Lecture Outline Tutoring Option. Strenthening Induction Hypothesis. - PowerPoint PPT Presentation

Lecture Outline Tutoring Option. Strenthening Induction Hypothesis. Theorem: The sum of the first n odd numbers is a perfect square. How does tutoring work? Theorem: The sum of the first n odd numbers is k 2 . 1. (Ideally) You work on homework and solve (most of) it. Base Case 1 (0th odd number) is perfect square. Strengthening Induction Hypothesis. 2. You do not need to write-up or turn in. Induction Hypothesis Sum of first k odds is perfect square a 2 = k 2 . Strong Induction 3. You read and understand homework solutions. Induction Step 1. The ( k + 1)st odd number is 2 k + 1. 4. Yofu see a tutor, who gives you a short oral quiz. Well ordered principle. 2. Sum of the first k + 1 odds is 4.1 If you do well. Full points. a 2 + 2 k + 1 = k 2 + 2 k + 1 4.2 Decent effort. Less than full points. k 2 + 2 k + 1 = ( k + 1 ) 2 3. ??? 4.3 Didn’t understand HW solutions. Uh oh. ... P(k+1)! 4.4 Can try again. Limit: 2 on average. Tiling Cory Hall Courtyard. Hole have to be there? Maybe just one? Hole in center? Theorem: Can tile the 2 n × 2 n square to leave a hole adjacent Theorem: Any tiling of 2 n × 2 n square has to have one hole. to the center. Proof: The remainder of 2 2 n divided by 3 is 1. Proof: C Base case: true for k = 0. 2 0 = 1 Base case: A single tile works fine. Ind Hyp: 2 2 k = 3 a + 1 for integer a . The hole is adjacent to the center of the 2 × 2 square. E Induction Hypothesis: Any 2 n × 2 n square can be tiled with a hole at the center. C E 2 2 k ∗ 2 2 2 2 ( k + 1 ) = 2 n + 1 B B 4 ∗ 2 2 k = = 4 ∗ ( 3 a + 1 ) B = 12 a + 3 + 1 D 2 n + 1 What to do now??? A = 3 ( 4 a + 1 )+ 1 2 n D a integer = ⇒ ( 4 a + 1 ) is an integer. 2 n

