Homework 7 Solutions, Math 55
|
|
- Randall Anderson
- 5 years ago
- Views:
Transcription
1 Homework 7 Solutions, Math (a) Since a is a positive integer, a = a 1 + b 0 is a positive integer of the form as + bt for some integers s and t, so a S. Thus S is nonempty. (b) Since S is nonempty, it has a least element c by the well-ordering principle. (c) Let d be a common divisor of a and b. Then a = dk and b = dl for some k, l Z. We know that c = ax + by for some x, y Z since c is an element of S. Since we see that d is also a divisor of c. c = ax + by = dkx + dly = d(kx + ly), (d) If c a, then by the division algorithm there exists a unique integer r such that a = qc+r and 0 < r < c. We know that c = ax + by since c is an element of S, so r = a qc = a q(ax + by) = a(1 qx) + b( qy) is an element of S also. But r < c, and c was supposed to be the smallest element of S. This is a contradiction. Thus c a. The proof that c b is identical. (e) The integer c must be the greatest common divisor of a and b. Indeed, it is a common divisor of a and b by part (d). Every other common divisor d is a divisor of c, so d c, which means that c is the greatest common divisor. If c were another common divisor that was at least as large as all other divisors, then by part (c) we would have c c, but c is maximal amongst common divisors so we must have c = c (c) (d) f(1) = f(0) f(0) = 9 6 = 1. f() = f(1) f(1) = 1 = 3. f(3) = f() f() = = 13. f(4) = f(3) f(3) = = 141. f() = f(1)/f(0) = 1. f(3) = f()/f(1) = 1. f(4) = f(3)/f() = 1. f(5) = f(4)/f(3) = 1. (a) This is well-defined, and f(n) = ( 1) n. We prove this by induction. Let P (n) be the statement that f(n) = ( 1) n. The base case P (0) is given to us. If P (k) is true for some k, then we know that f(k) = ( 1) k, which means that f(k + 1) = f(k) = ( 1) k = ( 1) k+1, so P (k + 1) is true too. Thus P (n) is true for all n. 1
2 (b) This is well-defined, and n/3 if n 0 mod 3 f(n) = 0 if n 1 mod 3 (n+1)/3 if n mod 3. We prove this by strong induction. The base cases n = 0, 1, are given to us. Suppose that we know this formula for all nonnegative integers up to some integer k. We have three cases. If k mod 3, then k 0 mod 3 also, so f(k + 1) = f(k ) = (k )/3 = (k+1)/3. If k mod 3, then k 0 mod 3 also, so f(k + 1) = f(k ) = 0. If k + 1 mod 3, then k mod 3 also, so This completes the induction. (c) This is not well-defiend. f(k + 1) = f(k ) = (k 1)/3 = (k+)/3. (d) This is well-defined, and we have f(n) = n 1 for all n 1. We prove this by induction. The base case n = 1 is given to us, and if f(k) = k 1 for some k 1, then completing the induction. f(k + 1) = f(k) = k 1 = k (e) This is well-defined, and we have f(n) = n/ +1. We prove this by strong induction. The base case is given to us, so suppose that, for some k, we have f(j) = j/ +1 for all 0 j k. If k + 1 is even, then k 1 = (k + 1) is even also, and which means that k = k 1 + = k + 1, f(k + 1) = f(k 1) = (k 1)/ +1 = (k+1)/ = (k+1)/ +1. If, on the other hand, k +1 is odd, then we need to verify that f(k) = f(k 1) and that both of these quantities are equal to (k+1)/ +1. Since k+1 is odd, we see that k is even and (k + 1)/ = k/. Moreover k 1 is odd and (k 1)/ = (k )/ = (k/) 1, so f(k) = k/ +1 = k/ = (k/) 1 = (k 1)/ +1 = f(k 1). Moreover, completing the induction. (a) a 1 = and a n+1 = a n + 4 for all n 1. (b) a 1 = and a n+1 = a n. (c) a 1 = and a n+1 = a n + (n + 1). f(k + 1) = f(k) = k/ +1 = (k+1)/ +1
3 (d) a 1 = 1 and a n+1 = a n + n We prove that f f n = f n f n+1 for all n 1 by induction. When n = 1, this just says that f 1 = 1 is equal to f 1 f = 1, which is true. Suppose that f 1 + +f k = f kf k+1 for some k. Then f f k + f k+1 = f k f k+1 + f k+1 = f k+1 (f k + f k+1 ) = f k+1 f k+ and this completes the induction Note that f f 0 f 1 = = 1 = ( 1) 1, so that kicks off the induction. Suppose f k+1 f k f k = ( 1)k for some positive integer k. Then and that completes the induction. f k+ f k f k+1 = (f k + f k+1 f k f k+1 = f k + f k f k+1 f k+1 = f k + f k+1 (f k f k+1 ) = f k + f k+1 ( f k 1 ) = (f k 1 f k+1 f k ) = ( 1) k = ( 1) k Note that f 0 f 1 + f = = 0 and f = f 1 1 = 0 also, so this kicks off the induction. If f 0 + f k = f k 1 1 for some positive integer k, then f 0 + f k f k+1 + f k+ = (f k 1 1) f k+1 + f k+ = (f k 1 1) f k+1 + (f k + f k+1 ) and that completes the induction. = (f k 1 1) + f k = f k We define and then max{a, b} := { a if a b b if b > a. max{a 1,..., a n } := max{a 1, max{a,..., a n }} for all n. Similar definitions work for min also First, we show that Z + S by usual induction. Let P (n) be the statement that n S. Then P (1) is the statement that 1 S, which we know is true. If P (k) is true for some k, then we know that k S. We know that 1 S, so then k + 1 S also, proving that P (k + 1) is also true. Thus P (n) is true for all n, which means that Z + S. Next, we show that S Z + by structural induction. Notice that 1 Z +. Moreover, if s, t Z +, then s + t Z + also clearly. This shows that S Z +, which proves that S = Z (a) 1 S and if s S, then s + S. (b) 3 S and if s S, then 3s S. 3
4 (c) 1, x S, and if p, q S then p + q S, p q S and pq S (a) First step: (, 3), (3, ). Second step: (4, 6), (5, 5), (6, 4). Third step: (6, 9), (7, 8), (8, 7), (9, 6). Fourth step: (8, 1), (9, 11), (10, 10), (11, 9), (1, 8). (b) Let P (n) be the statement that, if (a, b) S is obtained from (0, 0) after n applications of the recursive definition, then 5 a+b. Then P (0) is clear, so suppose P (0),..., P (k) are all true. If (a, b) S is obtained using k +1 applications of the recursive definition, then either (a, b) = (a +, b + 3) for some (a, b ) S that is obtained using k applications, or (a, b) = (a + 3, b + ) for some (a, b ) S that is obtained using k applications. In either case, a + b = (a + b ) + 5 so a + b is divisible by 5 because a + b is. This completes the induction. (c) Clearly 5 (0 + 0). Inductively, if 5 (a + b), then 5 ((a + ) + (b + 3)) and also 5 ((a + 3) + (b + )) The reversal of the empty string is the empty string. If w is a string of length n+1, then we can write w as xy where x has length n and y is just a single character. Then we define w R := yx R Notice that (1+1)+1 = 5, so we have a m,n = (m+n)+1 for (m, n) = 1. Suppose that for some (m, n) (1, 1) we have a m,n = (m + n ) + 1 for all (m, n ) < (m, n), where means the lexicographic order on Z + Z +. If n = 1, then (m 1, 1) < (m, 1), so a m,n = a m 1,n + = (((m 1) + n) + 1) + = (m + n) +. If n > 1, then (m, n 1) < (m, n) also, so again (a) (b) a m,n = a m,n 1 + = ((m + (n 1)) + 1) + = (m + n) ( 6 ) n 1 = ( 7 1) 1 = 7. n= If n = 1, there is only one such string. If n, we get to choose all but the first and last bits freely, so we have n options There are 6 4 total strings, and 5 4 strings that do not have x in them, so the number of strings that do have an x in them is (b) 999/7 of them are divisible by 7. Of those, the ones that are also divisible by 11 are the ones that are multiples of 77, and there are 999/77 of those. So the number that are divisible by 7 but not 11 is 999/7 999/77. 4
5 (f) /7 999/ /77. (g) There are 9 ways of choosing a 1-digit positive integer, 9 9 ways of choosing a -digit positive integer with distinct digits, and ways of choosing a 3-digit positive integer with distinct digits. So we just add these numbers up. (h) There are 4 ways of choosing a 1-digit positive integer that is even. There are 5 5 ways of choosing a -digit integer where the first (tens) digit is odd and the second digit is even (the first can be 1,3,5,7,9 and the second can be 0,,4,6,8, and notice that the end result will necessarily have two distinct digits and will be even). Also, there are 4 4 ways of choosing a -digit integer where both digits are even and distinct (the first digit can be,4,6,8 and the second can 0,,4,6,8 except whatever the first digit was). Similarly, there are ways of choosing 3-digit numbers (a) There are 9999/9 = 1111 integers less than or equal to 9999 that are divisible by 9, and of those 999/9 = 111 of them are less than So there are 1000 such integers. (b) There are = 9000 total integers in this range, and half of them will be even, so (g) There are 9999/5 integers that are divisible by 5 less than or equal to 9999, and of them 999/5 are less than Also, a multiple of 5 is also a multiple of 7 if and only if its a multiple of 35, and there are 9999/35 integers less than or equal to 9999 that are divisible by 35, and of them 999/35 are less than So we get ( 9999/5 999/5 ) ( 9999/35 999/35 ). (h) An integer is divisible by 5 and 7 if and only if it is divisible by /35 999/35, like we saw above (d) (6 7). So we have (e) (d) There are 100 total subsets. Of them, 1 has no elements and 100 have 1 element. So have more than 1 element There are 6! ways of arranging the people around the table, but each arrangement is being considered equivalent to 6 arrangements, so we get (6!)