COMP232 - Mathematics for Computer Science

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1 COMP3 - Mathematics for Computer Science Tutorial 10 Ali Moallemi moa Iraj Hedayati h Concordia University, Winter 016 Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 1 / 11

2 Table of Contents Mathematical Induction Exercise 4 Exercise 6 Exercise 9 Exercise 18 Exercise 33 Exercise 4 Exercise 45 Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science / 11

3 Exercise 4 Let P (n) be the statement that n 3 = ( n(n+1) ) for the positive integer n. a) What is the statement P (1)? 1 3 = ( 1(1+1) ) b) Show that P (1) is true, completing the basis step of the proof. Left side: 1 3 = 1 Right side: ( 1(1+1) ) = ( ) = 1 = 1 c) What is the inductive hypothesis? P (k) holds for an arbitrary positive integer k 1, k 3 = ( k(k+1) ) d) What do you need to prove in the inductive step? We have to prove that P (k) P (k + 1) Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 3 / 11

4 Exercise 4 - Cont... e) Complete the inductive step, identifying where you use the inductive hypothesis. P (k + 1) : k 3 + (k + 1) 3 = ( (k+1)(k+) ) Under assumption of P (k) is true: k 3 + (k + 1) 3 = ( k(k+1) ) + (k + 1) 3 = k (k+1) 4 + (k + 1) 3 = k (k+1) +4(k+1) 3 4 = (k+1) (k +4(k+1)) 4 = (k+1) (k +4k+4) 4 = (k+1) (k+) 4 = ( (k+1)(k+) ) This shows that P (k + 1) is true. f) Explain why these steps show that this formula is true whenever n is a positive integer. Since we proved correctness of basis step (a and b) and also completed inductive step (e), using mathematical induction method, P (n) holds for every positive integer n Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 4 / 11

5 Exercise 6 Prove that 1 1! +! + + n n! = (n + 1)! 1 whenever n is a positive integer. Using mathematical induction; Suppose P (n) : 1 1! +! + + n n! = (n + 1)! 1 1 P (1) : 1 1! = 1 and (n + 1)! 1 = (1 + 1)! 1 =! 1 = 1 (TRUE) P (k) is true. Hence 1 1! +! + + k k! = (k + 1)! 1 3 P (k) P (k + 1) P (k + 1) : 1 1! +! + + k k! + (k + 1) (k + 1)! = (k + )! 1 Using we can rewrite left side as: 1 1! +! + + k k! + (k + 1) (k + 1)! = (k + 1)! 1 + (k + 1) (k + 1)! = (k + ) (k + 1)! 1 = (k + )! 1 Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 5 / 11

6 Exercise 9 a) Find a formula for the sum of the first n even positive integers n = n(n + 1) b) Prove the formula that you conjectured in part (a). Basis step: = 1 (1 + 1) is true. Inductive Hypothesis: k = k(k + 1) Inductive step: P (k) P (k + 1) ( k)+(k +1) = k(k +1)+(k +1) = (k +1)(k +). Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 6 / 11

7 Exercise 18 Let P (n) be the statement that n! < n n, where n is an integer greater than 1. a) What is the statement P ()? Basis step because it is the first integer positive greater than 1. b) Show that P () is true, completing the basis step of the proof.! = and = 4.Also < 4 hence P () is true c) What is the inductive hypothesis? P (k) is true and k! < k k d) What do you need to prove in the inductive step? k > 1 (P (k) P (k + 1)) Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 7 / 11

8 Exercise 18 - Cont... Let P (n) be the statement that n! < n n, where n is an integer greater than 1. e) Complete the inductive step. P (k) is true means k! < k k k!(k + 1) < (k + 1)k k < (k + 1)(k + 1) k (k + 1)! < (k + 1) (k+1) f) Explain why these steps show that this inequality is true whenever n is an integer greater than 1. Since we proved correctness of basis step (a and b) and also completed inductive step (e), using mathematical induction method, P (n) holds for every positive integer n greater than or equal to. Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 8 / 11

9 Exercise 33 Prove that 5 divides n 5 n whenever n is a non-negative integer. Let P (n) be n 5 n is divisible by 5. Basis step: P (0) is true because = 0 is divisible by 5. Inductive step: Assume that P (k) is true, that is, k 5 k is divisible by 5. Then (k + 1) 5 (k + 1) = (k 5 + 5k k k + 5k + 1) (k + 1) = (k 5 k) + 5(k 4 + k 3 + k + k) is also divisible by 5, because both terms in this sum are divisible by 5. Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 9 / 11

10 Exercise 4 Prove that if A 1, A,..., A n and B are sets, then (A 1 B) (A B) (A n B) = (A 1 A A n ) B answer: P (n) : (A 1 B) (A B) (A n B) = (A 1 A A n ) B Basis Step:p(1) is trivial: A 1 B = A 1 B Inductive Step: Assume P (k) holds. We show P (k + 1) also holds. (A 1 B) (A B) (A k B) (A k+1 B) = ( ) (A 1 B) (A B) (A k B) (A k+1 B) = ( ) (A 1 A A k ) B (A k+1 B) = ( ) (A 1 A A k ) B (A k+1 B) = (A 1 A A k ) A k+1 B = (A 1 A A k A k+1 ) B Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 10 / 11

11 Exercise 45 Prove that a set with n elements has n(n 1) subsets containing exactly two elements whenever n is an integer greater than or equal to. Let P (n) be the statement that a set with n elements has n(n 1) two-element subsets. Basis step: P (), is true, because a set with two elements has one subset with two elements (itself), and ( 1) = = 1. Inductive step: Now assume that P (k) is true. Let S be a set with k + 1 elements.choose an element a in S and let T = S {a}. A two-element subset of S either contains a or does not. Those subsets not containing a are the subsets of T with two elements; by the inductive hypothesis there are k(k 1)/ of these. There are k subsets of S with two elements that contain a, because such a subset contains a and one of the k elements in T.Hence, there are k(k 1) + k = k k + k = k + k = (k + 1)k two-element subsets of S. This completes the inductive proof. Ali Moallemi, Iraj Hedayati COMP3 - Mathematics for Computer Science 11 / 11

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