DOCSLIB.ORG

3 Mathematical Induction

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
3 Mathematical Induction 3 Mathematical Induction Reading: Metalogic Part II, 23,27 Contents 3.1 Mathematical Induction . 35 3.1.1 Peano's Axioms . 35 3.1.2 **Example: Parentheses . 39 3.1.3 The Principle of Induction on the Construction ofaW®.................................... 40 3.1.4 **The Induction Principle . 41 3.2 Unique Decomposition . 43 3.2.1 Unique Readability Theorem . 43 3.2.2 Finding the Main Connnective . 45 3.3 Some Theorems Involving Induction . 46 3.3.1 Unique Valuation . 46 3.3.2 The Interpolation Theorem . 46 3.3.3 Disjunctive Normal Form . 47 3.1 Mathematical Induction Our objective in this section is to familiarize the reader with the technique of Proof by Mathematical Induction. This proof technique is used throughout mathematical logic, and so we should learn it right away. 3.1.1 Peano's Axioms We think of the natural numbers as forming a sequence beginning with 0 followed by its successor, and then its successor, and so on. We begin with 0 and then generate the sequence by adding on n0 whenever we have generated n. n0 is said to be the successor of n, and it may be thought of as n + 1. We thus get the sequence 0; 00; 000; 0000; 00000;::: 36 3 Mathematical Induction This informal characterization of the set of natural numbers N, is formally characterized by Peano's Postulates1 P1 0 2 N. 0 P2 If n 2 N, then there is a unique n 2 N. 0 0 P3 For all m; n 2 N, if m = n , then m = n. 0 P4 For each n 2 N, n 6= 0. 0 P5 If M ⊆ N such that 0 2 M and m 2 M whenever m 2 M, then M = N. 0 P1 is straightforward. P2 tells us that is a function, so that if x = y, then 0 0 0 x = y .P3 tells us that the function is 1-1, i.e., that no number can be the successor of two distinct numbers. P4 tells us that 0 is not the successor of any number, so that, in e®ect, the Range of the function 0 is the set N ¡ f0g. And P5 is the basis for what is known as Weak Induction Principle 3.1.1 (The Principle of Mathematical Induction) Let P be a property of natural numbers. Suppose that: (1) 0 has the property P. (2) If any natural number n has the property P, then its successor n0 has the property P. Then, every natural number n has the property P. The Principle of Mathematical Induction can be drawn from P5 by letting M = fm 2 NjP (m)g. The following terminology is fairly standard in conjunction with proof by mathematical induction. The part of the proof which consists in establishing (1), i.e., that 0 has P, is called the Basis step. The part of the proof which consists in establishing (2), i.e., that n0 has the property P whenever n has the property P, is called the Induction Step. The induction step is carried out by assuming that n has P, and showing that on this assumption n0 has P. This assumption that n has P is called the Induction Hypothesis. We will use mathematical induction to prove the following elementary consequence of these postulates, namely, that every natural number other than 0 is the successor of a natural number2: Proposition 20 Every n 2 N is such that either n = 0 or else there exists an m 2 N and n = m0 Proof By Mathematical Induction. Basis Step. n = 0. The disjunction follows immediately by addition: either n = 0 or else there exists an m 2 N and n = m0. Induction Step. Assume the theorem holds for n 2 N, i.e., assume that either n = 0 or else there exists an m 2 N and n = m0. (This is the Induction 1 These are actually due to Dedekind 2 This establishes that the function 0 is not merely 1-1, but also onto N ¡ f0g 3.1 Mathematical Induction 37 Hypothesis.) We have to show that this also holds for n0. Clearly n0 2 N, 0 by P2. And, clearly n 6= 0, by P4. So, we have to show that there exists an m 2 N such that n0 = m0. But, of course, there is: let m = n. ¤ There are a number of variants of this principle that mathematicians rely on in proofs by induction. Kleene (1950) also acknowledges Course-of-Values Induction, on which the induction hypothesis is not simply that n has P, but that m has P, for all m · n. So, we have Principle 3.1.2 (The Principle of Course-of-Values Induction:) Let P be a property of natural numbers. Suppose that: (1) 0 has the property P. (2) If