Numbers (8A) Young Won Lim 5/24/17
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1 Numbers (8A
2 Copyright (c 2017 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled "GNU Free Documentation License". Please send corrections (or suggestions to youngwlim@hotmail.com. This document was produced by using OpenOffice and Octave.
3 GCD A 24-by-60 rectangle is covered with ten 12-by-12 square tiles, where 12 is the GCD of 24 and = mod 12 = 0 60 = mod 12 = 0 More generally, an a-by-b rectangle can be covered with square tiles of side-length d only if d is a common divisor of a and b d a d b d : common divisor the largest d : gcd (greatest common divisor Numbers (8A 3
4 LCM What is the LCM of 4 and 6? Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76,... and the multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,... Common multiples of 4 and 6 are simply the numbers that are in both lists: 12, 24, 36, 48, 60, 72,... So, from this list of the first few common multiples of the numbers 4 and 6, their least common multiple is Numbers (8A 4
5 GCD * LCM a = p 1 a 1 p2 a 2 pn a n Prime Factorization b = p 1 b 1 p2 b 2 pn b n gcd(a,b = p 1 min(a 1, b 1 p 2 min(a 2, b 2 p n min(a n,b n lcm(a,b = p 1 Max(a 1, b 1 p 2 Max (a 2, b 2 p n Max(a n, b n gcd(a, b lcm(a,b = p 1 a 1 +b 1 p2 a 2 +b 2 pn a n +b n = a b Numbers (8A 5
6 Finding common unit length Euclid's method for finding the greatest common divisor (GCD of two starting lengths BA and DC, both defined to be multiples of a common "unit" length. The length DC being shorter, it is used to "measure" BA, but only once because remainder EA is less than DC. EA now measures (twice the shorter length DC, with remainder FC shorter than EA. Then FC measures (three times length EA. Because there is no remainder, the process ends with FC being the GCD. On the right Nicomachus' example with numbers 49 and 21 resulting in their GCD of 7 (derived from Heath 1908:300. Numbers (8A 6
7 Euclid Algorithm Steps remainder remainder zero divisor divisor divisor gcd Numbers (8A 7
8 Euclid Algorithm 1071 = = gcd (1071,462 = 3 7 = = = = Numbers (8A 8
9 Common Divisor 1071 = = = common divisor d d 1071 and d mod d = 0 d mod d = 0 d Numbers (8A 9
10 Common Divisor Properties common divisor d? d 1071 and d mod d = 0 and 462 mod d = = remainder ( mod d = 0 2 (462 mod d mod d = mod d = mod d = mod d = mod d = mod d = 0 d 462 and d Numbers (8A 10
11 Reducing GCD Problems 1071 = d 1071 and d 462 gcd (1071, = d 462 and d 147 gcd (462, = d 147 and d 21 gcd (147, Numbers (8A 11
12 Linear Combination of gcd(1071, 462= = = = = ( = = = gcd(1071, 462 = 21 = gcd (a,b = sa + t b Numbers (8A 12
13 Linear Combination of gcd(252, 198= = = = = = = = 2 18 ( (198 3 ( = ( = = 18 gcd (a,b = sa + t b Numbers (8A 13
14 Bezout s Identity a, b Z + x, y Z x a + y b = gcd (a,b Bezout s coefficients (not unique Bezout s identities Generally, a linear combination of a & b must be unique and its coefficients x & y need not be integers. Numbers (8A 14
