence sequences. For a three term recurrence we would need to check three initial cases, n =,, 3, and in the induction step restrict k to values 3 or greater. 9. Prove that n f i = f n+ for all n Z +. 4

5 Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand n f i = f n+. Base case: When n =, the left side of is f =, and the right side is f 3 = =, so both sides are equal and is true for n =. Induction step: Let k Z + be given and suppose is true for n = k. Then f i = f i + f = f k+ + f (by ind. hyp. with n = k) = f k+3 (by recurrence for f n ) Thus, holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of induction, it follows that is true for all n Z +. Remark: Here standard induction was sufficient, since we were able to relate the n = k + case directly to the n = k case, in the same way as in the induction proofs for summation formulas like n i = n(n + )/. Hence, a single base case was sufficient. 0. Let the Tribonacci sequence be defined by T = T = T 3 = and T n = T n + T n + T n 3 for n 4. Prove that T n < n for all n Z +. Proof: We will prove by strong induction that, for all n Z +, T n < n Base case: We will need to check directly for n =,, 3 since the induction step (below) is only valid when k 3. For n =,, 3, T n is equal to, whereas the right-hand side of is equal to =, = 4, and 3 = 8, respectively. Thus, holds for n =,, 3. Induction step: Let k 3 be given and suppose is true for all n =,,..., k. Then T = T k + T k + T k (by recurrence for T n ) < k + k + k (by strong ind. hyp. with n = k, k, and k ) ( = ) 8 = 7 8 <. Thus, holds for n = k +, and the proof of the induction step is complete. Conclusion: By the principle of strong induction, it follows that is true for all n Z +.. Let a n be the sequence defined by a =, a = 8, a n = a n + a n (n 3). Prove that a n = 3 n + ( ) n for all n Z +. Proof: We will prove by strong induction that, for all n Z +, a n = 3 n + ( ) n. Base case: When n =, the left side of is a =, and the right side is ( ) =, so both sides are equal and is true for n =. 5

6 Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand When n =, the left and right sides of are a = 8 and 3 + ( ) = 8, so holds in this case as well. Induction step: Let k Z + with k be given and suppose is true for n =,,..., k. Then a = a k + a k (by recurrence for a n ) = 3 k + ( ) k + ( 3 k + ( ) k ) (by for n = k and n = k ) = 3 ( k + k ) + ( ( ) k + ( ) k ) (by algebra) = 3 k + ( ) (more algebra). Thus, holds for n = k +, and the proof of the induction step is complete. Conclusion: By the strong induction principle, it follows that is true for all n Z +.. Number of subsets: Show that a set of n elements has n subsets. Proof: P (n) We will prove by induction that, for all n Z +, the following holds: Any set of n elements has n subsets. Base case: Since any -element set has subsets, namely the empty set and the set itself, and =, the statement P (n) is true for n =. Induction step: Let k Z + be given and suppose P (k) is true, i.e., that any k-element set has k subsets. We seek to show that P (k + ) is true as well, i.e., that any (k + )-element set has subsets. Let A be a set with (k + ) elements. Let a be an element of A, and let A = A {a} (so that A is a set with k elements). We classify the subsets of A into two types: (I) subsets that do not contain a, and (II) subsets that do contain a. The subsets of type (I) are exactly the subsets of the set A. Since A has k elements, the induction hypothesis can be applied to this set and we get that there are k subsets of type (I). The subsets of type (II) are exactly the sets of the form B = B {a}, where B is a subset of A. By the induction hypothesis there are k such sets B, and hence k subsets of type (II). Since there are k subsets of each of the two types, the total number of subsets of A is k + k =. Since A was an arbitrary (k + )-element set, we have proved that any (k + )-element set has subsets. Thus P (k + ) is true, completing the induction step. Conclusion: By the principle of induction, P (n) is true for all n Z Number of -element subsets: Show that a set of n elements has n(n )/ subsets with elements. (Take n = as the base case.) Proof: (Sketch) This can be proved in the same way as the formula for the number of all subsets. For the number of Type I subsets (i.e., those not containing the element a), the induction hypothesis can be used as before. The Type II subsets of A can be counted directly: the latter subsets are those of the form {a, b}, where a is the selected element and b is an arbitrary element in A {a}. There are k choices for b, and hence k subsets of Type II. 4. Generalization of De Morgan s Law to unions of n sets. Show that if A,..., A n are sets, then (A A n ) = A A n. 6

