Mathematical Induction. Part Three


 Joanna McDonald
 1 years ago
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1 Mathematical Induction Part Three
2 Announcements Problem Set Two due Friday, October 4 at 2:5PM (just before class) Stop by office hours with questions! us at with questions! Problem session tonight in at 7:00PM. Hope to see you there! Francisco's office hours moved to Hewlett 20 (here!), 3:30 5:30 today.
3 Strong Induction
4 The Unstacking Game You are given a stack of n boxes. Each turn, you take a stack of at least one box, then split it into two smaller stacks. You then earn points equal to the product of the height of the two stacks. Once every stack has height one, the game is over. Adapted from material by Prof. Dieter van Melkebeek of the University of Wisconsin
5 The Unstacking Game: An Example Score: 0
6 The Unstacking Game: An Example Score: 0
7 The Unstacking Game: An Example Score: 0
8 The Unstacking Game: An Example Score: = 6
9 The Unstacking Game: An Example Score: = 6
10 The Unstacking Game: An Example Score: 6
11 The Unstacking Game: An Example Score: 6
12 The Unstacking Game: An Example Score: 6
13 The Unstacking Game: An Example Score: 6 =
14 The Unstacking Game: An Example Score: 7 =
15 The Unstacking Game: An Example Score: 7
16 The Unstacking Game: An Example Score: 7
17 The Unstacking Game: An Example Score: 7
18 The Unstacking Game: An Example Score: 7 2 = 2
19 The Unstacking Game: An Example Score: 9 2 = 2
20 The Unstacking Game: An Example Score: 9
21 The Unstacking Game: An Example Score: 9
22 The Unstacking Game: An Example Score: 9
23 The Unstacking Game: An Example Score: 9 =
24 The Unstacking Game: An Example Score: 0 =
25 The Unstacking Game: An Example Score: 0
26 Score: 0 Another Example
27 Score: 0 Another Example
28 Score: 0 Another Example
29 Another Example Score: 4 4 = 4
30 Score: 4 Another Example
31 Score: 4 Another Example
32 Score: 4 Another Example
33 Another Example Score: 4 3 = 3
34 Another Example Score: 7 3 = 3
35 Score: 7 Another Example
36 Score: 7 Another Example
37 Score: 7 Another Example
38 Another Example Score: 7 2 = 2
39 Another Example Score: 9 2 = 2
40 Score: 9 Another Example
41 Score: 9 Another Example
42 Score: 9 Another Example
43 Another Example Score: 9 =
44 Another Example Score: 0 =
45 Score: 0 Another Example
46 What is the optimal strategy for the unstacking game?
47 Try it! Play the unstacking game with 6 boxes in the stack. How many points can you get?
48 An Observation It looks like every strategy seems to get fifteen points. The two strategies we adopted for the game with five boxes both ended up giving ten points. Do you get the same number of points, regardless of your strategy, if you start with the same number of boxes?
49
50 0 Points
51 0 Points
52 = 0 Points
53 0 Points Point =
54 0 Points Point =
55 0 Points Point = 2 = 2 =
56 0 Points Point = 3 Points 2 = 2 =
57 0 Points Point = 3 Points 2 = 2 =
58 0 Points Point = 3 Points 2 = 2 = 3 = 3 2 = 2 =
59 0 Points Point = 3 Points 2 = 2 = 3 = 3 2 = 2 =
60 0 Points Point = 3 Points 2 = 2 = 3 = 3 2 = 2 = 2 2 = 4 = =
61 0 Points Point = 3 Points 2 = 2 = 6 Points 3 = 3 2 = 2 = 2 2 = 4 = =
62 Unstacking Game Scores n =, score is 0 n = 2, score is n = 3, score is 3 n = 4, score is 6 n = 5, score is 0 n = 6, score is 5
63 Unstacking Game Scores n =, score is 0 n = 2, score is = n = 3, score is 3 = + 2 n = 4, score is 6 = n = 5, score is 0 = n = 6, score is 5 =
64 Unstacking Game Scores n =, score is 0 n = 2, score is = n = 3, score is 3 = + 2 n = 4, score is 6 = n = 5, score is 0 = n = 6, score is 5 = n = k, score is (?) k
65 Unstacking Game Scores n =, score is 0 n = 2, score is = n = 3, score is 3 = + 2 n = 4, score is 6 = n = 5, score is 0 = n = 6, score is 5 = n = k, score is (?) k(k ) / 2
