am_lect_18 - 1 P(0 2 ∀ k ∈ N ∀ m ≤ k P m → P k 1...

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Induction: A geometric example A triomino is a 2 × 2 square with one square missing: Claim : n 1, any 2 n × 2 n checkerboard with one square missing can be tiled with triominos Proof by induction on n : Base case ( n = 1): A 2 1 × 2 1 checkerboard with one square missing is a triomino, so it can obviously be tiled Inductive step: Let k N such that any 2 k × 2 k checkerboard with one square missing can be tiled with triominos [IH] Consider a 2 k +1 × 2 k +1 checkerboard with one square missing. This is clearly made up of three complete 2 k × 2 k checkerboards and one 2 k × 2 k checkerboard with one square missing. Place one triomino so that it covers one square of each complete checkerboard. We are left with the problem of tiling four 2 k × 2 k checkerboard each “missing” one square, which can by done by the IH. This completes the proof by induction.
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Strong Induction Normal induction: In order to prove: n N , P ( n ) … we prove two things: 1) P (0) 2) k N , P ( k ) P ( k +1) Strong induction: In order to prove: n N , P ( n ) … we prove two things:
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Unformatted text preview: 1) P (0) 2) ∀ k ∈ N , ( ∀ m ≤ k , P ( m )) → P ( k +1) Inductive hypothesis Strong inductive hypothesis Example Claim : Every postage value n ≥ 12 cents can be made up of 4-cent and 5-cent stamps — Mathematically, ∀ n ≥ 12, ∃ a ∈ N , ∃ b ∈ N , n = 4 a + 5 b Proof by induction on n : Base cases ( n = 12, 13, 14, 15): Since 12 = 4 × 3+5 × 0, 13 = 4 × 2+5 × 1, 14 = 4 × 1+5 × 2, 15 = 4 × 0+5 × 3, the base cases all hold. Inductive step: Let k ≥ 15 such that ∀ m ≤ k and m ≥ 12, ∃ a ∈ N , ∃ b ∈ N , m = 4 a + 5 b [strong IH] We need to show that ∃ p ∈ N , ∃ q ∈ N , k +1 = 4 p + 5 q Now k +1 = ( k − 3)+4. Since k ≥ 15, k − 3 ≥ 12 and hence by the strong IH, ∃ a ∈ N , ∃ b ∈ N , k − 3 = 4 a + 5 b Hence k +1 = 4 a + 5 b + 4 = 4( a +1) + 5 b . Hence the statement holds for k +1, which completes the proof by (strong) induction....
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am_lect_18 - 1 P(0 2 ∀ k ∈ N ∀ m ≤ k P m → P k 1...

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