7 Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand Proof: P (n): We seek to prove by induction on n the following statement: For all sets A,..., A n we have (A A n ) = A A n. The key to the argument is two set version of De Morgan s Law: ( ) (A B) = A B, which holds for any sets A and B. Base case: For n =, the left and right sides of are both equal to A, so holds trivially in this case. Hence P () is true. Though not absolutely necessary, we can also easily verify the next case, n = : In this case, the left and right sides of are (A A ) and A A, respectively, so the identity is just the two set version of De Morgan s Law, i.e., ( ) with A = A and B = A. Induction step: Let k, and suppose P (k) is true, i.e., suppose that holds for n = k and any sets A,..., A k. We seek to show that P (k + ) is true, i.e., that for any sets A,..., A, holds. Let A,..., A be given sets. Then (A A ) = ((A A k ) A ) = (A A k ) A (by ( ) with A = (A A k ) and B = A ) = ( A A k ) A (by induction hypothesis applied to A,..., A k ) = A A k A. Thus, holds for n = k +, and since the A,..., A were arbitrary sets, we have obtained statement P (k + ). Hence, the proof of the induction step is complete. Conclusion: By the principle of induction, it follows that P (n) is true for all n Z +. 7

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)

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

Math 115 Spring 2011 Written Homework 5 Solutions

. Evaluate each series. a) 4 7 0... 55 Math 5 Spring 0 Written Homework 5 Solutions Solution: We note that the associated sequence, 4, 7, 0,..., 55 appears to be an arithmetic sequence. If the sequence

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

Solutions for Practice problems on proofs

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

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

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,

Mathematical Induction. Mary Barnes Sue Gordon

Mathematics Learning Centre Mathematical Induction Mary Barnes Sue Gordon c 1987 University of Sydney Contents 1 Mathematical Induction 1 1.1 Why do we need proof by induction?.... 1 1. What is proof by

Full and Complete Binary Trees

Full and Complete Binary Trees Binary Tree Theorems 1 Here are two important types of binary trees. Note that the definitions, while similar, are logically independent. Definition: a binary tree T is full

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

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

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

Math 55: Discrete Mathematics

Math 55: Discrete Mathematics UC Berkeley, Spring 2012 Homework # 9, due Wednesday, April 11 8.1.5 How many ways are there to pay a bill of 17 pesos using a currency with coins of values of 1 peso, 2 pesos,

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

1(a). How many ways are there to rearrange the letters in the word COMPUTER?

CS 280 Solution Guide Homework 5 by Tze Kiat Tan 1(a). How many ways are there to rearrange the letters in the word COMPUTER? There are 8 distinct letters in the word COMPUTER. Therefore, the number of

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

MATH 4330/5330, Fourier Analysis Section 11, The Discrete Fourier Transform

MATH 433/533, Fourier Analysis Section 11, The Discrete Fourier Transform Now, instead of considering functions defined on a continuous domain, like the interval [, 1) or the whole real line R, we wish

arxiv:math/0202219v1 [math.co] 21 Feb 2002

RESTRICTED PERMUTATIONS BY PATTERNS OF TYPE (2, 1) arxiv:math/0202219v1 [math.co] 21 Feb 2002 TOUFIK MANSOUR LaBRI (UMR 5800), Université Bordeaux 1, 351 cours de la Libération, 33405 Talence Cedex, France

A Little Set Theory (Never Hurt Anybody)

A Little Set Theory (Never Hurt Anybody) Matthew Saltzman Department of Mathematical Sciences Clemson University Draft: August 21, 2013 1 Introduction The fundamental ideas of set theory and the algebra

Answer Key for California State Standards: Algebra I

Algebra I: Symbolic reasoning and calculations with symbols are central in algebra. Through the study of algebra, a student develops an understanding of the symbolic language of mathematics and the sciences.

Discrete Mathematics: Homework 7 solution. Due: 2011.6.03

EE 2060 Discrete Mathematics spring 2011 Discrete Mathematics: Homework 7 solution Due: 2011.6.03 1. Let a n = 2 n + 5 3 n for n = 0, 1, 2,... (a) (2%) Find a 0, a 1, a 2, a 3 and a 4. (b) (2%) Show that

CONTRIBUTIONS TO ZERO SUM PROBLEMS

CONTRIBUTIONS TO ZERO SUM PROBLEMS S. D. ADHIKARI, Y. G. CHEN, J. B. FRIEDLANDER, S. V. KONYAGIN AND F. PAPPALARDI Abstract. A prototype of zero sum theorems, the well known theorem of Erdős, Ginzburg

We can express this in decimal notation (in contrast to the underline notation we have been using) as follows: 9081 + 900b + 90c = 9001 + 100c + 10b

In this session, we ll learn how to solve problems related to place value. This is one of the fundamental concepts in arithmetic, something every elementary and middle school mathematics teacher should