66 Empirically, it looks like Score for n = n(n ) / 2 How would we prove this?
67 The Intuition
68 The Intuition n blocks
69 The Intuition
70 k blocks The Intuition
71 The Intuition
72 The Intuition
73 The Intuition We are now playing two independent games of the unstacking game!
74 The Intuition
75 The Intuition n k blocks in this game k blocks in this game
76 n k blocks in this game The Intuition Total score: k (n k) for making the split + (score for n k block game) + (score for k block game) k blocks in this game
77 Proof Idea Proceed by induction! Want to show: Score for n blocks is n(n ) / 2. Show that the score for one block is ( )/2 = 0 as a base case. Show that when we split a stack into two pieces, the sum of the score for that move, the total score for the left tower subgame, and the total score for the right tower subgame comes out to n(n ) / 2 regardless of what split we make.
78 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n the theorem holds and consider a stack with n + blocks in it. Suppose we take k boxes off of the stack, leaving two stacks, one of size k and one of size n + k. The score for this move is k(n + k). By the inductive hypothesis, our score for the two smaller stacks are k(k )/2 and (n + k)(n k) / 2, respectively.
79 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n the theorem holds and consider a stack with n + blocks in it. Suppose we take k boxes off of the stack, leaving two stacks, one of size k and one of size n + k. The score for this move is k(n + k). By the inductive hypothesis, our score for the two smaller stacks are k(k )/2 and (n + k)(n k) / 2, respectively.
80 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n the theorem holds and consider a stack with n + blocks in it. Suppose we take k boxes off of the stack, leaving two stacks, one of size k and one of size n + k. The score for this move is k(n + k). By the inductive hypothesis, our score for the two smaller stacks are k(k )/2 and (n + k)(n k) / 2, respectively.
81 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n the theorem holds and consider a stack with n + blocks in it. Suppose we take k boxes off of the stack, leaving two stacks, one of size k and one of size n + k. The score for this move is k(n + k). By the inductive hypothesis, our score for the two smaller stacks are k(k )/2 and (n + k)(n k) / 2, respectively.
82 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n the theorem holds and consider a stack with n + blocks in it. Suppose we take k boxes off of the stack, leaving two stacks, one of size k and one of size n + k. The score for this move is k(n + k). By the inductive hypothesis, our score for the two smaller stacks are k(k )/2 and (n + k)(n k) / 2, respectively.
83 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n the theorem holds and consider a stack with n + blocks in it. Suppose we take k boxes off of the stack, leaving two stacks, one of size k and one of size n + k. The score for this move is k(n + k). By the inductive hypothesis, our score for the two smaller stacks are k(k )/2 and (n + k)(n k) / 2, respectively.
84 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n the theorem holds and consider a stack with n + blocks in it. Suppose we take k boxes off of the stack, leaving two stacks, one of size k and one of size n + k. The score for this move is k(n + k). By the inductive hypothesis, our score for the two smaller stacks are k(k )/2 and (n + k)(n k) / 2, respectively.
85 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n the theorem holds and consider a stack with n + blocks in it. Suppose we take k boxes off of the stack, leaving two stacks, one of size k and one of size n + k. The score for this move is k(n + k). By the inductive hypothesis, our score for the two smaller stacks are k(k )/2 and (n + k)(n k) / 2, respectively.
86 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n the theorem holds and consider a stack with n + blocks in it. Suppose we take k boxes off of the stack, leaving two stacks, one of size k and one of size n + k. The score for this move is k(n + k). By the inductive hypothesis, our score for the two smaller stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Is this legal?
