CS70: Lecture 2. Outline. Quick Background and Notation. Direct - PowerPoint PPT Presentation

CS70: Lecture 2. Outline. Quick Background and Notation. Direct Proof (Forward Reasoning). Theorem: For any a , b , c ∈ Z , if a | b and a | c then a | b − c . Integers closed under addition. Proof: Assume a | b and a | c a , b ∈ Z = ⇒ a + b ∈ Z b = aq and c = aq ′ where q , q ′ ∈ Z Today: Proofs!!! a | b means “a divides b”. b − c = aq − aq ′ = a ( q − q ′ ) Done? 1. By Example (or Counterexample). 2 | 4? Yes! ( b − c ) = a ( q − q ′ ) and ( q − q ′ ) is an integer so 2. Direct. (Prove P = ⇒ Q . ) 7 | 23? No! a | ( b − c ) 3. by Contraposition (Prove P = ⇒ Q ) 4 | 2? No! 4. by Contradiction (Prove P .) Works for ∀ a , b , c ? Formally: a | b ⇐ ⇒ ∃ q ∈ Z where b = aq . Argument applies to every a , b , c ∈ Z . 5. by Cases 3 | 15 since for q = 5, 15 = 3 ( 5 ) . Direct Proof Form: Goal: P = ⇒ Q A natural number p > 1, is prime if it is divisible only by 1 and Assume P . itself. ... Therefore Q. Another direct proof. The Converse Another Direct Proof. Let D 3 be the 3 digit natural numbers. Theorem: For n ∈ D 3 , if the alternating sum of digits of n is divisible Theorem: ∀ n ∈ D 3 , ( 11 | n ) = ⇒ ( 11 | alt. sum of digits of n ) by 11, than 11 | n . Proof: Assume 11 | n . n = 100 a + 10 b + c = 11 k = ⇒ ∀ n ∈ D 3 , ( 11 | alt. sum of digits of n ) = ⇒ 11 | n 99 a + 11 b +( a − b + c ) = 11 k = ⇒ Thm: ∀ n ∈ D 3 , ( 11 | alt. sum of digits of n ) = ⇒ 11 | n Examples: a − b + c = 11 k − 99 a − 11 b = ⇒ n = 121 Alt Sum: 1 − 2 + 1 = 0. Divis. by 11. As is 121. Is converse a theorem? a − b + c = 11 ( k − 9 a − b ) = ⇒ a − b + c = 11 ℓ where ℓ = ( k − 9 a − b ) ∈ Z n = 605 Alt Sum: 6 − 0 + 5 = 11 Divis. by 11. As is 605 = 11 ( 55 ) ∀ n ∈ D 3 , ( 11 | n ) = ⇒ ( 11 | alt. sum of digits of n ) That is 11 | alternating sum of digits. Proof: For n ∈ D 3 , n = 100 a + 10 b + c , for some a , b , c . Example: n = 264. 11 | n ? 11 | 2 − 6 + 4? Note: similar proof to other. In this case every = ⇒ is ⇐ ⇒ Assume: Alt. sum: a − b + c = 11 k for some integer k . Often works with arithmetic properties except when multiplying by 0. Add 99 a + 11 b to both sides. We have. 100 a + 10 b + c = 11 k + 99 a + 11 b = 11 ( k + 9 a + b ) Left hand side is n , k + 9 a + b is integer. = ⇒ 11 | n . Theorem: ∀ n ∈ N , ( 11 | alt. sum of digits of n ) ⇐ ⇒ ( 11 | n ) Direct proof of P = ⇒ Q : Assumed P : 11 | a − b + c . Proved Q : 11 | n .

