solution5

# solution5 - = 2 m-3 (4 + 2 + 1) = 2 m-3 · 7 < 2 m-3 ·...

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Assignment 5 Solution October 2, 2009 1. Let n be a positive number. (a) Give an expression (using summation notation) for the sum of the ﬁrst n odd integers. (Note: this is not the sum of the odd integers n .) n X k =1 2 k - 1 (b) Prove by mathematical induction that this sum is equal to n 2 for all n 0. Proof of n X k =1 2 k - 1 = n 2 by induction on n for all n 1. i) n = 0: sum of 0 odd integers is 0. 0 2 = 0. ii) Let m X k =1 2 k - 1 = m 2 m +1 X k =1 2 k - 1 = ± m X k =1 2 k - 1 ! + 2( m + 1) - 1 = m 2 + 2( m + 1) - 1 = m 2 + 2 m + 1 = ( m + 1)( m + 1) = ( m + 1) 2 1

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3. Prove by mathematical induction that for every integer n 4 there exist integers p and q such that n = 2 p + 5 q . Proof by induction on n 4. i) n = 4: 4 = 2 · 2 + 5 · 0 = 4 p = 2, q = 0 ii) Assume m = 2 p + 5 q for p,q Z . m + 1 = 2 p + 5 q + 1 = 2 p + 5 q + 1 + 4 - 4 + 5 - 5 = (2 p - 4) + (5 q + 5) + (1 + 4 - 5) = 2( p - 2) + 5( q + 1) + 0 = 2( p - 2) + 5( q + 1) 5. Consider the following sequence: a 1 = 1; a 2 = 2; a 3 = 3; and for k 4, a k = a k - 1 + a k - 2 + a k - 3 . Prove using strong induction that a n < 2 n for all n > 0. Proof of a n < 2 n for n > 0 by induction on n . i) n = 1: a 1 = 1 < 2 = 2 1 a 1 < 2 1 ii) Suppose that a i < 2 i for all i < m . Then, we have a m - 1 < 2 m - 1 a m - 2 < 2 m - 2 a m - 3 < 2 m - 3 . . . a 1 < 2 1 2
a m = a m - 1 + a m - 2 + a m - 3 < 2 m - 1 + 2 m - 3 + 2 m - 3 = 2 m - 3 (2 2 + 2 1 + 2 0 )

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Unformatted text preview: = 2 m-3 (4 + 2 + 1) = 2 m-3 · 7 < 2 m-3 · 8 = 2 m-3 · 2 3 = 2 m a m < 2 m 7. Prove using strong induction that for any propositional formula P that is in negation-normal form, ¬ P is equivalent to a formula in negation-normal form. Proof by induction on P ≥ NNF i) P ≡ q : ¬ P = ¬ q ∈ NNF by deﬁnition of NNF . P ≡ ¬ q : ¬ P = ¬¬ q = q ∈ NNF by deﬁnition of NNF . ii) Assume that S and T satisﬁes the property. We have S ∈ NNF , ¬ S ≡ S where S ∈ NNF , and T ∈ NNF , ¬ T ≡ T where T ∈ NNF P ≡ S ∧ T : ¬ P ≡ ¬ ( S ∧ T ) ≡ ¬ S ∨ ¬ T ≡ S ∨ T ∈ NNF P ≡ S ∨ T : ¬ P ≡ ¬ ( S ∨ T ) ≡ ¬ S ∧ ¬ T ≡ S ∧ T ∈ NNF 3 P ≡ S → T : ¬ P ≡ ¬ ( S → T ) ≡ ¬ ( ¬ S ∨ T ) ≡ ¬¬ S ∧ ¬ T ≡ S ∧ T ∈ NNF 4...
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## This note was uploaded on 10/04/2009 for the course CMPSC 360 taught by Professor Haullgren during the Fall '08 term at Penn State.

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solution5 - = 2 m-3 (4 + 2 + 1) = 2 m-3 · 7 < 2 m-3 ·...

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