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Chapter 3: Elementary Number Theory and Methods of Proofs 3.1-.3.4 Direct Methods and Counterexamples • Introduction • Rational Numbers • Divisibility • Division Algorithm Instructor: Hayk Melikya Introduction to Abstract Mathematics [email protected] 1 Basic Definitions Definition: An integer n is an even number if there exists an integer k such that n = 2k. Symbolically: Let Even(n) := “an integer n is even”: E(n) = ( k Z)( n = 2k) . Def: An integer n is an odd number if there exists an integer k such that n = 2k+1. Symbolically: Let O(n) := “an integer is odd”: Odd(n) = ( k Z)( n = 2k +1) . Def: An integer n is a prime number if and only if n>1 and if n=rs for some positive integers r and s then r=1 or s=1. Introduction to Abstract Mathematics 2 Primes and Composites Def: An integer n is a prime number if and only if n>1 and if n=rs for some positive integers r and s then r=1 or s=1. Symbolically: Prime(n):= n is prime positive integers r and s, if n = rs then r =1 or s =1 Def: A positive integer n is a composite if and only if n=rs for some positive integers r and s then r ≠ 1 and s ≠ 1. Symbolically: Cpmposite(n):= n is compoeite positive integers r and s, such that n = rs and r ≠ 1 and s ≠ 1 The RSA Challenge (up to US$200,000) http://www.rsasecurity.com/rsalabs/challenges/factoring/numbers.html Examples: Find the truth value of the following prpopositions E(6), P(12), C(17) Introduction to Abstract Mathematics 3 Existential Statements x P(x) Proofs: – Constructive Construct an example of such a such that P(a) is true – Non-constructive By contradiction – Show that if such x does NOT exist than a contradiction can be derived Introduction to Abstract Mathematics 4 Example Let G(n):= a b ((a+b=n) Prime(a) Prime(b)) Prove that (nN)G(n) Proof: – n=210 – a=113 – b=97 Piece of cake… What about (nN) G(n) ( many Million $ baby) Introduction to Abstract Mathematics 5 Universal Statements x P(x) x [Q(x) R(x)] Proof techniques: – Exhaustion – By contradiction Assume the statement is not true Arrive at a contradiction – Direct Generalizing from an arbitrary particular member Introduction to Abstract Mathematics 6 ├ (xU)P(x) To prove a theorem of the form (xU)P(x) (same as ├ (xU)P(x)) which states “for all elements x in a given universe U, the proposition P(x) is true” we select an arbitrary aU from the universe, and then prove the assertion P(a). Then by Universal generalization we conclude P(a)├ (xU)P(x) For arguments of the form├ x [Q(x) R(x)] Introduction to Abstract Mathematics 7 Example 1 Exhaustion: – Any even number between 4 and 30 can be written as a sum of two primes: – 4=2+2 – 6=3+3 – 8=3+5 – … – 30=11+19 Works for finite domains only What if I want to prove that for any integer n the product of n and n+1 is even? Can I exhaust all integer values of n? Introduction to Abstract Mathematics 8 Example 2 Theorem: (nZ)( even(n*(n+1)) ) Proof: – Consider a particular but arbitrary chosen integer n – n is odd or even – Case 1: n is odd Then n=2k+1, n+1=2k+2 n(n+1) = (2k+1)(2k+2) = 2(2k+1)(k+1) = 2p for some integer p So n(n+1) is even Case 2: n is even Then n=2k, n+1=2k+1 n(n+1) = 2k(2k+1) = 2p for some integer p So n(n+1) is even Introduction to Abstract Mathematics 9 Fallacy Generalizing from a particular but NOT arbitrarily chosen example I.e., using some additional properties of n Example: – “all odd numbers are prime” – “Proof”: Consider odd number 3 It is prime Thus for any odd n prime(n) holds Such “proofs” can be given for correct statements as well! Introduction to Abstract Mathematics 10 Prevention Try to stay away from specific instances (e.g., 3) Make sure that you are not using any additional properties of n considered Challenge your proof – Try to play the devil’s advocate and find holes in it… • • • • Using the same letter to mean different things Jumping to a conclusion Insufficient justification Begging