Number theoryตัวจริง

Number Theory

by

Nittaya Noinan Kanchanapisekwittayalai

Phetchabun

Definitions.

Let a,b be integers with b ≠ 0. if there exists an integer c such that a = bc,

we say b divides a which is denoted by b|a

b is called divisor of a a is called multiple of b

Symbol .

b|a instead of “b divides a “

b|a instead of “b can’t divides a “

Example

1. 3 | 12 because 12 = 3 42. -5 | 15 because 15 = (-5) (-3)3. 8 | 0 because 0 = 8 04. 9 | 30 because there is not integer c to make 30 = 9c5. 11 | 66 and 66 | 198 then 11 | 1986. 21 | 126 and 126 | 882 then 21 | 882

Theorem 1.Let a,b be integers with a ≠ 0 and b ≠ 0 .

if a | b and b there exists an integer x such that b = ax c then a | c

Proof.assume that a | b and b | c there exists an integer x such that b = ax there exists an integer y such that c = by therefore c = (ax)y

c = a(xy) and xy be integerthus a | c

Theorem 2.If a,b be positive integers such a | b then a b

Proof.assume that a | b there exists an integer x such that b = axsince a and b be positive integer then x be

positive integer therefore x ≥ 1then ax ≥ a ( because a be positive integer)

thus b ≥ a

Theorem 3.If a,b and c be integers with a | b and a | c then a | (bx + cy) when x and y be each integer.

Proof.assume that a | b and a | c there exists an integer d such that b = adthere exists an integer e such that c = aetherefore bx + cy = (ad)x + (ae)y (when x and y be integer)

= a (dx + ey) (since dx + ey be integer)

thus a | (bx + cy)

bx + cy is called linear combination of b and c

Prime numbersDefinition. Let positive integer P is prime numbers

if and only if p 1 and if there exist integer x divided p then x { 1 , -1 , p , -p}

Example. The following is a prime numbers2, 3 , 5, 7, 11 , 13 , 17 , 19 , 23 etc.

A number that has only 1 and itself as factors is a prime number.

A number like 24 that has factors other than 1 and itself is called a composite number.

Example. Composite number

24 = 4 6= 2 2 2 3= 23 3 (composite numbers can write

be factor a prime number)

120 = 10 12= 2 5 2 2 3= 23 3 5 (composite numbers can

write be factor a prime number)

Theorem 4. Fundamental Theorem of Arithmetic

The fundamental theorem of arithmetic states that any integer n > 1 can be expressed uniquely as a product of prime numbers apart from the order of primes.

6 3 2 36 2 2 3 3or

Express 430 as a product of prime numbers.

430 2 215

215 5 43

43 43 1

430 2 5 43

Division Algorithm

Theorem 5. Division Algorithm. if a and b be integers with b 0 then There exist uniquely determined integers q and r such that

a = bq + r and 0 r < b

q is called quotient and

r is called remainder

Example.

1) 4 = 9(0) + 4 2) 24 = 4(6) + 03) 62 = 12(5) + 24) -22 = 6(-4) + 25) -34 = 5(7) + 1

Denifition.

integer a be even number if and only if there exit integer k such that a = 2k

integer a be odd number if and only if there exit integer k such that a = 2k + 1

Example. 1Prove that “Squares of odd numbers is odd”

Proof. Give a is odd numbers then there exist integer k with a = 2k + 1

a2 = (2K + 1)2

= 4K2 + 4K + 1= 2(2K2 + 2K) + 1

When K be integer then 2(2K2 + 2K) be even numbersThen 2(2K2 + 2K) + 1 be odd numbersHence a2 be odd numbers

Example. 2 Prove that “when 6 divides integer x then it hasremainder be 3. find the remainder when 6 divides 3x “Proof. when 6 divides integer x then it hasremainder be 3 can write following

x = 6n + 3 when n be integer3x = 18n + 93x = 18n + 6 + 33x = 6(3n + 1) + 3

Hence when 3x is divided by 6 has remainder is 3

Example.3Give n be positive integers .Find the remainder

when n(n+1)(n+2) is divided by 3 .Proof. When n is divided by 3 .it will probability has remainder are 0 , 1 and 2 .Case 1. when the remainder is 0 can be write n

following n = 3k when k is integer .then n(n+1)(n+2) = 3k(3k+1)(3k+2)

