Problem Set 5 Solutions

Similar documents
Discrete Mathematics. Spring 2017

Mathematical Induction Assignments

MTH 3318 Solutions to Induction Problems Fall 2009

Math 324 Summer 2012 Elementary Number Theory Notes on Mathematical Induction

Homework #2 Solutions Due: September 5, for all n N n 3 = n2 (n + 1) 2 4

Climbing an Infinite Ladder

Discrete Math, Spring Solutions to Problems V

MATH 324 Summer 2011 Elementary Number Theory. Notes on Mathematical Induction. Recall the following axiom for the set of integers.

Climbing an Infinite Ladder

Sums of Squares. Bianca Homberg and Minna Liu

Induction. Induction. Induction. Induction. Induction. Induction 2/22/2018

Mathematical Induction

Midterm Preparation Problems

Mathematical Induction

Note that r = 0 gives the simple principle of induction. Also it can be shown that the principle of strong induction follows from simple induction.

Chapter 5.1: Induction

201-1A5-MT - Mathematics Summer 2015 HOMEWORK 2 Deadline : Sunday, August 30, 2015 at 12 :30.

MATH 215 Final. M4. For all a, b in Z, a b = b a.

Solutions to Assignment 1

MATH 271 Summer 2016 Practice problem solutions Week 1

CSC 344 Algorithms and Complexity. Proof by Mathematical Induction

Consider an infinite row of dominoes, labeled by 1, 2, 3,, where each domino is standing up. What should one do to knock over all dominoes?

Mathematical Induction

Chapter Summary. Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms

PRIMALITY TEST FOR FERMAT NUMBERS USING QUARTIC RECURRENCE EQUATION. Predrag Terzic Podgorica, Montenegro

Basic Proof Examples

THE ISLAMIC UNIVERSITY OF GAZA ENGINEERING FACULTY DEPARTMENT OF COMPUTER ENGINEERING DISCRETE MATHMATICS DISCUSSION ECOM Eng. Huda M.

Math 3000 Section 003 Intro to Abstract Math Homework 6

MATH CSE20 Homework 5 Due Monday November 4

Final Exam Review. 2. Let A = {, { }}. What is the cardinality of A? Is

1.1 Inductive Reasoning filled in.notebook August 20, 2015

Fall 2017 Test II review problems

PRINCIPLE OF MATHEMATICAL INDUCTION

Lesson 3.5 Exercises, pages

Mathematical Induction. Section 5.1

CSCE 222 Discrete Structures for Computing. Dr. Hyunyoung Lee

PRINCIPLE OF MATHEMATICAL INDUCTION

Mathematical Fundamentals

n(n 1) = (n + 1)n(n 1)/ ( 1) n(n 1)(n 2) = (n + 1)n(n 1)(n 2)/4

Sequences, their sums and Induction

(3,1) Methods of Proof

The following techniques for methods of proofs are discussed in our text: - Vacuous proof - Trivial proof

Solutions to Practice Final

Mathematical Induction Again

1 Examples of Weak Induction

2301 Assignment 1 Due Friday 19th March, 2 pm

Mathematical Induction Again

Exercises. Template for Proofs by Mathematical Induction

MATH 55 - HOMEWORK 6 SOLUTIONS. 1. Section = 1 = (n + 1) 3 = 2. + (n + 1) 3. + (n + 1) 3 = n2 (n + 1) 2.

Recitation 7: Existence Proofs and Mathematical Induction

Appendix G: Mathematical Induction

9/5/17. Fermat s last theorem. CS 220: Discrete Structures and their Applications. Proofs sections in zybooks. Proofs.

Math 210 Exam 2 - Practice Problems. 1. For each of the following, determine whether the statement is True or False.