the question assuming the claim first Introduction to Abstract Mathematics 11 Rational Numbers A real number is rational iff it can be represented as a ratio/quotient/fraction of integers a and b(b0) – rR [rQ a,bZ [r=a/b & b0]] Notes: – a is numerator – b is denominator – Any rational number can be represented in infinitely many ways – The fractional part of any rational number written in any natural radix has a period in it Introduction to Abstract Mathematics 12 Rational or not? -12 – -12/1 3.1459 – 3+1459/10000 0.56895689568956895689… – 5689/9999 1+1/2+1/4+1/8+… – 2 0 – 0/1 Introduction to Abstract Mathematics 13 Theorem 1 Any number with a periodic fractional part in a natural radix representation is rational Proof: – Constructive: – x=0.n1…nmn1…nm… – x=0.(n1…nm) – x*10m-x=n1…nm – x=n1…nm/(10m-1) Introduction to Abstract Mathematics 14 Theorem 2 Any geometric series: – S=q0+q1+q2+q3+… – where -1<q<1 – evaluates to S=1/(1-q) Proof – Proof idea – More formal proof – Definitions of limits and partial sums Introduction to Abstract Mathematics 15 ZQ Every integer is a rational number Proof : set the denominator to 1 Book : page 127 Q The set of rational numbers is closed with respect to arithmetic operations +, -, *, / Partial proofs : textbook pages 121-131 Formal proof Introduction to Abstract Mathematics 16 Irrational Numbers So far all the examples were of rational numbers How about some irrationals? – – e – sqrt(2) Introduction to Abstract Mathematics 17 Simple Exercises The sum of two even numbers is even. The product of two odd numbers is odd. direct proof. Introduction to Abstract Mathematics 18 Divisibility a “divides” b or is b divisible by a (a|b ): b = ak for some integer k Also we say that 5|15 because 15 = 35 b is multiple of a n|0 because 0 = n0 a is a factor of b 1|n because n = 1n b is divisor for a n|n because n = n1 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. Introduction to Abstract Mathematics 19 Simple Divisibility Facts 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. Proof of (??) Introduction to Abstract Mathematics direct proof. 20 Divisibility by a Prime Theorem. Any integer n > 1 is divisible by a prime number. Introduction to Abstract Mathematics 21 Fundamental Theorem of Arithmetic Every integer, n>1, has a unique factorization into primes: p0 ≤ p1 ≤ ··· ≤ pk p0 p1 ··· pk = n Example: 61394323221 = 3·3·3·7·11·11·37·37·37·53 Introduction to Abstract Mathematics 22 Prime Products Claim: Every integer > 1 is a product of primes. Proof: (by contradiction) Suppose not. Then set of non-products is nonempty. There is a smallest integer n > 1 that is not a product of primes. In particular, n is not prime. So n = k·m for integers k, m where n > k,m >1. Since k,m smaller than the least nonproduct, both are prime products, eg., k = p1 p2 p94 m = q1 q2 q214 Introduction to Abstract Mathematics 23 Prime Products Claim: Every integer > 1 is a product of primes. …So n = k m = p1 p2 p94 q1 q2 q214 is a prime product, a contradiction. The set of nonproducts > 1 must be empty. QED (The proof of the fundamental theorem will be given later.) Introduction to Abstract Mathematics 24 The Quotient-Reminder Theorem For b > 0 and any a, there are unique numbers q : quotient(a,b), r : remainder(a,b), such that a = qb + r and We also say 0 r < b. q = a div b r = a mod b. When b=2, this says that for any a, there is a unique q such that a=2q or a=2q+1. When b=3, this says that for any a, there is a unique q such that a=3q or a=3q+1 or a=3q+2. Introduction to Abstract Mathematics 25 The Division Theorem For b > 0 and any a, there are unique numbers q : quotient(a,b), r : remainder(a,b), such that a = qb + r and 0 r < b. Given any b, we can divide the integers into many blocks of b numbers. For any a, there is a unique “position” for a in this line. q = the block where a is in -b 0 b 2b r = the offset in this block kb a (k+1)b Clearly, given a and b, q and r are uniquely defined. Introduction to Abstract Mathematics 26 The Square of an Odd Integer Idea 0: find counterexample. 