Lecture 6 Integers

Lecture 6Integers

CSCI – 1900 Mathematics for Computer ScienceFall 2014Bill Pine

CSCI 1900 Lecture 6 - 2

Lecture Introduction

• Reading– Rosen Section 4.1

• Remainder Theorem• Divisibility of integers• Prime numbers• GCD• LCM• Representing integers in different bases


Remainder Theorem

• Given two integers, n and m with n > 0,– Perform the integer division of m by n

• q is the quotient and r is the remainder• q and r are unique because we require 0 <= r < n

• Therefore, we can write m = q*n + r

• This result is known as the Remainder Theorem


Examples of m = qn + r

• If n is 3 and m is 16– 16 = 5*3 + 1 q = 5; r =

1

• If n is 10 and m is 3– 3 = 0*10 + 3 q = 0; r = 3

• If n is 5 and m is –11– 11 = – 3*5 + 4 q = – 3; r = 4


Divisibility

• If one integer, n, divides into a second integer, m, without a remainder, then we say that – n divides m – Denoted n | m

• If one integer, n, does not divide evenly into a second integer, m, then we say that – n does not divide m– Denoted n | m


Some Properties of Divisibility• If n | m,

– There exists an integer k such that m = k * n• The absolute values of both k and n are less than

the absolute value of m, i.e., |n| < |m| and |k| < |m|• Examples:

4 | 2424 = 4 * 6 both 4 and 6 are less than 24

5 | 135135 = 5 * 27 both 5 and 27 are less than 135


Prime Numbers

• A number p is called prime if the only positive integers that divide p are p and 1

• Examples of prime numbers: 2, 3, 5, 7, 11 • There are many computer algorithms that can

be used to determine if a number n>1 is prime, with greater or lesser efficiency

• Who cares ?– Anyone who buys anything online or has a wireless

network they do not want to share !– Cryptography involves prime number in some manner


Basic Prime Number Algorithm

Function IsPrime( n )nIsPrime = Truefor i= 2 to n-1

if( i | n)nIsPrime = FalseExit Loop

return (nIsPrime)


Factoring a Number into its Primes

• Repeatedly dividing a number into its multiples until the multiples no longer can be divided, shows us that any number can be expressed a a product prime numbers

• Examples: 9 = 3 * 3 = 32

24 = 8 * 3 = 2 * 2 * 2 * 3 = 23 * 3315 = 3*105 = 3*3*35 = 3*3*5*7 = 32 * 5 * 7

• Any number can be expressed as a product of prime numbers– This factorization is unique


Modulus

• The mod n operator is a direct consequence of the Remainder Theorem

• m mod n is defined to be the remainder when m is divided by n– The divisor n is called the modulus

• Given m = q * n + r then we say m mod n = r• If m mod n = 0 then m | n


Modulus(cont)

• Examples 13 mod 3 = 1 => 4*3+1=1332 mod 5 = 2 => 6*5+2=32a mod 7 = 1 => 7*k+1=a, for some integer k


Greatest Common Divisor

• If a, b, and c are in Z+, and c | a and c | b, we say that c is a common divisor of a and b

• If d is the largest such c, d is called the greatest common divisor (GCD)

• d is a multiple of every c, i.e., every c divides d

• If the GCD(a, b) = 1 then a and b are relatively prime


GCD Example

Find the GCD of 540 and 315:• First find the prime factors of each

540 = 22 * 33 * 5315 = 32 *5* 7

• 540 and 315 share the divisors 3 and 5,540 has 33 and 5 315 has 32 and 5

• So each is equal 32 * 5 times some different primesSo the largest is the GCD 32 * 5 = 45– 315 45 = 7 and 540 45=12


Euclid’s Algorithm

• Inputs: two positive integers a and b, a > b

• Output: gcd(a, b) – the greatest common divisor of a and b

• Procedure:r = a mod bwhile ( r > 0 )

a = bb = rr = a mod b

return b


Euclid’s Algorithm Example

• For two integers a= 846 and b = 212 846 = 3 * 212 + 210 k1 = 3; r1 = 210212 = 1 * 210 + 2 k2 = 1; r2 = 2210 = 105 * 2 + 0 k3 = 105; r3 = 0

GCD=2• For two integers a= 555 and b = 296

555 = 1 * 296 + 259 k1 = 1; r1 = 259296 = 1 * 259 + 37 k2 = 1; r2 = 37259 = 7 * 37 + 0 k3 = 7; r3 = 0

GCD = 37

846 = 47 * 32 * 2 212 = 53 * 22

555 = 37 * 5 * 3296 = 37 * 23


Least Common Multiple

• If a, b, and k are in Z+, and a | k, b | k, we say that k is a common multiple of a and b

• The smallest such k, call it c, is called the least common multiple or LCM of a and b

• We write c = LCM(a, b)• An important result is

– GCD(a, b)*LCM(a, b) = a*b– This provides a convenient way to calculate

LCM(a, b)


Representation of Integers

• In day-to-day life, we use decimal (base 10) arithmetic , but it is only one of many ways to express an integer value

• We say that a decimal value is the “base 10 expansion of n” or the “decimal expansion of n”

• If b > 1 is an integer, then every positive integer n can be uniquely expressed in the form:n = dkbk + dk-1bk-1 + dk-2bk-2 + … + d1b1 + d0b0

where 0 < di < b, i = 0, 1, …, k

Algorithm: Base 10 to Base b

• Input:two positive integers, base b and number n in base 10

• Output:the value of n in base b

• Procedure:See Handout



Example: Decimal 482 to Base 5

482 = 96*5 + 2 (remainder (2) is d0 digit)96 = 19*5 + 1 (remainder (1) is d1 digit)19 = 3*5 + 4 (remainder (4) is d2 digit)3 = 0*5 + 3 (remainder (3) is d3 digit)48210 = 34125


Example: Decimal 704 to Base 8 (Octal)

• 704 = 88*8 + 0 (remainder (0) is d0 digit)• 88 = 11*8 + 0 (remainder (0) is d1 digit)• 11 = 1*8 + 3 (remainder (3) is d2 digit)• 1 = 0*8 + 1 (remainder (1) is d3 digit)• 70410 = 13008

Algorithm: Base b to Base 10

• Input:two positive integers, base b and number n in base b

• Output:the value of n in base 10

• Procedure:See Handout



Example: 32125 to Base 10

34125 = 3 * 53 + 4 * 52 + 1 * 51 + 2 * 50 = 3 * 125 + 4 * 25 + 1 * 5 + 2 * 1= 375 + 100 + 5 + 2 = 48210

Example: 13008 to Base 10

13008= 1 * 83 + 3 * 82 + 0 * 81 + 0 * 80 = 1 * 512 + 3 * 64 + 0 * 8 + 0 * 1= 512 + 192 = 70410


Nota Bene

• The two conversion algorithms are pairs• If you convert a number n from base 10 to

base b– You can check your result by converting the

result back to base 10• If you convert a number n from base b to

base 10– You can check your result by converting the

result back to base b CSCI 1900 Lecture 6 - 25


Key Concepts Summary

• Divisibility of integers• Prime numbers• Remainder Theorem• GCD• LCM• Expansion into different base

Lecture 6 Integers

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