The Mean Value Theorem

The Mean Value Theorem THEOREM (The Extreme Value Theorem): If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers

I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

Connectivity and cuts

Math 104, Graph Theory February 19, 2013 Measure of connectivity How connected are each of these graphs? > increasing connectivity > I G 1 is a tree, so it is a connected graph w/minimum # of edges. Every

1. Prove that the empty set is a subset of every set.

1. Prove that the empty set is a subset of every set. Basic Topology Written by Men-Gen Tsai email: b89902089@ntu.edu.tw Proof: For any element x of the empty set, x is also an element of every set since

Math 319 Problem Set #3 Solution 21 February 2002

Math 319 Problem Set #3 Solution 21 February 2002 1. ( 2.1, problem 15) Find integers a 1, a 2, a 3, a 4, a 5 such that every integer x satisfies at least one of the congruences x a 1 (mod 2), x a 2 (mod

Chapter 3. Distribution Problems. 3.1 The idea of a distribution. 3.1.1 The twenty-fold way

Chapter 3 Distribution Problems 3.1 The idea of a distribution Many of the problems we solved in Chapter 1 may be thought of as problems of distributing objects (such as pieces of fruit or ping-pong balls)

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

Chapter 17. Orthogonal Matrices and Symmetries of Space

Chapter 17. Orthogonal Matrices and Symmetries of Space Take a random matrix, say 1 3 A = 4 5 6, 7 8 9 and compare the lengths of e 1 and Ae 1. The vector e 1 has length 1, while Ae 1 = (1, 4, 7) has length

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a

Exponential time algorithms for graph coloring

Exponential time algorithms for graph coloring Uriel Feige Lecture notes, March 14, 2011 1 Introduction Let [n] denote the set {1,..., k}. A k-labeling of vertices of a graph G(V, E) is a function V [k].

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,

Cartesian Products and Relations

Cartesian Products and Relations Definition (Cartesian product) If A and B are sets, the Cartesian product of A and B is the set A B = {(a, b) :(a A) and (b B)}. The following points are worth special

CHAPTER 3. Methods of Proofs. 1. Logical Arguments and Formal Proofs

CHAPTER 3 Methods of Proofs 1. Logical Arguments and Formal Proofs 1.1. Basic Terminology. An axiom is a statement that is given to be true. A rule of inference is a logical rule that is used to deduce

Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

Fibonacci Numbers and Greatest Common Divisors. The Finonacci numbers are the numbers in the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,...

Fibonacci Numbers and Greatest Common Divisors The Finonacci numbers are the numbers in the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,.... After starting with two 1s, we get each Fibonacci number

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

Chapter 3. Cartesian Products and Relations. 3.1 Cartesian Products

Chapter 3 Cartesian Products and Relations The material in this chapter is the first real encounter with abstraction. Relations are very general thing they are a special type of subset. After introducing

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

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

Notes on Determinant

ENGG2012B Advanced Engineering Mathematics Notes on Determinant Lecturer: Kenneth Shum Lecture 9-18/02/2013 The determinant of a system of linear equations determines whether the solution is unique, without

Section IV.1: Recursive Algorithms and Recursion Trees

Section IV.1: Recursive Algorithms and Recursion Trees Definition IV.1.1: A recursive algorithm is an algorithm that solves a problem by (1) reducing it to an instance of the same problem with smaller

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).

I. Pointwise convergence

MATH 40 - NOTES Sequences of functions Pointwise and Uniform Convergence Fall 2005 Previously, we have studied sequences of real numbers. Now we discuss the topic of sequences of real valued functions.

v w is orthogonal to both v and w. the three vectors v, w and v w form a right-handed set of vectors.

3. Cross product Definition 3.1. Let v and w be two vectors in R 3. The cross product of v and w, denoted v w, is the vector defined as follows: the length of v w is the area of the parallelogram with

MATH10212 Linear Algebra. Systems of Linear Equations. Definition. An n-dimensional vector is a row or a column of n numbers (or letters): a 1.

MATH10212 Linear Algebra Textbook: D. Poole, Linear Algebra: A Modern Introduction. Thompson, 2006. ISBN 0-534-40596-7. Systems of Linear Equations Definition. An n-dimensional vector is a row or a column

Lecture 1: Course overview, circuits, and formulas

Lecture 1: Course overview, circuits, and formulas Topics in Complexity Theory and Pseudorandomness (Spring 2013) Rutgers University Swastik Kopparty Scribes: John Kim, Ben Lund 1 Course Information Swastik

The thing that started it 8.6 THE BINOMIAL THEOREM

476 Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers (b) Based on your results for (a), guess the minimum number of moves required if you start with an arbitrary number of n disks.