/6 = 5! (a) There are = (10!)/4! ways of choosing some permutation of 6 people out of the 10. There are 9!/3! ways where the bride is excluded. So there are 10!/4! 9!/3! ways. (b) Using the same idea as above, 10!/4! 8!/!. (c) We can count the ways of choosing a permutation of 6 people of of the 9 non-groom people that include the bride using the idea above. We then repeat the idea for the 9 non-bride people. These two sets of permutations are disjoint, so we get a total of (9!/3! 8!/!) + (9!/3! 8!/!). 5
6 It s sufficient to calculate the number N of bit strings with 5 consecutive 0 s. Indeed, the number of bit strings with 5 consecutive 1 s is also N since the bijection which swaps 1 s for 0 s and conversely maps the set of bit strings with 5 consecutive 0 s to the set of bit strings with 5 consecutive 1 s. Notice that there are bit strings that have both 5 consecutive 0 s and 5 consecutive 1 s, so the total number we re looking for is N. The bit strings with 5 consecutive 0 s are all of one the following possible forms, where x stands for an arbitrary bit. (i) 00000xxxxx. (ii) xxxx. (iii) x100000xxx. (iv) xx100000xx. (v) xxx100000x. (vi) xxxx Notice that a string cannot simultaneously be of two of these types. There are 5 strings of type (i), and 4 strings of types (ii)-(vi). So N = There are students in the class. 6
MATH 55 - HOMEWORK 6 SOLUTIONS. 1. Section = 1 = (n + 1) 3 = 2. + (n + 1) 3. + (n + 1) 3 = n2 (n + 1) 2.
MATH 55 - HOMEWORK 6 SOLUTIONS Exercise Section 5 Proof (a) P () is the statement ( ) 3 (b) P () is true since ( ) 3 (c) The inductive hypothesis is P (n): ( ) n(n + ) 3 + 3 + + n 3 (d) Assuming the inductive
More informationPRACTICE PROBLEMS: SET 1
PRACTICE PROBLEMS: SET MATH 437/537: PROF. DRAGOS GHIOCA. Problems Problem. Let a, b N. Show that if gcd(a, b) = lcm[a, b], then a = b. Problem. Let n, k N with n. Prove that (n ) (n k ) if and only if
More informationMath 324 Summer 2012 Elementary Number Theory Notes on Mathematical Induction
Math 4 Summer 01 Elementary Number Theory Notes on Mathematical Induction Principle of Mathematical Induction Recall the following axiom for the set of integers. Well-Ordering Axiom for the Integers If
More informationHomework #2 Solutions Due: September 5, for all n N n 3 = n2 (n + 1) 2 4
Do the following exercises from the text: Chapter (Section 3):, 1, 17(a)-(b), 3 Prove that 1 3 + 3 + + n 3 n (n + 1) for all n N Proof The proof is by induction on n For n N, let S(n) be the statement
More informationMATH 220 (all sections) Homework #12 not to be turned in posted Friday, November 24, 2017
MATH 220 (all sections) Homework #12 not to be turned in posted Friday, November 24, 2017 Definition: A set A is finite if there exists a nonnegative integer c such that there exists a bijection from A
More informationFall 2017 Test II review problems
Fall 2017 Test II review problems Dr. Holmes October 18, 2017 This is a quite miscellaneous grab bag of relevant problems from old tests. Some are certainly repeated. 1. Give the complete addition and
More informationa + b = b + a and a b = b a. (a + b) + c = a + (b + c) and (a b) c = a (b c). a (b + c) = a b + a c and (a + b) c = a c + b c.
Properties of the Integers The set of all integers is the set and the subset of Z given by Z = {, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, }, N = {0, 1, 2, 3, 4, }, is the set of nonnegative integers (also called
More informationProperties of the Integers
Properties of the Integers The set of all integers is the set and the subset of Z given by Z = {, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, }, N = {0, 1, 2, 3, 4, }, is the set of nonnegative integers (also called
More information0 Sets and Induction. Sets
0 Sets and Induction Sets A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set
More informationMATH 324 Summer 2011 Elementary Number Theory. Notes on Mathematical Induction. Recall the following axiom for the set of integers.