87 Induction, Graphically
88 Induction, Graphically
89 Induction, Graphically block n blocks
90 Induction, Graphically block n blocks We assume P(n) as our inductive hypothesis. This means that if we ever have a stack of exactly n blocks, we know what the total score will be.
91 Induction, Graphically
92 Induction, Graphically
93 Induction, Graphically Neither of these stacks have size n, so we can't assume anything about the total score we'll get!
94 An Important Detail In a proof by induction, the inductive step works by showing, for some n, that P(n) P(n + ). There is no guarantee that P(n') holds for any n' < n. Induction only lets you use the immediately previous result to prove the next result. This is often not desirable.
95 The principle of strong induction states that if for some property P(n), we have that P(0) is true and For any natural number n, if P(n') is true for all n' n, then P(n + ) is true then For any natural number n, P(n) is true.
96 The principle of strong induction states that if for some property P(n), we have that Assume that P(n) holds for n and all smaller n. P(0) is true and For any natural number n, if P(n') is true for all n' n, then P(n + ) is true then For any natural number n, P(n) is true.
97 Induction and Dominoes Weak Induction: The zeroth domino falls. If the nth domino falls, the (n + )st domino falls. Strong induction: The zeroth domino falls. If dominoes 0,, 2,, n all fall, then the (n + )st domino falls.
98 Weak and Strong Induction Weak induction works by assuming that P(n) holds and showing P(n + ). Strong induction works by assuming P(n) is true and that P(n') is true for any n' < n, then showing P(n + ). Weak induction is good for proofs that work by peeling off one piece of a large structure. Fall down from the n + case to the n case. Strong induction is good for proofs that work by splitting a large structure down into one or more smaller pieces. Fall down from the n + case to some number of cases no larger than n.
99 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. The score for this move is k(n + k). This leaves two stacks, one of size k and one of size (n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
100 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. The score for this move is k(n + k). This leaves two stacks, one of size k and one of size (n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
101 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' Note in the range that we n' explicitly n. Consider mention a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the we're using strong induction here, score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. The just score as we for this would move say is k(n we + were k). This leaves two stacks, one of size k and proceeding one of size (n + by induction, k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n contradiction, + k)(n k) / 2, or respectively. contrapositive. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
102 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. The score for this move is k(n + k). This leaves two stacks, one of size k and one of size (n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
103 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. The score for this move is k(n + k). This leaves two stacks, one of size k and one of size (n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
104 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. The score for this move is k(n + k). This leaves two stacks, one of size k and one of size (n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, Note respectively. the strong Then the version total score of for the game is the inductive hypothesis. We assume that the theorem is true for all n' from up to some value n, not just that it's true for n. Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
105 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. The score for this move is k(n + k). This leaves two stacks, one of size k and one of size (n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
106 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. The score for this move is k(n + k). This leaves two stacks, one of size k and one of size (n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
107 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. The score for this move is k(n + k). This leaves two stacks, one of size k and one of size (n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
108 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). and one of size (n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
109 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
110 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
111 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Now we can use the stronger version of the inductive hypothesis to assume we know this is already true. A weak Since our choice of k was arbitrary, the total score for n boxes is the same induction would fail here. regardless of what move we make. So the claim holds for n +, completing the proof.