Real Roots of Univariate Polynomials with Real Coefficients

Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials

8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express

Lecture 17 : Equivalence and Order Relations DRAFT

CS/Math 240: Introduction to Discrete Mathematics 3/31/2011 Lecture 17 : Equivalence and Order Relations Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT Last lecture we introduced the notion

Elements of probability theory

2 Elements of probability theory Probability theory provides mathematical models for random phenomena, that is, phenomena which under repeated observations yield di erent outcomes that cannot be predicted

Graph Theory Problems and Solutions

raph Theory Problems and Solutions Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles November, 005 Problems. Prove that the sum of the degrees of the vertices of any finite graph is

If n is odd, then 3n + 7 is even.

Proof: Proof: We suppose... that 3n + 7 is even. that 3n + 7 is even. Since n is odd, there exists an integer k so that n = 2k + 1. that 3n + 7 is even. Since n is odd, there exists an integer k so that

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

LEARNING OBJECTIVES FOR THIS CHAPTER

CHAPTER 2 American mathematician Paul Halmos (1916 2006), who in 1942 published the first modern linear algebra book. The title of Halmos s book was the same as the title of this chapter. Finite-Dimensional

INCIDENCE-BETWEENNESS GEOMETRY

INCIDENCE-BETWEENNESS GEOMETRY MATH 410, CSUSM. SPRING 2008. PROFESSOR AITKEN This document covers the geometry that can be developed with just the axioms related to incidence and betweenness. The full

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

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

a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4

Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2

CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2 Proofs Intuitively, the concept of proof should already be familiar We all like to assert things, and few of us

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 1. SYSTEMS OF EQUATIONS AND MATRICES 1.1. Representation of a linear system. The general system of m equations in n unknowns can be written a 11 x 1 + a 12 x 2 +

Student Outcomes. Lesson Notes. Classwork. Discussion (10 minutes)

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 8 Student Outcomes Students know the definition of a number raised to a negative exponent. Students simplify and write equivalent expressions that contain

RECURSIVE ENUMERATION OF PYTHAGOREAN TRIPLES

RECURSIVE ENUMERATION OF PYTHAGOREAN TRIPLES DARRYL MCCULLOUGH AND ELIZABETH WADE In [9], P. W. Wade and W. R. Wade (no relation to the second author gave a recursion formula that produces Pythagorean

Linear Algebra. A vector space (over R) is an ordered quadruple. such that V is a set; 0 V ; and the following eight axioms hold:

Linear Algebra A vector space (over R) is an ordered quadruple (V, 0, α, µ) such that V is a set; 0 V ; and the following eight axioms hold: α : V V V and µ : R V V ; (i) α(α(u, v), w) = α(u, α(v, w)),

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights

THE DIMENSION OF A VECTOR SPACE

THE DIMENSION OF A VECTOR SPACE KEITH CONRAD This handout is a supplementary discussion leading up to the definition of dimension and some of its basic properties. Let V be a vector space over a field

Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field

Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will

NIM with Cash. Abstract. loses. This game has been well studied. For example, it is known that for NIM(1, 2, 3; n)

NIM with Cash William Gasarch Univ. of MD at College Park John Purtilo Univ. of MD at College Park Abstract NIM(a 1,..., a k ; n) is a -player game where initially there are n stones on the board and the

Similarity and Diagonalization. Similar Matrices

MATH022 Linear Algebra Brief lecture notes 48 Similarity and Diagonalization Similar Matrices Let A and B be n n matrices. We say that A is similar to B if there is an invertible n n matrix P such that

Representation of functions as power series

Representation of functions as power series Dr. Philippe B. Laval Kennesaw State University November 9, 008 Abstract This document is a summary of the theory and techniques used to represent functions

Practice with Proofs

Practice with Proofs October 6, 2014 Recall the following Definition 0.1. A function f is increasing if for every x, y in the domain of f, x < y = f(x) < f(y) 1. Prove that h(x) = x 3 is increasing, using

The Determinant: a Means to Calculate Volume

The Determinant: a Means to Calculate Volume Bo Peng August 20, 2007 Abstract This paper gives a definition of the determinant and lists many of its well-known properties Volumes of parallelepipeds are

1. Give the 16 bit signed (twos complement) representation of the following decimal numbers, and convert to hexadecimal:

Exercises 1 - number representations Questions 1. Give the 16 bit signed (twos complement) representation of the following decimal numbers, and convert to hexadecimal: (a) 3012 (b) - 435 2. For each of