MATH 4 Summer 011 Elementary Number Theory Notes on Mathematical Induction Principle of Mathematical Induction Recall the following axiom for the set of integers. Well-Ordering Axiom for the Integers If
More informationHomework 1 (revised) Solutions
Homework 1 (revised) Solutions 1. Textbook, 1.1.1, # 1.1.2 (p. 24) Let S be an ordered set. Let A be a non-empty finite subset. Then A is bounded and sup A, inf A A Solution. The hint was: Use induction,
More informationQuantitative Aptitude
WWW.UPSCMANTRA.COM Quantitative Aptitude Concept 1 1. Number System 2. HCF and LCM 2011 Prelims Paper II NUMBER SYSTEM 2 NUMBER SYSTEM In Hindu Arabic System, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7,
More informationQ 1 Find the square root of 729. 6. Squares and Square Roots Q 2 Fill in the blank using the given pattern. 7 2 = 49 67 2 = 4489 667 2 = 444889 6667 2 = Q 3 Without adding find the sum of 1 + 3 + 5 + 7
More informationProofs. Chapter 2 P P Q Q
Chapter Proofs In this chapter we develop three methods for proving a statement. To start let s suppose the statement is of the form P Q or if P, then Q. Direct: This method typically starts with P. Then,
More information1. Consider the conditional E = p q r. Use de Morgan s laws to write simplified versions of the following : The negation of E : 5 points
Introduction to Discrete Mathematics 3450:208 Test 1 1. Consider the conditional E = p q r. Use de Morgan s laws to write simplified versions of the following : The negation of E : The inverse of E : The
More informationSolutions to Homework Set 1
Solutions to Homework Set 1 1. Prove that not-q not-p implies P Q. In class we proved that A B implies not-b not-a Replacing the statement A by the statement not-q and the statement B by the statement
More informationChapter 2. Divisibility. 2.1 Common Divisors
Chapter 2 Divisibility 2.1 Common Divisors Definition 2.1.1. Let a and b be integers. A common divisor of a and b is any integer that divides both a and b. Suppose that a and b are not both zero. By Proposition
More information(1) Which of the following are propositions? If it is a proposition, determine its truth value: A propositional function, but not a proposition.
Math 231 Exam Practice Problem Solutions WARNING: This is not a sample test. Problems on the exams may or may not be similar to these problems. These problems are just intended to focus your study of the
More informationInduction and Recursion
. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Induction and Recursion
More informationName (please print) Mathematics Final Examination December 14, 2005 I. (4)
Mathematics 513-00 Final Examination December 14, 005 I Use a direct argument to prove the following implication: The product of two odd integers is odd Let m and n be two odd integers Since they are odd,
More informationMathathon Round 1 (2 points each)
Mathathon Round ( points each). A circle is inscribed inside a square such that the cube of the radius of the circle is numerically equal to the perimeter of the square. What is the area of the circle?
More informationBinomial Coefficient Identities/Complements
Binomial Coefficient Identities/Complements CSE21 Fall 2017, Day 4 Oct 6, 2017 https://sites.google.com/a/eng.ucsd.edu/cse21-fall-2017-miles-jones/ permutation P(n,r) = n(n-1) (n-2) (n-r+1) = Terminology
More informationMath 370 Homework 2, Fall 2009
Math 370 Homework 2, Fall 2009 (1a) Prove that every natural number N is congurent to the sum of its decimal digits mod 9. PROOF: Let the decimal representation of N be n d n d 1... n 1 n 0 so that N =
More informationRecursive Definitions
Recursive Definitions Example: Give a recursive definition of a n. a R and n N. Basis: n = 0, a 0 = 1. Recursion: a n+1 = a a n. Example: Give a recursive definition of n i=0 a i. Let S n = n i=0 a i,
More informationInduction and recursion. Chapter 5
Induction and recursion Chapter 5 Chapter Summary Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms Mathematical Induction Section 5.1
More informationProofs. Chapter 2 P P Q Q
Chapter Proofs In this chapter we develop three methods for proving a statement. To start let s suppose the statement is of the form P Q or if P, then Q. Direct: This method typically starts with P. Then,
More informationChapter Summary. Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms
1 Chapter Summary Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Recursive Algorithms 2 Section 5.1 3 Section Summary Mathematical Induction Examples of
More informationFour Basic Sets. Divisors
Four Basic Sets Z = the integers Q = the rationals R = the real numbers C = the complex numbers Divisors Definition. Suppose a 0 and b = ax, where a, b, and x are integers. Then we say a divides b (or
More informationLecture Notes 1 Basic Concepts of Mathematics MATH 352
Lecture Notes 1 Basic Concepts of Mathematics MATH 352 Ivan Avramidi New Mexico Institute of Mining and Technology Socorro, NM 87801 June 3, 2004 Author: Ivan Avramidi; File: absmath.tex; Date: June 11,
More informationCSE 20 DISCRETE MATH. Fall
CSE 20 DISCRETE MATH Fall 2017 http://cseweb.ucsd.edu/classes/fa17/cse20-ab/ Today's learning goals Determine whether a relation is an equivalence relation by determining whether it is Reflexive Symmetric
More informationCSC B36 Additional Notes sample induction and well-ordering proofs. c Nick Cheng
CSC B36 Additional Notes sample induction and well-ordering proofs c Nick Cheng Introduction We present examples of induction proofs here in hope that they can be used as models when you write your own
More informationDiscrete Structures, Final Exam
Discrete Structures, Final Exam Monday, May 11, 2009 SOLUTIONS 1. (40 pts) Short answer. Put your answer in the box. No partial credit. [ ] 0 1 (a) If A = and B = 1 0 [ ] 0 0 1. 0 1 1 [ 0 1 1 0 0 1 ],
More informationChapter 5.1: Induction
Chapter.1: Induction Monday, July 1 Fermat s Little Theorem Evaluate the following: 1. 1 (mod ) 1 ( ) 1 1 (mod ). (mod 7) ( ) 8 ) 1 8 1 (mod ). 77 (mod 19). 18 (mod 1) 77 ( 18 ) 1 1 (mod 19) 18 1 (mod
More informationIntermediate Math Circles February 14, 2018 Contest Prep: Number Theory
Intermediate Math Circles February 14, 2018 Contest Prep: Number Theory Part 1: Prime Factorization A prime number is an integer greater than 1 whose only positive divisors are 1 and itself. An integer
More informationSolutions to Homework Problems
Solutions to Homework Problems November 11, 2017 1 Problems II: Sets and Functions (Page 117-118) 11. Give a proof or a counterexample of the following statements: (vi) x R, y R, xy 0; (x) ( x R, y R,
More informationDefinition: Let S and T be sets. A binary relation on SxT is any subset of SxT. A binary relation on S is any subset of SxS.