Discrete Structures - CM0246 Mathematical Induction

8. Given a rational number r, prove that there exist coprime integers p and q, with q 0, so that r = p q. . For all n N, f n = an b n 2

21 Induction. Tom Lewis. Fall Term Tom Lewis () 21 Induction Fall Term / 14

Intro to Algebra Today. We will learn names for the properties of real numbers. Homework Next Week. Due Tuesday 45-47/ 15-20, 32-35, 40-41, *28,29,38

Proof Terminology. Technique #1: Direct Proof. Learning objectives. Proof Techniques (Rosen, Sections ) Direct Proof:

Generating Functions (Revised Edition)

C241 Homework Assignment 7

Chapter 5: Sequences, Mathematic Induction, and Recursion

Math 283 Spring 2013 Presentation Problems 4 Solutions

M381 Number Theory 2004 Page 1

Definition: A sequence is a function from a subset of the integers (usually either the set

CS1802 Week 11: Algorithms, Sums, Series, Induction

Solutions Quiz 9 Nov. 8, Prove: If a, b, m are integers such that 2a + 3b 12m + 1, then a 3m + 1 or b 2m + 1.

1.2 The Well-Ordering Principle

Legendre s Equation. PHYS Southern Illinois University. October 18, 2016

Induction and recursion. Chapter 5

We want to show P (n) is true for all integers

Introduction to Abstract Mathematics

MATH10040: Numbers and Functions Homework 1: Solutions

Infinite Continued Fractions

CISC-102 Fall 2017 Week 3. Principle of Mathematical Induction

CHAPTER 4 SOME METHODS OF PROOF

Induction and Recursion

Sum of Squares. Defining Functions. Closed-Form Expression for SQ(n)

Systems Simulation Chapter 7: Random-Number Generation

With Question/Answer Animations

A Level. A Level Mathematics. Proof by Contradiction (Answers) AQA, Edexcel, OCR. Name: Total Marks:

IB Mathematics HL Year 2 Unit 11: Completion of Algebra (Core Topic 1)

Review of Power Series

1. Given the public RSA encryption key (e, n) = (5, 35), find the corresponding decryption key (d, n).

Name CMSC203 Fall2008 Exam 2 Solution Key Show All Work!!! Page (16 points) Circle T if the corresponding statement is True or F if it is False.

Mathematics 228(Q1), Assignment 3 Solutions

1 Direct Proofs Technique Outlines Example Implication Proofs Technique Outlines Examples...

CS 220: Discrete Structures and their Applications. Mathematical Induction in zybooks

Solution Set 2. Problem 1. [a] + [b] = [a + b] = [b + a] = [b] + [a] ([a] + [b]) + [c] = [a + b] + [c] = [a + b + c] = [a] + [b + c] = [a] + ([b + c])

2k n. k=0. 3x 2 7 (mod 11) 5 4x 1 (mod 9) 2 r r +1 = r (2 r )

Outline. We will cover (over the next few weeks) Induction Strong Induction Constructive Induction Structural Induction

Use mathematical induction in Exercises 3 17 to prove summation formulae. Be sure to identify where you use the inductive hypothesis.

DR.RUPNATHJI( DR.RUPAK NATH )

Solutions for Homework Assignment 2

12x + 18y = 50. 2x + v = 12. (x, v) = (6 + k, 2k), k Z.

Notes on Continued Fractions for Math 4400

Math 11 Foundations: Unit 1 - Logic. Unit 1: Logic

INDUCTION AND RECURSION. Lecture 7 - Ch. 4

Mathematical Induction

Seunghee Ye Ma 8: Week 2 Oct 6

Introduction to Number Theory

Transcription:

Problem Set 5 Solutions Section 4.. Use mathematical induction to prove each of the following: a) For each natural number n with n, n > + n. Let P n) be the statement n > + n. The base case, P ), is true because 9 > +. We assume that P k) is true, which states that k > + k for some natural number k. We will show P k + ) is true. Now, k+ k > + k ) + k > + k + k+ and hence, k+ > + k+. That is, if P k) is true, then P k + ) is true. Therefore, for all natural numbers n, n > + n. b) For each natural number b with n 6, n > n + ). Let P n) be the statement n > n + ). The base case, P 6), is true because 6 64 > 49 6 + ). Assume P k) is true, which says that k > k + ). We will show that P k + ) is true as well. Consider Now, for k 6, k > k +, so we have k+ k k + ) k + 4k + k + 4k + > k + + 4k + k + 4k + 4 k + ) as desired. So, P k + ) is true wheneverp k) is true. Therefore, by mathematical induction, n > n + ) for all natural numbers n 6. c) For each natural number n with n, + n) n < n. Let P n) be the statement that + n) < n. The base case, P ), is true because + ) 4 ) 64 7 <. If P k) is true, then + ) k k < k. We will show P k + ) is true as well. Consider + ) k+ < k + ) k k k +