all natural numbers m · n have the property P, then n0 has the property P. Then, every natural number n has the property P. [Andrews, Mendelson (1987) identi¯es a Principle of Complete Induction,3 This is also known as Strong Induction: Principle 3.1.3 (The Principle of Complete Induction (Strong Induction):) Let P be a property of natural numbers. Suppose that: If all natural numbers m < n have the property P, then n has the property P. Then, every natural number n has the property P. These variations need not upset us, however, for they are provably equiv- alent. As such, we can choose whichever form makes the proof easiest. We will show that the Principle of Mathematical Induction is equivalent to the Principle of Complete Induction in two steps. Lemma 3.1.1 The Principle of Mathematical Induction is a consequence of The Principle of Complete Induction. Proof [Andrews] We want to show that if P (0) and if, for all n, if P (n) then P (n0), then for all n, P (n). So, (as with any conditional proof), we assume the antecedent and try to derive the consequent. So, we assume (*) P (0) and for all n, if P (n) then P (n0) We want to show that P (n) for all n. We do so using the Principle of Complete Induction, i.e., we take as our induction hypothesis that P (j) for all j < n, and try to derive P (n). The Principle of Complete Induction will then entitle 3 Hunter (1996) has a di®erent terminology. \In the Induction Step we prove that if the Theorem holds for all cases up to an arbitrarily given point, then it holds also for all cases at the next higher point. (This will be a strong induction. A weak induction is one in which the Induction Step shows that if the Theorem holds for all cases at an arbitrarily given point, then it holds also for all cases at the next higher point." By these lights, our Principle of Mathematical Induction would be a weak induction and both our Course-of-Values Induction and our Principle of Complete Induction would be strong inductions|assuming, of course, that Hunter would count the Induction Hypothesis in Complete Induction as the Induction Step even though there is no Basis Step. 38 3 Mathematical Induction us to conclude that P (n) for all n. Induction Step. Assume P (j) for all j < n. If n = 0, then by (*), P (n). Otherwise n > 0. So, n = m0. In that case, m < n, so, P (m) by the induction hypothesis. And by (*), for every number k, if P (k) then P (k0), so P (m0). So, for all n, P (n). ¤ Lemma 3.1.2 The Principle of Complete Induction is a consequence of The Principle of Mathematical Induction Proof [Andrews] We want to show that if, for every number n, P (n) whenever all number j < n are such that P (j), then P (n) for all n. So, we make the following assumption: (**) For all n, P (n) whenever all number j < n are such that P (j) We must show that P (n) for all n. We prove this using The Principle of Mathematical Induction. Basis Step. n = 0. Since there are no j < 0, then by (**) it is trivial that P (0). Induction Step. The Induction hypothesis is that P (n). We want to show P (n0). But, by (**), since P (n), then for all j < n, P (j). So, since P (n) and for all j < n, P (j), it follows that, for all k < n0, P (k). By (**), P (n0). So, for all n, P (n). QED From these two lemmas, we derive Lemma 3.1.3 The Principle of Mathematical Induction is equivalent to the Principle of Complete Induction Problems 3.1. The Principle of Mathematical Induction is a consequence of Course-of- Values Induction. 3.2. Course-of-Values Induction is a consequence of The Principle of Mathe- matical Induction. 3.3. The Principle of Complete Induction is a consequence of Course- of- Values Induction. 3.4. Course-of-Values Induction is a consequence of The Principle of Complete Induction. 3.5. Suppose that the Basis step were eliminated from Course-of Values In- duction to get: Let P be a property of natural numbers. Suppose that: If all natural numbers m · n have the property P, then n0 has the property P. Then, every natural number n has the property P. Show that this is equivalent to The Principle of Mathematical Induction. 3.1 Mathematical Induction 39 3.1.2 **Example: Parentheses We take a very liberal attitude about what is to count as a property of num- bers.