4 Functions Before studying functions we will first quickly define a more general idea, namely the notion of a relation. A function turns out to be a special type of relation. Definition: Let S and T be
More informationMATH 215 Final. M4. For all a, b in Z, a b = b a.
MATH 215 Final We will assume the existence of a set Z, whose elements are called integers, along with a well-defined binary operation + on Z (called addition), a second well-defined binary operation on
More informationCSE 20 DISCRETE MATH. Winter
CSE 20 DISCRETE MATH Winter 2017 http://cseweb.ucsd.edu/classes/wi17/cse20-ab/ Today's learning goals Determine whether a relation is an equivalence relation by determining whether it is Reflexive Symmetric
More informationExam Practice Problems
Math 231 Exam Practice Problems WARNING: This is not a sample test. Problems on the exams may or may not be similar to these problems. These problems are just intended to focus your study of the topics.
More informationWith Question/Answer Animations
Chapter 5 With Question/Answer Animations Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Chapter Summary
More informationMAT115A-21 COMPLETE LECTURE NOTES
MAT115A-21 COMPLETE LECTURE NOTES NATHANIEL GALLUP 1. Introduction Number theory begins as the study of the natural numbers the integers N = {1, 2, 3,...}, Z = { 3, 2, 1, 0, 1, 2, 3,...}, and sometimes
More informationMathematical Background
Chapter 1 Mathematical Background When we analyze various algorithms in terms of the time and the space it takes them to run, we often need to work with math. That is why we ask you to take MA 2250 Math
More informationLecture 4: Counting, Pigeonhole Principle, Permutations, Combinations Lecturer: Lale Özkahya
BBM 205 Discrete Mathematics Hacettepe University http://web.cs.hacettepe.edu.tr/ bbm205 Lecture 4: Counting, Pigeonhole Principle, Permutations, Combinations Lecturer: Lale Özkahya Resources: Kenneth
More informationDISCRETE MATH: FINAL REVIEW
DISCRETE MATH: FINAL REVIEW DR. DANIEL FREEMAN 1) a. Does 3 = {3}? b. Is 3 {3}? c. Is 3 {3}? c. Is {3} {3}? c. Is {3} {3}? d. Does {3} = {3, 3, 3, 3}? e. Is {x Z x > 0} {x R x > 0}? 1. Chapter 1 review
More information11 Division Mod n, Linear Integer Equations, Random Numbers, The Fundamental Theorem of Arithmetic
11 Division Mod n, Linear Integer Equations, Random Numbers, The Fundamental Theorem of Arithmetic Bezout s Lemma Let's look at the values of 4x + 6y when x and y are integers. If x is -6 and y is 4 we
More informationCOMBINATORIAL COUNTING
COMBINATORIAL COUNTING Our main reference is [1, Section 3] 1 Basic counting: functions and subsets Theorem 11 (Arbitrary mapping Let N be an n-element set (it may also be empty and let M be an m-element
More informationComplete Induction and the Well- Ordering Principle
Complete Induction and the Well- Ordering Principle Complete Induction as a Rule of Inference In mathematical proofs, complete induction (PCI) is a rule of inference of the form P (a) P (a + 1) P (b) k
More informationPolynomials. In many problems, it is useful to write polynomials as products. For example, when solving equations: Example:
Polynomials Monomials: 10, 5x, 3x 2, x 3, 4x 2 y 6, or 5xyz 2. A monomial is a product of quantities some of which are unknown. Polynomials: 10 + 5x 3x 2 + x 3, or 4x 2 y 6 + 5xyz 2. A polynomial is a
More informationEDULABZ INTERNATIONAL NUMBER SYSTEM