Thus n(n+1)(n+2) divided by 3 has remainder is 0 because 3 3k

Case 2. when the remainder is 1 can be write n following n = 3k + 1 when k is integer .then n(n+1)(n+2) = (3k+1)(3k+2)(3k+3)

Thus n(n+1)(n+2) divided by 3 has remainder is 0 because 3 (3k + 3)

Case 3. when the remainder is 2 can be write n following n = 3k + 2 when k is integer .then n(n+1)(n+2) = (3k+2)(3k+3)(3k+4)

Thus n(n+1)(n+2) divided by 3 has remainder is 0 because 3 (3k + 3)

Hence from all 3 cases we can said n(n+1)(n+2) divided by 3 has remainder is 0

Example.4

Find all positive integer that divide between 417 and 390 with the remainder equally.Proof. Give x be positive integer that divide between 417 and 390 with remainder is r Then 417 = kx + r when k be integer ..(1)

390 = mx + r when m be integer..(2)

(1) – (2) then 27 = (k – m)x …(3)

Since k – m be integers thus x 27 Thus the possible value of x is 1, 3 , 9 , 27Of checking see that 1 and 3 divide between 419 and 390 and the remainder is 0 equally , 9 divide between 419 and 390 and the remainder is 3 equally and 27 divide between 419 and 390 and the remainder is 12 equally .Hence all positive integer that divide between 417 and 390 with the remainder equally are 1 , 3 , 9 and 27

Theorem 6.Let b be integer more than 1 .each integer n can write in distributed base b such that

n = akbk + ak-1bk-1 +…+ a1b + a0

When k is integer and a0 , a1 , a2 ,…,ak be

not negative integer and less than with b and ak 0

Proof. By division algorithm.When n is divided by b then the quotient q0

and remainder a0 such that

n = bq0 + a0 0 a0 b ….1)

q0 is divided by b such that

q0 = bq1 + a1 0 a1 b ….2)

To do it as same the upper until the quotient be zero q1 = bq2 + a2 0 a2 b

q2 = bq3 + a3 0 a3 b

.

.

.

qk-2 = bqk-1 + ak-1 0 ak-1 b

qk-1 = b(0) + ak 0 ak b

Since n q0q1q2…≥ 0 is not negative integer

Sequence that is decreased. Thus the division algolithm have to finished by the last quotience is zero.

From 1) n = bq0 + a0

Instead q0 by 2) then n = b(bq1 + a1 ) + a0

Then = b2q1 + a1b + a0

Do it as same the upper then n = b3q2 + a2b2 + a1b + a0

.

.

.

n = bk-1qk-2 + ak-2bk-2 +..+ a1b + a0

n = bkqk-1 + ak-1bk-1 +..+ a1b + a0

Since 3) qk-1 = ak

Thus n = akbk + ak-1bk-1 +..+ a1b + a0

Example. Write 96 in distributed base 5.

Solve. 96 = (5 19) + 119 = (5 3) + 43 = ( 5 0) + 3

Then 96 = (5 19) + 1

= [5 (5 3) + 4 ] + 1= (52 3)+( 5 4) + 1

Hence 96 = (3 52)+(4 5) + 1We use symbol ( akak-1 …a0 )b instead of

akbk + ak-1bk-1 +..+ a1b + a0

Hence 96 = (3 52)+(4 5) + 1 = (341) 5


Solve. 25 = (2 24) + 124 = (2 12) + 012 = ( 2 6) + 0 6 = ( 2 3) + 0 3 = ( 2 1) + 1 1 = ( 2 0) + 1

Hence 25 = (11001) 2


Solve. 12345 = (8 1543) + 11543 = (8 192) + 7192 = ( 8 24) + 0 24 = ( 8 3) + 0 3 = ( 8 0) + 3

Hence 12345 = (30071) 8

Greatest Common DivisorRelatively Prime

Definition. Let a,b be integers. Integer c which divide of a and b is called common divisor

Definition. Let a,b be integers, not both zero. The greatest common divisor of a and b (or gcd(a,b) ) is the biggest number d which divides both a and b.