112 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
113 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is k (n+ k)+ k (k ) (n+ k )(n k) Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
114 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is k (n+ k )+ k (k ) 2 + (n+ k)(n k) 2 = 2 k (n+ k) 2 + k (k ) + (n+ k)(n k) 2 2 Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
115 = Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is k (n+ k )+ k (k ) 2 + (n+ k)(n k) 2 = 2 k (n+ k) 2 + k (k ) + (n+ k)(n k) k (n+ k)+k (k )+(n+ k )(n k ) = 2 kn+2 k 2 k 2 +k 2 k+n 2 +n nk kn k+k Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
116 = Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is k (n+ k )+ k (k ) 2 + (n+ k)(n k) 2 = 2 k (n+ k) 2 + k (k ) + (n+ k)(n k) k (n+ k)+k (k )+(n+ k )(n k ) = 2 kn+2 k 2 k 2 +k 2 k+n 2 +n nk kn k+k Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
117 = Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is k (n+ k )+ k (k ) 2 + (n+ k)(n k) 2 = 2 k (n+ k) 2 + k (k ) + (n+ k)(n k) k (n+ k )+k (k )+(n+ k)(n k) = 2 kn+2 k 2 k 2 +k 2 k+n 2 +n nk kn k+k Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
118 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is k (n+ k )+ k (k ) 2 + (n+ k)(n k) 2 = 2 k (n+ k) 2 + k (k ) + (n+ k)(n k) k (n+ k)+k (k )+(n+ k )(n k ) = = 2 kn+2 k 2 k 2 +k 2 k+n 2 +n nk kn k+k = n2 +n = (n+)n 2 2 Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
119 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is k (n+ k )+ k (k ) 2 + (n+ k)(n k) 2 = 2 k (n+ k) 2 + k (k ) + (n+ k)(n k) k (n+ k)+k (k )+(n+ k )(n k ) = = 2 kn+2 k 2 k 2 +k 2 k+n 2 +n nk kn k+k = n2 +n = (n+)n 2 2 Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
120 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is k (n+ k )+ k (k ) 2 + (n+ k)(n k) 2 = 2 k (n+ k) 2 + k (k ) + (n+ k)(n k) k (n+ k)+k (k )+(n+ k )(n k ) = = 2 kn+2 k 2 k 2 +k 2 k+n 2 +n nk kn k+k = n2 +n = (n+)n 2 2 Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
121 Theorem: For any n, any sequence of moves in the unstacking game gives a score of n(n ) / 2. Proof: By strong induction. As a base case, if n =, the game is over because all stacks have size one. Our score is then 0 = ( )/2, so the theorem holds for n =. For the inductive step, assume that for some n that the theorem holds for any n' in the range n' n. Consider a stack with n + blocks in it. We need to show that no matter what sequence of moves we make on this stack, the score is (n + )n / 2. Suppose that for arbitrary k, we take k boxes off of the stack. This leaves two stacks, one of size k and one of size (n + k). The score for this move is k(n + k). By the inductive hypothesis, no matter what moves we make on the two stacks, the score for the two stacks are k(k )/2 and (n + k)(n k) / 2, respectively. Then the total score for the game is k (n+ k )+ k (k ) 2 + (n+ k)(n k) 2 = 2 k (n+ k) 2 + k (k ) + (n+ k)(n k) k (n+ k)+k (k )+(n+ k )(n k ) = = 2 kn+2 k 2 k 2 +k 2 k+n 2 +n nk kn k+k = n2 +n = (n+)n 2 2 Since our choice of k was arbitrary, the total score for n boxes is the same regardless of what move we make. So the claim holds for n +, completing the proof.
122 Induction and Intuition Often, a proof by induction can be used to prove a result without giving any intuition for why the result is true. To gain an understanding of why the result is true, you often need to look for a different proof. Extra Credit Opportunity (worth 5 points on the next problem set): Come up with an intuitive explanation for the previous result, then formalize it with a proof. You don't need to use induction.
123 Induction and Intuition Often, a proof by induction can be used to prove a result without giving any intuition for why the result is true. To gain an understanding of why the result is true, you often need to look for a different proof. Extra Credit Opportunity (worth 5 points on the next problem set): Come up with an intuitive explanation for the previous result, then formalize it with a proof. You don't need to use induction if it isn't necessary.
124 Application: Continued Fractions
125 Continued Fractions
126 Continued Fractions
127 Continued Fractions
128 Continued Fractions
129 Continued Fractions
130 Continued Fractions
131 Continued Fractions 4 3
132 Continued Fractions 4 3
133 Continued Fractions 3 4
134 Continued Fractions
135 Continued Fractions
136 Continued Fractions
137 Continued Fractions
138 Continued Fractions
139 Continued Fractions
140 Continued Fractions
141 Continued Fractions
142 Continued Fractions
143 Continued Fractions
144 Continued Fractions
145 Continued Fractions
146 Continued Fractions
147 Continued Fractions
148 Continued Fractions 37 42
149 Continued Fractions A continued fraction is an expression of the form More formally, a continued fraction is either An integer n, or a 0 + a + a 2 + n + /F, where n is an integer and F is a continued fraction. Continued fractions have numerous applications in number theory and computer science. (They're also really fun to write!) a
150 Fun with Continued Fractions Any rational number (including negative numbers) has a continued fraction. Any irrational number has an (infinite) continued fraction. If we truncate an infinite continued fraction for an irrational number, we can get progressively better approximations of that number.
