Practice Questions Solutions for Exam - Mathematical Induction | CS 3653, Study notes of Computer Science

Download Practice Questions Solutions for Exam - Mathematical Induction | CS 3653 and more Study notes Computer Science in PDF only on Docsity! CS 3653 Fall 2002 Lecture 13-1 T. Menhaj Lecture 13 Mathematical Induction (Cont’d) Example 5: Knowing that 2 is irrational, prove that p(n): ( ) 1' 1 1 1 n s P n = + + +… is irrational, { }1, 2,3,n∀ ∈ … . 1. Basis step: ( )2 1 1 2P = + = is irrational, so p(2) is true. 2. Assumption: Assume p(n) is true; i.e P(n)= 1 1 1 ...+ + + (n 1’s) is irrational. 3. Induction: Prove p(n+1) is true; need to show that P(n+1)= 1 1 1 ...+ + + (n+1 1’s) is irrational P(n+1)= ( )1 1' 1 1 1 ... n s+ + + + = 1 ( )P n+ Note, 1+ P(n) is irrational. CS 3653 Fall 2002 Lecture 13-2 T. Menhaj Why? Prove by contradiction. Assume 1+ P(n) is rational. Therefore 1+ P(n)= for some , , , relatively prime .q q r q r r ∈ Z Thus, P(n)=1- q r = r q r − . But this cannot be true since P(n) is irrational. Therefore, 1+ P(n) is irrational. Now, show 1 ( )P n+ is irrational. Prove by contraction. Assume 1 ( )P n+ is rational. Therefore 1 ( )P n+ = for some , , , relatively prime.q q r q r r ∈ Z 2 2 2 2 21 ( ) ( ) q r qP n P n r r − + = ⇒ = which is rational, contradiction. Back to the initial assumption of 2 is irrational. Check if 2 is irrational. Prove by contradiction. Assume it is rational. Therefore CS 3653 Fall 2002 Lecture 13-5 T. Menhaj Example 7: Prove that 2 1 2,n n n n> + ∀ ≥ ∈N Let p(n): the statement P(n)=2n>n+1. 1. Basis step: P(2)= 22 4 2 1 3= > + = , so p(2) is true. 2. Assume p(n) is true; P(n)=2 1 2,n n n n> + ∀ ≥ ∈N 3. Prove p(n+1) is true or prove that 12 ( 1) 1n n+ > + + . 12 2 2 2 ( 1)n n n+ = ⋅ > ⋅ + by the assumption. now show 2 ( 1)n⋅ + >( 1) 1n + + ? 2 ( 1) ( 1) 1 2 2 since 2 n n n n n n ⋅ + > + + + + > + ≥ Therefore: 12 2 2 2 ( 1)n n n+ = ⋅ > ⋅ + >n+2=( 1) 1n + + 12n+ >( 1) 1n + + ; hence p(n+1) is true. CS 3653 Fall 2002 Lecture 13-6 T. Menhaj Example 8: Show that 1 1 n i i i n i α αβ β= = ∑ = ∏ Let p(n): P(n)= 1 1 n i i i n i α αβ β= = ∑ = ∏ 1. Basis step: 1 1 1 2 1 2 1 1 2 1 Knowing that (1) and i i i i i i i i P α α α α αα α α α β β β β β β β β = = + = ∑ = ⇒ = =∏ = ⋅ =∏ This says that p(1) and p(2) are true. 2. Assume that p(k) is true for all 1 k n≤ ≤ ; i.e., 1 1 ( ) , 1,2, , k i i i k i P k k n α αβ β= = ∑ = = ∏ ∀ = … 3. Prove that p(n+1) is true; show that 1 1 1 1 ( 1) n i i i n i P n α αβ β + = + = ∑ + = = ∏ Proof: CS 3653 Fall 2002 Lecture 13-7 T. Menhaj 1 1 1 1 1 1 1 1 ? 1 1 ( ) ( ) 1from p(2) from assumption 1 1 1 n i i i n n i n i i i n i n i n i n i n i n n i i α α α α α α α α α α α β β β β β β β β β β + = + = = + + + + = + = + = = ∑ =∏ ∑ ∑ = ⋅ = ∏ ⋅ ∏ ⋅ =∏ This says that p(n+1) is true. Example 9: Prove: 1 1 1 ( ) 1 n n n i i i i i i P n x x x n = = =   = ≤ ≤ + −         ∑ ∑ ∑ Let p(n) denote the proposition that 1 1 1 ( ) 1 n n n i i i i i i P n x x x n = = =   = ≤ ≤ + −         ∑ ∑ ∑ is true. 1. Basis: First we show that p(2) is true. 2 2 2 1 1 1 1 2 1 2 1 2 2 1 2 1 i i i i i i x x x x x x x x x = = =   ≤ ≤ + −         ≡ + ≤ + ≤ + + −                   ∑ ∑ ∑ CS 3653 Fall 2002 Lecture 13-10 T. Menhaj Example 10: Find summation: 1 1 1! 