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## Elementary Number Theory and Methods of Proof

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**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. 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.**Simple Exercises**The sum of two even numbers is even. The product of two odd numbers is odd. direct proof.**Rational Number**R is rational there are integers a and b such that numerator and b ≠ 0. 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 0.12121212…… a rational number?**Divisibility**a “divides” b (a|b): b = ak for some integer k 5|15 because 15 = 35 n|0 because 0 = n0 1|n because n = 1n 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.**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 (1) direct proof.**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 (2) direct proof.**Divisibility by a Prime**Theorem. Any integer n > 1 is divisible by a prime number. Idea of induction.**Fundamental Theorem of Arithmetic**Every integer, n>1, has a unique factorization into primes: p0≤ p1 ≤ ··· ≤ pk p0p1 ··· pk = n Example: 61394323221 = 3·3·3·7·11·11·37·37·37·53**Prime Products**Claim: Every integer > 1 is a product of primes. Proof:(by contradiction) Supposenot. 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 leastnonproduct, • both are prime products, eg., • k = p1 p2p94 • m = q1 q2 q214**Prime Products**Claim: Every integer > 1 is a product of primes. …So n = k m = p1 p2p94q1q2 q214 is a prime product, a contradiction. The set of nonproducts > 1 must be empty. QED Idea of induction (or smallest counterexample). (The proof of the fundamental theorem will be given later.)**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 0 r < b. We also say q = a div b and 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.**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 r = the offset in this block a (k+1)b kb 2b b -b 0 Clearly, given a and b, q and r are uniquely defined.**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**The Square of an Odd Integer**Idea 3: Use quotient-remainder theorem. Proof by cases.**Trial and Error Won’t Work!**Fermat (1637): If an integer n is greater than 2, then the equation an + bn = cn has no solutions in non-zero integers a, b, and c. has no solutions in non-zero integers a, b, and c. Claim: False. But smallest counterexample has more than 1000 digits. Euler conjecture: has no solution for a,b,c,d positive integers. Open for 218 years, until Noam Elkies found**The Square Root of an Even Square**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. Proof by contrapositive.**Irrational Number**Theorem:is irrational. Proof (by contradiction): • 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!**so m is even.**Irrational Number Theorem:is irrational. Proof (by contradiction): Want to prove both m and n are even. so can assume so n is even. Proof by contradiction.**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. Proof by contradiction.**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. • If a=kb, then gcd(a,b)=b, and we are done. • Otherwise, by the Division Theorem, a = qb + r for r>0.**Greatest Common Divisors**• Let’s say a>b. • If a=kb, then gcd(a,b)=b, and we are done. • Otherwise, by the Division Theorem, a = qb + r for r>0. gcd(8,4) = 4 gcd(12,8) = 4 a=12, b=8 => 12 = 8 + 4 gcd(9,3) = 3 a=21, b=9 => 21 = 2x9 + 3 gcd(21,9) = 3 gcd(99,27) = 9 a=99, b=27 => 99 = 3x27 + 18 gcd(27,18) = 9 Euclid: gcd(a,b) = gcd(b,r)!**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)**Example 1**gcd(a,b) if b = 0, then answer = a. else write a = qb + r answer = gcd(b,r) GCD(102, 70) 102 = 70 + 32 = GCD(70, 32) 70 = 2x32 + 6 = GCD(32, 6) 32 = 5x6 + 2 = GCD(6, 2) 6 = 3x2 + 0 = GCD(2, 0) Returnvalue:2.**Example 2**gcd(a,b) if b = 0, then answer = a. else write a = qb + r answer = gcd(b,r) GCD(252, 189) 252 = 1x189 + 63 = GCD(189, 63) 189 = 3x63 + 0 = GCD(63, 0) Returnvalue:63.**Example 3**gcd(a,b) if b = 0, then answer = a. else write a = qb + r answer = gcd(b,r) GCD(662, 414) 662 = 1x414 + 248 = GCD(414, 248) 414 = 1x248 + 166 = GCD(248, 166) 248 = 1x166 + 82 = GCD(166, 82) 166 = 2x82 + 2 = GCD(82, 2) 82 = 41x2 + 0 = GCD(2, 0) Returnvalue:2.**Correctness of Euclid’s GCD Algorithm**a = qb + r Euclid: gcd(a,b) = gcd(b,r) When r = 0:**Correctness of Euclid’s GCD Algorithm**a = qb + r Euclid: gcd(a,b) = gcd(b,r) When r > 0: Let d be a common divisor of b, r**Correctness of Euclid’s GCD Algorithm**a = qb + r Euclid: gcd(a,b) = gcd(b,r) When r > 0: Let d be a common divisor of a, b.