NUMBER SYSTEM 1. Find the product of the place value of 8 and the face value of 7 in the number 7801. Ans. Place value of 8 in 7801 = 800, Face value of 7 in 7801 = 7 Required product = 800 7 = 00. How
More informationWe want to show P (n) is true for all integers
Generalized Induction Proof: Let P (n) be the proposition 1 + 2 + 2 2 + + 2 n = 2 n+1 1. We want to show P (n) is true for all integers n 0. Generalized Induction Example: Use generalized induction to
More informationMATH 114 Fall 2004 Solutions to practice problems for Final Exam
MATH 11 Fall 00 Solutions to practice problems for Final Exam Reminder: the final exam is on Monday, December 13 from 11am - 1am. Office hours: Thursday, December 9 from 1-5pm; Friday, December 10 from
More informationClimbing an Infinite Ladder
Section 5.1 Section Summary Mathematical Induction Examples of Proof by Mathematical Induction Mistaken Proofs by Mathematical Induction Guidelines for Proofs by Mathematical Induction Climbing an Infinite
More informationIn N we can do addition, but in order to do subtraction we need to extend N to the integers
Chapter The Real Numbers.. Some Preliminaries Discussion: The Irrationality of 2. We begin with the natural numbers N = {, 2, 3, }. In N we can do addition, but in order to do subtraction we need to extend
More information1 Predicates and Quantifiers
1 Predicates and Quantifiers We have seen how to represent properties of objects. For example, B(x) may represent that x is a student at Bryn Mawr College. Here B stands for is a student at Bryn Mawr College
More informationPUTNAM PROBLEMS SEQUENCES, SERIES AND RECURRENCES. Notes
PUTNAM PROBLEMS SEQUENCES, SERIES AND RECURRENCES Notes. x n+ = ax n has the general solution x n = x a n. 2. x n+ = x n + b has the general solution x n = x + (n )b. 3. x n+ = ax n + b (with a ) can be
More informationLecture Notes on DISCRETE MATHEMATICS. Eusebius Doedel
Lecture Notes on DISCRETE MATHEMATICS Eusebius Doedel c Eusebius J. Doedel, 009 Contents Logic. Introduction............................................................................... Basic logical
More informationMATH 580, midterm exam SOLUTIONS
MATH 580, midterm exam SOLUTIONS Solve the problems in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page or use the Extra space empty page
More informationALGEBRA. 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers
ALGEBRA CHRISTIAN REMLING 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers by Z = {..., 2, 1, 0, 1,...}. Given a, b Z, we write a b if b = ac for some
More informationMath Homework # 4
Math 446 - Homework # 4 1. Are the following statements true or false? (a) 3 5(mod 2) Solution: 3 5 = 2 = 2 ( 1) is divisible by 2. Hence 2 5(mod 2). (b) 11 5(mod 5) Solution: 11 ( 5) = 16 is NOT divisible
More informationHMMT February 2018 February 10, 2018
HMMT February 018 February 10, 018 Algebra and Number Theory 1. For some real number c, the graphs of the equation y = x 0 + x + 18 and the line y = x + c intersect at exactly one point. What is c? 18
More informationNOTES ON INTEGERS. 1. Integers
NOTES ON INTEGERS STEVEN DALE CUTKOSKY The integers 1. Integers Z = {, 3, 2, 1, 0, 1, 2, 3, } have addition and multiplication which satisfy familar rules. They are ordered (m < n if m is less than n).
More informationD-MATH Algebra I HS18 Prof. Rahul Pandharipande. Solution 1. Arithmetic, Zorn s Lemma.
