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

Save this PDF as:

Size: px
Start display at page:

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

## Transcription

1 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 the sum of the first n integers: Claim 1 For any positive integer n, Σ n i=1 i = n(n+1). At that point, we didn t prove this formula correct, because this is most easily done using a new proof technique: induction. Mathematical induction is a technique for showing that a statement P(n) is true for all natural numbers n, or for some infinite subset of the natural numbers (e.g. all positive even integers). It s a nice way to produce quick, easy-to-read proofs for a variety of fact that would be awkward to prove with the techniques you ve seen so far. It is particularly well suited to analyzing the performance of recursive algorithms. Most of you have seen a few of these in previous programming classes and you ll see a lot more of them as you finish the rest of the major. Induction is very handy, but it may strike you as a bit weird. It may take you some time to get used to it. So, we ve got two tasks which are a bit independent: 1

2 Learn how to write an inductive proof. Understand why inductive proofs are legitimate. Everyone needs to be able to write an inductive proof to pass this course. However, some of you might finish the term feeling a bit shaky on whether you believe induction works correctly. That s ok. You can still use the technique, relying on the fact that we say it s ok. You ll see induction (and its friend recursion) in later terms and will gradually come to feel more confident about its validity. An Example A simple proof by induction has the following outline: Claim: P(n) is true for all positive integers n. Proof: We ll use induction on n. Base: We need to show that P(1) is true. Induction: Suppose that P(k) is true, for some positive integer k. We need to show that P(k + 1) is true. The part of the proof labelled induction is a conditional statement. We assume that P(k) is true. This assumption is called the inductive hypothesis. We use this assumption to show that P(k + 1) is true. For our formula example, our proposition P(n) is Σ n i=1 i = n(n+1). Substituting in the definition of P for our example, we get the following outline: Proof: We will show that Σ n i=1 i = n(n+1) for any positive integer n, using induction on n. Base: We need to show that the formula holds for n = 1, i.e. Σ 1 i=1 i = 1. Induction: Suppose that Σ k i=1 i = k(k+1) for some positive integer k. We need to show that Σ k+1 i=1 i = (k+1)(k+).

3 So, in the part marked induction, we re assuming the formula works for some k and we ll use that formula for k to work out the formula for k +1. For our specific example, the induction proof might look as follows: Proof: We will show that Σ n i=1 i = n(n+1) for any positive integer n, using induction on n. Base: We need to show that the formula holds for n = 1. Σ 1 i=1 i = 1. And also 1 = 1. So the two are equal for n = 1. Induction: Suppose that Σ k i=1 i = k(k+1) for some positive integer k. We need to show that Σ k+1 i=1 i = (k+1)(k+). By the definition of summation notation, Σ k+1 i=1 i = (Σk i=1 i)+(k+ 1) Substituting in the formula from our inductive hypothesis, we get that (Σ k i=1 i) + (k + 1) = (k(k+1)) + (k + 1). But ( k(k+1) ) + (k + 1) = k(k+1) + (k+1) = (k+)(k+1) = (k+1)(k+). So, combining these equations, we get that Σ k+1 i=1 i = (k+1)(k+) which is what we needed to show. One way to think of a proof by induction is that it s a template for building direct proofs. If I give you the specific value n = 47, you could write a direct proof by starting with the base case and using the inductive step 46 times to work your way from the n = 1 case up to the n = 47 case. 3 Why is this legit? There are several ways to think about mathematical induction, and understand why it s a legitimate proof technique. Different people prefer different motivations at this point, so I ll offer several. A proof by induction of that P(k) is true for all positive integers k involves showing that P(1) is true (base case) and that P(k) P(k + 1) (inductive step). 3

4 Domino Theory: Imagine an infinite line of dominoes. The base step pushes the first one over. The inductive step claims that one domino falling down will push over the next domino in the line. So dominos will start to fall from the beginning all the way down the line. This process continues forever, because the line is infinitely long. However, if you focus on any specific domino, it falls after some specific finite delay. Recursion fairy: The recursion fairy is the mathematician s version of a programming assistant. Suppose you tell her how to do the proof for P(1) and also why P(k) implies P(k + 1). Then suppose you pick any integer (e.g. 1034) then she can take this recipe and use it to fill in all the details of a normal direct proof that P holds for this particular integer. That is, she takes P(1), then uses the inductive step to get from P(1) to P(), and so on up to P(1034). Smallest counter-example: Let s assume we ve established that P(1) is true and also that P(k) implies P(k+1). Let s prove that P(j) is true for all positive integers j, by contradiction. That is, we suppose that P(1) is true, also that P(k) implies P(k + 1), but there is a counter-example to our claim that P(j) is true for all j. That is, suppose that P(m) was not true for some integer m. Now, let s look at the set of all counter-examples. We know that all the counter-examples are larger than 1, because our induction proof established explicitly that P(1) was true. Suppose that the smallest counter-example is s. So P(s) is not true. We know that s > 1, since P(1) was true. Since s was supposed to be the smallest counter-example, s 1 must not be a counter-example, i.e. P(s 1) is true. But now we know that P(s 1) is true but P(s) is not true. This directly contradicts our assumption that P(k) implies P(k + 1) for any k. The smallest counter-example explanation is a formal proof that induction works, given how we ve defined the integers. If you dig into the mathematics, 4

