lect14_1 - Induction proofs To prove that some proposition...

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1 Induction proofs To prove that some proposition P ( n ) is true for all n n 0 by induction on n n 0 1. Basis : Check that P ( n 0 ) is true. 2. Inductive step . Assume that P ( n ) is true for n = k , where k is some integer, k n 0 Prove that P ( n ) is true for n = k +1 So, in inductive step we prove that for any n n 0 P ( n ) P ( n +1) Usually this is simpler than prove directly that for any n n 0 P ( n ).
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2 Example. Let P ( n ) : n 3 - n is divisible by 6 To prove that P ( n ) is true for all n 0: 1. Basis. P (0): “0 3 - 0=0 is divisible by 6”. P (0) is true 2. Inductive step P ( n = k ) P ( n = k +1), k 0. Assume P ( n = k ) is true, where k is some integer, k 0. So, we assume that k 3 - k is divisible by 6 i.e. k 3 - k = 6 s ( s is any integer) Goal: to prove that P ( n = k +1) is true, i. e. that ( k +1) 3 - ( k +1) is divisible by 6. ( k +1) 3 - ( k +1) = k 3 +3 k 2 + 3 k +1 - k - 1 = = ( k 3 - k ) + 3 k 2 + 3 k = = 6 s + 3 k ( k + 1)
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3 Suppose we want to prove the formula for the sum of ‘geometric progression’ q q q ... q q q n n - - = + + + + + + 1 1 1 1 3 2 where 1 0 < < q , and n is any integer, n 0 How can we prove that the formula is correct for any n 0 ? Notation: = = + + + + = n i i n n q q ... q q S 0 2 1 q q S n n - - = + 1 1 1 Show it for arbitrary fixed n 0 (prove directly, without induction) Use induction on n 0 Two alternatives:
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4 Show it for some n 0 (prove P ( n ) for any n directly) 1 2 1 2 1 ) (1 + + + - + + + + = + + + = n n n n q q ... q q q ... q q S q n S 1 1 - + = + n n n q S S q 1 1 ) (1 + - = - n n q q S q q S n n - - = + 1 1 1
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5 1) Base: n =0 1; 0 0 0 0 = = = = q q S i i 1 1 1 1 1 1 0 = - - = - - + q q q q 2) Induction Hypothesis : Assume the formula is correct for n = k , where k is some integer, k 0 q q S k k - - = + 1 1 1 Induction Step : prove the formula for n = k +1, i. e. q q S k k - - = + + + 1 1 1 1) ( 1 1 1 2 1 1 + + + + = + + + + + = k k k k k q S q q ... q q S q q q q q q q q q k k k k k k - - = - - + - = + - - = + + + + + + + + 1 1 1 1 1 1 1 1) ( 1 1) ( 1 1 1 1 Proof by induction on n 0 that q q S n n - - = + 1 1 1
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6 ( 2200 n n 0 ) P ( n ) P ( n 0 ) ( 2200 n n 0 ) [ P ( n ) P ( n +1)] ? Induction Principle . If A is a subset of N that satisfies two properties: 1) n 0 A 2) 2200 n n 0 [ n A ( n +1) A ], then A = N We just can think about A as a set of integers, that satisfy P : A ={ n N | P ( n )} Then, 1) n 0 A P ( n 0 ) 2) 2200 n n 0 [ n A ( n +1) A ] 2200 n n 0 ( P ( n ) P ( n +1))
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7 Although it sounds very obvious, the induction
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lect14_1 - Induction proofs To prove that some proposition...

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