Lecture 6 Integers

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# Lecture 6 Integers - PowerPoint PPT Presentation

Lecture 6 Integers. CSCI – 1900 Mathematics for Computer Science Spring 2014 Bill Pine. Lecture Introduction. Reading Kolman Section 1.4 Remainder Theorem Divisibility of integers Prime numbers GCD LCM Representing integers in different bases. Remainder Theorem.

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### Lecture 6Integers

CSCI – 1900 Mathematics for Computer Science

Spring 2014

Bill Pine

Lecture Introduction
• Kolman Section 1.4
• Remainder Theorem
• Divisibility of integers
• Prime numbers
• GCD
• LCM
• Representing integers in different bases

CSCI 1900

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

CSCI 1900

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

CSCI 1900

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

CSCI 1900

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 | 24

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

5 | 135

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

CSCI 1900

Simple properties of divisibility
• Given three integers a, b, c with a | b and a | c, then
• a | (b + c)
• a | (b - c)
• a | bc
• Given three integers a, b, c with a | b and b | c, then
• a | c

CSCI 1900

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

CSCI 1900

Basic Prime Number Algorithm

Function IsPrime( n )

nIsPrime = True

for i= 2 to n-1

if( i | n)

nIsPrime = False

Exit Loop

return (nIsPrime)

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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 = 3224 = 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

CSCI 1900

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

CSCI 1900

Modulus(cont)
• Examples

13 mod 3 = 1 => 4*3+1=13

32 mod 5 = 2 => 6*5+2=32

a mod 7 = 1 => 7*k+1=a, for some integer k

CSCI 1900

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

CSCI 1900

GCD Example

Find the GCD of 540 and 315:

• First find the prime factors of each

540 = 22* 33 * 5

315 = 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 primes

So the largest is the GCD  32 * 5 = 45

• 315  45 = 7 and 540  45=12

CSCI 1900

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=amodb

while ( r> 0 )

a=b

b=r

r=amodb

returnb

CSCI 1900

Euclid’s Algorithm Example
• For two integers a= 846 and b = 212

846 = 3 * 212 + 210 k1 = 3; r1 = 210

212 = 1 * 210 + 2 k2 = 1; r2 = 2

210 = 105 * 2 + 0 k3 = 105; r3 = 0

GCD=2

• For two integers a= 555 and b = 296

555 = 1 * 296 + 259 k1 = 1; r1 = 259

296 = 1 * 259 + 37 k2 = 1; r2 = 37

259 = 7 * 37 + 0 k3 = 7; r3 = 0

GCD = 37

846 = 47 * 32 * 2

212 = 53 * 22

555 = 37 * 5 * 3

296 = 37 * 23

CSCI 1900

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)

CSCI 1900

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 + d0b0where 0 < di< b, i = 0, 1, …, k

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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

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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

CSCI 1900

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

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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

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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

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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

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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

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

CSCI 1900