151 Pi as a Continued Fraction π=
152 Approximating Pi
153 Approximating Pi π= =
154 Approximating Pi π= = And he made the Sea of cast bronze, ten cubits from one brim to the other; it was completely round. [ A] line of thirty cubits measured its circumference. Kings 7:23, New King James Translation
155 Approximating Pi π= = /7 =
156 Approximating Pi π= = /7 = Archimedes knew of this approximation, circa 250 BCE
157 Approximating Pi π= = /7 = /06 =
158 Approximating Pi π= = /7 = /06 = /3 =
159 Approximating Pi π= = /7 = /06 = /3 = Chinese mathematician Zu Chongzhi discovered this approximation in the early fifth century; this was the best approximation of pi for over a thousand years
160 Approximating Pi π= = /7 = /06 = /3 = /3302 =
161 More Continued Fractions 4 3
162 More Continued Fractions 4 3
163 More Continued Fractions 2 3
164 More Continued Fractions 2 3
165 More Continued Fractions 2
166 More Continued Fractions 2
167 More Continued Fractions 4 3
168 More Continued Fractions =
169 More Continued Fractions =
170 An Interesting Continued Fraction x=
171 An Interesting Continued Fraction x= /
172 An Interesting Continued Fraction x= / 2 /
173 An Interesting Continued Fraction x= / 2 / 3 / 2
174 An Interesting Continued Fraction x= / 2 / 3 / 2 5 / 3
175 An Interesting Continued Fraction x= / 2 / 3 / 2 5 / 3 8 / 5
176 An Interesting Continued Fraction x= / 2 / 3 / 2 5 / 3 8 / 5 3 / 8
177 An Interesting Continued Fraction x= / 2 / 3 / 2 5 / 3 8 / 5 3 / 8 2 / 3
178 An Interesting Continued Fraction x= / 2 / 3 / 2 5 / 3 8 / 5 3 / 8 2 / 3 34 / 2
179 An Interesting Continued Fraction x= / 2 / 3 / 2 5 / 3 8 / 5 3 / 8 2 / 3 34 / 2 Each fraction is the ratio of consecutive Fibonacci numbers!