Notice that k+ < k for k, and so + ) k+ < + ) k+ < k + k + k 4. Thus, if P k) is true, then P k + ) is true. Therefore, for all natural numbers n, + n) n < n. a) Verify that 4) 4 and that ) ) 4 9 4 6. b) Verify that 4) 9 6 0. ) ) 4 9 4 4 4 4 4 ) 9 4 9 ) 9 ) ) 6 5 8 and that 4 ) ) ) 9 6 4 6 ) ) ) ) 4 9 6 5 ) ) 8 4 4 9 6 ) 9 ) ) 5 5 6 8 5 8 ) ) 6 5) ) ) 4 6 5 0 c) For n N with n, make a conjecture for the formula for the product ) ) ) ) 4 9 6 n n k ) k n + n d) Based on your work in Parts 4a) and 4b), state a proposition and then use the Extended Principle of Mathematical Induction to prove your proposition. Let P n) be the proposition that for natural numbers n, n i ) i First, for n, we have that P ) is true as i ) i n + n ) 4 4 +

Now, suppose for all j k that P k) is true, where k is some natural number greater than. We will show P k + ) is true, which states j k+ j j ) Consider the following: k+ j ) k j ) j k + k + ) ) k + ) ) ) k + k k + ) ) k + k + ) ) k k + ) k + k k + k k + ) k + k + ) So, P k + ) is true whenever P k) is true. Thus, for natural numbers n, n k ) k n + n 9. Can each natural number greater than or equal to 6 be written as the sum of at least two natural numbers, each of which is a or a 5? Justify your conclusion. For example, 6 + +, 9 + + 5, and 7 + 5 + 5 + 5. Let P n) be the statement that n can be written as the sum of at least two natural numbers, each of which is a or a 5. For base cases, we have P 6) is true because 6 + + P 7) is true because 7 + 5 P 8) is true because 8 + + + P 9) is true because 9 + + 5 P 0) is true because 0 5 + 5 Now let k N with k 0. Assume that P 6),..., P k) are true. Since k + k 4) + 5 and 6 k 4 < k, we see that P k 4) is true. So, k 4 is a sum of at least natural numbers, each of which is a or a 5. So, the above equation implies that k + is the sum of at least natural numbers, each of which is a or a 5, and so P k + ) is true. Therefore, for all natural numbers n 6, we can write n as the sum of at least two natural numbers, each of which is a or a 5.

0. Use mathematical induction to prove the following proposition: Let x be a real number with x > 0. Then for each natural number n with n, + x) n > + nx. Explain why the assumption x > 0 was used in the proof. Let x be a real number with x > 0. Let P n) be + x) n > + nx. First, P ) is true. For any choice of x > 0, we have + x) > + x x + x + > + x And, since x R +, this last line is always true. x > 0 Let k N with k and assume P k) is true. Then, + x) k > + kx Since x + 0, we see that + x) > 0 as well, so we can multiply both sides of the above inequality by + x). Then, + x) k+ > + kx) + x) + kx + x + kx > + k + )x Hence, if P k) is true, then P k + ) is true. So, for all n N with n, + x) n > + nx when x > 0. In the case where x 0, we have equality in the statement for any choice of n. So by restricting to x > 0, the statement will be true for any choice of x R + and any choice of n N where n.. Prove or disprove each of the following propositions: a) For each n N, + + + nn + ) n n + This is a true statement. We will prove this with mathematical induction. Let n N. Let P n) be the statement P ) is true because +. Assume P k) is true, which says that + + + nn + ) + + + kk + ) n n + k k +

for some k N. We will show that P k + ) is true as well. Consider + + + kk + ) + k + )k + ) [ + ] + + + kk + ) k + )k + ) k k + + k + )k + ) kk + ) + k + )k + ) k + k + k + )k + ) k + ) k + )k + ) k + k + as desired. This shows that P k + ) is true whenever P k) is true. So, by mathematical induction, + + + nn + ) n n + for all natural numbers n. b) For each natural number n with n, 4 + 4 5 +... + nn + ) n n + This is a true statement. We will prove this with mathematical induction. Let n N, where n, and let P n) be that Now, P ) is true because that for some k N with k, 4 + 4 5 +... + nn + ) n n + 4 )+ 4 + 4 5 +... + kk + ) k k + We will show P k + ) is true as well. Consider. Next, assume P k) is true, which is to say 4 + 4 5 +... + kk + ) + k + )k + ) [ 4 + ] 4 5 +... + + kk + ) k + )k + ) k k + + k + )k + )

