1. 1 Introduction to Abstract Mathematics Chapter 3: Elementary Number Theory and Methods of Proofs Instructor: Hayk Melikya melikyan@nccu.edu 3.1-.3.4 Direct Methods and Counterexamples Introduction Rational Numbers Divisibility Division Algorithm

2. 2 Introduction to Abstract Mathematics Basic Definitions Definition: An integer n is an even number if there exists an integer k such that n = 2k. Def: An integer n is an odd number if there exists an integer k such that 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. Symbolically: Let Even(n) := “an integer n is even”: E(n) = (  k  Z)( n = 2k). Symbolically: Let O(n) := “an integer is odd”: Odd(n) = (  k  Z)( n = 2k +1).

3. 3 Introduction to Abstract Mathematics 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)

4. 4 Introduction to Abstract Mathematics Existential Statements  x P(x) Proofs: –Constructive u Construct an example of such a such that P(a) is true –Non-constructive u By contradiction –Show that if such x does NOT exist than a contradiction can be derived

5. 5 Introduction to Abstract Mathematics Example Let G(n):=  a  b ((a+b=n)  Prime(a)  Prime(b)) Prove that (  n  N)G(n) Proof: –n=210 –a=113 –b=97 Piece of cake… What about (  n  N) G(n) ( many Million $ baby)

6. 6 Introduction to Abstract Mathematics Universal Statements v  x P(x) v  x [Q(x)  R(x)] v Proof techniques: –Exhaustion –By contradiction u Assume the statement is not true u Arrive at a contradiction –Direct u Generalizing from an arbitrary particular member

7. 7 Introduction to Abstract Mathematics ├ (  x  U)P(x) To prove a theorem of the form (  x  U)P(x) (same as ├ (  x  U)P(x) ) which states “for all elements x in a given universe U, the proposition P(x) is true” we select an arbitrary a  U from the universe, and then prove the assertion P(a). Then by Universal generalization we conclude P(a) ├ (  x  U)P(x) For arguments of the form ├  x [Q(x)  R(x)]

8. 8 Introduction to Abstract Mathematics 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?

9. 9 Introduction to Abstract Mathematics Example 2 Theorem: (  n  Z)( even(n*(n+1)) ) Proof: –Consider a particular but arbitrary chosen integer n –n is odd or even –Case 1: n is odd u Then n=2k+1, n+1=2k+2 u n(n+1) = (2k+1)(2k+2) = 2(2k+1)(k+1) = 2p for some integer p u 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

10. 10 Introduction to Abstract Mathematics 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”: u Consider odd number 3 u It is prime u Thus for any odd n prime(n) holds Such “proofs” can be given for correct statements as well!

11. 11 Introduction to Abstract Mathematics Prevention v Try to stay away from specific instances (e.g., 3) v Make sure that you are not using any additional properties of n considered v 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

12. 12 Introduction to Abstract Mathematics Rational Numbers A real number is rational iff it can be represented as a ratio/quotient/fraction of integers a and b(b  0) –  r  R [r  Q   a,b  Z [r=a/b & b  0]] 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

13. 13 Introduction to Abstract Mathematics Rational or not? v -12 –-12/1 v 3.1459 –3+1459/10000 v 0.56895689568956895689… –5689/9999 v 1+1/2+1/4+1/8+… –2 v 0 –0/1

14. 14 Introduction to Abstract Mathematics Theorem 1 Any number with a periodic fractional part in a natural radix representation is rational Proof: –Constructive: –x=0.n 1 …n m n 1 …n m … –x=0.(n 1 …n m ) –x*10 m -x=n 1 …n m –x=n 1 …n m /(10 m -1)

15. 15 Introduction to Abstract Mathematics Theorem 2 Any geometric series: –S=q 0 +q 1 +q 2 +q 3 +… –where -1<q<1 –evaluates to S=1/(1-q) Proof –Proof idea –More formal proof –Definitions of limits and partial sums

16. 16 Introduction to Abstract Mathematics Z  Q Every integer is a rational number Proof : set the denominator to 1 u Book : page 127 The set of rational numbers is closed with respect to arithmetic operations +, -, *, / Partial proofs : textbook pages 121-131 Formal proof Q

17. 17 Introduction to Abstract Mathematics Irrational Numbers v So far all the examples were of rational numbers v How about some irrationals? – –e –sqrt(2)

18. 18 Introduction to Abstract Mathematics Simple Exercises The sum of two even numbers is even. The product of two odd numbers is odd. direct proof.

19. 19 Introduction to Abstract Mathematics a “divides” b or is b divisible by a (a|b ): b = ak for some integer k Also we say that b is multiple of a a is a factor of b b is divisor for a 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.