5 you ll find that it depends on the integers having what s called the wellordering property: any subset that has a lower bound also has a smallest element. Standard axioms used to define the integers include either a wellordering or an induction axiom. These arguments don t depend on whether our starting point is 1 or some other integer, e.g. 0 or or -47. All you need to do is ensure that your base case covers the first integer for which the claim is supposed to be true. 4 Building an inductive proof In constructing an inductive proof, you ve got two tasks. First, you need to set up this outline for your problem. This includes identifying a suitable proposition P and a suitable integer variable n. Notice that P(n) must be a statement, i.e. something that is either true or false. For example, it is never just a formula whose value is a number. Also, notice that P(n) must depend on an integer n. This integer n is known as our induction variable. The assumption at the start of the inductive step ( P(k) is true ) is called the inductive hypothesis. Your second task is to fill in the middle part of the induction step. That is, you must figure out how to relate a solution for a larger problem P(k +1) to a solution for a small problem P(k). Most students want to do this by starting with the small problem and adding something to it. For more complex situations, it s usually better to start with the larger problem and try to find an instance of the smaller problem inside it. 5 Another example of induction Let s do another example: Claim For every positive integer n 4, n < n!. First, you should try (on your own) some specific integers and verify that the claim is true. Since the claim specifies n 4, it s worth checking that 4 5

6 does work but the smaller integers don t. In this claim, the proposition P(n) is n < n!. So an outline of our inductive proof looks like: Proof: Suppose that n is an integer and n 4. We ll prove that n < n! using induction on n. Base: n = 4. [show that the formula works for n = 4] Induction: Suppose that the claim holds for n = k. That is, suppose that we have an integer k 4 such that k < k!. We need to show that the claim holds for n = k+1, i.e. that k+1 < (k+1)!. Notice that our base case is for n = 4 because the claim was specified to hold only for integers 4. Fleshing out the details of the algebra, we get the following full proof. When working with inequalities, it s especially important to write down your assumptions and what you want to conclude with. You can then work from both ends to fill in the gap in the middle of the proof. Proof: Suppose that n is an integer and n 4. We ll prove that n < n! using induction on n. Base: n = 4. In this case n = 4 = 16. Also n! = = 4. Since 16 < 4, the formula holds for n = 4. Induction: Suppose that the claim holds for n = k. That is, suppose that we have an integer k 4 such that k < k!. We need to show that k+1 < (k + 1)!. k+1 = k. By the inductive hypothesis, k < k!, so k < k!. Since k 4, < k + 1. So k! < (k + 1) k! = (k + 1)!. Putting these equations together, we find that k+1 < (k + 1)!, which is what we needed to show. 6 Some comments about style Notice that the start of the proof tells you which variable in your formula (n in this case) is the induction variable. In this formula, the choice of 6

8 Proof: By induction on n. Base: Let n = 0. Then n 3 n = = 0 which is divisible by 3. Induction: Suppose that k 3 k is divisible by 3, for some positive integer k. We need to show that (k + 1) 3 (k + 1) is divisible by 3. (k+1) 3 (k+1) = (k 3 +3k +3k+1) (k+1) = (k 3 k)+3(k +k) From the inductive hypothesis, (k 3 k) is divisible by 3. And 3(k + k) is divisible by 3 since (k + k) is an integer. So their sum is divisible by 3. That is (k + 1) 3 (k + 1) is divisible by 3. The zero base case is technically enough to make the proof solid, but sometimes a zero base case doesn t provide good intuition or confidence. So you ll sometimes see an extra base case written out, e.g. n = 1 in this example, to help the author or reader see why the claim is plausible. 8 A geometrical example Let s see another example of the basic induction outline, this time on a geometrical application. Tiling some area of space with a certain type of puzzle piece means that you fit the puzzle pieces onto that area of space exactly, with no overlaps or missing areas. A right triomino is a -by- square minus one of the four squares. I claim that 8

