a4s - Concordia University Department of CSE COMP 232...

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Unformatted text preview: Concordia University Department of CSE COMP 232 Mathematics for Computer Science Solutions to Assignment 4 1. Use strong induction to prove that for every integer n 8 there exist non-negative integers x,y such that n = 3 x + 5 y . SOLUTION: Basis : n = 8 : Take x = y = 1. n = 9 : Take x = 3 ,y = 0. n = 10 : Take x = 0; y = 2. Inductive hypothesis : Assume that for some n 10, and for all m such that 8 m n , there exist non-negative integers x,y such that m = 3 x + 5 y . Inductive step : Now consider n + 1. Since n + 1- 3 = n- 2 8, using the inductive hypothesis, there exist x and y such that n- 2 = 3 x + 5 y . Then n + 1 = 3( x + 1) + 5 y . This completes the inductive step. Remark: This problem can also be solved using standard induction. For the basis, if n = 8 then take x = y = 1. For the inductive step, suppose for some n ( n 8) we have n = 3 x + 5 y . Then n + 1 = 3 x + 5 y + 1 = 3 x + 5 y + 6- 5 = 3( x + 2) + 5( y- 1). Now y- 1 could be negative if y = 0. In this case n = 3 x , and we must have x > = 3. Thus in this case we can write n + 1 = 3 x + 5 y + 1 = 3 x + 5 y + 10- 9 = 3( x- 3) + 5( y + 2) where x- 3 0. 2. The Fibonacci numbers are defined as follows: f = 0 ,f 1 = 1, and f n +2 = f n + f n +1 whenever n 0. Prove that when n is a positive integer: f- f 1 + f 2 + ...- f 2 n- 1 + f 2 n = f 2 n- 1- 1 SOLUTION: Basis : n = 1. f- f 1 + f 2 = 0- 1 + 1 = 0 = f 1- 1, since f = 0 and f 1 = 1. Inductive hypothesis : Suppose for some positive integer n : f- f 1 + f 2- - f 2 n- 1 + f 2 n = f 2 n- 1- 1. Inductive step: We must show that f- f 1 + f 2- - f 2 n- 1 + f 2 n- f 2 n +1 + f 2 n +2 = f 2 n +1- 1. ( f- f 1 + f 2- - f 2 n- 1 + f 2 n )- f 2 n +1 + f 2 n +2 = f 2 n- 1- 1- f 2 n +1 + f 2 n +2 = f 2 n- 1- 1- ( f 2 n + f 2 n- 1 ) + f 2 n +1 + f 2 n = f 2 n +1- 1, as needed....
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a4s - Concordia University Department of CSE COMP 232...

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