Chapter 3 Basic Number Theory

Transcription

1 Chapter 3 Basic Number Theory

2 What is Number Theory? Well...

3 What is Number Theory? Well... Number Theory The study of the natural numbers (Z + ), especially the relationship between different sorts of numbers.

4 What is Number Theory? Well... Number Theory The study of the natural numbers (Z + ), especially the relationship between different sorts of numbers. Divisibility Suppose n, m Z, m 0, we say m divides n if n is a multiple of m. That is, k Z n = mk. If m divides n, we write m n. If not, m n.

5 Greatest Common Factors Basic Examples gcd(12,20)=

6 Greatest Common Factors Basic Examples gcd(12,20)= 4 gcd(18,30)=

7 Greatest Common Factors Basic Examples gcd(12,20)= 4 gcd(18,30)= 6 gcd(225,120)=

8 Greatest Common Factors Basic Examples gcd(12,20)= 4 gcd(18,30)= 6 gcd(225,120)= 15

9 Greatest Common Factors Basic Examples gcd(12,20)= 4 gcd(18,30)= 6 gcd(225,120)= 15 For the last one, we cant just look at it and know the answer - we need some technique, and prime factorization works here. 120 = , 225 = This is not practical for large numbers.

10 The Euclidean Algorithm Example Find (132, 36)

11 The Euclidean Algorithm Example Find (132, 36) 132 =

12 The Euclidean Algorithm Example Find (132, 36) 132 = =

13 The Euclidean Algorithm Example Find (132, 36) 132 = = =

14 The Euclidean Algorithm Example Find (132, 36) 132 = = = So, (132, 36) = 12.

15 Another Example Example Find ( , )

16 Another Example Example Find ( , ) The answer we seek is 1078.

17 Another Example Example Find ( , ) The answer we seek is Can you generalize this? (a, b) =

18 Another Example Example Find ( , ) The answer we seek is Can you generalize this? (a, b) = (b, r)

19 The Division Algorithm Before we analyze, where does this equation come from? The division algorithm, which states

20 The Division Algorithm Before we analyze, where does this equation come from? The division algorithm, which states The Division Algorithm a = bq + r, 0 r < b There are two parts to this proof, the existence of q and r, and uniqueness.

21 The Division Algorithm Before we analyze, where does this equation come from? The division algorithm, which states The Division Algorithm a = bq + r, 0 r < b There are two parts to this proof, the existence of q and r, and uniqueness. Proof Existence Consider the set S = {a nd n Z}. We claim S contains a non-negative integer. There are two cases to consider: 1 If a 0, choose n = 0 2 If a < 0, choose n = ad

22 The Division Algorithm Proof Proof In both cases, a nd is non-negative and thus S always contains at least one non-negative integer. This means we can apply the well-ordering principle.

23 The Division Algorithm Proof Proof In both cases, a nd is non-negative and thus S always contains at least one non-negative integer. This means we can apply the well-ordering principle. Every non-empty set of positive integers contains a smallest element.

24 The Division Algorithm Proof Proof In both cases, a nd is non-negative and thus S always contains at least one non-negative integer. This means we can apply the well-ordering principle. Every non-empty set of positive integers contains a smallest element. and we can deduce that S contains a least non-negative integer r. By definition, r = a nd for some n. Let q be this n. Then, by rearranging, a = qd + r.

25 The Division Algorithm Proof Proof It remains to show 0 r < d. The first inequality holds as r was chosen to be non-negative. To show r < d, suppose r d. Since d 0, r > 0 but d > 0 or d < 0.

26 The Division Algorithm Proof Proof It remains to show 0 r < d. The first inequality holds as r was chosen to be non-negative. To show r < d, suppose r d. Since d 0, r > 0 but d > 0 or d < 0. If d > 0 then r d implies a qd d, further implying a qd d 0 a (q + 1)d 0. Therefore, a (q + 1)d S, and since a (q + 1)d = r d with d > 0, we know a (q + 1)d < r, contradicting that r was the least non-negative element in S.

27 The Division Algorithm Proof Proof It remains to show 0 r < d. The first inequality holds as r was chosen to be non-negative. To show r < d, suppose r d. Since d 0, r > 0 but d > 0 or d < 0. If d > 0 then r d implies a qd d, further implying a qd d 0 a (q + 1)d 0. Therefore, a (q + 1)d S, and since a (q + 1)d = r d with d > 0, we know a (q + 1)d < r, contradicting that r was the least non-negative element in S. If d < 0, then r d implies that a qd d. This implies that a qd + d 0 a (q 1)d 0. Therefore, a (q 1)d S and, since a (q 1)d = r + d with d < 0, we know a (q 1)d < r. So, r < d, completing the existence proof.