9 Claim 4 For any positive integer n, a n n checkerboard with any one square removed can be tiled using right triominoes. Proof: by induction on n. Base: Suppose n = 1. Then our n n checkerboard with one square removed is exactly one right triomino. Induction: Suppose that the claim is true for some integer k. That is a k k checkerboard with any one square removed can be tiled using right triominoes. Suppose we have a k+1 k+1 checkerboard C with any one square removed. We can divide C into four k k sub-checkerboards P, Q, R, and S. One of these sub-checkerboards is already missing a square. Suppose without loss of generality that this one is S. Place a single right triomino in the middle of C so it covers one square on each of P, Q, and R. Now look at the areas remaining to be covered. In each of the sub-checkerboards, exactly one square is missing (S) or already covered (P, Q, and R). So, by our inductive hypothesis, each of these sub-checkerboards minus one square can be tiled with right triominoes. Combining these four tilings with the triomino we put in the middle, we get a tiling for the whole of the larger checkerboard C. This is what we needed to construct. 9 Graph coloring We can also use induction to prove a useful general fact about graph colorability: Claim 5 If all vertices in a graph G have degree D, then G can be colored with D + 1 colors. The objects involved in this claim are graphs. To apply induction to objects like graphs, we organize our objects by their size. Each step in the induction process will show that the claim holds for all objects of a particular 9

10 (integer) size. For graphs, the size would typically be either the number of vertices or the number of edges. For this proof, it s most convenient to use the number of vertices. Proof: by induction on the number of vertices in G. Base: The graph with just one vertex has maximum degree 0 and can be colored with one color. Induction: Suppose that any graph with k vertices and maximum vertex degree D can be colored with D + 1 colors. Let G be a graph with k +1 vertices and maximum vertex degree D. Remove some vertex v (and its edges) from G to create a smaller graph G. G has k vertices. Also, the maximum vertex degree of G is no larger than D, because removing a vertex can t increase the degree. So, by the inductive hypothesis, G can be colored with D + 1 colors. Because v has at most D neighbors, its neighbors are only using D of the available colors, leaving a spare color that we can assign to v. The coloring of G can be extended to a coloring of G with D + 1 colors. We can use this idea to design an algorithm (called the greedy algorithm) for coloring a graph. This algorithm walks through the vertices one-by-one, giving each vertex a color without revising any of the previouslyassigned colors. When we get to each vertex, we see what colors have been assigned to its neighbors. If there is a previously used color not assigned to a neighbor, we re-use that color. Otherwise, we deploy a new color. The above theorem shows that the greedy algorithm will never use more than D + 1 colors. Notice, however, that D + 1 is only an upper bound on the chromatic number of the graph. The actual chromatic number of the graph might be a lot smaller. For example, D + 1 would be 7 for the wheel graph W 6 but this graph actually has chromatic number only three: 10

11 G B B R G G B The performance of the greedy algorithm is very sensitive to the order in which the vertices are considered. For example, suppose we start coloring W 6 by coloring the center hub, then a vertex v on the outer ring, and then the vertex opposite v. Then our partial coloring might look as shown below. Completing this coloring will require using four colors. G R G Notice that whether we need to deploy a new color to handle a vertex isn t actually determined by the degree of the vertex but, rather, by how many of its neighbors are already colored. So a useful heuristic is to order vertices by their degrees and color higher-degree vertices earlier in the process. This tends to mean that, when we reach a high-degree vertex, some of its neighbors will not yet be colored. So we will be able to handle the high-degree vertices with fewer colors and then extend this partial coloring to all the low-degree vertices. 10 Strong induction The inductive proofs you ve seen so far have had the following outline: Proof: We will show P(n) is true for all n, using induction on n. 11

12 Base: We need to show that P(1) is true. Induction: Suppose that P(k) is true, for some integer k. We need to show that P(k + 1) is true. Think about building facts incrementally up from the base case to P(k). Induction proves P(k) by first proving P(i) for every i from 1 up through k 1. So, by the time we ve proved P(k), we ve also proved all these other statements. For some proofs, it s very helpful to use the fact that P is true for all these smaller values, in addition to the fact that it s true for k. This method is called strong induction. A proof by strong induction looks like this: Proof: We will show P(n) is true for all n, using induction on n. Base: We need to show that P(1) is true. Induction: Suppose that P(n) is true for n = 1,,..., k. We need to show that P(k + 1) is true. The only new feature about this proof is that, superficially, we are assuming slightly more in the hypothesis of the inductive step. The difference is actually only superficial, and the two proof techniques are equivalent. However, this difference does make some proofs much easier to write. 11 Postage example Strong induction is useful when the result for n = k+1 depends on the result for some smaller value of n, but it s not the immediately previous value (k). Here s a classic example: Claim 6 Every amount of postage that is at least 1 cents can be made from 4-cent and 5-cent stamps. For example, 1 cents uses three 4-cent stamps. 13 cents of postage uses two 4-cent stamps plus a 5-cent stamp. 14 uses one 4-cent stamp plus two 1