Load more

Recommended publications
  • Chapter 3 Induction and Recursion
    Chapter 3 Induction and Recursion 3.1 Induction: An informal introduction This section is intended as a somewhat informal introduction to The Principle of Mathematical Induction (PMI): a theorem that establishes the validity of the proof method which goes by the same name. There is a particular format for writing the proofs which makes it clear that PMI is being used. We will not explicitly use this format when introducing the method, but will do so for the large number of different examples given later. Suppose you are given a large supply of L-shaped tiles as shown on the left of the figure below. The question you are asked to answer is whether these tiles can be used to exactly cover the squares of an 2n × 2n punctured grid { a 2n × 2n grid that has had one square cut out { say the 8 × 8 example shown in the right of the figure. 1 2 CHAPTER 3. INDUCTION AND RECURSION In order for this to be possible at all, the number of squares in the punctured grid has to be a multiple of three. It is. The number of squares is 2n2n − 1 = 22n − 1 = 4n − 1 ≡ 1n − 1 ≡ 0 (mod 3): But that does not mean we can tile the punctured grid. In order to get some traction on what to do, let's try some small examples. The tiling is easy to find if n = 1 because 2 × 2 punctured grid is exactly covered by one tile. Let's try n = 2, so that our punctured grid is 4 × 4.
    [Show full text]
  • Mathematical Induction
    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 the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we can reach the next rung. From (1), we can reach the first rung. Then by applying (2), we can reach the second rung. Applying (2) again, the third rung. And so on. We can apply (2) any number of times to reach any particular rung, no matter how high up. This example motivates proof by mathematical induction. Principle of Mathematical Induction Principle of Mathematical Induction: To prove that P(n) is true for all positive integers n, we complete these steps: Basis Step: Show that P(1) is true. Inductive Step: Show that P(k) → P(k + 1) is true for all positive integers k. To complete the inductive step, assuming the inductive hypothesis that P(k) holds for an arbitrary integer k, show that must P(k + 1) be true. Climbing an Infinite Ladder Example: Basis Step: By (1), we can reach rung 1. Inductive Step: Assume the inductive hypothesis that we can reach rung k. Then by (2), we can reach rung k + 1. Hence, P(k) → P(k + 1) is true for all positive integers k. We can reach every rung on the ladder. Logic and Mathematical Induction • Mathematical induction can be expressed as the rule of inference (P(1) ∧ ∀k (P(k) → P(k + 1))) → ∀n P(n), where the domain is the set of positive integers.
    [Show full text]
  • Chapter 4 Induction and Recursion
    Chapter 4 Induction and Recursion 4.1 Induction: An informal introduction This section is intended as a somewhat informal introduction to The Principle of Mathematical Induction (PMI): a theorem that establishes the validity of the proof method which goes by the same name. There is a particular format for writing the proofs which makes it clear that PMI is being used. We will not explicitly use this format when introducing the method, but will do so for the large number of different examples given later. Suppose you are given a large supply of L-shaped tiles as shown on the left of the figure below. The question you are asked to answer is whether these tiles can be used to exactly cover the squares of an 2n × 2n punctured grid { a 2n × 2n grid that has had one square cut out { say the 8 × 8 example shown in the right of the figure. 1 2 CHAPTER 4. INDUCTION AND RECURSION In order for this to be possible at all, the number of squares in the punctured grid has to be a multiple of three. By direct calculation we can see that it is true when n = 1; 2 or 3, and these are the cases we're interested in here. It turns out to be true in general; this is easy to show using congruences, which we will study later, and also can be shown using methods in this chapter. But that does not mean we can tile the punctured grid. In order to get some traction on what to do, let's try some small examples.