Definition. a and b are said to be relatively prime if gcd(a,b) = 1, so no prime common divisors.

The first ways to find the greatest common divisor.

1) The common divisor of 4 and 8 is 1, 2, 4 thus The greatest common divisor of 4 and 8 is 42) The common divisor of 8 and 12 is 1, 2, 4 thus The greatest common divisor of 4 and 8 is 4 3) The common divisor of 12 and 35 is 1 thus The greatest common divisor of 4 and 8 is 14) The common divisor of 24 and 32 is 1, 2, 4,

6, 8 thus The greatest common divisor of 4 and 8 is 8

The second ways to find the greatest common divisor.

1) Since 4 = 2 28 = 2 2 2

Thus The greatest common divisor of 4 and 8 is 2 2 = 4

2) Since 8 = 2 2 212 = 2 2 3

Thus The greatest common divisor of 8 and 12 is

2 2 = 4

The second ways to find the greatest common divisor.

3) Since 12 = 2 2 335 = 5 7

Thus there is no number is the same product of prime numbers between 12 and 35 The greatest common divisor of 12 and 35 is 1

4) Since 24 = 2 2 2 332 = 2 2 2 2 2

Thus The greatest common divisor of 4 and 8 is 2 2 2 = 8

The third ways to find the greatest common divisor.

1) Since 2 4 , 8 2 2 , 4

1 , 2



2) Since 2 8 , 12 2 4 , 6

2 , 3



3) Since 1 12 , 35 12 , 35

Thus there is no number divide 12 and 35 The greatest Common divisor of 12 and 35 is 1


4) Since 2 24 , 32 2 12 , 16

2 6 , 8 3 , 4

Thus The greatest common divisor of 24 and 32 is

2 2 2 = 8

Euclidean Algorithm. Given integers a and b 1, use the division algorithm repeatedly:

a = q1b + r1 0 r1 < a

b = q2r1 + r2 0 r2 < r1

r1 = q3r2 + r3 0 r3 < r2

...

...

r k-2= qkrk−1 + rk 0 rk < rk-1

rk−1 = qk+1rk

where in each equation the divisor at the preceding stage is divided by the remainder. These remainders decrease

r1 > r2 > · · · 0

The fourth ways to find the greatest common divisor.Theorem 7. Euclidean Algorithm.

so the process eventually stops when the remainder becomes zero. If r1 = 0, then gcd(a, b) = n. Otherwise, rk = gcd(a, b), where rk is the last nonzero remainder and can be expressed as a linear combination of a and b by eliminating remainders.

Proof. Express rk as a linear combination of a and b by eliminating remainders in the equations from the second last equation up. Hence every common

divisor of a and b divides rk. But rk is itself a common divisor of a and b (it divides every ri—work up through the equations). Hence rk = gcd(a, b).

Two integers a and b are called relatively prime if gcd(a, b) = 1.Hence 12 and 35 are relatively prime, but this is not true for 12 and 15Because gcd(12, 15) = 3. Note that 1 is relatively prime to every integer a. The following theorem collects three basic properties of relatively prime integers.

Example. 1. Find the gcd of 3240 and 2484

Solve.3240 = 1(2484) + 756 2484 = 3(756) + 216 756 = 3(216) + 108 216 = 2(108) + 0

Hence (3240 , 2484) = 108

Example. 2. Find the gcd of 450 and 840

Solve. 840 = 1(450) + 390 450 = 1(390) + 60 390 = 6(60) + 30 60 = 2(30) + 0

Hence (450 , 840) = 30

Find the greatest common factor (GCF).

