Methods_of_Proof In Discrete Mathematics
- 1. Methods of Proof Lecture 4: September 09 Rosen 1.7, 1.8
- 2. Some terminology • Theorem: a mathematical statement that can be shown to be true • Proposition: less important theorem • Axiom (postulate): a statement that is assumed to be true • Lemma: less important theorem that is helpful in the proof of other results • Corollary: a theorem that can be established directly from a theorem that has been proved • Conjecture: a statement proposed to be true, but not proven yet
- 3. Basic Definitions An integer n is an even number if there exists an integer k such that n = 2k. An integer n is an odd number if there exists an integer k such that n = 2k+1.
- 4. Proving an Implication Goal: If P, then Q. (P implies Q) Method 1: Write assume P, then show that Q logically follows. If Claim: , then Reasoning: When x=0, it is true. When x grows, 4x grows faster than x3 in that range. Proof: When
- 5. Direct Proofs The sum of two even numbers is even. The product of two odd numbers is odd. x = 2m, y = 2n x+y = 2m+2n = 2(m+n) x = 2m+1, y = 2n+1 xy = (2m+1)(2n+1) = 4mn + 2m + 2n + 1 = 2(2mn+m+n) + 1. Proof Proof
- 6. a “divides” b (a|b): b = ak for some integer k Divisibility 5|15 because 15 = 3∙5 n|0 because 0 = n∙0 1|n because n = 1∙n n|n because n = n∙1 A number p > 1 with no positive integer divisors other than 1 and itself is called a prime. Every other number greater than 1 is called composite. 2, 3, 5, 7, 11, and 13 are prime, 4, 6, 8, and 9 are composite.
- 7. 1. If a | b, then a | bc for all c. 2. If a | b and b | c, then a | c. 3. If a | b and a | c, then a | sb + tc for all s and t. 4. For all c ≠ 0, a | b if and only if ca | cb. Simple Divisibility Facts Proof of (1) a | b b = ak bc = ack bc = a(ck) a|bc a “divides” b (a|b): b = ak for some integer k
- 8. 1. If a | b, then a | bc for all c. 2. If a | b and b | c, then a | c. 3. If a | b and a | c, then a | sb + tc for all s and t. 4. For all c ≠ 0, a | b if and only if ca | cb. Simple Divisibility Facts Proof of (2) a | b => b = ak1 b | c => c = bk2 => c = ak1k2 => a|c a “divides” b (a|b): b = ak for some integer k
- 9. 1. If a | b, then a | bc for all c. 2. If a | b and b | c, then a | c. 3. If a | b and a | c, then a | sb + tc for all s and t. 4. For all c ≠ 0, a | b if and only if ca | cb. Simple Divisibility Facts Proof of (3) a | b => b = ak1 a | c => c = ak2 sb + tc = sak1 + tak2 = a(sk1 + tk2) => a|(sb+tc) a “divides” b (a|b): b = ak for some integer k
- 10. Proving an Implication Claim: If r is irrational, then √r is irrational. How to begin with? What if I prove “If √r is rational, then r is rational”, is it equivalent? Yes, this is equivalent; proving “if P, then Q” is equivalent to proving “if not Q, then not P”. Goal: If P, then Q. (P implies Q) Method 1: Write assume P, then show that Q logically follows.
- 11. Rational Number R is rational there are integers a and b such that and b ≠ 0. numerator denominator Is 0.281 a rational number? Is 0 a rational number? If m and n are non-zero integers, is (m+n)/mn a rational number? Is the sum of two rational numbers a rational number? Is x=0.12121212…… a rational number? Yes, 281/1000 Yes, 0/1 Yes Yes, a/b+c/d=(ad+bc)/bd Note that 100x-x=12, and so x=12/99.
- 12. Proving an Implication Claim: If r is irrational, then √r is irrational. Method 2: Prove the contrapositive, i.e. prove “not Q implies not P”. Proof: We shall prove the contrapositive – “if √r is rational, then r is rational.” Since √r is rational, √r = a/b for some integers a,b. So r = a2 /b2 . Since a,b are integers, a2 ,b2 are integers. Therefore, r is rational. Goal: If P, then Q. (P implies Q) Q.E.D.
- 13. Proving an “if and only if” Goal: Prove that two statements P and Q are “logically equivalent”, that is, one holds if and only if the other holds. Example: An integer is even if and only if the its square is even. Method 1: Prove P implies Q and Q implies P. Method 1’: Prove P implies Q and not P implies not Q. Method 2: Construct a chain of if and only if statement.
- 14. Proof by Contraposition Statement: If m2 is even, then m is even Statement: If m is even, then m2 is even m = 2k m2 = 4k2 Proof: Proof: m2 = 2k m = √(2k) ?? An integer is even if and only if the its square is even. Method 1: Prove P implies Q and Q implies P.