    [Show full text]
  • Chapter 7. Induction and Recursion
    Chapter 7. Induction and Recursion Part 1. Mathematical Induction The principle of mathematical induction is this: to establish an infinite sequence of propositions P1,P2,P3,...,Pn,... (or, simply put, Pn (n 1)), it is enough to verify the following two things ≥ (1) P1, and (2) Pk Pk ; (that is, assuming Pk vaild, we can rpove the validity of Pk ). ⇒ + 1 + 1 These two things will give a “domino effect” for the validity of all Pn (n 1). Indeed, ≥ step (1) tells us that P1 is true. Applying (2) to the case n = 1, we know that P2 is true. Knowing that P2 is true and applying (2) to the case n = 2 we know that P3 is true. Continue in this manner, we see that P4 is true, P5 is true, etc. This argument can go on and on. Now you should be convinced that Pn is true, no matter how big is n. We give some examples to show how this induction principle works. Example 1. Use mathematical induction to show 1 + 3 + 5 + + (2n 1) = n2. ··· − (Remember: in mathematics, “show” means “prove”.) Answer: For n = 1, the identity becomes 1 = 12, which is obviously true. Now assume the validity of the identity for n = k: 1 + 3 + 5 + + (2k 1) = k2. ··· − For n = k + 1, the left hand side of the identity is 1 + 3 + 5 + + (2k 1) + (2(k + 1) 1) = k2 + (2k + 1) = (k + 1)2 ··· − − which is the right hand side for n = k + 1. The proof is complete. 1 Example 2. Use mathematical induction to show the following formula for a geometric series: 1 xn+ 1 1 + x + x2 + + xn = − .
    [Show full text]
  • 3 Mathematical Induction
    Hardegree, Metalogic, Mathematical Induction page 1 of 27 3 Mathematical Induction 1. Introduction.......................................................................................................................................2 2. Explicit Definitions ..........................................................................................................................2 3. Inductive Definitions ........................................................................................................................3 4. An Aside on Terminology.................................................................................................................3 5. Induction in Arithmetic .....................................................................................................................3 6. Inductive Proofs in Arithmetic..........................................................................................................4 7. Inductive Proofs in Meta-Logic ........................................................................................................5 8. Expanded Induction in Arithmetic.....................................................................................................7 9. Expanded Induction in Meta-Logic...................................................................................................8 10. How Induction Captures ‘Nothing Else’...........................................................................................9 11. Exercises For Chapter 3 .................................................................................................................11
    [Show full text]
  • Proofs and Mathematical Reasoning
    Proofs and Mathematical Reasoning University of Birmingham Author: Supervisors: Agata Stefanowicz Joe Kyle Michael Grove September 2014 c University of Birmingham 2014 Contents 1 Introduction 6 2 Mathematical language and symbols 6 2.1 Mathematics is a language . .6 2.2 Greek alphabet . .6 2.3 Symbols . .6 2.4 Words in mathematics . .7 3 What is a proof? 9 3.1 Writer versus reader . .9 3.2 Methods of proofs . .9 3.3 Implications and if and only if statements . 10 4 Direct proof 11 4.1 Description of method . 11 4.2 Hard parts? . 11 4.3 Examples . 11 4.4 Fallacious \proofs" . 15 4.5 Counterexamples . 16 5 Proof by cases 17 5.1 Method . 17 5.2 Hard parts? . 17 5.3 Examples of proof by cases . 17 6 Mathematical Induction 19 6.1 Method . 19 6.2 Versions of induction. 19 6.3 Hard parts? . 20 6.4 Examples of mathematical induction . 20 7 Contradiction 26 7.1 Method . 26 7.2 Hard parts? . 26 7.3 Examples of proof by contradiction . 26 8 Contrapositive 29 8.1 Method . 29 8.2 Hard parts? . 29 8.3 Examples . 29 9 Tips 31 9.1 What common mistakes do students make when trying to present the proofs? . 31 9.2 What are the reasons for mistakes? . 32 9.3 Advice to students for writing good proofs . 32 9.4 Friendly reminder . 32 c University of Birmingham 2014 10 Sets 34 10.1 Basics . 34 10.2 Subsets and power sets . 34 10.3 Cardinality and equality .