13 5-cent stamps. If you experiment with small values, you quickly realize that the formula for making k cents of postage depends on the one for making k 4 cents of postage. That is, you take the stamps for k 4 cents and add another 4-cent stamp. We can make this into an inductive proof as follows: Proof: by induction on the amount of postage. Base: If the postage is 1 cents, we can make it with three 4-cent stamps. If the postage is 13 cents, we can make it with two 4-cent stamps. plus a 5-cent stamp. If it is 14, we use one 4-cent stamp plus two 5-cent stamps. If it is 15, we use three 5-cent stamps. Induction: Suppose that we have show how to construct postage for every value from 1 up through k. We need to show how to construct k +1 cents of postage. Since we ve already proved base cases up through 15 cents, we ll assume that k Since k+1 16, (k+1) 4 1. So by the inductive hypothesis, we can construct postage for (k + 1) 4 cents using m 4-cent stamps and n 5-cent stamps, for some natural numbers m and n. In other words (k + 1) 4) = 4m + 5n. But then k + 1 = 4(m + 1) + 5n. So we can construct k + 1 cents of postage using m+1 4-cent stamps and n 5-cent stamps, which is what we needed to show. Notice that we needed to directly prove four base cases, since we needed to reach back four integers in our inductive step. It s not always obvious how many base cases are needed until you work out the details of your inductive step. 1 Nim In the parlour game Nim, there are two players and two piles of matches. At each turn, a player removes some (non-zero) number of matches from one of the piles. The player who removes the last match wins. 1 1 Or, in some variations, loses. There seem to be several variations of this game. 13

14 Claim 7 If the two piles contain the same number of matches at the start of the game, then the second player can always win. Here s a winning strategy for the second player. Suppose your opponent removes m matches from one pile. In your next move, you remove m matches from the other pile, thus evening up the piles. Let s prove that this strategy works. Proof by induction on the number of matches (n) in each pile. Base: If both piles contain 1 match, the first player has only one possible move: remove the last match from one pile. The second player can then remove the last match from the other pile and thereby win. Induction: Suppose that the second player can win any game that starts with two piles of n matches, where n is any value from 1 through k. We need to show that this is true if n = k + 1. So, suppose that both piles contain k + 1 matches. A legal move by the first player involves removing j matches from one pile, where 1 j k + 1. The piles then contain k + 1 matches and k + 1 j matches. The second player can now remove j matches from the other pile. This leaves us with two piles of k+1 j matches. If j = k+1, then the second player wins. If j < k + 1, then we re now effectively at the start of a game with k + 1 j matches in each pile. Since j 1, k + 1 j k. So, by the induction hypothesis, we know that the second player can finish the rest of the game with a win. The induction step in this proof uses the fact that our claim P(n) is true for a smaller value of n. But since we can t control how many matches the first player removes, we don t know how far back we have look in the sequence of earlier results P(1)...P(k). Our previous proof about postage can be rewritten so as to avoid strong induction. It s less clear how to rewrite proofs like this Nim example. 14

15 13 Prime factorization Early in this course, we saw the Fundamental Theorem of Arithmetic, which states that every positive integer n, n, can be expressed as the product of one or more prime numbers. Let s prove that this is true. Recall that a number n is prime if its only positive factors are one and n. n is composite if it s not prime. Since a factor of a number must be no larger than the number itself, this means that a composite number n always has a factor larger than 1 but smaller than n. This, in turn, means that we can write n as ab, where a and b are both larger than 1 but smaller than n. Proof by induction on n. Base: can be written as the product of a single prime number,. Induction: Suppose that every integer between and k can be written as the product of one or more primes. We need to show that k + 1 can be written as a product of primes. There are two cases: Case 1: k + 1 is prime. Then it is the product of one prime, i.e. itself. Case : k + 1 is composite. Then k + 1 can be written as ab, where a and b are integers such that a and b lie in the range [, k]. By the induction hypothesis, a can be written as a product of primes p 1 p...p i and b can be written as a product of primes q 1 q...q j. So then k + 1 can be written as the product of primes p 1 p...p i q 1 q...q j. In both cases k + 1 can be written as a product of primes, which is what we needed to show. Again, the inductive step needed to reach back some number of steps in our sequence of results, but we couldn t control how far back we needed to go. We ll leave the details of proving this as an exercise for the reader. 15

16 14 Variation in notation Certain details of the induction outline vary, depending on the individual preferences of the author and the specific claim being proved. Some folks prefer to assume the statement is true for k and prove it s true for k + 1. Other assume it s true for k 1 and prove it s true for k. You can pick either one, though one of the two choices may yield a slightly simpler proof for some specific problem. Some authors prefer to write strong induction hypotheses all the time, even when a weak hypothesis would be sufficient. This saves mental effort, because you don t have to figure out in advance whether a strong hypothesis was really required. However, for some problems, a strong hypothesis may be more complicated to state than a weak one. Many beginners make technical mistakes writing strong hypotheses, even if they can correct write weak ones. You ll probably use the strong form more often as you become more fluent with mathematical writing. Authors writing for more experienced audiences may abbreviate the outline somewhat, e.g. packing an entirely short proof into one paragraph without labelling the base and inductive steps separately. However, being careful about the outline is important when you are still getting used to the technique. 16

### New Zealand Mathematical Olympiad Committee. Induction

New Zealand Mathematical Olympiad Committee Induction Chris Tuffley 1 Proof by dominoes Mathematical induction is a useful tool for proving statements like P is true for all natural numbers n, for example