Additional Example 1: Using a List to Find the GCF

Course 2

Greatest Common Factor

12, 36, 54

12: 1, 2, 3, 4, 6, 12

36: 1, 2, 3, 4, 6, 9, 12, 18, 36

54: 1, 2, 3, 6, 9, 18, 27, 54

The GCF is 6.

List all of the factors of each number.

Circle the greatest factor that is in allthe lists.

Try This: Example 1

Insert Lesson Title Here

Course 2



14, 28, 63

14: 1, 2, 7, 14

28: 1, 2, 4, 7, 14, 28

63: 1, 3, 7, 9, 21, 63

The GCF is 7.

List all of the factors ofeach number.

Circle the greatest factor that is in allthe lists.


Additional Example 2A: Using Prime Factorization to Find the GCF

Course 2


A. 40, 56

40 = 2 · 2 · 2 · 5

56 = 2 · 2 · 2 · 7

2 · 2 · 2 = 8

The GFC is 8.

Write the prime factorization of each number and circle the common factors.

Multiply the common prime factors.


Additional Example 2B: Using Prime Factorization to Find the GCF

Course 2


B. 252, 180, 96, 60

252 = 2 · 2 · 3 · 3 · 7

180 = 2 · 2 · 3 · 3 · 5

96 = 2 · 2 · 2 · 2 · 2 · 3

60 = 2 · 2 · 3 · 5

2 · 2 · 3 = 12

The GCF is 12.

Write the prime factorizationof each number and circlethe common prime factors.

Multiply the common primefactors.

Try This: Example 2A


Course 2



A. 72, 84

72 = 2 · 2 · 2 · 9

84 = 2 · 2 · 7 · 3

2 · 2 = 4

The GCF is 4.

Write the prime factorization of each number and circle the common factors.Multiply the common prime factors.

Try This: Example 2B


Course 2



B. 360, 250, 170, 40

360 = 2 · 2 · 2 · 9 · 5

250 = 2 · 5 · 5 · 5

170 = 2 · 5 · 17

40 = 2 · 2 · 2 · 5

2 · 5 = 10

The GCF is 10.


Multiply the common primefactors.

Find the GCM of 48 and 80.

48 80224 40212 2026 1023 5

Common FactorsOnce

Remaining Factors

2

Greastest Common divisor Method # 3

2 · 2 · 2 · 2 = 16

The GCF is 16.

Find the LCM of 108 , 45 and 90.

108 45 90336 15 30218 15 153

6 5 556 1 1

Common FactorsOnce

Remaining Factors

Greastest Common divisor Method # 3

3 · 2 · 3 · 5 = 90

The GCF is 90.

You have 120 red beads, 100 white beads, and 45 blue beads. You want to use all the beads to make bracelets that have red, white, and blue beads on each. What is the greatest number of matching bracelets you can make?

Additional Example 3: Problem Solving Application

Course 2


Additional Example 3 Continued

Course 2


1 Understand the Problem

Rewrite the question as a statement.

• Find the greatest number of matching bracelets you can make.

List the important information:• There are 120 red beads, 100 white beads, and 45 blue beads.

• Each bracelet must have the same number of red, white, and blue beads.

The answer will be the GCF of 120, 100, and 45.

Course 2


2 Make a Plan

You can list the prime factors of 120, 100,and 45 to find the GFC.

Solve3

120 = 2 · 2 · 2 · 3 · 5

100 = 2 · 2 · 5 · 5

45 = 3 · 3 · 5

The GFC of 120, 100, and 45 is 5.

You can make 5 bracelets.


Course 2


Look Back4

If you make 5 bracelets, each one will have 24 red beads, 20 white beads, and 9 bluebeads, with nothing left over.


Try This: Example 3

Nathan has made fishing flies that he plans to give away as gift sets. He has 24 wet flies and 18 dry flies. Using all of the flies, how many sets can he make?


Course 2


Try This: Example 3 Continued


Course 2


1 Understand the Problem

Rewrite the question as a statement.

• Find the greatest number of sets of flies he can make.