28 The Division Algorithm Proof Proof. Uniqueness Suppose there exists q, q, r, r with 0 r, r < d a = dq + r and a = dq + r. Without loss of generality, assume q q. Subtracting the two equations yields d(q q) = r r.

29 The Division Algorithm Proof Proof. Uniqueness Suppose there exists q, q, r, r with 0 r, r < d a = dq + r and a = dq + r. Without loss of generality, assume q q. Subtracting the two equations yields d(q q) = r r. If d > 0 then r r and r < d d + r, so (r r ) < d. Similarly, if d < 0 then r r and r < d d + r, so (r r ) < d. Combining these yields r r < d.

30 The Division Algorithm Proof Proof. Uniqueness Suppose there exists q, q, r, r with 0 r, r < d a = dq + r and a = dq + r. Without loss of generality, assume q q. Subtracting the two equations yields d(q q) = r r. If d > 0 then r r and r < d d + r, so (r r ) < d. Similarly, if d < 0 then r r and r < d d + r, so (r r ) < d. Combining these yields r r < d. The original equation implies d divides r r. So, d r r or r r = 0. But, we established that r r < d, so r = r. Substituting into the original equation yields dq = dq and since d 0, q = q, proving uniqueness.

31 The Division Algorithm Back to the Euclidean Algorithm. The general method looks like: a = q 1 b + r 1 b = q 2 r 1 + r 2 r 1 = q 3 r 2 + r 3 r 2 = q 4 r 3 + r 4 r n 2 = q n r n 1 + r n r n 1 = q n+1 r n + 0.

32 The Division Algorithm Back to the Euclidean Algorithm. The general method looks like: a = q 1 b + r 1 b = q 2 r 1 + r 2 r 1 = q 3 r 2 + r 3 r 2 = q 4 r 3 + r 4 r n 2 = q n r n 1 + r n r n 1 = q n+1 r n + 0 Why is this last nonzero remainder r n a common divisor of a and b?.

33 The Division Algorithm Back to the Euclidean Algorithm. The general method looks like: a = q 1 b + r 1 b = q 2 r 1 + r 2 r 1 = q 3 r 2 + r 3 r 2 = q 4 r 3 + r 4 r n 2 = q n r n 1 + r n r n 1 = q n+1 r n + 0 Why is this last nonzero remainder r n a common divisor of a and b? Start from the bottom and work upwards..

34 Justification The last line r n 1 = q n+1 r n shows that r n r n 1.

35 Justification The last line r n 1 = q n+1 r n shows that r n r n 1. The line above that r n 2 = q n r n 1 + r n shows that r n divides r n 2 since it divides r n and r n 1.

36 Justification The last line r n 1 = q n+1 r n shows that r n r n 1. The line above that r n 2 = q n r n 1 + r n shows that r n divides r n 2 since it divides r n and r n 1. By continuing in this fashion, we see it also divides r n 3, r n 4,, all the way through to a and b.

37 Justification But, why is r n the greatest common divisor of a and b?

38 Justification But, why is r n the greatest common divisor of a and b? Suppose d is any common divisor of a and b. We will work our way back down the list. Consider a = q 1 b + r 1. Since d divides a and b, it must also divide r 1. The second equation b = q 2 r 1 + r 2 shows d must divide r 2.

39 Justification But, why is r n the greatest common divisor of a and b? Suppose d is any common divisor of a and b. We will work our way back down the list. Consider a = q 1 b + r 1. Since d divides a and b, it must also divide r 1. The second equation b = q 2 r 1 + r 2 shows d must divide r 2. Continuing down the line, at each stage we know d divides the previous two remainders, r i 1 and r i, and the current line r i 1 = q i+1 r i + r i+1 will tell us d also divides r i+1.

40 Justification But, why is r n the greatest common divisor of a and b? Suppose d is any common divisor of a and b. We will work our way back down the list. Consider a = q 1 b + r 1. Since d divides a and b, it must also divide r 1. The second equation b = q 2 r 1 + r 2 shows d must divide r 2. Continuing down the line, at each stage we know d divides the previous two remainders, r i 1 and r i, and the current line r i 1 = q i+1 r i + r i+1 will tell us d also divides r i+1. Eventually, we reach r n 2 = q n r n 1 + r n, at which point we conclude d r n. So, if we have any common divisor of a and b in d then d r n. Therefore r n must be the greatest common divisor of a and b.