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 3: Use quotient-remainder theorem. Introduction to Abstract Mathematics 27 Contrapositive Proof Statement: If m2 is even, then m is even Contrapositive: If m is odd, then m2 is odd. Proof (the contrapositive): Since m is an odd number, m = 2l+1 for some natural number l. So m2 = (2l+1)2 = (2l)2 + 2(2l) + 1 So m2 is an odd number. Introduction to Abstract Mathematics Proof by contrapositive. 28 Irrational Number Theorem: 2 is irrational. Proof (by contradiction): • Suppose 2 was rational. • Choose m, n integers without common prime factors (always possible) such that m 2 n • Show that m and n are both even, thus having a common factor 2, a contradiction! Introduction to Abstract Mathematics 29 Irrational Number Theorem: 2 is irrational. Proof (by contradiction): Want to prove both m and n are even. so can assume m 2 n m 2l m 2 4l 2 2n m 2n 4l 2 2n m 2 2 2 n 2 2l 2 so m is even. so n is even. Introduction to Abstract Mathematics Proof by contradiction. 30 Infinitude of the Primes Theorem. There are infinitely many prime numbers. Let p1, p2, …, pN be all the primes. Consider p1p2…pN + 1. Claim: if p divides a, then p does not divide a+1. Introduction to Abstract Mathematics Proof by contradiction. 31 Floor and Ceiling Def: For any real number x, the floor of x, written x, is the unique integer n such that n x < n + 1. It is the largest integer not exceeding x ( x). Def: For any real number x, the ceiling of x, written x, is the unique integer n such that n – 1 < x n. What is n? If k is an integer, what are x and x + 1/2 ? Is x + y = x + y? ( what if x = 0.6 and y = 0.7) For all real numbers x and all integers m, x + m = x + m For any integer n, n/2 is n/2 for even n and (n–1)/2 for odd n Introduction to Abstract Mathematics 32 Exercises Is it true that for all real numbers x and y: x – y = x - y x – 1 = x - 1 x + y = x + y x + 1 = x + 1 For positive integers n and d, n = d * q + r, where d = n / d and r = n – d * n / d with 0 r < d Introduction to Abstract Mathematics 33 Greatest Common Divisors Given a and b, how to compute gcd(a,b)? Can try every number, but can we do it more efficiently? Let’s say a>b. 1. If a=kb, then gcd(a,b)=b, and we are done. 2. Otherwise, by the Division Theorem, a = qb + r for r>0. Introduction to Abstract Mathematics 34 Greatest Common Divisors Let’s say a>b. 1. If a=kb, then gcd(a,b)=b, and we are done. 2. Otherwise, by the Division Theorem, a = qb + r for r>0. a=12, b=8 => 12 = 8 + 4 gcd(12,8) = 4 gcd(8,4) = 4 a=21, b=9 => 21 = 2x9 + 3 gcd(21,9) = 3 gcd(9,3) = 3 a=99, b=27 => 99 = 3x27 + 18 gcd(99,27) = 9 gcd(27,18) = 9 Euclid: gcd(a,b) = gcd(b,r)! Introduction to Abstract Mathematics 35 Euclid’s GCD Algorithm a = qb + r Euclid: gcd(a,b) = gcd(b,r) gcd(a,b) if b = 0, then answer = a. else write a = qb + r answer = gcd(b,r) Introduction to Abstract Mathematics 36 Example 1 gcd(a,b) if b = 0, then answer = a. GCD(102, 70) 102 = 70 + 32 = GCD(70, 32) 70 = 2x32 + 6 write a = qb + r = GCD(32, 6) 32 = 5x6 + 2 answer = gcd(b,r) = GCD(6, 2) 6 = 3x2 + 0 else = GCD(2, 0) Return value: 2. Example 3 Example 2 GCD(662, 414) 662 = 1x414 + 248 = GCD(414, 248) 414 = 1x248 + 166 = GCD(189, 63) = GCD(248, 166) 248 = 1x166 + 82 = GCD(63, 0) = GCD(166, 82) 166 = 2x82 + 2 = GCD(82, 2) 82 = 41x2 + 0 GCD(252, 189) 252 = 1x189 + 63 189 = 3x63 + 0 Return value: 63. = GCD(2, 0) Return value: 2. Introduction to Abstract Mathematics 37 Practice problems Study the Sections 3.1- 3.4 from your textbook. 2. Be sure that you understand all the examples discussed in class and in textbook. 3. Do the following problems from the textbook: Exercise 3.1 # 13, 16, 32, 36, 45 Exercise 3.2 # 15, 19, 21, 32, Exercise 3.3 # 13, 16, 25, 26, Exercise 3.4 # 4, 6, 8, 10, 18, 33 1. Introduction to Abstract Mathematics 39