20. 20 Introduction to Abstract Mathematics 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 (??) direct proof.

21. 21 Introduction to Abstract Mathematics Divisibility by a Prime Theorem. Any integer n > 1 is divisible by a prime number.

22. 22 Introduction to Abstract Mathematics Every integer, n>1, has a unique factorization into primes: p 0 ≤ p 1 ≤ ··· ≤ p k p 0 p 1 ··· p k = n Fundamental Theorem of Arithmetic Example: 61394323221 = 3·3·3·7·11·11·37·37·37·53

23. 23 Introduction to Abstract Mathematics Claim: Every integer > 1 is a product of primes. Prime Products 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 = p 1  p 2    p 94 m = q 1  q 2    q 214

24. 24 Introduction to Abstract Mathematics Prime Products …So n = k  m = p 1  p 2    p 94  q 1  q 2    q 214 is a prime product, a contradiction.  The set of nonproducts > 1 must be empty. QED Claim: Every integer > 1 is a product of primes. (The proof of the fundamental theorem will be given later.)

25. 25 Introduction to Abstract Mathematics 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. The Quotient-Reminder Theorem 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. We also say q = a div b r = a mod b.

26. 26 Introduction to Abstract Mathematics 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. The Division Theorem 0 b 2b kb (k+1)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 a r = the offset in this block Clearly, given a and b, q and r are uniquely defined. -b

27. 27 Introduction to Abstract Mathematics The Square of an Odd Integer 3 2 = 9 = 8+1, 5 2 = 25 = 3x8+1 …… 131 2 = 17161 = 2145x8 + 1, ……… Idea 1: prove that n 2 – 1 is divisible by 8. Idea 2: consider (2k+1) 2 Idea 0: find counterexample. Idea 3: Use quotient-remainder theorem.

28. 28 Introduction to Abstract Mathematics Since m is an odd number, m = 2l+1 for some natural number l. So m 2 is an odd number. Contrapositive Proof Statement: If m 2 is even, then m is even Contrapositive: If m is odd, then m 2 is odd. So m 2 = (2l+1) 2 = (2l) 2 + 2(2l) + 1 Proof (the contrapositive): Proof by contrapositive.

29. 29 Introduction to Abstract Mathematics 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! Theorem: is irrational. Proof (by contradiction): Irrational Number

30. 30 Introduction to Abstract Mathematics so can assume so n is even. so m is even. Theorem: is irrational. Proof (by contradiction):Want to prove both m and n are even. Proof by contradiction. Irrational Number

31. 31 Introduction to Abstract Mathematics Infinitude of the Primes Theorem. There are infinitely many prime numbers. Claim: if p divides a, then p does not divide a+1. Let p 1, p 2, …, p N be all the primes. Proof by contradiction. Consider p 1 p 2 …p N + 1.

32. 32 Introduction to Abstract Mathematics Floor and Ceiling 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 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?

33. 33 Introduction to Abstract Mathematics 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

34. 34 Introduction to Abstract Mathematics 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.

35. 35 Introduction to Abstract Mathematics 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. Euclid: gcd(a,b) = gcd(b,r)! a=12, b=8 => 12 = 8 + 4 gcd(12,8) = 4 a=21, b=9 => 21 = 2x9 + 3gcd(21,9) = 3 a=99, b=27 => 99 = 3x27 + 18 gcd(99,27) = 9 gcd(8,4) = 4 gcd(9,3) = 3 gcd(27,18) = 9

36. 36 Introduction to Abstract Mathematics Euclid’s GCD Algorithm 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) a = qb + r

37. 37 Introduction to Abstract Mathematics gcd(a,b) if b = 0, then answer = a. else write a = qb + r answer = gcd(b,r) Example 1 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) Return value: 2. Example 2 GCD(252, 189) 252 = 1x189 + 63 = GCD(189, 63) 189 = 3x63 + 0 = GCD(63, 0) Return value: 63. 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) Return value: 2. Example 3

38. 38 Introduction to Abstract Mathematics Euclid: gcd(a,b) = gcd(b,r)a = qb + r Correctness of Euclid’s GCD Algorithm Let d be a common divisor of b, r  b = k 1 d and r = k 2 d for some k 1, k 2.  a = qb + r = qk 1 d + k 2 d = (qk 1 + k 2 )d => d is a divisor of a Let d be a common divisor of a, b  r = a – qb = k 3 d – qk 1 d = (k 3 – qk 1 )d => d is a divisor of r So d is a common factor of a, b iff d is a common factor of b, r  d = gcd(a, b) iff d = gcd(b, r) When r > 0:

39. 39 Introduction to Abstract Mathematics Practice problems 1. 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