This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CS 2800 – Fall 2008 Homework #4 Solution Some general comments on writing inductive proofs • Clearly state your base case and inductive hypothesis • When proving the inductive case, state clearly where you use your inductive hypothesis. • It is important to distinguish fact (i.e. what is known or assumed to be true) from conjecture (for example, what you are trying to show). For example, if you are trying to show ( k + 1) 2 ≤ ( k + 1)! and you choose to simplify this expression, you have to either preface each simplification step with the phrase ‘To show/prove:’ or indicate that all the steps are reversible. If not, then in effect, your argument is backwards. Section 4.1 Problem 24 [5pts] Let P ( n ) be the statement: 1 2 n ≤ 1 · 3 · 5 . . . (2 n 1) 2 · 4 · 6 . . . 2 n for n ≥ 1. We proceed by induction. • Base case: When n = 1, P(1) is the proposition 1 2 ≤ 1 2 which is clearly true. • Inductive case: Assume that P ( n ) holds for some n ≥ 1, i.e. 1 2 n ≤ 1 · 3 · 5 . . . (2 n 1) 2 · 4 · 6 . . . 2 n We then need to show that P ( n + 1) is true, i.e. To Prove: 1 2( n + 1) ≤ 1 · 3 · 5 . . . (2 n 1) · (2 n + 1) 2 · 4 · 6 . . . 2 n · (2 n + 2) Multiplying the inductive hypothesis by 2 n/ 2( n + 1), we get: 1 2( n + 1) ≤ 1 · 3 · 5 . . . (2 n 1) · 2 n 2 · 4 · 6 . . . 2 n · (2 n + 2) ≤ 1 · 3 · 5 . . . (2 n 1) · (2 n + 1) 2 · 4 · 6 . . . 2 n · (2 n + 2) (since 2 n ≤ 2 n + 1) Thus, P ( n + 1) is true. 1 Problem 26 [5pts] Let P ( n ) be the statement a n b n ≤ na n 1 ( a b ), 0 < b < a , for n ≥ 1. We proceed by induction. • Base case: When n = 1, we have a b ≤ a b which is true. • Inductive case: Assume P ( k ) holds for some k ≥ 1 i.e. a k b k ≤ ka k 1 ( a b ). We need to show that P ( k + 1) is true, i.e. To Prove: a k +1 b k +1 ≤ ( k + 1) a k ( a b ) i.e. TP: a k +1 b k +1 ≤ ka k ( a b ) + a k ( a b ) i.e. TP: a k +1 b k +1 ≤ ka k ( a b ) + a k +1 ba k i.e. TP: ba k b k +1 ≤ a ( ka k 1 ( a b )) i.e. TP: b ( a k b k ) ≤ a ( ka k 1 ( a b )) Now, we already know from the inductive hypothesis that a k b k ≤ ka k 1 ( a b ). Thus, to prove the above inequality, all we need to show is that b ≤ a — but this is given to us as well. Thus, P ( k + 1) is true. Problem 46 [5pts] We proceed by induction. • Base case: When n = 3, then there is exactly one subset containing 3 elements. Substituting n = 3 into the formula n ( n 1)( n 2) / 6, we arrive at 1 as well. Thus, the base case is established. • Inductive case: Assume that the given statement is true for some k ≥ 3 i.e. for a set with k elements, the number of subsets of size 3 is k ( k 1)( k 2) / 6. We need to show that a set S with k + 1 elements has ( n + 1) n ( n 1) / 6 subsets with exactly 3 elements. Fix an element a in set S . Let T be the set ( S { a } ) i.e. the set of all elements in S excluding a . Now the subsets of S that contain exactly three elements come in two flavors — there are those that consist of elements drawn entirely from...
View
Full Document
 Fall '07
 SELMAN
 Mathematical Induction, Recursion, Inductive Reasoning, Natural number, Structural induction, inductive hypothesis

Click to edit the document details