Transcript Discrete Mathematics

이산수학 (Discrete Mathematics)
2.4 정수와 나눗셈
(The Integers and Division)
2006년 봄학기
문양세
강원대학교 컴퓨터과학과
Introduction
2.4 The Integers and Division
Of course you already know what the integers are, and
what division is…
(정수와 나누기… 다 아는거~)
But: There are some specific notations, terminology, and
theorems associated with these concepts which you may
not know.
(몇 가지 주요 표기, 용어, 정리 등을 배웁시다.)
These form the basics of number theory.
(정수론의 기초..)
• Vital in many important algorithms today (hash functions,
cryptography, digital signatures).
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Divides, Factor, and Multiple
2.4 The Integers and Division
Let a,bZ with a0.
a|b  “a divides b” : “cZ: b=ac”
“There is an integer c such that c times a equals b.”
(b가 a로 나누어짐(a로 b를 나눌 수 있음)을 의미하며, 이때 몫이 c이다.)
• Example: 312  True, but 37  False.
인수
Iff a divides b, then we say a is a factor or a divisor of b,
and b is a multiple of a.
배수
“b is even” :≡ 2b. Is 0 even? Is −4?
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Some Facts of Division
2.4 The Integers and Division
a,b,c  Z:
1. a|0
(2|0, 3|0, …)
2. (a|b  a|c)  a | (b + c)
(2|4  2|6  2|10)
3. a|b  a|bc
(2|4  2|4∙3)
4. (a|b  b|c)  a|c
(2|4  4|8  2|8)
Proof of (2):
• a|b means there is an s such that b=as,
• and a|c means that there is a t such that c=at,
• so b+c = as+at = a(s+t), so a|(b+c) also.□
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More Detailed Version of Proof
2.4 The Integers and Division
Show a,b,c  Z: (a|b  a|c)  a | (b + c).
Let a, b, c be any integers such that a|b and a|c, and
show that a | (b + c).
By definition of |, we know s: b=as, and t: c=at.
Let s, t, be such integers.
Then b+c = as + at = a(s+t)
so u: b+c=au, namely u=s+t.
Thus a|(b+c).
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Prime Numbers (소수)
2.4 The Integers and Division
An integer p>1 is prime iff it is not the product of any two
integers greater than 1:
p>1  a,bN: a>1, b>1, ab=p.
The only positive factors of a prime p are 1 and p itself.
Some primes: 2,3,5,7,11,13...
(양의 인수(divisor)가 1과 자기 자신뿐이면 소수(prime)라 한다.)
Non-prime integers greater than 1 are called composite,
because they can be composed by multiplying two integers
greater than 1.
합성수
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Fundamental Theorem of Arithmetic
(산술의 기본 정리)
2.4 The Integers and Division
Every positive integer has a unique representation as the
product of a non-decreasing series of zero or more primes.
(모든 양의 정수는 오름차순으로 정렬된 소수들의 곱으로 유일하게 표현된다.
결국 Prime Factorization(소인수분해)를 이야기한다고 볼 수 있다.)
• 1 = (product of empty series) = 1
• 2 = 2 (product of series with one element 2)
• 4 = 2·2 (product of series 2,2)
• 2000 = 2·2·2·2·5·5·5 = 24·53; 2001 = 3·23·29;
2002 = 2·7·11·13; 2003 = 2003
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소수의 개수?
2.4 The Integers and Division
“무한히 많은 소수가 존재함”을 증명하라.
1. 유한 개의 소수 p1, p2, …, pn이 있다고 가정하자.
2. Q = p1·p2·…·pn + 1이라 하자.
3. “산술의 기본 정리”에 의하여 Q는 소수들의 곱으로 표현된다.
4. 따라서, pj | Q(1  j  n)를 만족하는 소수 pj가 존재한다.
5. 그러면, 다음과 같은 전개가 이루어진다.
5.1 (pj | p1·p2·…·pn)  (pj |-(p1·p2·…·pn))
[a|b  a|bc]
5.2 (pj | Q) = (pj | (p1·p2·…·pn + 1))
[by step 4]
5.3 (pj | 1)?
[(a|b  a|c)  a | (b + c)]
6. pj는1을 나눌 수 없으므로, 결국 가정(유한 개의 소수가 있다)이 잘못된 것이다.
즉, 소수의 개수는 무한하다. □
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The Division “Algorithm”
2.4 The Integers and Division
Really just a theorem, not an algorithm…
• The name is used here for historical reasons.
For any integer dividend a and divisor d≠0, there is a
unique integer quotient q and remainder rN
such that a = dq + r and 0  r < |d|.
• dividend = “피제수”, divisor = “제수”
• quotient = “몫”, remainder = “나머지”
a,dZ, d>0: !q,rZ: 0r<|d|, a=dq+r. (! means “unique”)
We can find q and r by: q=ad, r=aqd.