41 The Division Algorithm Statement of the Euclidean Algorithm To compute (a, b), let r 1 = a and r 0 = b and compute successive quotients and remainders of r i 1 = q i+1 r i + r i+1 until some remainder r n+1 = 0. Then r n is the greatest common divisor. Note: Although the notation may look odd, we let a = r 1 so that we have a mechanism to refer to all values in the algorithm in the form r i.

42 Primes Theorem There are infinitely many primes.

43 Primes Theorem There are infinitely many primes. Proof. Suppose we start with a list of all primes p 1, p 2,, p n. Let A = p 1 p 2 p n + 1. A is larger than any number on our list. So, if A is prime then we are done.

44 Primes Theorem There are infinitely many primes. Proof. Suppose we start with a list of all primes p 1, p 2,, p n. Let A = p 1 p 2 p n + 1. A is larger than any number on our list. So, if A is prime then we are done. If A is composite, then there must be a q 1 q 1 A where q 1 p i for all i = 1, n. Because q 1 1 we have a contradiction, So, it must be so that q 1 is a prime not on our original list.

45 Primes Theorem There are infinitely many primes. Proof. Suppose we start with a list of all primes p 1, p 2,, p n. Let A = p 1 p 2 p n + 1. A is larger than any number on our list. So, if A is prime then we are done. If A is composite, then there must be a q 1 q 1 A where q 1 p i for all i = 1, n. Because q 1 1 we have a contradiction, So, it must be so that q 1 is a prime not on our original list. Let B = q 1 p 1 p n + 1 and repeat this process.

46 Helpful Tools for Number Theory Theorem If x 0, y 0 is a solution of ax + by = c, then so is x 0 + bt, y 0 at. Proof. We know ax 0 + by 0 = c. Consider the following: a(x 0 + bt) + b(y 0 at) = ax 0 + abt + by 0 abt = ax 0 + by 0 = c

47 Helpful Tools for Number Theory Theorem If (a, b) c then ax + by = c has no solution and if (a, b) c then ax + by = c has a solution. Proof. Suppose x 0, y 0 Z ax 0 + by 0 = c. Since (a, b) ax 0 and (a, b) by 0, (a, b) c. Conversely, suppose (a, b) c. then c = m(a, b) for some m. We know ar + bs = (a, b) for some r, s Z. Then, and x = rm, y = sm is a solution. a(rm) + b(sm) = m(a, b) = c

48 One More Theorem Theorem Suppose (a, b) = 1 and x 0, y 0 is a solution of ax + by = c. Then all solutions are given by x = x 0 + bt, y = y 0 at for t Z.

49 Congruences Definition Let a, b, n be integers with n 0, We say a b(mod n) if a b is a multiple of n.

50 Congruences Definition Let a, b, n be integers with n 0, We say a b(mod n) if a b is a multiple of n. The Linear Congruence Theorem Suppose g = (a, m). a) If g b then ax b(mod m) has no solutions. b) If g b then ax b(mod m) has exactly g incongruent solutions.

51 Proof of (a) Proof. (by contrapositive) If ax b(mod m) has a solution then (a, m) b. Suppose r is a solution. Then ar b(mod m) by definition, and from the definition, m (ar b), or ar b = km for some k. Since (a, m) a and (a, m) km, it follows that (a, m) b.

52 Proof of (b) Proof. Since g = (a, m) and g b then we can rewrite our congruence as a g x b ( mod m ) g g

53 Proof of (b) Proof. Since g = (a, m) and g b then we can rewrite our congruence as a g x b ( mod m ) g g ( ) But, a g, m g =1, so the right hand side has a unique solution modulo ( ) m g, say x x m 1 g.

54 Proof of (b) Proof. Since g = (a, m) and g b then we can rewrite our congruence as a g x b ( mod m ) g g ( ) But, a g, m g =1, so the right hand side has a unique solution modulo ( ) m g, say x x m 1 g. So, the integers x which satisfy ax b(mod m) are exactly those of the form x = x 1 + k m g for some k.