k )k + ) + k + )k + ) k k + )k + ) k k + 6 as desired. This shows that P k + ) is true whenever P k) is true. So, by mathematical induction, 4 + 4 5 +... + nn + ) n n + for all natural numbers n. c) For each n N, + + 4 + + nn + ) nn+)n+). This is a true statement. We will prove this with mathematical induction. Let P n) be that for natural numbers n, + + 4+ +nn+) nn+)n+). P ) is true because +)+). Assume P k) is true, which says that + + 4 + + kk + ) kk + )k + ) for some natural number n. We will show that P k + ) is true, which says that Now, + + 4 + + k + )k + ) k + )k + )k + ) + + 4 + + k + )k + ) [ + + 4 + kk + )] + k + )k + ) kk + )k + ) + k + )k + ) kk + )k + ) + k + )k + ) k + )k + )k + ) as desired. This shows that P k + ) is true whenever P k) is true. So, by mathematical induction, nn + )n + ) + + 4 + + nn + ) for all n N.

Section 4.. Assume that f, f,..., f n,... are the Fibonacci numbers. Prove each of the following: a) For each n N, f 4n is a multiple of. Let P n) be f 4n is a multiple of. Since f 4, P ) is true. Now, if P k) is true, then there exists m N such that f 4k m. Then, f 4k+) f 4k+4 f 4k+ + f 4k+ f 4k+ + f 4k+ + f 4k+ + f 4k f 4k+ + f 4k + f 4k+ + f 4k f 4k+ + f 4k f 4k+ + m) f 4k+ + m) This proves that if P k) is true, then P k + ) is true. So, for all n N, divides f 4n. b) For each n N, f 5n is a multiple of 5. Let P n) be f 5n is divisible by 5 where n is any natural number. Since f 5 5, P ) is true. Now, if P k) is true, then there exists an integer m such that f 5k 5m. Then, f 5k+) 5 5k+5 f 5k+4 + f 5k+ f 5k+ + f 5k+ + f 5k+ + f 5k+ f 5k+ + f 5k+ + f 5k+ f 5k+ + f 5k+ + f 5k+ + f 5k ) + f 5k+ f 5k+ + 4f 5k+ + f 5k 5f 5k+ + f 5k 5f 5k+ + 5m) 5f 5k+ + m) This proves that P k + ) is true whenever P k) is true. So, for all n N, 5 divides f 5n. c) OMIT d) For each n N, f + f +... + f n f n. Let n N and let P n) be the statement f + f +... + f n f n. Since f f, P ) is true. For k N, if P k) is true, then f + f +... + f k f k. Then f + f +... + f k ) + f k+) f + f +... + f k ) + f k+ f k + f k+ f k+ f k+) This proves that if P k) is true, then P k + ) is true. So, for all n N, f + f +... + f n f n.

e) For each n N, f + f 4 +... f n f n+. Let n N and let P n) be f + f 4 +... f n f n+. P ) is true because f f. For k N, if P k) is true, then f + f 4 +... f k f k+. Then, f + f 4 +... + f k ) + f k+) f + f 4 +... + f k ) + f k+ f k+ + f k+ f k+ f k+)+ This shows that P k +) is true whenever P k) is true. So, for n N, f +f 4 +... f n f n+. f) For each n N, f + f +... + f n f n f n+. For n N, let P n) be f + f +... + f n f n f n+. P ) is true because f f f. For k N, if P k) is true, then f + f +... + f k f kf k+. Then, f + f +... + f k ) + f k+ f kf k+ + f k+ f k+ f k + f k+ ) f k+ f k+ This proves that if P k) is true, then P k + ) is true as well. So, for n N, f + f +... + f n f n f n+. g) For each n N such that n 0 mod ), f n is an odd integer. Let P n) be f n+ and f n+ are both odd. Now, P ) is true because f 4 and f 5 5. For k N, if P k) is true, then f k+ and f k+ are both odd. Consider f k+)+ f k+4 f k+ + f k+ Now, by a proposition from the text, Proposition 4.), f k+ is even. And, f k+ is odd by our inductive hypothesis. Therefore, f k+)+ is odd. Also, f k+)+ f k+5 f k+4 + f k+ Again, by proposition 4., f k+ is even. And, we have just proven that f k+4 is odd. Therefore, f k+)+ is odd. This proves that if P k) is true, then P k + ) is true. So, by mathematical induction, if n is a natural number such that n 0 mod ), f n is an odd integer. 4. The quadratic formula can be used to show that α + 5 and β 5 are the two real number solutions to the quadratic equation x x 0. Notice that this implies α α+ and β β +. It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers

a) Verify that f α β α β and that f α β α β. f + 5 5 + 5 5 5 5 f + 5 ) ) 5 + 5 5 + 5+5 4 5+5 4 5 6+ 5 6+ 5 4 5 b) Work with the relation f f + f and substitute the expressions for f and f from part a). Rewrite the expression as a single fraction and then in the numerator use α + α αα + ) and a similar equation involving β. Now prove that f α β α β. 4 5 4 5 f f + f + α + α) β + β) αα + ) ββ + ) αα ) ββ) α β c) Use induction to prove that for each natural number n, if α + 5 and β 5, then f n αn β n α β. Let P n) be f n αn β n α β. We have previously shown that P ) and P ) are true. Now, f k+ f k + f k αk β k + αk β k

αk + α k β k + β k ) ak α + ) β k β + ) ak α ) β k β ) αk+ β k+ This shows that P k + ) is true whenever P k) is true. So, for n N, f n αn β n α β, where α + 5 and β 5. 9. For the sequence a, a,..., a n,..., assume that a and that for each n N, a n+ a n + 5. a) Calculate a through a 6. a a + 5 7 a a + 5 a 4 a + 5 7 a 5 a 4 + 5 a 6 a 5 + 5 7 b) Make a conjecture for a formula for a n for each n N. Conjecture: For n N, a n + 5n ). c) Prove your conjecture in Exercise 9b) is correct. Let P n) be a n + 5n ) for n N. P ) is true because a + 5 ). Now, if P k) is true then a k + 5k ). Consider a k+ a k + 5 + 5k ) + 5 + 5k This shows that P k+) is true whenever P k) is true. So, for all n N, a n +5n ).. For each sequence a, a,..., a n,..., assume that a, a, and that for each n N, a n+ a n+ a n. a) Calculate a through a 6. a a a ) ) 7 a 4 a a 7) ) 5 a 5 a 4 a 5) 7) a 6 a 5 a 4 ) 5) 6 b) Make a conjecture for a formula for a n for each n N. Conjecture: For all natural numbers n, a n n.

c) Prove your conjecture in Exercise b) is correct. Let P n) be a n n for n N. P ) is true because a. If P k) is true, then a k k for some natural number k. We want to show that a k+ k+. Consider a k+ a k a k k ) k ) k k + k k k k+ This proves that P k + ) is true whenever P k) is true. So, for all natural numbers n, a n n. 6. For each sequence a, a,..., a n,..., assume that a and that for each natural number n, a n+ a n + n n! a) Compute n! for the first 0 natural numbers.!!! 6 4! 4 5! 0 6! 70 7! 5040 8! 400 9 6880 0! 68800 b) Compute a n for the first 0 natural numbers. a a +! a +! 6 a 4 6 +! 4 a 5 4 + 4 4! 0 a 6 0 + 5 5! 70 a 7 70 + 6 6! 5040 a 8 5040 + 7 7! 400 a 9 400 + 8 8! 6880 a 0 6880 + 9 9! 68800

c) Make a conjecture about a formula for a n in terms of n that does not involve a summation or a recursion. Conjecture: for all natural numbers n, a n n!. d) Prove your conjecture in Part c). Let P n) be the statement that a n n! for any natural number n. P is true because a. Now, if P k) is true, then a k k! for some natural number k. Then, a k+ a k + k k! k! + k k! k! + k) k + )! as desired. So, P k + ) is true whenever P k) is true. Therefore, by mathematical induction, a n n! for all natural numbers n.

2024 © sciencedocbox.com Privacy Policy | Terms of Service | Feedback