    [Show full text]
  • MATHEMATICAL INDUCTION SEQUENCES and SERIES
    MISS MATHEMATICAL INDUCTION SEQUENCES and SERIES John J O'Connor 2009/10 Contents This booklet contains eleven lectures on the topics: Mathematical Induction 2 Sequences 9 Series 13 Power Series 22 Taylor Series 24 Summary 29 Mathematician's pictures 30 Exercises on these topics are on the following pages: Mathematical Induction 8 Sequences 13 Series 21 Power Series 24 Taylor Series 28 Solutions to the exercises in this booklet are available at the Web-site: www-history.mcs.st-andrews.ac.uk/~john/MISS_solns/ 1 Mathematical induction This is a method of "pulling oneself up by one's bootstraps" and is regarded with suspicion by non-mathematicians. Example Suppose we want to sum an Arithmetic Progression: 1+ 2 + 3 +...+ n = 1 n(n +1). 2 Engineers' induction € Check it for (say) the first few values and then for one larger value — if it works for those it's bound to be OK. Mathematicians are scornful of an argument like this — though notice that if it fails for some value there is no point in going any further. Doing it more carefully: We define a sequence of "propositions" P(1), P(2), ... where P(n) is "1+ 2 + 3 +...+ n = 1 n(n +1)" 2 First we'll prove P(1); this is called "anchoring the induction". Then we will prove that if P(k) is true for some value of k, then so is P(k + 1) ; this is€ c alled "the inductive step". Proof of the method If P(1) is OK, then we can use this to deduce that P(2) is true and then use this to show that P(3) is true and so on.
    [Show full text]
  • Logicism, Interpretability, and Knowledge of Arithmetic
    ZU064-05-FPR RSL-logicism-final 1 December 2013 21:42 The Review of Symbolic Logic Volume 0, Number 0, Month 2013 Logicism, Interpretability, and Knowledge of Arithmetic Sean Walsh Department of Logic and Philosophy of Science, University of California, Irvine Abstract. A crucial part of the contemporary interest in logicism in the philosophy of mathematics resides in its idea that arithmetical knowledge may be based on logical knowledge. Here an implementation of this idea is considered that holds that knowledge of arithmetical principles may be based on two things: (i) knowledge of logical principles and (ii) knowledge that the arithmetical principles are representable in the logical principles. The notions of representation considered here are related to theory-based and structure- based notions of representation from contemporary mathematical logic. It is argued that the theory-based versions of such logicism are either too liberal (the plethora problem) or are committed to intuitively incorrect closure conditions (the consistency problem). Structure-based versions must on the other hand respond to a charge of begging the question (the circularity problem) or explain how one may have a knowledge of structure in advance of a knowledge of axioms (the signature problem). This discussion is significant because it gives us a better idea of what a notion of representation must look like if it is to aid in realizing some of the traditional epistemic aims of logicism in the philosophy of mathematics. Received February 2013 Acknowledgements: This is material coming out of my dissertation, and I would thus like to thank my advisors, Dr. Michael Detlefsen and Dr.
    [Show full text]
  • Why Proofs by Mathematical Induction Are Generally Not Explanatory MARC LANGE
    Why proofs by mathematical induction are generally not explanatory MARC LANGE Philosophers who regard some mathematical proofs as explaining why the- orems hold, and others as merely proving that they do hold, disagree sharply about the explanatory value of proofs by mathematical induction. I offer an argument that aims to resolve this conflict of intuitions without making any controversial presuppositions about what mathematical explanations would be. 1. The problem Proposed accounts of scientific explanation have long been tested against certain canonical examples. Various arguments are paradigmatically expla- natory, such as Newton’s explanations of Kepler’s laws and the tides, Darwin’s explanations of various biogeographical and anatomical facts, Wegener’s explanation of the correspondence between the South American and African coastlines, and Einstein’s explanation of the equality of inertial and gravitational mass. Any promising comprehensive theory of scientific explanation must deem these arguments to be genuinely explanatory. By the same token, there is widespread agreement that various other arguments are not explanatory – examples so standard that I need only give their famil- iar monikers: the flagpole, the eclipse, the barometer, the hexed salt and so forth. Although there are some controversial cases, of course (such as ‘expla- nations’ of the dormitive-virtue variety), philosophers who defend rival accounts of scientific explanation nevertheless agree to a large extent on the phenomena that they are trying to save. Alas, the same cannot be said when it comes to mathematical explanation. Philosophers disagree sharply about which proofs of a given theorem explain why that theorem holds and which merely prove that it holds.1 This kind of disagreement makes it difficult to test proposed accounts of mathematical explanation without making controversial presuppositions about the phe- nomena to be saved.