55 Proof of (b) Proof. Consider the set of integers { x 1, x 1 + m g, x m g,, x 1 + (g 1) m } g

56 Proof of (b) Proof. Consider the set of integers { x 1, x 1 + m g, x m g,, x 1 + (g 1) m } g None of these are congruent modulo m and none differ by as much as m. further, for any k Z, we have that x 1 + k m g is congruent modulo m to one of them.

57 Proof of (b) Proof. To see this, write k = gq + r where 0 r < d from the Division Algorithm. then, x 1 + k m g = x 1 + (gq + r) m g = x 1 + mq + r m g x 1 + r m (mod m) g So, these are the g solutions of ax b(mod m).

58 Congruence Rules 1 a a(mod m)

59 Congruence Rules 1 a a(mod m) 2 a 0(mod m) iff m a

60 Congruence Rules 1 a a(mod m) 2 a 0(mod m) iff m a 3 a b(mod m) iff b a(mod m)

61 Congruence Rules 1 a a(mod m) 2 a 0(mod m) iff m a 3 a b(mod m) iff b a(mod m) 4 If a b(mod m) and c d(mod m) then 1 a + c b + d(mod m)

62 Congruence Rules 1 a a(mod m) 2 a 0(mod m) iff m a 3 a b(mod m) iff b a(mod m) 4 If a b(mod m) and c d(mod m) then 1 a + c b + d(mod m) 2 ac bd(mod m)

63 Congruence Rules 1 a a(mod m) 2 a 0(mod m) iff m a 3 a b(mod m) iff b a(mod m) 4 If a b(mod m) and c d(mod m) then 1 a + c b + d(mod m) 2 ac bd(mod m) 5 (a, n) = 1 then ab ac(mod n) implies that b c(mod n)

64 Congruence Rules 1 a a(mod m) 2 a 0(mod m) iff m a 3 a b(mod m) iff b a(mod m) 4 If a b(mod m) and c d(mod m) then 1 a + c b + d(mod m) 2 ac bd(mod m) 5 (a, n) = 1 then ab ac(mod n) implies that b c(mod n) 6 If (a, n) = d 1 then a b(mod n) implies a d b d (mod n d )

65 Congruence Rules 1 a a(mod m) 2 a 0(mod m) iff m a 3 a b(mod m) iff b a(mod m) 4 If a b(mod m) and c d(mod m) then 1 a + c b + d(mod m) 2 ac bd(mod m) 5 (a, n) = 1 then ab ac(mod n) implies that b c(mod n) 6 If (a, n) = d 1 then a b(mod n) implies a d b d (mod n d ) 7 If (a, m) = 1 then ax b(mod m) has one solution.

66 Using These Properties Example Solve for x in 4x 3(mod 19).

67 Using These Properties Example Solve for x in 4x 3(mod 19). Since (5, 19) = 1, we can multiply both sides by 5. This gives 4x 3(mod 19) 20x 15(mod 19) and since 20 1(mod 19), we have that x 15(mod 19).

68 And Another One Example Solve for x in 6x 15(mod 514)

69 And Another One Example Solve for x in 6x 15(mod 514) Since 6x 15 is always odd, it can never be divisible by 514. So, there is no solution.

70 Fermat s Little Theorem Theorem Let p be a prime which does not divide the integer a, then a p 1 1(mod p). Proof. Start by listing the first p 1 positive multiples of a: a, 2a, 3a,, (p 1)a.

71 Fermat s Little Theorem Theorem Let p be a prime which does not divide the integer a, then a p 1 1(mod p). Proof. Start by listing the first p 1 positive multiples of a: a, 2a, 3a,, (p 1)a. Suppose that ra and sa are the same modulo p, then we have r s(mod p), so the p 1 multiples of a above are distinct and nonzero; that is, they must be congruent to 1, 2, 3,, p 1 in some order.

72 Fermat s Little Theorem Theorem Let p be a prime which does not divide the integer a, then a p 1 1(mod p). Proof. Start by listing the first p 1 positive multiples of a: a, 2a, 3a,, (p 1)a. Suppose that ra and sa are the same modulo p, then we have r s(mod p), so the p 1 multiples of a above are distinct and nonzero; that is, they must be congruent to 1, 2, 3,, p 1 in some order. Multiply all these congruences together and we find a 2a 3a (p 1)a (p 1)(mod p) or better, a p 1 (p 1)! (p 1)!(mod p). Divide both sides by (p 1)! to complete the proof.