Lecture Notes. Advanced Discrete Structures COT S

Transcription

1 Lecture Notes Advanced Discrete Structures COT S

2 Recap Divisibility Prime Number Theorem Euclid s Lemma Fundamental Theorem of Arithmetic Euclidean Algorithm

3 Basic Notions - Section 3.2 SOLVING ax + by = d

4 Definition A Diophantine equation is a polynomial equation (in two or more unknowns) such that only the integer solutions are searched or studied. Examples: 3 x + 3y = 2 No integer solutions 2 x 2 + 5y 2 = 1 No integer solutions 3x + 5y = 2 x = 4 + 5k y = 2 3k, k Z

5 Definition A set of integers is computably enumerable if there is an algorithm such that: For each integer input n, if n S, then the algorithm eventually halts; otherwise it runs forever. Example: S: there is a run of exactly n 6 s in the decimal expansion of τ = 2π = S = {1,2,?, }

6 Hilbert s 10 th Problem (1900) Is there a general algorithm for solving all types of Diophantine equations? Answer (Matiyasevich-Robinson-Davis-Putnam, 1977): No, every computably enumerable set is Diophantine. Corresponding to any given consistent axiomatization of number theory, one can explicitly construct a Diophantine equation which has no solutions, but this fact cannot be proved within the given axiomatization.

7 Linear Diophantine Equations Theorem: Let a, b Z with d = gcd(a, b). The equation ax + by = c has no integral solutions if d c. If d c, then there are infinitely many integral solutions. Moreover, if X = x and Y = y is a particular solution of the equation, then all solutions are given by: X = x + b d k, Y = y + a d k where k Z.

8 Linear Diophantine Equations Consider: ax + by = c and d = gcd(a, b) If d c but c d, solve instead: a Now, gcd a d, b d = c d. d X + b d Y = c d

9 Solving ax + by = gcd(a, b) 1. Extended Euclidean Algorithm Form the following sequence: x 0 = 0, x 1 = 1, x j = q j 1 x j 1 + x j 2, y 0 = 1, y 1 = 0, y j = q j 1 y j 1 + y j 2, Then a x n + b y n = gcd(a, b)

10 Recall from last Lecture Use the Euclidean algorithm to find gcd(600, 252): 600 = q 1 = = q 2 = 2 96 = q 3 = 1 60 = q 4 = 1 36 = q 5 = 2 24 = q 6 = 2 gcd(600, 252) = 12

11 Solving Linear Diophantine Equations Example: 252 x y = gcd 252,600 = 12 q 1 = 2, q 2 = 2, q 3 = 1, q 4 = 1, q 5 = 1, q 6 = 2 x 0 = 0, x 1 = 1, y 0 = 1, y 1 = 0, x j = q j 1 x j 1 + x j 2 y j = q j 1 y j 1 + y j 2 x 2 = 2 x 1 + x 0 = 2, y 2 = 2 y 1 + y 0 = 1, x 3 = 2 x 2 + x 1 = 5, y 3 = 2 y 2 + y 1 = 2, x 4 = 1 x 3 + x 2 = 7, y 4 = 1 y 3 + y 2 = 3, x 5 = 1 x 4 + x 3 = 12, y 5 = 1 y 4 + y 3 = 5, x 6 = 1 x 5 + x 4 = 19, y 6 = 1 y 5 + y 4 = 8, = 12 = gcd(252,600)

12 Solving Linear Diophantine Equations Recursive Form: x 0 = 0, x 1 = 1, x j = q j 1 x j 1 + x j 2, y 0 = 1, y 1 = 0, y j = q j 1 y j 1 + y j 2, Matrix Form: 0 1 x j 2 y j 2 1 q j 1 x j 1 y j 1 = x j 1 y j 1 x j 2 q j 1 x j 1 y j 2 q j 1 y j 1 = x j 1 y j 1 x j y j

13 Solving Linear Diophantine Equations =

14 Two was to solve ax + by = gcd(a, b) Reverse Euclidean Algorithm (Substitution) 600 = = = = = = = = = = = = = = =

15 Basic Notions - Section 3.3 CONGRUENCES

16 Definition Let a, b, n Z with n 0. We say that a b mod n (read: a is congruent to b mod n) if n a b, i.e., there exists a k Z such that a b = k n or a = b + k n.

17 Examples 2 9 mod = 11 = (mod 10) = 120 = (mod 10) = 0 = (mod 21) 21 0 = 21 = 1 21

18 Proposition Let a, n Z with n 0. a 0 (mod n) if and only if n a. Proof. ( ) Suppose a 0 (mod n), i.e., there exists a k Z such that a 0 = k n. By definition, n a. ( ) Likewise, if n a there exists a k Z such that a 0 = k n which means a 0 (mod n). a a (mod n). Proof. Note that a a = 0 = 0 n, thus a a (mod n).

19 Proposition Let a, b, n Z with n 0. a b (mod n) if and only if b a (mod n). Proof. ( ) Suppose a b (mod n), i.e., there exists a k Z such that a b = k n. Multiply both sides by 1 to get: b a = a b = k n = k n which means b a (mod n). ( ) Proof is similar; interchange a and b.

20 Proposition Let a, b, c, n Z with n 0. If a b (mod n) and b c (mod n), then a c mod n. Proof. By definition, a = b + k 1 n and b = c + k 2 n for some integers k 1, k 2 Z. Thus, a = b + k 1 n = c + k 2 n + k 1 n = c + k 2 + k 1 n which means a c (mod n).