Proof by Contraposition Another Contrapostion... Proof by Contradiction Lemma: For every n in N , n 2 is even = ⇒ n is even. ( P = ⇒ Q ) Thm: For n ∈ Z + and d | n . If n is odd then d is odd. √ √ n 2 is even, n 2 = 2 k , ... Theorem: 2 is irrational. 2 k even? n = 2 k + 1 what do we know about d ? b ) 2 � = 2. Must show: For every a , b ∈ Z , ( a Proof by contraposition: ( P = ⇒ Q ) ≡ ( ¬ Q = ⇒ ¬ P ) What to do? P = ’ n 2 is even.’ ........... ¬ P = ’ n 2 is odd’ A simple property (equality) should always “not” hold. Goal: Prove P = ⇒ Q . Proof by contradiction: Q = ’n is even’ ........... ¬ Q = ’n is odd’ Assume ¬ Q ⇒ n 2 is odd. Theorem: P . ...and prove ¬ P . Prove ¬ Q = ⇒ ¬ P : n is odd = ¬ P = ⇒ P 1 ··· = ⇒ R Conclusion: ¬ Q = ⇒ ¬ P equivalent to P = ⇒ Q . n = 2 k + 1 Proof: Assume ¬ Q : d is even. d = 2 k . n 2 = 4 k 2 + 4 k + 1 = 2 ( 2 k + k )+ 1. ¬ P = ⇒ P 1 ··· = ⇒ ¬ R d | n so we have n 2 = 2 l + 1 where l is a natural number.. ¬ P = ⇒ False n = qd = q ( 2 k ) = 2 ( kq ) ... and n 2 is odd! Contrapositive: True = ⇒ P . Theorem P is proven. n is even. ¬ P ¬ Q = ⇒ ¬ P so P = ⇒ Q and ... Contradiction Proof by contradiction: example Product of first k primes.. √ Theorem: 2 is irrational. Theorem: There are infinitely many primes. √ Assume ¬ P : 2 = a / b for a , b ∈ Z . Proof: Did we prove? Reduced form: a and b have no common factors. ◮ Assume finitely many primes: p 1 ,..., p k . ◮ “The product of the first k primes plus 1 is prime.” √ ◮ Consider ◮ No. 2 b = a q = p 1 × p 2 ×··· p k + 1 . ◮ The chain of reasoning started with a false statement. 2 b 2 = a 2 = 4 k 2 Consider example.. ◮ q cannot be one of the primes as it is larger than any p i . a 2 is even = ◮ 2 × 3 × 5 × 7 × 11 × 13 + 1 = 30031 = 59 × 509 ⇒ a is even. ◮ q has prime divisor p (” p > 1” = R ) which is one of p i . ◮ There is a prime in between 13 and q = 30031 that divides ◮ p divides both x = p 1 · p 2 ··· p k and q , and divides q − x , a = 2 k for some integer k q . ◮ = ⇒ p | q − x = ⇒ p ≤ q − x = 1. b 2 = 2 k 2 ◮ Proof assumed no primes in between. ◮ so p ≤ 1. ( Contradicts R . ) b 2 is even = ⇒ b is even. The original assumption that “the theorem is false” is false, a and b have a common factor. Contradiction. thus the theorem is proven.

Proof by cases. (“divide-and-conquer” strategy) Proof by cases. Be careful. Theorem: x 5 − x + 1 = 0 has no solution in the rationals. Theorem: There exist irrational x and y such that x y is rational. Theorem: 3 = 4 √ Proof: First a lemma... Let x = y = 2. Lemma: If x is a solution to x 5 − x + 1 = 0 and x = a / b for a , b ∈ Z , Proof: Assume 3 = 4. Start with 12 = 12. Divide one side by 3 and √ √ 2 is rational. Done! then both a and b are even. Case 1: x y = the other by 4 to get 4 = 3. By commutativity theorem holds. 2 √ Reduced form a √ b : a and b can’t both be even! + Lemma 2 is irrational. Don’t assume what you want to prove! Case2: 2 = ⇒ no rational solution. Theorem: 1 = 2 √ √ √ 2 , y = Proof of lemma: Assume a solution of the form a / b . Proof: For x = y , we have ◮ New values: x = 2 2. ( x 2 − xy ) = x 2 − y 2 � 5 � a ◮ 2 � √ − a / b + 1 = 0 √ √ √ x ( x − y ) = ( x + y )( x − y ) � √ 2 √ √ 2 = 2 = 2 . b 2 ∗ x y = 2 = 2 2 x = ( x + y ) multiply by b 5 , x = 2 x a 5 − ab 4 + b 5 = 0 1 = 2 Thus, in this case, we have irrational x and y with a rational x y (i.e., Case 1: a odd, b odd: odd - odd +odd = even. Not possible. Dividing by zero is no good. 2). Case 2: a even, b odd: even - even +odd = even. Not possible. Also: Multiplying inequalities by a negative. Case 3: a odd, b even: odd - even +even = even. Not possible. One of the cases is true so theorem holds. Case 4: a even, b even: even - even +even = even. Possible. P = ⇒ Q does not mean Q = ⇒ P . Question: Which case holds? Don’t know!!! The fourth case is the only one possible, so the lemma follows. Summary Direct Proof: To Prove: P = ⇒ Q . Assume P . reason forward, Prove Q . By Contraposition: To Prove: P = ⇒ Q Assume ¬ Q . Prove ¬ P . By Contradiction: To Prove: P Assume ¬ P . Prove False . By Cases: informal. Universal: show that statement holds in all cases. Existence: used cases where one is true. √ √ Either 2 and 2 worked. √ √ √ 2 worked. or 2 and 2 Careful when proving! Don’t assume the theorem. Divide by zero. Watch converse. ...