### 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)

### Mathematical Induction

Mathematical S 0 S 1 S 2 S 3 S 4 S 5 S 6 S 7 S 8 S 9 S 10 Like dominoes! Mathematical S 0 S 1 S 2 S 3 S4 S 5 S 6 S 7 S 8 S 9 S 10 Like dominoes! S 4 Mathematical S 0 S 1 S 2 S 3 S5 S 6 S 7 S 8 S 9 S 10

### Oh Yeah? Well, Prove It.

Oh Yeah? Well, Prove It. MT 43A - Abstract Algebra Fall 009 A large part of mathematics consists of building up a theoretical framework that allows us to solve problems. This theoretical framework is built

### Mathematical Induction

Mathematical Induction MAT30 Discrete Mathematics Fall 016 MAT30 (Discrete Math) Mathematical Induction Fall 016 1 / 19 Outline 1 Mathematical Induction Strong Mathematical Induction MAT30 (Discrete Math)

### Chapter 6 Finite sets and infinite sets. Copyright 2013, 2005, 2001 Pearson Education, Inc. Section 3.1, Slide 1

Chapter 6 Finite sets and infinite sets Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 1 Section 6. PROPERTIES OF THE NATURE NUMBERS 013 Pearson Education, Inc.1 Slide Recall that denotes

### 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,

### Lecture 3. Mathematical Induction

Lecture 3 Mathematical Induction Induction is a fundamental reasoning process in which general conclusion is based on particular cases It contrasts with deduction, the reasoning process in which conclusion

### Induction and Recursion

Induction and Recursion Winter 2010 Mathematical Induction Mathematical Induction Principle (of Mathematical Induction) Suppose you want to prove that a statement about an integer n is true for every positive

### Introduction to mathematical arguments

Introduction to mathematical arguments (background handout for courses requiring proofs) by Michael Hutchings A mathematical proof is an argument which convinces other people that something is true. Math

### Planar Graphs and Graph Coloring

Planar Graphs and Graph Coloring Margaret M. Fleck 1 December 2010 These notes cover facts about graph colorings and planar graphs (sections 9.7 and 9.8 of Rosen) 1 Introduction So far, we ve been looking

### 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

### COMP 250 Fall Mathematical induction Sept. 26, (n 1) + n = n + (n 1)

COMP 50 Fall 016 9 - Mathematical induction Sept 6, 016 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

### Mathematical induction & Recursion

CS 441 Discrete Mathematics for CS Lecture 15 Mathematical induction & Recursion Milos Hauskrecht milos@cs.pitt.edu 5329 Sennott Square Proofs Basic proof methods: Direct, Indirect, Contradiction, By Cases,

### Students in their first advanced mathematics classes are often surprised

CHAPTER 8 Proofs Involving Sets Students in their first advanced mathematics classes are often surprised by the extensive role that sets play and by the fact that most of the proofs they encounter are

### 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

### Discrete Mathematics, Chapter 5: Induction and Recursion

Discrete Mathematics, Chapter 5: Induction and Recursion Richard Mayr University of Edinburgh, UK Richard Mayr (University of Edinburgh, UK) Discrete Mathematics. Chapter 5 1 / 20 Outline 1 Well-founded

### 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

### Mathematical Induction

Chapter 2 Mathematical Induction 2.1 First Examples Suppose we want to find a simple formula for the sum of the first n odd numbers: 1 + 3 + 5 +... + (2n 1) = n (2k 1). How might we proceed? The most natural

### 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

### 8.7 Mathematical Induction

8.7. MATHEMATICAL INDUCTION 8-135 8.7 Mathematical Induction Objective Prove a statement by mathematical induction Many mathematical facts are established by first observing a pattern, then making a conjecture

### Mathematical Induction. Rosen Chapter 4.1 (6 th edition) Rosen Ch. 5.1 (7 th edition)

Mathematical Induction Rosen Chapter 4.1 (6 th edition) Rosen Ch. 5.1 (7 th edition) Mathmatical Induction Mathmatical induction can be used to prove statements that assert that P(n) is true for all positive

### Discrete Mathematics and Probability Theory Fall 2016 Seshia and Walrand Note 3

CS 70 Discrete Mathematics and Probability Theory Fall 06 Seshia and Walrand Note 3 Mathematical Induction Introduction. In this note, we introduce the proof technique of mathematical induction. Induction

### Principle of (Weak) Mathematical Induction. P(1) ( n 1)(P(n) P(n + 1)) ( n 1)(P(n))

Outline We will cover (over the next few weeks) Mathematical Induction (or Weak Induction) Strong (Mathematical) Induction Constructive Induction Structural Induction Principle of (Weak) Mathematical Induction

### CS 2336 Discrete Mathematics

CS 2336 Discrete Mathematics Lecture 5 Proofs: Mathematical Induction 1 Outline What is a Mathematical Induction? Strong Induction Common Mistakes 2 Introduction What is the formula of the sum of the first