Hole can be anywhere! Strong Induction. Induction = ⇒ Strong Induction. Theorem: Every natural number n > 1 can be written as a Theorem: Can tile the 2 n × 2 n to leave a hole adjacent Let Q ( k ) = P ( 0 ) ∧ P ( 1 ) ··· P ( k ) . product of primes. anywhere. By the induction principle: Fact: A prime n has exactly 2 factors 1 and n . “If Q ( 0 ) , and ( ∀ k ∈ N )( Q ( k ) = ⇒ Q ( k + 1 )) then Base Case: n = 2. Better theorem ...better induction hypothesis! ( ∀ k ∈ N )( Q ( k )) ” Induction Step: Base case: Sure. A tile is fine. P ( n ) = “ n can be written as a product of primes. “ Also, Q ( 0 ) ≡ P ( 0 ) , and ( ∀ k ∈ N )( Q ( k )) ≡ ( ∀ k ∈ N )( P ( k )) Flipping the orientation can leave hole anywhere. Either n + 1 is a prime or n + 1 = a · b where 1 < a , b < n + 1 . Induction Hypothesis: P ( n ) says nothing about a , b ! ( ∀ k ∈ N ) ( Q ( k ) = ⇒ Q ( k + 1 )) “Any 2 n × 2 n square can be tiled with a hole anywhere. ” Strong Induction Principle: If P ( 0 ) and ≡ ( ∀ k ∈ N )(( P ( 0 ) ···∧ P ( k )) = ⇒ ( P ( 0 ) ··· P ( k ) ∧ P ( k + 1 ))) Consider 2 n + 1 × 2 n + 1 square. ( ∀ k ∈ N )(( P ( 0 ) ∧ ... ∧ P ( k )) = ⇒ P ( k + 1 )) , ≡ ( ∀ k ∈ N )(( P ( 0 ) ···∧ P ( k )) = ⇒ P ( k + 1 )) then ( ∀ k ∈ N )( P ( k )) . Use induction hypothesis in each. P ( 0 ) = ⇒ P ( 1 ) = ⇒ P ( 2 ) = ⇒ P ( 3 ) = Strong Induction Principle: If P ( 0 ) and ⇒ ··· Strong induction hypothesis: “ a and b are products of primes” ( ∀ k ∈ N )(( P ( 0 ) ∧ ... ∧ P ( k )) = ⇒ P ( k + 1 )) , = ⇒ “ n + 1 = a · b is the product of the prime factors” Use L-tile and ... we are done. then ( ∀ k ∈ N )( P ( k )) . i.e., the product of the factors of a and the factors of b . Well Ordering Principle. Tournaments have short cycles Tournament has a cycle of length 3 if at all. If ∀ n . P ( n ) is not true, then ∃ n . ¬ P ( n ) . Assume the the smallest cycle is of length k . Consider smallest m , with ¬ P ( m ) , Case 1: Of length 3. Done. P ( m − 1 ) = ⇒ P ( m ) must be false (assuming P ( 0 ) holds.) Case 2: Of length larger than 3. Def: A round robin tournament on n players : each player p This is a proof of the induction principle! p 2 plays each player q , either p → q ( p beats q ) or q → q ( q beats I.e., q .) ¬∀ n . P ( n ) = ⇒ ( ∃ n , ¬ ( P ( n − 1 ) = ⇒ P ( n )) . “ p 3 → p 1 ” = ⇒ 3 cycle Def: A cycle : a sequence of p 1 ,..., p k , p i → p i + 1 and p k → p 1 . (Contrapositive of Induction principle.) Theorem: Any tournament that has a cycle has a cycle of p 1 p 3 Contradiction. But it assumes that there is a smallest m where P ( m ) does not length 3. hold. p k p 4 “ p 1 → p 3 ” = ⇒ k − 1 length cycle! The Well ordering principle states that for any subset of the natural numbers there is a smallest element. ··· ··· Contradiction! ··· ··· Smallest may not be what you expect: the well ordering ··· principal holds for rationals but with different ordering!!

Horses of the same color... Summary: principle of induction. Theorem: All horses have the same color. Base Case: P ( 1 ) - trivially true. ( P ( 0 ) ∧ (( ∀ k ∈ N )( P ( k ) = ⇒ P ( k + 1 )))) = ⇒ ( ∀ n ∈ N )( P ( n )) New Base Case: P ( 2 ) : there are two horses with same color. Variations: Induction Hypothesis: P ( k ) Any k horses have the same color. ( P ( 0 ) ∧ (( ∀ n ∈ N )( P ( n ) = ⇒ P ( n + 1 )))) = ⇒ ( ∀ n ∈ N )( P ( n )) Induction step P ( k + 1 ) ? First k have same color by P ( k ) . 1 , 2 1 , 2 , 3 ,..., k , k + 1 Second k have same color by P ( k ) . 1 , 2 1 , 2 , 3 ,..., k , k + 1 Statement to prove: P ( n ) for n starting from n 0 A horse in the middle in common! 1 , 2 1 , 2 , 3 ,..., k , k + 1 Base Case: Prove P ( n 0 ) . All k must have the same color. No horse in common! 1 , 2 , 3 ,..., k , k + 1 Ind. Step: Prove. For all values, n ≥ n 0 , P ( n ) = ⇒ P ( n + 1 ) . How about P ( 1 ) = ⇒ P ( 2 ) ? Statement is proven! Fix base case. Strong Induction: ...Still doesn’t work!! ( P ( 1 ) ∧ (( ∀ n ∈ N )(( n ≥ 1 ) ∧ P ( n )) = ⇒ P ( n + 1 )))) (There are two horses is �≡ For all two horses!!!) = ⇒ ( ∀ n ∈ N )(( n ≥ 1 ) = ⇒ P ( n )) Of course it doesn’t work. As we will see, it is more subtle to catch errors in proofs of correct theorems!!