- 15. Since m is an odd number, m = 2k+1 for some integer k. So m2 is an odd number. Proof by Contraposition Statement: If m2 is even, then m is even Contrapositive: If m is odd, then m2 is odd. So m2 = (2k+1)2 = (2k)2 + 2(2k) + 1 Proof (the contrapositive): Method 1’: Prove P implies Q and not P implies not Q. An integer is even if and only if the its square is even. = 2(2k2 + 2k) + 1
- 16. F P P Proof by Contradiction To prove P, you prove that not P would lead to ridiculous result, and so P must be true. You are working as a clerk. If you have won Mark 6, then you would not work as a clerk. You have not won Mark 6.
- 17. • Suppose was rational. • Choose m, n integers without common prime factors (always possible) such that • Show that m and n are both even, thus having a common factor 2, a contradiction! n m 2 Theorem: is irrational. 2 Proof (by contradiction): Proof by Contradiction 2
- 18. l m 2 so can assume 2 2 4 m l 2 2 2l n so n is even. n m 2 m n 2 2 2 2 m n so m is even. 2 2 2 4 n l Proof by Contradiction Theorem: is irrational. 2 Proof (by contradiction): Want to prove both m and n are even.
- 19. Divisibility by a Prime Theorem. Any integer n > 1 is divisible by a prime number. Idea of induction. • Let n be an integer. • If n is a prime number, then we are done. • Otherwise, n = ab, both are smaller than n. • If a or b is a prime number, then we are done. • Otherwise, a = cd, both are smaller than a. • If c or d is a prime number, then we are done. • Otherwise, repeat this argument, since the numbers are getting smaller and smaller, this will eventually stop and we have found a prime factor of n.
- 20. Infinitude of the Primes Theorem. There are infinitely many prime numbers. Claim: if p divides a, then p does not divide a+1. Let p1, p2, …, pN be all the primes. Consider p1p2…pN + 1. Proof (by contradiction): Proof (by contradiction): a = cp for some integer c a+1 = dp for some integer d => 1 = (d-c)p, contradiction because p>=2. So none of p1, p2, …, pN can divide p1p2…pN + 1, a contradiction.
- 21. Proof by Cases x is positive or x is negative e.g. want to prove a nonzero number always has a positive square. if x is positive, then x2 > 0. if x is negative, then x2 > 0. x2 > 0. A proof by cases must cover all possible cases that arise in a theorem
- 22. The Square of an Odd Integer 32 = 9 = 8+1, 52 = 25 = 3x8+1 …… 1312 = 17161 = 2145x8 + 1, ……… Idea 1: prove that n2 – 1 is divisible by 8. Idea 2: consider (2k+1)2 Idea 0: find counterexample. n2 – 1 = (n-1)(n+1) = ??… (2k+1)2 = 4k2 +4k+1 If k is even, then both k2 and k are even, and so we are done. If k is odd, then both k2 and k are odd, and so k2 +k even, also done. = 4(k2 +k)+1 So n2 = (2k+1)2 = 4(2m)+1
- 23. Proof by Cases Show that there are no solutions in integers x and y of x2 + 3y2 = 8. • When |x| 3, x2 8 • When |y| 2, 3y2 8 • This leaves possible x = {−2, −1, 0, 1, 2} and possible y = {−1, 0, 1} • Possible x2 are 0, 1, 4 and possible 3y2 are 0 and 3 • Largest sum of possible values for x2 and 3y2 is 7 • Consequently, it is impossible for x2 + 3y2 = 8 to hold when x and y are integers
- 24. Mistakes in proofs • What is wrong with this proof “1=2”? 1. a=b (given) 2. a2 =ab (multiply both sides of 1 by a) 3. a2 -b2 =ab-b2 (subtract b2 from both sides of 2) 4. (a-b)(a+b)=b(a-b) (factor both sides of 3) 5. a+b=b (divide both sides of 4 by a-b) 6. 2b=b (replace a by b in 5 as a=b and simply) 7. 2=1 (divide both sides of 6 by b) 24
- 25. What is wrong with this proof? • “Theorem”: If n2 is positive, then n is positive “Proof”: Suppose n2 is positive. As the statement “If n is positive, then n2 is positive” is true, we conclude that n is positive • The mistake occurs when one or more steps of a proof are based on the truth of the statement being proved • Counterexample: n = -1 25
- 26. In Summary Direct Proofs (cases/induction) H1 ∧ H2 ∧ ⋯ ∧ Hn ⟹ C Proofs of the Contrapositive ¬C ⟹ (H1 ∧ H2 ∧ ⋯ ∧ Hn) Proofs by Contradiction H1 ∧ H2 ∧ ⋯ ∧ Hn ∧ ¬C ⟹ a contradiction
- 27. Challenge for the Bored – Rational vs Irrational Question: If a and b are irrational, can ab be rational?? We know that √2 is irrational, what about √2√2 ?