    [Show full text]
  • Proofs by Induction Maths Delivers! Proofs by Induction
    1 Maths delivers! 2 3 4 5 6 12 11 7 10 8 9 A guide for teachers – Years 11 and 12 Proofs by induction Maths delivers! Proofs by induction Professor Brian A. Davey, La Trobe University Editor: Dr Jane Pitkethly, La Trobe University Illustrations and web design: Catherine Tan, Michael Shaw For teachers of Secondary Mathematics These notes are based in part on lecture notes and printed notes prepared by the author for use in teaching mathematical induction to first-year students at Latrobe University. © The University of Melbourne on behalf of the Australian Mathematical Sciences Institute (AMSI), 2013 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. 2011. http://creativecommons.org/licenses/by-nc-nd/3.0/ Australian Mathematical Sciences Institute Building 161 The University of Melbourne VIC 3010 Email: [email protected] Website: www.amsi.org.au The principle of mathematical induction ........................ 4 A typical proof by induction ................................ 6 Two non-proofs by induction ................................ 7 Modified or strong induction ................................ 9 The Tower of Hanoi puzzle ................................. 10 A two-colour theorem .................................... 14 If you can’t prove it, prove something harder! ..................... 17 Answers to exercises ..................................... 20 maths delivers! Proofs by induction The principle of mathematical induction The idea of induction: falling dominoes There are two steps involved in knocking over a row of dominoes. Step 1. You have to knock over the first domino. Step 2. You have to be sure that when domino k falls, it knocks over domino k 1. Å These are the same as the steps in a proof by induction.
    [Show full text]
  • Mathematical Induction Sept
    COMP 250 Fall 2016 9 - Mathematical induction Sept. 26, 2016 You will see many examples in this course and upcoming courses of algorithms for solving various problems. It many cases, it will be obvious that the algorithms do what you think. However, it many cases it will not be obvious, and one needs to argue very convincingly (or "prove") why the algorithms are correct. To do so, and to understand the arguments of others, we need to learn about how such proofs work. In particular, the next core topic in the course is recursion. We will look at a number of recursive algorithm and analyze how long they take to run. Recursion can be a bit confusing when one first learns about it. One way to understand recursion is to relate it to a proof technique in mathematics called mathematical induction. In a nutshell, you can use mathematical induction to prove that recursive algorithms are correct. Today I will introduce the basics of mathematical induction, and in the following few lectures we'll look at some recursive algorithms. Before I introduce induction, let me give an example of a convincing proof of a statement you have seen many times before:. n(n + 1) 1 + 2 + 3 + ··· + (n − 1) + n = : 2 One trick for showing this statement is true is to add up two copies of the left hand side, one forward and one backward: 1 + 2 + 3 + ··· + (n − 1) + n n + (n − 1) + :::: + 3 + 2 + 1: Then pair up terms: 1 + n; 2 + (n − 1); 3 + (n − 2); :::(n − 1) + 2; n + 1 and note that each pair sums to n + 1 and there are n pairs, which gives (n + 1) ∗ n.
    [Show full text]
  • Peano Axioms to Present a Rigorous Introduction to the Natural Numbers Would Take Us Too Far Afield
    Peano Axioms To present a rigorous introduction to the natural numbers would take us too far afield. We will however, give a short introduction to one axiomatic approach that yields a system that is quite like the numbers that we use daily to count and pay bills. We will consider a set, N,tobecalledthenatural numbers, that has one primitive operation that of successor which is a function from N to N. We will think of this function as being given by Sx x 1(this will be our definition of addition by 1). We will only assume the following rules. 1.IfSx Sy then x y (we say that S is one to one). 2. There exists exactly one element, denoted 1,inN that is not of the form Sx for any x N. 3.IfX N is such that 1 X and furthermore, if x X then Sx X, then X N. These rules comprise the Peano Axioms for the natural numbers. Rule 3. is usually called the principle of mathematical induction As above, we write x 1 for Sx.We assert that the set of elements that are successors of successors consists of all elements of N except for 1 and S1 1 1. We will now prove this assertion. If z SSx then since 1 Su and S is one to one (rule 1.) z S1 and z 1.If z N 1,S1 then z Su by 2. and u 1 by 1. Thus u Sv by 2. Hence, z SSv. This small set of rules (axioms) allow for a very powerful system which seems to have all of the properties of our intuitive notion of natural numbers.
    [Show full text]