(e.g., if a = 14 and d = 3, then q = 143 = 4 and r = 14 - 3·4 = 2.
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Greatest Common Divisor (최대공약수)
2.4 The Integers and Division
The greatest common divisor gcd(a,b) of integers a,b (not
both 0) is the largest (most positive) integer d that is a
divisor both of a and of b.
(최대공약수는 두 수의 공약수 중에서 가장 큰 공약수이다.)
d = gcd(a,b) = max(d: d|a  d|b)
Example: gcd(24,36)=?
Positive common divisors: 1,2,3,4,6,12…
Greatest is 12.
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GCD shortcut
2.4 The Integers and Division
If the prime factorizations are written as
a  p1a1 p2a2  pnan and b  p1b1 p2b2  pnbn ,
then the GCD is given by:
gcd(a, b)  p1min(a1 ,b1 ) p2min(a2 ,b2 )  pnmin(an ,bn ) .
Example:
• a=84=2·2·3·7
= 22·31·71
• b=96=2·2·2·2·2·3 = 25·31·70
• gcd(84,96)
= 22·31·70 = 2·2·3 = 12.
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Relatively Primality (서로 소, 상대 소수)
2.4 The Integers and Division
Integers a and b are called relatively prime or coprime iff
their gcd = 1.
(두 수의 최대공약수가 1인 경우, 두 수는 서로 소(상대 소수)라 한다.)
Example: Neither 21 and 10 are prime, but they are
coprime. 21=3·7 and 10=2·5, so they have no common
factors > 1, so their gcd = 1.
A set of integers {a1,a2,…} is (pairwise) relatively prime if
all pairs ai, aj, ij, are relatively prime.
(주어진 정수들의 모든 쌍이 서로 소이면, 해당 정수들은 서로 소라 한다.)
• E.g., 10, 17, and 21 are relatively primes.
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Least Common Multiple (최소공배수)
2.4 The Integers and Division
lcm(a,b) of positive integers a, b, is the smallest positive
integer that is a multiple both of a and of b.
m = lcm(a,b) = min(m: a|m  b|m)
(최대공배수는 두 수의 공배수 중에서 가장 작은 공배수이다.)
• E.g. lcm(6,10)=30
If the prime factorizations are written as
a  p1a1 p2a2  pnan and b  p1b1 p2b2  pnbn ,
then the LCM is given by
lcm(a, b)  p1max(a1 ,b1 ) p2max(a2 ,b2 )  pnmax(an ,bn ) .
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A Useful Rule
2.4 The Integers and Division
If a and b are positive integers,
then ab = gcd(a,b)·lcm(a,b)
Prove it by yourself!
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The mod Operator
2.4 The Integers and Division
An integer “division remainder” operator.
Let a,dZ with d>1, then
• a mod d denotes the remainder r;
• i.e. the remainder when a is divided by d.
• r = a mod d
q
We can compute (a mod d) by: a  d·a/d.
In C programming language, “%” = mod.
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Modular Congruence (모듈로 합동?)
2.4 The Integers and Division
Let Z+={nZ | n>0}, the positive integers.
Let a,bZ, mZ+.
Then a is congruent to b modulo m, written “ab(mod m)”,
iff m|(ab). Also equivalent to: (ab) mod m = 0.
•
m으로 나누었을 때, a와 b의 나머지가 동일(합동)하다.
•
m으로 나누어 나머지가 동일한 수의 차를 m으로 나누면 나머지가 0이다.
•
두 수의 차를 m으로 나눌 수 있으면, 두 수를 m으로 나눈 나머지는 동일하다.
Also equivalent to: (ab) mod m = 0.
Example: 175(mod 6)
6|(17–5), (17–5)%6 = 0
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Spiral Visualization of mod
Example shown:
modulo-5
arithmetic
2.4 The Integers and Division
≡ 0 (mod 5)
20
15
10
≡ 4 (mod 5)
19
14
≡ 1 (mod 5)
5
9
0
4
3
8
6
1
11
16
21
2
7
13
12
18
17
22
≡ 3 (mod 5)
≡ 2 (mod 5)
5로 나누어 나머지가 3인 수
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Useful Congruence Theorems
Let a,bZ, mZ+. Then:
ab (mod m)  kZ a=b+km.
2.4 The Integers and Division
나머지
몫
Let a,b,c,dZ, mZ+. Then if
ab (mod m) and cd (mod m), then:
• a+c  b+d (mod m), and
• ac  bd (mod m)
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Applications of Congruence (합동의 응용)
2.4 The Integers and Division
The mod operator is widely used in hash functions.
• h(key) = key mod m
Linear congruential methods is used to generate pseudo
random numbers.
• x[n+1] = (a·x[n] + c) mod m
Also, in cryptography, encryption, …
• In C,
char shift_three_char(char c) /* assume c is upper */
{
return (char)(‘A’ + (c + 3)%26);
}
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