Strong Mathematical Induction

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1 Strong Mathematical Induction Lecture 23 Section 5.4 Robb T. Koether Hampden-Sydney College Mon, Feb 24, 2014 Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

2 1 The Principle 2 The Method 3 Examples Prime Factorization The Checkerboard Puzzle Trees Binary Strings 4 Assignment Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

3 Outline 1 The Principle 2 The Method 3 Examples Prime Factorization The Checkerboard Puzzle Trees Binary Strings 4 Assignment Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

4 The Principle of Strong Mathematical Induction Let P(n) be a predicate defined for integers n. Let a be an integer. If it is true that P(a), P(a + 1),..., P(b) are true, and For all integers k b, if P(a), P(a + 1),..., P(k) are true, then P(k + 1) is true, then it follows that P(n) is true for all n a. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

5 Outline 1 The Principle 2 The Method 3 Examples Prime Factorization The Checkerboard Puzzle Trees Binary Strings 4 Assignment Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

6 The Method The basic step. Show that P(a), P(a + 1),..., P(b) are true. The inductive step. Show that for all integers k b, if P(a), P(a + 1),..., P(k) are true, then P(k + 1) is true, The conclusion. Conclude that P(n) is true for all n a. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

7 The Method The power of the method is that we are allowed to assume that the statement is true for all integers less than or equal to k, not just k itself. Depending on the nature of the statement, this can be a tremendous advantage. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

8 Outline 1 The Principle 2 The Method 3 Examples Prime Factorization The Checkerboard Puzzle Trees Binary Strings 4 Assignment Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

9 Outline 1 The Principle 2 The Method 3 Examples Prime Factorization The Checkerboard Puzzle Trees Binary Strings 4 Assignment Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

10 Prime Factorization Theorem Every integer n 2 can be factored into a product of primes. The basis step. Let n = 2. 2 is prime, so the statement is true. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

11 Prime Factorization The inductive step. Suppose that the statement is true for all n k for some integer k 2. That is, suppose that every integer n from 2 through k factors into a product of primes, for some integer k 2. Consider the integer k + 1. Either it factors or it does not factor. If it does not factor, then it is prime and we are done. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

12 Prime Factorization So suppose that it does factor, say n = rs for some integers r and s with 2 r < k + 1 and 2 s < k + 1. Then, by the induction hypothesis, r and s factor into products of primes. Therefore, k + 1 factors into a product of primes. Therefore, all integers n 2 factor into a product of primes. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

13 Outline 1 The Principle 2 The Method 3 Examples Prime Factorization The Checkerboard Puzzle Trees Binary Strings 4 Assignment Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

14 The Generalized Checkboard Puzzle Theorem Given a checkboard with an even number of squares, if we remove any two squares of opposite color, the remaining squares can be covered with 1 2 and 2 1 tiles. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

15 A Lemma Lemma For any integers m and n, if the diagonally opposite corners of an m n checkerboard are of opposite colors, then either m is even and n is odd or m is odd and n is even. If m and n are both odd, then each row and column will end with the same color that it started with. If m and n are both even, then each row and column will end with the opposite color that it started with. Either way, diagonally opposite corners would be the same color. Therefore, if diagonally opposite corners are of opposite colors, then one of m and n must be even and the other one odd. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

16 Inductive Proof of the Theorem Let m = 2 and n = 1 and remove two squares of opposite from a 2 1 checkerboard. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

17 Inductive Proof of the Theorem There are no squares left, so they can be covered (vacuously) by 2 1 and 1 2 tiles. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

18 Inductive Proof of the Theorem Let m and n be two integers, at least one of which is even, and suppose that the theorem is true for all smaller checkerboards with an even number of squares Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

19 Inductive Proof of the Theorem Remove any two squares of opposite colors. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

20 Inductive Proof of the Theorem Consider the bounding rectangle of the removed squares. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

21 Inductive Proof of the Theorem It either equals the original checkerboard or it is smaller. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

22 Inductive Proof of the Theorem Case 1: Suppose it is smaller. Then, by the induction hypothesis, it can be tiled. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

23 Inductive Proof of the Theorem We must show that the remainder of the checkerboard can also be tiled. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

24 Inductive Proof of the Theorem We may assume that the original checkerboard has an even number of columns. (Why?) Even Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

25 Inductive Proof of the Theorem The smaller checkerboard must have one even dimension and one odd dimension. Even Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

26 Inductive Proof of the Theorem Case 1-A: Suppose that it has an even number of rows and an odd number of columns. Even Odd Even Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

27 Inductive Proof of the Theorem Then we can tile the checkerboard as shown. Even Odd Even Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