21 Proposition Let a, b, c, d, n Z with n 0, and suppose a b (mod n) and c d (mod n). Then Proof. a + c b + d mod n. We have that a = b + n k 1 and c = d + n k 2 for some k 1, k 2 Z. Thus, a + c = b + n k 1 + d + n k 2 = b + d + n (k 1 + k 2 ), i.e., a + c b + d (mod n).

22 Proposition Let a, b, c, d, n Z with n 0, and suppose a b (mod n) and c d (mod n). Then Proof. a c b d mod n. We have that a = b + n k 1 and c = d + n k 2 for some k 1, k 2 Z. Thus, a c = b + n k 1 d + n k 2 = b d + n d k 1 + b k 2 + n k 1 k 2, i.e., a c b d (mod n).

23 Proposition Let a, b, c, d, n Z with n 0, and suppose a b (mod n) and c d (mod n). Then Proof. a c b d mod n. We have that a = b + n k 1 and c = d + n k 2 for some k 1, k 2 Z. Thus, a c = b + n k 1 d + n k 2 = b d + n d k 1 + b k 2 + n k 1 k 2, i.e., a c b d (mod n).

24 Examples We can do algebra modulo n: x (mod 11) x ( 7) (mod 11) x 7 (mod 11) Remember, this means that, x = 18, x = 7, x = 4, x = 15, are all solutions.

25 Basic Notions - Section DIVISION

26 Subtraction and Division Subtraction: Addition of negative numbers 7 4 = 7 + ( 4) Division: Multiplication of fractions Negative numbers: Negative powers: 3 4 = = is the number k which solves 4 + k = is the number k which solves 4 k = 1

27 Proposition Let a, b, c, n Z with n 0 and gcd a, n = 1. If ab ac (mod n), then b c (mod n). Proof. Recall that gcd a, n = 1 implies that there exist x, y Z such that a x + n y = 1 b c a x + n y = (b c) or a b a c x + b y c y n = b c. Since n ab ac and n by cy n, n b c also. Thus, b c (mod n).

28 Example Solve: 2x (mod 11) 2x ( 5) (mod 11) 2x 4 (mod 11) x 2 (mod 11) since gcd(2,11) = 1.

29 Example Solve: 2x (mod 11) 2x ( 5) (mod 11) 2x 3 (mod 11) 2 1 2x (mod 11) since gcd(2,11) = 1. But what does 2 1 (mod 11) mean?

30 2 1 (mod 11) Want to find a k Z with the following property: this would mean k 2 1 mod 11, 2k 1 = 11j for some j Z. Rewritten, we are looking for j, k Z such that 2k + 11j = 1. Using the extended Euclidean algorithm: k = 6, j = 1 Check: 6 2 = 12 1 (mod 11)

31 Example Solve: 2x (mod 11) 2x ( 5) (mod 11) 2x 3 (mod 11) 2 1 2x (mod 11) x 6 3 = 18 7 (mod 11)

32 Proposition Suppose gcd(a, n) = 1. Let s, t Z s.t. as + nt = 1. Then a s 1 mod n. [The integer s is said to be the multiplicative inverse of a modulo n and written a 1.] Proof. Since a s 1 = n t, a s 1 (mod n). What can happen when gcd a, n 1? 3k 1 (mod 6)

33 Example However, 15 x 21 (mod 39) 3 5 x 3 7 mod 39 5 x 7 (mod 39) Here we re looking for 5 1 (mod 39). Find j, k such that 39j + 5k = 1 by the extended Euclidean algorithm: j = 1, k = 8 works, so x 8 7 = (mod 39) Other solutions: x 4 (mod 39) and x 30 (mod 39).

34 Solving ax c (mod n) If gcd a, n = 1: 1. Use the extended Euclidean algorithm to find s, t Z such that a s + n t = The solution is x s c (mod n). If gcd(a, n) > 1: 1. If d does not divide b, there is no solution. 2. Assume d b. Consider the new congruence a/d x b/d mod n/d and obtain the new solution x = x Solutions to original congruence are: x 0, x 0 + n d, x n d,, x 0 + d 1 n d (mod n)

35 Non-linear Equations An important congruence: x 2 a mod n. Example Check: x 2 1 mod = 0 0 (mod 5) 1 2 = 1 1 (mod 5) 2 2 = 4 4 (mod 5) 3 2 = 9 4 (mod 5) 4 2 = 16 1 (mod 5) Solutions: x ±1 (mod 5)

36 Non-linear Equations Example Check: x 2 1 mod = 1 1 (mod 15) 8 2 = 64 4 (mod 15) 2 2 = 4 4 (mod 15) 9 2 = 81 6 (mod 15) 3 2 = 9 9 (mod 15) 10 2 = (mod 15) 4 2 = 16 1 (mod 15) 11 2 = (mod 15) 5 2 = (mod 15) 12 2 = (mod 15) 6 2 = 36 6 (mod 15) 13 2 = (mod 15) 7 2 = 49 4 (mod 15) 14 2 = (mod 15) Solutions: x ±1, ±4 (mod 5)

37 Basic Notions - Section WORKING WITH FRACTIONS

38 Don t Use multiplicative inverse notation: a 1 Always check that the integer a 1 actually exists, i.e., gcd a, n = 1.