15 Pairs of Bézout Coefficients Examples 42 = = = 6 ( = 6 ( = 6 x a + y b = gcd (a,b x 12 + y 42 = gcd (12, 42 (x, y not unique Numbers (8A 15
16 Pairs of Bézout Coefficients not unique /6=7-3 < 7 4 < 7 12/6=2 1 < 2-1 < 2 Numbers (8A 16
17 Pairs of Bézout Coefficients 2 minimal pairs x a + y b = gcd (a,b 42/6=7 12/6=2 Among these pairs of Bézout coefficients, exactly two of them satisfy -3 < 7 4 < 7 1 < 2-1 < 2 The Extended Euclidean Algorithm always produces one of these two minimal pairs. Numbers (8A 17
18 Pairs of Bézout Coefficients all pairs x a + y b = gcd (a,b all pairs can be represented in the form 42/6=7 12/6= k 1 + 2k The Extended Euclidean Algorithm always produces one of these two minimal pairs. Numbers (8A 18
19 Extended Euclid Algorithm Given a & b, the extended Euclid algorithm produce the same coefficients. Uniquely, one is chosen among many possible Bézout s coefficients Numbers (8A 19
20 Congruence Etymology Middle English, from Latin congruentia ( agreement, from congruēns, present active participle of congruō ( meet together, agree. Noun: congruence (plural congruences The quality of agreeing or corresponding; being suitable and appropriate. (mathematics, number theory A relation between two numbers indicating they give the same remainder when divided by some given number. (mathematics, geometry The quality of being isometric roughly, the same measure and shape. (algebra More generally: any equivalence relation defined on an algebraic structure which is preserved by operations defined by the structure. Numbers (8A 20
21 Congruence in Geometry congruent congruent similar Numbers (8A 21
22 Congruent modulo n a b (mod n a is congruent to b modulo n n (a b n divides (a-b (a b mod n = 0 (a mod n = (b mod n the same remainder A remainder is positive (0,.. n-1 Numbers (8A 22
23 Congruence Relation Numbers (8A 23
24 Properties of a Congruence Relation Numbers (8A 24
25 Remainders Numbers (8A 25
26 Linear Congruence Problems a x b (mod n find x =? A linear congruence a x = b find x =? A linear equation A remainder is positive (0,.. n-1 Numbers (8A 26
27 Modular Multiplicative Inverse A linear congruence a x b (mod n āa x āb (mod n x ā b (mod n A linear equation a x = b a 1 a x = a 1 b x = a 1 b āa 1 (mod n a 1 a = 1 A remainder is positive (0,.. n-1 Numbers (8A 27
28 Relatively Prime Numbers gcd (a,n = 1 Relatively prime numbers a & n s a + t n = 1 s a + t n 1 (mod n t n mod n = 0 s a 1 (mod n āa 1 (mod n the inverse of a exists : s linear combination of gcd(a, n=1 Numbers (8A 28
29 Chinese Remainder Theorem x 2 (mod 3 and x 3 (mod 5 and x 2 (mod 7 Sunzi's original formulation: x 2 (mod 3 3 (mod 5 2 (mod 7 with the solution x = k where k Z Numbers (8A 29
30 Chinese Remainder Theorem x a 1 x a 2 (mod m 1 and (mod m 2 and m 1, m 2, m n pairwise relatively coprime x a n (mod m n x b (mod m 1 m 2 m n has a unique solution Numbers (8A 30