D-MATH Algebra I HS18 Prof. Rahul Pandharipande Solution 1 Arithmetic, Zorn s Lemma. 1. (a) Using the Euclidean division, determine gcd(160, 399). (b) Find m 0, n 0 Z such that gcd(160, 399) = 160m 0 +
More informationWORKSHEET MATH 215, FALL 15, WHYTE. We begin our course with the natural numbers:
WORKSHEET MATH 215, FALL 15, WHYTE We begin our course with the natural numbers: N = {1, 2, 3,...} which are a subset of the integers: Z = {..., 2, 1, 0, 1, 2, 3,... } We will assume familiarity with their
More informationBasic Proof Examples
Basic Proof Examples Lisa Oberbroeckling Loyola University Maryland Fall 2015 Note. In this document, we use the symbol as the negation symbol. Thus p means not p. There are four basic proof techniques
More informationDiscrete Mathematics & Mathematical Reasoning Chapter 6: Counting
Discrete Mathematics & Mathematical Reasoning Chapter 6: Counting Kousha Etessami U. of Edinburgh, UK Kousha Etessami (U. of Edinburgh, UK) Discrete Mathematics (Chapter 6) 1 / 39 Chapter Summary The Basics
More information1 Total Gadha s Complete Book of NUMBER SYSTEM TYPES OF NUMBERS Natural Numbers The group of numbers starting from 1 and including 1,,, 4, 5, and so on, are known as natural numbers. Zero, negative numbers,
More informationRecitation 7: Existence Proofs and Mathematical Induction
Math 299 Recitation 7: Existence Proofs and Mathematical Induction Existence proofs: To prove a statement of the form x S, P (x), we give either a constructive or a non-contructive proof. In a constructive
More informationChapter 5: The Integers
c Dr Oksana Shatalov, Fall 2014 1 Chapter 5: The Integers 5.1: Axioms and Basic Properties Operations on the set of integers, Z: addition and multiplication with the following properties: A1. Addition
More informationPreparing for the CS 173 (A) Fall 2018 Midterm 1
Preparing for the CS 173 (A) Fall 2018 Midterm 1 1 Basic information Midterm 1 is scheduled from 7:15-8:30 PM. We recommend you arrive early so that you can start exactly at 7:15. Exams will be collected
More informationMaths Book Part 1. By Abhishek Jain
Maths Book Part 1 By Abhishek Jain Topics: 1. Number System 2. HCF and LCM 3. Ratio & proportion 4. Average 5. Percentage 6. Profit & loss 7. Time, Speed & Distance 8. Time & Work Number System Understanding
More informationHomework 3 Solutions, Math 55
Homework 3 Solutions, Math 55 1.8.4. There are three cases: that a is minimal, that b is minimal, and that c is minimal. If a is minimal, then a b and a c, so a min{b, c}, so then Also a b, so min{a, b}
More informationChapter Generating Functions
Chapter 8.1.1-8.1.2. Generating Functions Prof. Tesler Math 184A Fall 2017 Prof. Tesler Ch. 8. Generating Functions Math 184A / Fall 2017 1 / 63 Ordinary Generating Functions (OGF) Let a n (n = 0, 1,...)
More information1.4 Equivalence Relations and Partitions
24 CHAPTER 1. REVIEW 1.4 Equivalence Relations and Partitions 1.4.1 Equivalence Relations Definition 1.4.1 (Relation) A binary relation or a relation on a set S is a set R of ordered pairs. This is a very
More informationWeek Some Warm-up Questions
1 Some Warm-up Questions Week 1-2 Abstraction: The process going from specific cases to general problem. Proof: A sequence of arguments to show certain conclusion to be true. If... then... : The part after
More informationNumbers. 2.1 Integers. P(n) = n(n 4 5n 2 + 4) = n(n 2 1)(n 2 4) = (n 2)(n 1)n(n + 1)(n + 2); 120 =
2 Numbers 2.1 Integers You remember the definition of a prime number. On p. 7, we defined a prime number and formulated the Fundamental Theorem of Arithmetic. Numerous beautiful results can be presented
More informationFinal Exam Review. 2. Let A = {, { }}. What is the cardinality of A? Is
1. Describe the elements of the set (Z Q) R N. Is this set countable or uncountable? Solution: The set is equal to {(x, y) x Z, y N} = Z N. Since the Cartesian product of two denumerable sets is denumerable,
More informationHW2 Solutions Problem 1: 2.22 Find the sign and inverse of the permutation shown in the book (and below).
Teddy Einstein Math 430 HW Solutions Problem 1:. Find the sign and inverse of the permutation shown in the book (and below). Proof. Its disjoint cycle decomposition is: (19)(8)(37)(46) which immediately
More informationEGMO 2016, Day 1 Solutions. Problem 1. Let n be an odd positive integer, and let x1, x2,..., xn be non-negative real numbers.
EGMO 2016, Day 1 Solutions Problem 1. Let n be an odd positive integer, and let x1, x2,..., xn be non-negative real numbers. Show that min (x2i + x2i+1 ) max (2xj xj+1 ), j=1,...,n i=1,...,n where xn+1
More information1 Sequences and Summation
1 Sequences and Summation A sequence is a function whose domain is either all the integers between two given integers or all the integers greater than or equal to a given integer. For example, a m, a m+1,...,
More informationLecture 17: Trees and Merge Sort 10:00 AM, Oct 15, 2018
CS17 Integrated Introduction to Computer Science Klein Contents Lecture 17: Trees and Merge Sort 10:00 AM, Oct 15, 2018 1 Tree definitions 1 2 Analysis of mergesort using a binary tree 1 3 Analysis of
More informationMATH 4400 SOLUTIONS TO SOME EXERCISES. 1. Chapter 1
MATH 4400 SOLUTIONS TO SOME EXERCISES 1.1.3. If a b and b c show that a c. 1. Chapter 1 Solution: a b means that b = na and b c that c = mb. Substituting b = na gives c = (mn)a, that is, a c. 1.2.1. Find
More information. As the binomial coefficients are integers we have that. 2 n(n 1).
