# Sample Induction Proofs

Save this PDF as:

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

### Worksheet on induction Calculus I Fall 2006 First, let us explain the use of for summation. The notation

Worksheet on induction MA113 Calculus I Fall 2006 First, let us explain the use of for summation. The notation f(k) means to evaluate the function f(k) at k = 1, 2,..., n and add up the results. In other

### 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

### 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

### 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

### 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

### 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

### 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

### Notes: Chapter 2 Section 2.2: Proof by Induction

Notes: Chapter 2 Section 2.2: Proof by Induction Basic Induction. To prove: n, a W, n a, S n. (1) Prove the base case - S a. (2) Let k a and prove that S k S k+1 Example 1. n N, n i = n(n+1) 2. Example

### 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

### 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

### 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

### 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

### 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

### 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

### CMPS 102 Solutions to Homework 1

CMPS 0 Solutions to Homework Lindsay Brown, lbrown@soe.ucsc.edu September 9, 005 Problem..- p. 3 For inputs of size n insertion sort runs in 8n steps, while merge sort runs in 64n lg n steps. For which

### 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

### Applications of Methods of Proof

CHAPTER 4 Applications of Methods of Proof 1. Set Operations 1.1. Set Operations. The set-theoretic operations, intersection, union, and complementation, defined in Chapter 1.1 Introduction to Sets are

### Discrete Mathematics: Homework 6 Due:

Discrete Mathematics: Homework 6 Due: 2011.05.20 1. (3%) How many bit strings are there of length six or less? We use the sum rule, adding the number of bit strings of each length to 6. If we include the

### 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,

### 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

### Sets and Subsets. Countable and Uncountable

Sets and Subsets Countable and Uncountable Reading Appendix A Section A.6.8 Pages 788-792 BIG IDEAS Themes 1. There exist functions that cannot be computed in Java or any other computer language. 2. There

### 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

### Sets and Counting. Let A and B be two sets. It is easy to see (from the diagram above) that A B = A + B A B

Sets and Counting Let us remind that the integer part of a number is the greatest integer that is less or equal to. It is denoted by []. Eample [3.1] = 3, [.76] = but [ 3.1] = 4 and [.76] = 6 A B Let A

### 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

### 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

### 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

### Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

### 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

### 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

### Notes on counting finite sets

Notes on counting finite sets Murray Eisenberg February 26, 2009 Contents 0 Introduction 2 1 What is a finite set? 2 2 Counting unions and cartesian products 4 2.1 Sum rules......................................

### Math 313 Lecture #10 2.2: The Inverse of a Matrix

Math 1 Lecture #10 2.2: The Inverse of a Matrix Matrix algebra provides tools for creating many useful formulas just like real number algebra does. For example, a real number a is invertible if there is

### Homework MA 725 Spring, 2012 C. Huneke SELECTED ANSWERS

Homework MA 725 Spring, 2012 C. Huneke SELECTED ANSWERS 1.1.25 Prove that the Petersen graph has no cycle of length 7. Solution: There are 10 vertices in the Petersen graph G. Assume there is a cycle C

### MTH6109: Combinatorics. Professor R. A. Wilson

MTH6109: Combinatorics Professor R. A. Wilson Autumn Semester 2011 2 Lecture 1, 26/09/11 Combinatorics is the branch of mathematics that looks at combining objects according to certain rules (for example:

### 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

### 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

### 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

### 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.

### 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

### 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].

### Math 2602 Finite and Linear Math Fall 14. Homework 9: Core solutions

Math 2602 Finite and Linear Math Fall 14 Homework 9: Core solutions Section 8.2 on page 264 problems 13b, 27a-27b. Section 8.3 on page 275 problems 1b, 8, 10a-10b, 14. Section 8.4 on page 279 problems

Mathematics 0N1 Solutions 1 1. Write the following sets in list form. 1(i) The set of letters in the word banana. {a, b, n}. 1(ii) {x : x 2 + 3x 10 = 0}. 3(iv) C A. True 3(v) B = {e, e, f, c}. True 3(vi)

### 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

### Foundations of Computing Discrete Mathematics Solutions to exercises for week 2

Foundations of Computing Discrete Mathematics Solutions to exercises for week 2 Agata Murawska (agmu@itu.dk) September 16, 2013 Note. The solutions presented here are usually one of many possiblities.