2 2! 3 3! ... ! ! n n i S n n S i i = = ⋅ + ⋅ + ⋅ + + ⋅ = ⋅∑ n 1 2 3 4 5 6 Sn 1 5 23 119 719 5039 1 n n S S − 5 4.8 5.1739 6.042 7.0083 (n+1)! 2 6 24 120 720 5040 It is reasonable to compare Sn with (n+1)! and guess ( )1 ! 1.nS n= + − Check: for n=7, ( ) 7 7 1 ! 40319, and 7+1 ! 40320. i S i i = = ⋅ = =∑ Now we need to prove it. Denote p(n): 1 ! n n i S i i = = ⋅∑ equals (n+1)!-1 and proceed as follows. 1- Basis step: 1 1 1 ! 1 2 1 (1) is true. i S i i p = = ⋅ = = − ⇒∑ 2- Assumption: ( ) 1 ( ) is true, i.e, ! 1 !-1. n n i p n S i i n = = ⋅ = +∑ CS 3653 Fall 2002 Lecture 13-11 T. Menhaj 3- Inductive Step: Show that p(n+1) is true, or prove that ( ) 1 1 1 ! 2 !-1. n n i S i i n + + = = ⋅ = +∑ Proof: ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 By assumption 1 ! ! 1 1 ! 1 !-1+ 1 1 != 1 ! n+1+1 1 1 ! n+2 1 2 !-1 n n n i i n S i i i i n n S n n n n n n + + = = + = ⋅ = ⋅ + + + ⇒ = + + + + − = = + − = + ∑ ∑ ∵ Example 11: Prove 2 2 1, 3,4,...n n n≥ + = Denote p(n): 2( ) 2 1, is true. 3,4,...P n n n n= ≥ + ∀ = 1. Basis: 2(3) : 3 2 3 1 9 7P ≥ ⋅ + ⇒ ≥ ; so p(3) is true. 2. Assumption: p(n) is true; 2 2 1.n n≥ + 3. Inductive step: show that p(n+1) is true; need to show that ( ) ( )21 2 1 1.n n+ ≥ + + Proof: 2 2( 1) 2 1 2 1 2 1: By assumption 4 2 2 2( 1) 2( 1) 1 n n n n n n n n n + = + + ≥ + + + ≥ + ≥ + + ≥ + + ∵ CS 3653 Fall 2002 Lecture 13-12 T. Menhaj Example 12: prove 22 , 4,5,...n n n≥ = Denote p(n): 2( ) 2nP n n= ≥ is true. 1. Basis: 4 2(4) : 2 4 16 16P ≥ ⇒ ≥ ; p(4) is true. 2. Assume: p(n) is true; 22 , 4,5,...n n n≥ = 3. Prove: p(n+1) is true; ( )212 1 , 4,5,...n n n+ ≥ + = ( ) 1 2 22 2 2 2 2 2 1 1 : By assumption and exmple 11 n n n n n n + = ⋅ ≥ ⋅ ≥ + + ≥ + ∵ Example 13: Prove that 1 ! 1, 1,2,... 2n n n− ≥ = p(n): 1 ! 1, 1,2,... 2n n n− ≥ = is true. 1) 0 1 1, (1) is true. 2 p≥ ⇒ 2) Assume: p(n) is true; 1 ! 1, 1,2,... 2n n n− ≥ = 3) Prove that p(n+1) is true; need to show that ( ) 2 1 ! 1. 2 n + ≥ Proof: CS 3653 Fall 2002 Lecture 13-15 T. Menhaj Example 16: By experimenting with small values of n, guess a formula for the given sum: 1 1 1... 1 2 2 3 ( 1)n S n n = + + + ⋅ ⋅ ⋅ + Then use induction to verify your formula. Let 1 1 1 ( 1) 1n a n n n n = = − + + 1 1 1 1 1 1 2 2 2 1 1 1 1 1 ( 1) 1 1 1 1 1 1 11 1 11 1 1 n n n n n i i i i n n n n i j j j S i i i i i i i j j j n n n n = = = = + = = = =      = = − = −     + + +            = − = + − −       +        = − = + + ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ Verification: Let p(n): 1 1 equals . ( 1) 1 n n i nS i i n= = + +∑ 1. Basis step: CS 3653 Fall 2002 Lecture 13-16 T. Menhaj 1 1 1 1 2 1 1 1 1 2 2 S = = ⋅ + = So, p(1) is true. 2. Assume p(n) is true; 1 1 ( 1) 1 n n i nS i i n= = = + +∑ 3. Inductive step: Prove p(n+1) is true; 1 1 1 1 1: ( 1) ( 1) 1 n n i nS i i n + + = + = + + +∑ Proof: 1 1 1 2 2 since n -1 1 1 From assumption ( 1) ( 1)( 2) 1 ( 2) 1 1 ( 1)( 2) ( 1)( 2) 2 1 ( 1) ( 1)( 2) ( 1)( 2) 1 1 ( 2) ( 1) 1 n n n i S S i i n n n n n n n n n n n n n n n n n n n n n + + = ≠ = = + + + + + + = + = + + + + + + + + = = + + + + + + = = + + + ∑ ∵