Math 580 Homework. 1. Divisibility. Definition 1. Let a, b be integers with a 0. Then b divides b iff there is an integer k such that b = ka. In the case we write a b. In this case we also say a is a factor
More informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of
More informationADVANCED CALCULUS - MTH433 LECTURE 4 - FINITE AND INFINITE SETS
ADVANCED CALCULUS - MTH433 LECTURE 4 - FINITE AND INFINITE SETS 1. Cardinal number of a set The cardinal number (or simply cardinal) of a set is a generalization of the concept of the number of elements
More informationLECTURE NOTES DISCRETE MATHEMATICS. Eusebius Doedel
LECTURE NOTES on DISCRETE MATHEMATICS Eusebius Doedel 1 LOGIC Introduction. First we introduce some basic concepts needed in our discussion of logic. These will be covered in more detail later. A set is
More informationChapter 1. Sets and Numbers
Chapter 1. Sets and Numbers 1. Sets A set is considered to be a collection of objects (elements). If A is a set and x is an element of the set A, we say x is a member of A or x belongs to A, and we write
More informationDiscrete Mathematics (CS503)
Discrete Mathematics (CS503) Module I Suggested Questions Day 1, 2 1. Translate the following statement into propositional logic using the propositions provided: You can upgrade your operating system only
More informationExecutive Assessment. Executive Assessment Math Review. Section 1.0, Arithmetic, includes the following topics:
Executive Assessment Math Review Although the following provides a review of some of the mathematical concepts of arithmetic and algebra, it is not intended to be a textbook. You should use this chapter
More information34 th United States of America Mathematical Olympiad
34 th United States of America Mathematical Olympiad 1. Determine all composite positive integers n for which it is possible to arrange all divisors of n that are greater than 1 in a circle so that no
More informationMATH HL OPTION - REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis
MATH HL OPTION - REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis PART B: GROUPS GROUPS 1. ab The binary operation a * b is defined by a * b = a+ b +. (a) Prove that * is associative.
More informationDepartment of Computer Science University at Albany, State University of New York Solutions to Sample Discrete Mathematics Examination II (Fall 2007)
Department of Computer Science University at Albany, State University of New York Solutions to Sample Discrete Mathematics Examination II (Fall 2007) Problem 1: Specify two different predicates P (x) and
More informationRecurrence Relations and Recursion: MATH 180
Recurrence Relations and Recursion: MATH 180 1: Recursively Defined Sequences Example 1: The sequence a 1,a 2,a 3,... can be defined recursively as follows: (1) For all integers k 2, a k = a k 1 + 1 (2)
More informationIn N we can do addition, but in order to do subtraction we need to extend N to the integers
Chapter 1 The Real Numbers 1.1. Some Preliminaries Discussion: The Irrationality of 2. We begin with the natural numbers N = {1, 2, 3, }. In N we can do addition, but in order to do subtraction we need
More informationSets. We discuss an informal (naive) set theory as needed in Computer Science. It was introduced by G. Cantor in the second half of the nineteenth
Sets We discuss an informal (naive) set theory as needed in Computer Science. It was introduced by G. Cantor in the second half of the nineteenth century. Most students have seen sets before. This is intended
More informationPUTNAM TRAINING NUMBER THEORY. Exercises 1. Show that the sum of two consecutive primes is never twice a prime.
PUTNAM TRAINING NUMBER THEORY (Last updated: December 11, 2017) Remark. This is a list of exercises on Number Theory. Miguel A. Lerma Exercises 1. Show that the sum of two consecutive primes is never twice
More informationMath 2070BC Term 2 Weeks 1 13 Lecture Notes
Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic
More informationCounting. Mukulika Ghosh. Fall Based on slides by Dr. Hyunyoung Lee
Counting Mukulika Ghosh Fall 2018 Based on slides by Dr. Hyunyoung Lee Counting Counting The art of counting is known as enumerative combinatorics. One tries to count the number of elements in a set (or,
More informationDefinitions, Theorems and Exercises. Abstract Algebra Math 332. Ethan D. Bloch
Definitions, Theorems and Exercises Abstract Algebra Math 332 Ethan D. Bloch December 26, 2013 ii Contents 1 Binary Operations 3 1.1 Binary Operations............................... 4 1.2 Isomorphic Binary
More informationPractice Exam 1 CIS/CSE 607, Spring 2009
Practice Exam 1 CIS/CSE 607, Spring 2009 Problem 1) Let R be a reflexive binary relation on a set A. Prove that R is transitive if, and only if, R = R R. Problem 2) Give an example of a transitive binary
More informationDefinition 6.1 (p.277) A positive integer n is prime when n > 1 and the only positive divisors are 1 and n. Alternatively
6 Prime Numbers Part VI of PJE 6.1 Fundamental Results Definition 6.1 (p.277) A positive integer n is prime when n > 1 and the only positive divisors are 1 and n. Alternatively D (p) = { p 1 1 p}. Otherwise
More information