180 The Golden Ratio ϕ= = ϕ
181 The Golden Ratio 2 34
182 The Golden Ratio 2 34
183 The Golden Ratio 2 3
184 The Golden Ratio 2 3
185 The Golden Ratio 8 3
186 The Golden Ratio 8 3
187 The Golden Ratio 8 5
188 The Golden Ratio 8 5
189 The Golden Ratio 3 5
190 The Golden Ratio 3 5
191 The Golden Ratio 3 2
192 The Golden Ratio 3 2
193 The Golden Ratio 2
194 The Golden Ratio 2
195 The Golden Ratio
196 The Golden Spiral
197 The Golden Spiral
198 How do we prove all rational numbers have continued fractions?
199 Constructing a Continued Fraction 25 9
200 Constructing a Continued Fraction 25 9 =
201 Constructing a Continued Fraction 25 9 =
202 Constructing a Continued Fraction 25 9 =
203 Constructing a Continued Fraction 25 9 = = + 2 7
204 Constructing a Continued Fraction 25 9 = = + 7 2
205 Constructing a Continued Fraction 25 9 = =
206 Constructing a Continued Fraction 25 9 = = = 3 + 2
207 Constructing a Continued Fraction 25 9 = = = 3 + 2
208 Constructing a Continued Fraction 25 9 = =
209 Constructing a Continued Fraction 25 9 = =
210 Constructing a Continued Fraction 25 9 =
211 Constructing a Continued Fraction =
212 Constructing a Continued Fraction =
213 Constructing a Continued Fraction = > 7 > 2 > 2
214 The Golden Ratio 2 34
215 The Golden Ratio 2 34
216 The Golden Ratio 2 3
217 The Golden Ratio 2 3
218 The Golden Ratio 8 3
219 The Golden Ratio 8 3
220 The Golden Ratio 8 5
221 The Golden Ratio 8 5
222 The Golden Ratio 3 5
223 The Golden Ratio 3 5
224 The Golden Ratio 3 2
225 The Golden Ratio 3 2
226 The Golden Ratio 2
227 The Golden Ratio 2
228 The Golden Ratio
229 The Division Algorithm For any integers a and b, with b 0, there exists unique integers q and r such that and a = qb + r 0 r < b q is called the quotient and r is called the remainder. If both a and b are nonnegative, then both q and r are nonnegative. Example: Given a = and b = 4: = Example: Given a = 37 and b = 42: 37 =
230 Theorem: Every nonnegative rational number can be expressed as a continued fraction. Proof: By strong induction. Let P(n) be any nonnegative rational number with denominator n can be represented as a continued fraction. We prove that P(n) is true for positive natural numbers. To show P(), note that any rational number a / can be written as the (trivial) continued fraction a. For the inductive step, assume that for all n' n, P(n') holds. We prove that P(n + ) holds, that is, that any nonnegative rational number a / (n + ) can be written as a continued fraction. Consider any nonnegative rational number a / (n + ). Using the division algorithm, write a = q(n + ) + r, with 0 r < n +. If r = 0, then we have that a = q(n + ), so a/(n + ) = q and the rational number a/(n + ) can be written as the trivial continued fraction q. Otherwise, r > 0. Notice that a q(n + ) = r, so Notice, therefore, that which is welldefined because r > 0. But we know that r < n +, so by the inductive hypothesis the nonnegative rational number (n + )/r can be written as a continued fraction. If we call this continued fraction F, then we have that a / (n + ) can be written as q + /F, which is a continued fraction. Thus P(n + ) holds, completing the proof by induction.
231 Theorem: Every nonnegative rational number can be expressed as a continued fraction. Proof: By strong induction. Let P(n) be any nonnegative rational number with denominator n can be represented as a continued fraction. We prove that P(n) is true for positive natural numbers. To show P(), note that any rational number a / can be written as the (trivial) continued fraction a. For the inductive step, assume that for all n' n, P(n') holds. We prove that P(n + ) holds, that is, that any nonnegative rational number a / (n + ) can be written as a continued fraction. Consider any nonnegative rational number a / (n + ). Using the division algorithm, write a = q(n + ) + r, with 0 r < n +. If r = 0, then we have that a = q(n + ), so a/(n + ) = q and the rational number a/(n + ) can be written as the trivial continued fraction q. Otherwise, r > 0. Notice that a q(n + ) = r, so Notice, therefore, that which is welldefined because r > 0. But we know that r < n +, so by the inductive hypothesis the nonnegative rational number (n + )/r can be written as a continued fraction. If we call this continued fraction F, then we have that a / (n + ) can be written as q + /F, which is a continued fraction. Thus P(n + ) holds, completing the proof by induction.
232 Theorem: Every nonnegative rational number can be expressed as a continued fraction. Proof: By strong induction. Let P(n) be any nonnegative rational number with denominator n can be represented as a continued fraction. We prove that P(n) is true for positive natural numbers. To show P(), note that any rational number a / can be written as the (trivial) continued fraction a. For the inductive step, assume that for all n' n, P(n') holds. We prove that P(n + ) holds, that is, that any nonnegative rational number a / (n + ) can be written as a continued fraction. Consider any nonnegative rational number a / (n + ). Using the division algorithm, write a = q(n + ) + r, with 0 r < n +. If r = 0, then we have that a = q(n + ), so a/(n + ) = q and the rational number a/(n + ) can be written as the trivial continued fraction q. Otherwise, r > 0. Notice that a q(n + ) = r, so Notice, therefore, that which is welldefined because r > 0. But we know that r < n +, so by the inductive hypothesis the nonnegative rational number (n + )/r can be written as a continued fraction. If we call this continued fraction F, then we have that a / (n + ) can be written as q + /F, which is a continued fraction. Thus P(n + ) holds, completing the proof by induction.