List the important information:

• There are 24 wet flies and 18 dry flies.

• He must use all of the flies.

The answer will be the GCF of 24 and 18.

Course 2


2 Make a Plan

You can list the prime factors of 24 and 18 to find the GCF.


Solve3

24 = 2 · 2 · 2 · 3 18 = 2 · 3 · 3

You can make 6 sets of flies.

2 · 3 = 6Multiply the prime factors that are common to both 24 and 18.



Course 2


Look Back4

If you make 6 sets, each set will have 3 dry flies and 4 wet flies.

Lesson Quiz: Part 1


1. 28, 40

2. 24, 56

3. 54, 99

4. 20, 35, 70

8

4


9

5

Course 2


Lesson Quiz: Part 2

5. The math clubs from 3 schools agreed to a competition. Members from each club must be divided into teams, and teams from all clubs must be equally sized. What is the greatest number of members that can be on a team if Georgia has 16 members, William has 24 members, and Fulton has 72 members?


Course 2


8

The Least Common MultipleDefinition. Let a,b be integers, not both zero.

The least common multiple of a and b (or lcm[a,b]) is the least number c which both a and b divides c.

Definition. Let a1 , a2 , a3 ,…,an be integers,not both zero. Integer C is the least number which a1 C , a2 C , …, an C is called The least common multiple of a1 , a2 , a3 ,…,an

symbol is [a1 , a2 , a3 ,…,an ]

EXAMPLE: Find the LCM of 12 and 18

The multiples of 12:

•12 x 1 = 12

•12 x 2 =24

•12 x 3 = 36

•12 x 4 = 48

•12 x 5 =60

The multiples of 18:

•18 x 1 = 18

•18 x 2 = 36

•18 x 3 = 54

•18 x 4 = 72

•18 x 5 = 90

12, 24, 36, 48, 60

18, 36, 54, 72, 90

The first number you see in both lists is 36.

The least common multiple of 12 and 18 is 36.

Example 2: Find the LCM of 9 and 10

9, 18, 27, 36, 45, 54, 63, 7210, 20, 30, 40, 50, 60, 70, 80If you don’t see a common multiple, make each list go further.

81, 90, 9990, 100, 110

The LCM of 9 and 10 is 90

Example 3:Find the LCM of 4 and 12

4, 8, 12, 16

12, 24, 36Answer: 12

Find the least common factor (LCF).

Course 2

Leastest Common Factor

A. 40, 56

40 = 2 · 2 · 2 · 5

56 = 2 · 2 · 2 · 7

2 · 2 · 2 · 5 · 7 = 280

The LCF is 280.

Write the prime factorization of each number and circle the common factors.

Multiply the common prime factors and remainder number.

Find the leatest common factor (LCF).

Course 2


B. 252, 180, 96, 60

252 = 2 · 2 · 3 · 3 · 7

180 = 2 · 2 · 3 · 3 · 5

96 = 2 · 2 · 2 · 2 · 2 · 3

60 = 2 · 2 · 3 · 5

2 · 2 · 3 · 3 · 5 · 2 · 2 · 2 · 7 = 10080 The LCF is 10080.


Multiply the common primefactors and remainder number.


Course 2


Find the leastest common factor (LCF).

A. 72, 84

72 = 2 · 2 · 2 · 9

84 = 2 · 2 · 7 · 3

2 · 2 · 2 · 3 · 7 · 9 = 1512

The LCF is 1512.

Write the prime factorization of each number and circle the common factors.Multiply the common prime factors and remainder number.

Find the LCM of 48 and 80.

48 80224 40212 2026 1023 5

Common FactorsOnce

Remaining Factors

LCM LCM 240

2 3222 5

Least Common Multiple Method # 3

Find the LCM of 108 , 45 and 90.

108 45 90336 15 30218 15 153

6 5 556 1 1

Common FactorsOnce

Remaining Factors

Least Common Multiple Method # 3

3 · 2 · 3 · 5 · 6 = 540

The LCF is 540.

Good Luck

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Number theoryตัวจริง