28 Inductive Proof of the Theorem Case 1-B: Suppose that it has an odd number of rows and an even number of columns. Odd Even Even Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

29 Inductive Proof of the Theorem Then we can tile the checkerboard as shown. Odd Even Even Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

30 Inductive Proof of the Theorem Case 2: Suppose the smaller checkerboard equals the original checkerboard. Then the two removed squares must be in diagonally opposite corners. Odd Even Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

31 Inductive Proof of the Theorem Then we can tile the checkerboard as shown. Odd Even Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

32 Inductive Proof of the Theorem There is a gap in the previous proof. Where is it? Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

33 Outline 1 The Principle 2 The Method 3 Examples Prime Factorization The Checkerboard Puzzle Trees Binary Strings 4 Assignment Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

34 Trees Definition (Tree) A graph is a tree with at least one vertex if it is connected and contains no cycles. A Tree Not a Tree Not a Tree Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

35 Trees Theorem If a connected graph with n 1 vertices is a tree, then it has exactly n 1 edges. When n = 1, there is only one vertex. If there were an edge, then it would connect that vertex to itself, creating a cycle. Therefore, there are 0 edges and the statement is true when n = 1. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

36 Trees Suppose that the statement is true for all n k for some k 1. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

37 Trees Let G be a graph with k + 1 vertices Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

38 Trees V Select any vertex V in the graph Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

39 Trees e 1 V e 2 e 4 e 3 Let m be the number of edges incident to V Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

40 Trees e 1 V e 2 e 4 e 3 m must be at least 1. Why? Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

41 Trees Remove V and the incident edges Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

42 Trees There are m separate component graphs G i. Why? Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

43 Trees Each component G i is a tree. Why? Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

44 Trees 9 vertices 8 edges 8 vertices 7 edges 7 vertices 6 edges 5 vertices 4 edges Each component G i has k i vertices and k i 1 edges Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

45 Trees The total number of edges in the components is (k 1 1) + + (k m 1) = (k 1 + k m ) m = k m. Now add back in the 1 vertex and m edges that we removed, and we have k + 1 vertices and k edges. Therefore, the statement is true when n = k + 1. Therefore, the statement is true for all n 1. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

46 Trees It is possible to give induction proof based on standard induction. The basic case is the same as before. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

47 Trees Suppose that the statement is true when n = k for some integer k 1. That is, suppose that any tree with k vertices has exactly k 1 edges. Let G be a tree with k + 1 edges. G must have a vertex of index 1. That is, there must be a vertex that is incident to only 1 edge. Remove that vertex and the incident edge, creating the graph G. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

48 Trees The graph G is a tree (why?) and it has k vertices. So G has exactly k 1 edges. Thus, G has exactly k edges. So the statement is true when n = k + 1. Therefore, it is true for all n 1. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

49 Outline 1 The Principle 2 The Method 3 Examples Prime Factorization The Checkerboard Puzzle Trees Binary Strings 4 Assignment Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

50 Binary Strings Theorem Let S be the set of all binary strings with an equal number of 0 s and 1 s. Then every string x S is of the form x = 0s1, where s S, x = 1s0, where s S, or x = st, where s, t S. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

51 Binary Strings Let x S have length n. Clearly, if x begins with 0 and ends with 1, or begins with 1 and ends with 0, then it must be in the form 0s1 or 1s0 for some s S. So, suppose that x begins and ends with 0 or begins and ends with 1. Without loss of generality, assume that x begins and ends with 0. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

52 Binary Strings Let x = x 1 x 2... x n 1 x n, where x 1 = 0 and x n = 0. Let k i be the number of 0 s minus the number of 1 s in the first i digits. Clearly, k 0 = 0, k 1 = 1, k n 1 = 1, k n = 0. For all i, the change from k i to k i+1 is either +1 or 1, depending on whether x i+1 = 0 or x i+1 = 1. Therefore, for some j, with 2 i n 2, we must have k j = 0. Then let s = x 1... x j and t = x j+1... x n. Then s, t S and x = st. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

53 Outline 1 The Principle 2 The Method 3 Examples Prime Factorization The Checkerboard Puzzle Trees Binary Strings 4 Assignment Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

54 Collected Collected Sec. 4.8: 16. Sec. 5.1: 15, 44. Sec. 5.2: 14, 26. Sec. 5.3: 10, 18. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

55 Assignment Assignment Read Section 5.4, pages Exercises 1, 6, 7, 8, 10, 11, 12, 17, page 277. Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, / 34

Direct Proof Division into Cases

Direct Proof Division into Cases Lecture 16 Section 4.4 Robb T. Koether Hampden-Sydney College Mon, Feb 10, 2014 Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 10, 2014