COMP 250 Fall 2012 lecture 2 binary representations Sept. 11, 2012

Binary numbers The reason humans represent numbers using decimal (the ten digits from 0,1,... 9) is that we have ten fingers. There is no other reason than that. There is nothing special otherwise about

Kevin James. MTHSC 412 Section 2.4 Prime Factors and Greatest Comm

MTHSC 412 Section 2.4 Prime Factors and Greatest Common Divisor Greatest Common Divisor Definition Suppose that a, b Z. Then we say that d Z is a greatest common divisor (gcd) of a and b if the following

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

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

1.2. Successive Differences

1. An Application of Inductive Reasoning: Number Patterns In the previous section we introduced inductive reasoning, and we showed how it can be applied in predicting what comes next in a list of numbers

God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886)

Chapter 2 Numbers God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886) God created the integers and the rest is the work

Page 331, 38.4 Suppose a is a positive integer and p is a prime. Prove that p a if and only if the prime factorization of a contains p.

Page 331, 38.2 Assignment #11 Solutions Factor the following positive integers into primes. a. 25 = 5 2. b. 4200 = 2 3 3 5 2 7. c. 10 10 = 2 10 5 10. d. 19 = 19. e. 1 = 1. Page 331, 38.4 Suppose a is a

10.2 Series and Convergence

10.2 Series and Convergence Write sums using sigma notation Find the partial sums of series and determine convergence or divergence of infinite series Find the N th partial sums of geometric series and

Lemma 5.2. Let S be a set. (1) Let f and g be two permutations of S. Then the composition of f and g is a permutation of S.

Definition 51 Let S be a set bijection f : S S 5 Permutation groups A permutation of S is simply a Lemma 52 Let S be a set (1) Let f and g be two permutations of S Then the composition of f and g is a

Math Common Core Sampler Test

High School Algebra Core Curriculum Math Test Math Common Core Sampler Test Our High School Algebra sampler covers the twenty most common questions that we see targeted for this level. For complete tests

WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT?

WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT? introduction Many students seem to have trouble with the notion of a mathematical proof. People that come to a course like Math 216, who certainly

1. Introduction circular deﬁnition Remark 1 inverse trigonometric functions

1. Introduction In Lesson 2 the six trigonometric functions were defined using angles determined by points on the unit circle. This is frequently referred to as the circular definition of the trigonometric

15. Symmetric polynomials

15. Symmetric polynomials 15.1 The theorem 15.2 First examples 15.3 A variant: discriminants 1. The theorem Let S n be the group of permutations of {1,, n}, also called the symmetric group on n things.

The Pointless Machine and Escape of the Clones

MATH 64091 Jenya Soprunova, KSU The Pointless Machine and Escape of the Clones The Pointless Machine that operates on ordered pairs of positive integers (a, b) has three modes: In Mode 1 the machine adds

Math Workshop October 2010 Fractions and Repeating Decimals

Math Workshop October 2010 Fractions and Repeating Decimals This evening we will investigate the patterns that arise when converting fractions to decimals. As an example of what we will be looking at,

IB Maths SL Sequence and Series Practice Problems Mr. W Name

IB Maths SL Sequence and Series Practice Problems Mr. W Name Remember to show all necessary reasoning! Separate paper is probably best. 3b 3d is optional! 1. In an arithmetic sequence, u 1 = and u 3 =

ON FIBONACCI NUMBERS WITH FEW PRIME DIVISORS

ON FIBONACCI NUMBERS WITH FEW PRIME DIVISORS YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK Abstract If n is a positive integer, write F n for the nth Fibonacci number, and ω(n) for the number

Sequences and Series

Sequences and Series Consider the following sum: 2 + 4 + 8 + 6 + + 2 i + The dots at the end indicate that the sum goes on forever. Does this make sense? Can we assign a numerical value to an infinite

5.1 Radical Notation and Rational Exponents

Section 5.1 Radical Notation and Rational Exponents 1 5.1 Radical Notation and Rational Exponents We now review how exponents can be used to describe not only powers (such as 5 2 and 2 3 ), but also roots

sin(x) < x sin(x) x < tan(x) sin(x) x cos(x) 1 < sin(x) sin(x) 1 < 1 cos(x) 1 cos(x) = 1 cos2 (x) 1 + cos(x) = sin2 (x) 1 < x 2

. Problem Show that using an ɛ δ proof. sin() lim = 0 Solution: One can see that the following inequalities are true for values close to zero, both positive and negative. This in turn implies that On the

THE BANACH CONTRACTION PRINCIPLE. Contents

THE BANACH CONTRACTION PRINCIPLE ALEX PONIECKI Abstract. This paper will study contractions of metric spaces. To do this, we will mainly use tools from topology. We will give some examples of contractions,