### Our goal first will be to define a product measure on A 1 A 2.

1. Tensor product of measures and Fubini theorem. Let (A j, Ω j, µ j ), j = 1, 2, be two measure spaces. Recall that the new σ -algebra A 1 A 2 with the unit element is the σ -algebra generated by the

### 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

### ARITHMETICAL FUNCTIONS II: CONVOLUTION AND INVERSION

ARITHMETICAL FUNCTIONS II: CONVOLUTION AND INVERSION PETE L. CLARK 1. Sums over divisors, convolution and Möbius Inversion The proof of the multiplicativity of the functions σ k, easy though it was, actually

### 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

### POWER SETS AND RELATIONS

POWER SETS AND RELATIONS L. MARIZZA A. BAILEY 1. The Power Set Now that we have defined sets as best we can, we can consider a sets of sets. If we were to assume nothing, except the existence of the empty

### GROUPS SUBGROUPS. Definition 1: An operation on a set G is a function : G G G.

Definition 1: GROUPS 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 the identity, such that (i) the

### 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).

### It is not immediately obvious that this should even give an integer. Since 1 < 1 5

Math 163 - Introductory Seminar Lehigh University Spring 8 Notes on Fibonacci numbers, binomial coefficients and mathematical induction These are mostly notes from a previous class and thus include some

### Chapter I Logic and Proofs

MATH 1130 1 Discrete Structures Chapter I Logic and Proofs Propositions A proposition is a statement that is either true (T) or false (F), but or both. s Propositions: 1. I am a man.. I am taller than

### Announcements. CS243: Discrete Structures. More on Cryptography and Mathematical Induction. Agenda for Today. Cryptography

Announcements CS43: Discrete Structures More on Cryptography and Mathematical Induction Işıl Dillig Class canceled next Thursday I am out of town Homework 4 due Oct instead of next Thursday (Oct 18) Işıl

### 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.

### Induction Problems. Tom Davis November 7, 2005

Induction Problems Tom Davis tomrdavis@earthlin.net http://www.geometer.org/mathcircles November 7, 2005 All of the following problems should be proved by mathematical induction. The problems are not necessarily

### MATHS 315 Mathematical Logic

MATHS 315 Mathematical Logic Second Semester, 2006 Contents 2 Formal Statement Logic 1 2.1 Post production systems................................. 1 2.2 The system L.......................................

### Math 317 HW #5 Solutions

Math 317 HW #5 Solutions 1. Exercise 2.4.2. (a) Prove that the sequence defined by x 1 = 3 and converges. x n+1 = 1 4 x n Proof. I intend to use the Monotone Convergence Theorem, so my goal is to show

### 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

### 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

### 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

### 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)

### 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

### 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

### 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

### 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,

### Solutions to Exercises 8

Discrete Mathematics Lent 2009 MA210 Solutions to Exercises 8 (1) Suppose that G is a graph in which every vertex has degree at least k, where k 1, and in which every cycle contains at least 4 vertices.

### Induction. Margaret M. Fleck. 10 October These notes cover mathematical induction and recursive definition

Induction Margaret M. Fleck 10 October 011 These notes cover mathematical induction and recursive definition 1 Introduction to induction At the start of the term, we saw the following formula for computing

### 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

### 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

### Section 5: Summary. Section 6. General Fourier Series

Section 5: Summary Periodic functions, (so far only with period π, can be represented ug the the Fourier series. We can use symmetry properties of the function to spot that certain Fourier coefficients

### 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

### 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

### Section 3 Sequences and Limits, Continued.

Section 3 Sequences and Limits, Continued. Lemma 3.6 Let {a n } n N be a convergent sequence for which a n 0 for all n N and it α 0. Then there exists N N such that for all n N. α a n 3 α In particular

### Sets and Cardinality Notes for C. F. Miller

Sets and Cardinality Notes for 620-111 C. F. Miller Semester 1, 2000 Abstract These lecture notes were compiled in the Department of Mathematics and Statistics in the University of Melbourne for the use

### 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

### 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

### Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note 11

CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note Conditional Probability A pharmaceutical company is marketing a new test for a certain medical condition. According

### 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

### Section 3.1 Worksheet NAME. f(x + h) f(x)

MATH 1170 Section 3.1 Worksheet NAME Recall that we have efine the erivative of f to be f (x) = lim h 0 f(x + h) f(x) h Recall also that the erivative of a function, f (x), is the slope f s tangent line

### CS 4310 HOMEWORK SET 1

CS 4310 HOMEWORK SET 1 PAUL MILLER Section 2.1 Exercises Exercise 2.1-1. Using Figure 2.2 as a model, illustrate the operation of INSERTION-SORT on the array A = 31, 41, 59, 26, 41, 58. Solution: Assume

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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.

### 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

### Generating Functions

Generating Functions If you take f(x =/( x x 2 and expand it as a power series by long division, say, then you get f(x =/( x x 2 =+x+2x 2 +x +5x 4 +8x 5 +x 6 +. It certainly seems as though the coefficient

### 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

### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on