31 m i, m, and M i x 2 (mod 3 x 3 (mod 5 x 2 (mod 7 m 1 = 3 m 2 = 5 m 3 = 7 m = = 105 = m/m 1 = /3 = 35 = m/m 2 = 3 5 7/5 = 21 = m/m 3 = 3 5 7/7 = 15 x a 1 (mod m 1 x a 2 (mod m 2 x a 3 (mod m 3 m = m 1 m 2 m 3 = m/m 1 = m 2 m 3 = m/m 2 = m 1 m 3 = m/m 3 = m 1 m 2 mod m 2 = mod m 3 = 0 mod m 1 = mod m 3 = 0 mod m 1 = mod m 2 = 0 M i mod m j = M j mod m i = 0 for i j Numbers (8A 31
32 Inverse of M i m 1, m 2, m 3 : pairwise relatively coprime gcd (, m 1 = 1 = 1 (mod m 1 y 1 : the inverse of m 2 m 3 gcd (, m 2 = 1 = 1 (mod m 2 y 2 : the inverse of m 1 m 3 gcd (, m 3 = 1 = 1 (mod m 3 y 3 : the inverse of m 1 m 2 = 1 (mod m 1 = 0 (mod m 1 = 0 (mod m 1 = 0 (mod m 2 = 1 (mod m 2 = 0 (mod m 2 = 0 (mod m 3 = 0 (mod m 3 = 1 (mod m 3 m 2 m 3 m 1 m 3 m 1 m 2 Numbers (8A 32
33 Sum of a i M i y i a 1 a 2 a 3 = 1 (mod m 1 = 0 (mod m 1 = 0 (mod m 1 = 0 (mod m 2 = 1 (mod m 2 = 0 (mod m 2 = 0 (mod m 3 = 0 (mod m 3 = 1 (mod m 3 a 1 = a 1 (mod m 1 a 2 = 0 (mod m 1 a 3 = 0 (mod m 1 a 1 = 0 (mod m 2 a 2 = a 2 (mod m 2 a 3 = 0 (mod m 2 a 1 = 0 (mod m 3 a 2 = 0 (mod m 3 a 3 = a 3 (mod m 3 a 1 + a 2 + a 3 = a 1 = a 1 (mod m 1 a 1 + a 2 + a 3 = a 2 = a 2 (mod m 2 a 1 + a 2 + a 3 = a 3 = a 3 (mod m 3 Numbers (8A 33
34 X = Sum of a i M i y i x a 1 (mod m 1 x a 2 (mod m 2 x a 3 (mod m 3 a 1 + a 2 + a 3 = a 1 = a 1 (mod m 1 a 1 + a 2 + a 3 = a 2 = a 2 (mod m 2 a 1 + a 2 + a 3 = a 3 = a 3 (mod m 3 x = a 1 + a 2 + a 3 Numbers (8A 34
35 Chinese Remainder Examples (1 x 2 (mod 3 x 3 (mod 5 x 2 (mod 7 m 1 = 3 m 2 = 5 m 3 = = 105 = m = m/m 1 = /3 = 35 = m/m 2 = 3 5 7/5 = 21 = m/m 3 = 3 5 7/7 = 15 m 2 m 3 m 1 m 3 m 1 m 2 = 2 (mod m 1 = 0 (mod m 1 = 0 (mod m 1 = 0 (mod m 2 = 1 (mod m 2 = 0 (mod m 2 = 0 (mod m 3 = 0 (mod m 3 = 1 (mod m 3 m 2 m 3 m 1 m 3 m 1 m 2 = 35 2 = 2 2 = 1 (mod 3 = 21 1 = 1 1 = 1 (mod 5 = 15 1 = 1 1 = 1 (mod 7 y 1 ( =2 : the inverse of ( =35 y 2 ( =1 : the inverse of ( =21 y 3 ( =1 : the inverse of ( =15 = 1 (mod m 1 = 0 (mod m 1 = 0 (mod m 1 = 0 (mod m 2 = 1 (mod m 2 = 0 (mod m 2 = 0 (mod m 3 = 0 (mod m 3 = 1 (mod m 3 Numbers (8A 35
36 Chinese Remainder Examples (2 a 1 a 2 a 3 = 1 (mod m 1 = 0 (mod m 1 = 0 (mod m 1 = 0 (mod m 2 = 1 (mod m 2 = 0 (mod m 2 = 0 (mod m 3 = 0 (mod m 3 = 1 (mod m 3 x = a 1 + a 2 + a 3 x = a 1 = a 1 (mod m 1 x = a 2 = a 2 (mod m 2 x = a 3 = a 3 (mod m 3 m 1 = 3 m 2 = 5 m 3 = 7 = 3 5 7/3 = 5 7 = 35 = /5 = 3 7 = 21 = 3 5 7/7 = 3 5 = 15 x = = 233 x = 233 = 23 (mod 105 m = = 105 Numbers (8A 36
37 Chinese Remainder Summary x a 1 (mod m 1 x a 2 (mod m 2 x a 3 (mod m 3 m = m 1 m 2 m 3 = m/m 1 = m 2 m 3 = m/m 2 = m 1 m 3 = m/m 3 = m 1 m 2 m 1, m 2, m 3 : pairwise relatively coprime gcd (, m 1 = 1 gcd (, m 2 = 1 gcd (, m 3 = 1 = 1 (mod m 1 = 1 (mod m 2 = 1 (mod m 3 y 1 : the inverse of y 2 : the inverse of y 3 : the inverse of x = a 1 + a 2 + a 3 Numbers (8A 37
38 Chinese Remainder Theorem Numbers (8A 38
39 Upper and Lower Bounds Numbers (8A 39
40 References [1] [2]
Numbers (8A) Young Won Lim 5/22/17
Numbers (8A Copyright (c 2017 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version
Numbers (8A) Young Won Lim 6/21/17
Numbers (8A Copyright (c 2017 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version
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