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Introduction to Number Theory Part B. - Congruence - Unique Factorisation Nikenasih B, M.Si Mathematics Educational Department Faculty of Mathematics and Natural Science State University of Yogyakarta
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Page 1: Introduction to Number Theory - Universitas Negeri Yogyakartastaff.uny.ac.id/sites/default/files/pendidikan/Nikenasih Binatari...Introduction to Number Theory Part B. ... bahwa setiap

Introduction to Number TheoryPart B.

- Congruence

- Unique Factorisation

Nikenasih B, M.Si

Mathematics Educational Department

Faculty of Mathematics and Natural Science

State University of Yogyakarta

Page 2: Introduction to Number Theory - Universitas Negeri Yogyakartastaff.uny.ac.id/sites/default/files/pendidikan/Nikenasih Binatari...Introduction to Number Theory Part B. ... bahwa setiap

Contents all part

Preliminary

Divisibility

Congruence

Unique Factorisation

Linear Diophantine Equation

Arithmetic Functions

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Congruence

Definition

Konsep kekongruenan bilangan dikembangkan berdasarkan konsepbahwa setiap bilangan bulat positif dapat dinyatakan ke dalambentuk N = pq + r atau N − r = pq dengan p, q, r adalah bilanganbulat dan r berada pada 0 ≤ r < p. Persamaan N = pq + r dengan pmenyatakan pembagi, q menyatakan hasil bagi dan r menyatakansisa.

Persamaan di atas sering pula ditulis N ≡ r (mod p)

(dibaca N kongruen modulo p terhadap r)

Dari hal tersebut didapat definisi bahwa a ≡ b (mod m)

jika m | (a − b) untuk bilangan bulat a, b dan m.

Contoh :

◦ (1) 25 ≡ 1 (mod 4) sebab 4|24

◦ (2) 1 ≡ −3 (mod 4) sebab 4|4

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Congruence

Properties 1

Beberapa sifat berkaitan dengan modulo adalah sebagai

berikut. Misalkan a, b, c, d dan m adalah bilangan-bilangan bulat

dengan d > 0 dan m > 0, berlaku :

i. a ≡ a (mod m)

ii. Jika a ≡ b (mod m) dan b ≡ c (mod m) maka a ≡ c (mod

m)

iii. Jika a ≡ b (mod m) dan d|m maka a ≡ b (mod d)

iv. Jika a ≡ b (mod m) maka ak ≡ bk (mod m) untuk semua k

bilangan asli

v. Jika a ≡ b (mod m) dan f(x) = anxn + an-1x

n-1 + ⋅⋅⋅ + ao

maka f(a) ≡ f(b) (mod m)

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Congruence

Properties 2

Beberapa sifat berkaitan dengan modulu adalah sebagai berikut.

Misalkan a, b, c, d dan m adalah bilangan-bilangan bulat dengan d > 0

dan m > 0, berlaku :

i. Jika a ≡ b (mod m) dan c ≡ d (mod m) maka a + c ≡ b + d (mod

m)

ii. Jika a ≡ b (mod m) dan c ≡ d (mod m) maka ac ≡ bd (mod m)

iii. (am + b)k ≡ bk (mod m) untuk semua k bilangan asli

iv. (am + b)k ⋅ (cm + d)n ≡ bk ⋅ dn (mod m) untuk semua k dan n

bilangan asli

v. Misalkan n ∈ N dan S(n) adalah penjumlahan digit-digit dari n

maka berlaku n ≡ S(n) (mod 9).

vi. n5 ≡ n (mod 10) untuk setiap n ∈ N.

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Congruence

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Unique Factorization

The Fundamental Theorem of Arithmetic

Every integer greater than 1 can be written

uniquely in the form

Where the pi are distinct primes and the

are positive integers.

k

kp...pp 21

21

i

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GCD and LCM

The greatest common divisor of two positiveintegers a and b is the greatest positive integerthat divides both a and b, which we denote bygcd(a, b), and similarly, the lowest commonmultiple of a and b is the least positive integerthat is a multiple of both a and b, which wedenote by lcm(a, b).

We say that a and b are relatively prime if gcd(a,b) = 1.

For integers a1, a2,. . . , an, gcd(a1, a2, . . . , an) is thegreatest positive integer that divides all of a1, a2, .. . , an, and lcm(a1, a2, . . . , an) is defined similarly.

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Theorem

Let b,n and r be positive integers, then

As we had learned on secondary school thatwe can use prime factorization method tofind the greatest common divisor of twointeger m and n.

Using this Theorem, we can find the greatestcommon divisor of two integer m and n withanother way.

rnGCDnrbnGCD ,,

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Theorem

Let m and n be positive integers where

0 < n < m. From division algorithm, we

know that there exist integers b and r

such that

Therefore

nrrbnm 0,

rnGCDnrbnGCDnmGCD ,,,

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Because n and r are positive integers

where 0 < r < n, we know that there

exist integers b1 and r1 such that

Therefore,

If we continu this process, then there

exist integers such that

rrrrbn 111 0,

111 ,,, rrGCDrrrbGCDrnGCD

srrr ,...,, 32

sss rrrGCDrrGCDrrGCDrnGCDnmGCD ,,,,, 1211