233 Theorem: Every nonnegative rational number can be expressed as a continued fraction. Proof: By strong induction. Let P(n) be any nonnegative rational number with denominator n can be represented as a continued fraction. We prove that P(n) is true for positive natural numbers. To show P(), note that any rational number a / can be written as the (trivial) continued fraction a. For the inductive step, assume that for all n' n, P(n') holds. We prove that P(n + ) holds, that is, that any nonnegative All nonnegative rational number rational a / (n + numbers ) can be written as a continued fraction. Consider any nonnegative rational number a / (n + ). Using the division algorithm, have write the a = form q(n + a/b ) + for r, with nonnegative 0 r < n +. If r = 0, then we have that a = q(n + ), a so and a/(n b. + ) The = q induction and the rational starts number with a/(n + ) can be written as the trivial continued the denominator fraction q. here Otherwise, because r > 0. this Notice that a q(n + ) = r, so choice works well. Notice, therefore, that which is welldefined because r > 0. But we know that r < n +, so by the inductive hypothesis the nonnegative rational number (n + )/r can be written as a continued fraction. If we call this continued fraction F, then we have that a / (n + ) can be written as q + /F, which is a continued fraction. Thus P(n + ) holds, completing the proof by induction.
234 Theorem: Every nonnegative rational number can be expressed as a continued fraction. Proof: By strong induction. Let P(n) be any nonnegative rational number with denominator n can be represented as a continued fraction. We prove that P(n) is true for positive natural numbers. To show P(), note that any rational number a / can be written as the (trivial) continued fraction a. For the inductive step, assume that for all n' n, P(n') holds. We prove that P(n + ) holds, that is, that any nonnegative rational number a / (n + ) can be written as a continued fraction. Consider any nonnegative rational number a / (n + ). Using the division algorithm, write a = q(n + ) + r, with 0 r < n +. If r = 0, then we have that a = q(n + ), so a/(n + ) = q and the rational number a/(n + ) can be written as the trivial continued fraction q. Otherwise, r > 0. Notice that a q(n + ) = r, so Notice, therefore, that which is welldefined because r > 0. But we know that r < n +, so by the inductive hypothesis the nonnegative rational number (n + )/r can be written as a continued fraction. If we call this continued fraction F, then we have that a / (n + ) can be written as q + /F, which is a continued fraction. Thus P(n + ) holds, completing the proof by induction.
235 Theorem: Every nonnegative rational number can be expressed as a continued fraction. Proof: By strong induction. Let P(n) be any nonnegative rational number with denominator n can be represented as a continued fraction. We prove that P(n) is true for positive natural numbers. To show P(), note that any rational number a / can be written as the (trivial) continued fraction a. For the inductive step, assume that for all n' n, P(n') holds. We prove that P(n + ) holds, that is, that any nonnegative rational number a / (n + ) can be written as a continued fraction. Consider any nonnegative rational number a / (n + ). Using the division algorithm, write a = q(n + ) + r, with 0 r < n +. If r = 0, then we have that a = q(n + ), so a/(n + ) = q and the rational number a/(n + ) can be written as the trivial continued fraction q. Otherwise, r > 0. Notice that a q(n + ) = r, so Notice, therefore, that which is welldefined because r > 0. But we know that r < n +, so by the inductive hypothesis the nonnegative rational number (n + )/r can be written as a continued fraction. If we call this continued fraction F, then we have that a / (n + ) can be written as q + /F, which is a continued fraction. Thus P(n + ) holds, completing the proof by induction.
SMMG December 2 nd, 2006 Dr. Edward Burger (Williams College) Discovering Beautiful Patterns in Nature and Number. Fun Fibonacci Facts
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