### Induction and the Division Theorem 2.1 The Method of Induction

2 Induction and the Division Theorem 2.1 The Method of Induction In the Introduction we discussed a mathematical problem whose solution required the verification of an infinite family of statements. We

### CHAPTER 3 Numbers and Numeral Systems

CHAPTER 3 Numbers and Numeral Systems Numbers play an important role in almost all areas of mathematics, not least in calculus. Virtually all calculus books contain a thorough description of the natural,

### Computer Science 211 Data Structures Mount Holyoke College Fall Topic Notes: Recursion and Mathematical Induction

Computer Science 11 Data Structures Mount Holyoke College Fall 009 Topic Notes: Recursion and Mathematical Induction Recursion An important tool when trying to solve a problem is the ability to break the

### 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

Section 5.1 Climbing an Infinite Ladder Suppose we have an infinite ladder and the following capabilities: 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder,

### This section demonstrates some different techniques of proving some general statements.

Section 4. Number Theory 4.. Introduction This section demonstrates some different techniques of proving some general statements. Examples: Prove that the sum of any two odd numbers is even. Firstly you

### 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

### WRITING PROOFS. Christopher Heil Georgia Institute of Technology

WRITING PROOFS Christopher Heil Georgia Institute of Technology A theorem is just a statement of fact A proof of the theorem is a logical explanation of why the theorem is true Many theorems have this

### 1.3 Induction and Other Proof Techniques

4CHAPTER 1. INTRODUCTORY MATERIAL: SETS, FUNCTIONS AND MATHEMATICAL INDU 1.3 Induction and Other Proof Techniques The purpose of this section is to study the proof technique known as mathematical induction.

### 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

### 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

### MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

### 3.3 MATHEMATICAL INDUCTION

Section 3.3 Mathematical Induction 3.3.1 3.3 MATHEMATICAL INDUCTION From modus ponens: p p q q basis assertion conditional assertion conclusion we can easily derive double modus ponens : p 0 p 0 p 1 p

### 2. Methods of Proof Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try.

2. METHODS OF PROOF 69 2. Methods of Proof 2.1. Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try. Trivial Proof: If we know q is true then

### 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

### 3. Recurrence Recursive Definitions. To construct a recursively defined function:

3. RECURRENCE 10 3. Recurrence 3.1. Recursive Definitions. To construct a recursively defined function: 1. Initial Condition(s) (or basis): Prescribe initial value(s) of the function.. Recursion: Use a

### Mathematics for Computer Science

mcs 01/1/19 1:3 page i #1 Mathematics for Computer Science revised Thursday 19 th January, 01, 1:3 Eric Lehman Google Inc. F Thomson Leighton Department of Mathematics and the Computer Science and AI Laboratory,

### It is time to prove some theorems. There are various strategies for doing

CHAPTER 4 Direct Proof It is time to prove some theorems. There are various strategies for doing this; we now examine the most straightforward approach, a technique called direct proof. As we begin, it

### CHAPTER 3. Sequences. 1. Basic Properties

CHAPTER 3 Sequences We begin our study of analysis with sequences. There are several reasons for starting here. First, sequences are the simplest way to introduce limits, the central idea of calculus.

### Direct Proofs. CS 19: Discrete Mathematics. Direct Proof: Example. Indirect Proof: Example. Proofs by Contradiction and by Mathematical Induction

Direct Proofs CS 19: Discrete Mathematics Amit Chakrabarti Proofs by Contradiction and by Mathematical Induction At this point, we have seen a few examples of mathematical proofs. These have the following

### Proof: A logical argument establishing the truth of the theorem given the truth of the axioms and any previously proven theorems.

Math 232 - Discrete Math 2.1 Direct Proofs and Counterexamples Notes Axiom: Proposition that is assumed to be true. Proof: A logical argument establishing the truth of the theorem given the truth of the

### Reading 7 : Program Correctness

CS/Math 240: Introduction to Discrete Mathematics Fall 2015 Instructors: Beck Hasti, Gautam Prakriya Reading 7 : Program Correctness 7.1 Program Correctness Showing that a program is correct means that

### Theorem 2. If x Q and y R \ Q, then. (a) x + y R \ Q, and. (b) xy Q.

Math 305 Fall 011 The Density of Q in R The following two theorems tell us what happens when we add and multiply by rational numbers. For the first one, we see that if we add or multiply two rational numbers

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

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

### Calculus for Middle School Teachers. Problems and Notes for MTHT 466

Calculus for Middle School Teachers Problems and Notes for MTHT 466 Bonnie Saunders Fall 2010 1 I Infinity Week 1 How big is Infinity? Problem of the Week: The Chess Board Problem There once was a humble

### Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that. a = bq + r and 0 r < b.

Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that a = bq + r and 0 r < b. We re dividing a by b: q is the quotient and r is the remainder,

### Basic Notions on Graphs. Planar Graphs and Vertex Colourings. Joe Ryan. Presented by

Basic Notions on Graphs Planar Graphs and Vertex Colourings Presented by Joe Ryan School of Electrical Engineering and Computer Science University of Newcastle, Australia Planar graphs Graphs may be drawn

### Chapter 4. Trees. 4.1 Basics

Chapter 4 Trees 4.1 Basics A tree is a connected graph with no cycles. A forest is a collection of trees. A vertex of degree one, particularly in a tree, is called a leaf. Trees arise in a variety of applications.

### Arithmetic Series. or The sum of any finite, or limited, number of terms is called a partial sum. are shorthand ways of writing n 1

CONDENSED LESSON 9.1 Arithmetic Series In this lesson you will learn the terminology and notation associated with series discover a formula for the partial sum of an arithmetic series A series is the indicated

### 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

### Induction. Mathematical Induction. Induction. Induction. Induction. Induction. 29 Sept 2015

Mathematical If we have a propositional function P(n), and we want to prove that P(n) is true for any natural number n, we do the following: Show that P(0) is true. (basis step) We could also start at

### COLORED GRAPHS AND THEIR PROPERTIES

COLORED GRAPHS AND THEIR PROPERTIES BEN STEVENS 1. Introduction This paper is concerned with the upper bound on the chromatic number for graphs of maximum vertex degree under three different sets of coloring

### SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION

CHAPTER 5 SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Alessandro Artale UniBZ - http://www.inf.unibz.it/ artale/ SECTION 5.2 Mathematical Induction I Copyright Cengage Learning. All rights reserved.

### PRINCIPLE OF MATHEMATICAL INDUCTION

Chapter 4 PRINCIPLE OF MATHEMATICAL INDUCTION Analysis and natural philosophy owe their most important discoveries to this fruitful means, which is called induction Newton was indebted to it for his theorem

### 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

### 1. R In this and the next section we are going to study the properties of sequences of real numbers.

+a 1. R In this and the next section we are going to study the properties of sequences of real numbers. Definition 1.1. (Sequence) A sequence is a function with domain N. Example 1.2. A sequence of real

### 6. GRAPH AND MAP COLOURING

6. GRPH ND MP COLOURING 6.1. Graph Colouring Imagine the task of designing a school timetable. If we ignore the complications of having to find rooms and teachers for the classes we could propose the following

### CS103X: Discrete Structures Homework Assignment 2: Solutions

CS103X: Discrete Structures Homework Assignment 2: Solutions Due February 1, 2008 Exercise 1 (10 Points). Prove or give a counterexample for the following: Use the Fundamental Theorem of Arithmetic to

### WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT?

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

### 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

### NIM Games: Handout 1

NIM Games: Handout 1 Based on notes by William Gasarch 1 One-Pile NIM Games Consider the following two-person game in which players alternate making moves. There are initially n stones on the board. During

### Math 140a - HW 1 Solutions

Math 140a - HW 1 Solutions Problem 1 (WR Ch 1 #1). If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Given that r is rational, we can write r = a b for some

### Topics in Number Theory

Chapter 8 Topics in Number Theory 8.1 The Greatest Common Divisor Preview Activity 1 (The Greatest Common Divisor) 1. Explain what it means to say that a nonzero integer m divides an integer n. Recall

### Summer High School 2009 Aaron Bertram. 1. Prime Numbers. Let s start with some notation:

Summer High School 2009 Aaron Bertram 1. Prime Numbers. Let s start with some notation: N = {1, 2, 3, 4, 5,...} is the (infinite) set of natural numbers. Z = {..., 3, 2, 1, 0, 1, 2, 3,...} is the set of

### An Innocent Investigation

An Innocent Investigation D. Joyce, Clark University January 2006 The beginning. Have you ever wondered why every number is either even or odd? I don t mean to ask if you ever wondered whether every number

### For the purposes of this course, the natural numbers are the positive integers. We denote by N, the set of natural numbers.

Lecture 1: Induction and the Natural numbers Math 1a is a somewhat unusual course. It is a proof-based treatment of Calculus, for all of you who have already demonstrated a strong grounding in Calculus

### 2.4 Mathematical Induction

2.4 Mathematical Induction What Is (Weak) Induction? The Principle of Mathematical Induction works like this: What Is (Weak) Induction? The Principle of Mathematical Induction works like this: We want

### Discrete Mathematics (2009 Spring) Induction and Recursion (Chapter 4, 3 hours)

Discrete Mathematics (2009 Spring) Induction and Recursion (Chapter 4, 3 hours) Chih-Wei Yi Dept. of Computer Science National Chiao Tung University April 17, 2009 4.1 Mathematical Induction 4.1 Mathematical

### Mathematical Induction. Part Three

Mathematical Induction Part Three Announcements Problem Set Two due Friday, October 4 at 2:5PM (just before class) Stop by office hours with questions! Email us at cs03@cs.stanford.edu with questions!

### Course notes on Number Theory

Course notes on Number Theory In Number Theory, we make the decision to work entirely with whole numbers. There are many reasons for this besides just mathematical interest, not the least of which is that

### Near Optimal Solutions

Near Optimal Solutions Many important optimization problems are lacking efficient solutions. NP-Complete problems unlikely to have polynomial time solutions. Good heuristics important for such problems.

### Mathematical induction: variants and subtleties

Mathematical induction: variants and subtleties October 29, 2010 Mathematical induction is one of the most useful techniques for solving problems in mathematics. I m assuming you re familiar with the basic

### 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

### Mathematical Induction

MCS-236: Graph Theory Handout #A5 San Skulrattanakulchai Gustavus Adolphus College Sep 15, 2010 Mathematical Induction The following three principles governing N are equivalent. Ordinary Induction Principle.

### CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

### Mathematical Induction with MS

Mathematical Induction 2008-14 with MS 1a. [4 marks] Using the definition of a derivative as, show that the derivative of. 1b. [9 marks] Prove by induction that the derivative of is. 2a. [3 marks] Consider

### CSE373: Data Structures and Algorithms Lecture 2: Proof by Induction. Linda Shapiro Winter 2015

CSE373: Data Structures and Algorithms Lecture 2: Proof by Induction Linda Shapiro Winter 2015 Background on Induction Type of mathematical proof Typically used to establish a given statement for all natural

### Finish Set Theory Nested Quantifiers

Finish Set Theory Nested Quantifiers Margaret M. Fleck 18 February 2009 This lecture does a final couple examples of set theory proofs. It then fills in material on quantifiers, especially nested ones,

### So let us begin our quest to find the holy grail of real analysis.

1 Section 5.2 The Complete Ordered Field: Purpose of Section We present an axiomatic description of the real numbers as a complete ordered field. The axioms which describe the arithmetic of the real numbers

### INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS

INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28

### Mathematical Induction

Chapter 4 Mathematical Induction 4.1 The Principle of Mathematical Induction Preview Activity 1 (Exploring Statements of the Form.8n N/.P.n//) One of the most fundamental sets in mathematics is the set

### 6.042/18.062J Mathematics for Computer Science October 24, 2006 Tom Leighton and Ronitt Rubinfeld. Recurrences I

6.042/8.062J Mathematics for Computer Science October 24, 2006 Tom Leighton and Ronitt Rubinfeld Lecture Notes Recurrences I This is the first of two lectures about solving recurrences and recurrent problems.

### (Refer Slide Time 1.50)

Discrete Mathematical Structures Dr. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras Module -2 Lecture #11 Induction Today we shall consider proof

### Fall 2015 Midterm 1 24/09/15 Time Limit: 80 Minutes

Math 340 Fall 2015 Midterm 1 24/09/15 Time Limit: 80 Minutes Name (Print): This exam contains 6 pages (including this cover page) and 5 problems. Enter all requested information on the top of this page,

### Graph Theory Problems and Solutions

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

### DEFINITIONS: 1. FACTORIAL: n! = n (n 1) (n 2) ! = 1. n! = n (n 1)! = n (n 1) (n 2)! (n + 1)! = (n + 1) n! = (n + 1) n (n 1)!

MATH 2000 NOTES ON INDUCTION DEFINITIONS: 1. FACTORIAL: n! = n (n 1) (n 2)... 3 2 1 0! = 1 n! = n (n 1)! = n (n 1) (n 2)! (n + 1)! = (n + 1) n! = (n + 1) n (n 1)! 2. SUMMATION NOTATION: f(i) = f(1) + f(2)

### MAT2400 Analysis I. A brief introduction to proofs, sets, and functions

MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take

### 31 is a prime number is a mathematical statement (which happens to be true).

Chapter 1 Mathematical Logic In its most basic form, Mathematics is the practice of assigning truth to welldefined statements. In this course, we will develop the skills to use known true statements to

### Congruences. Robert Friedman

Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition

### Mathematical Induction

Mathematical Induction Victor Adamchik Fall of 2005 Lecture 1 (out of three) Plan 1. The Principle of Mathematical Induction 2. Induction Examples The Principle of Mathematical Induction Suppose we have

### x if x 0, x if x < 0.

Chapter 3 Sequences In this chapter, we discuss sequences. We say what it means for a sequence to converge, and define the limit of a convergent sequence. We begin with some preliminary results about the

### Strong Induction. Catalan Numbers.

.. The game starts with a stack of n coins. In each move, you divide one stack into two nonempty stacks. + + + + + + + + + + If the new stacks have height a and b, then you score ab points for the move.

### Dedekind s forgotten axiom and why we should teach it (and why we shouldn t teach mathematical induction in our calculus classes)

Dedekind s forgotten axiom and why we should teach it (and why we shouldn t teach mathematical induction in our calculus classes) by Jim Propp (UMass Lowell) March 14, 2010 1 / 29 Completeness Three common