MUHAMMAD YUSRAN BASRI
MATH ICP B 2012
1211441023
FMIPA UNM
SUMMARY
THE FUNDAMENTAL THEOREM OF ARITHMETIC
Definition 1.1
The integer 𝑝 > 1 is called a prime number, if the divisor of it’s number only 𝑝 and
1.If a integer number more than 1 and i’s not prime number then it’s called composite
number.
Theorem 1.1
Let 𝑎 and 𝑏 is a integers. If prime 𝑝 divides 𝑎𝑏, then 𝑝 divides either 𝑎 or 𝑏.
Example: Find the largest n such that 2n || 31024 − 1.
Solution:
10
Note that 210 = 1024 and x2 − y2 = (x + y)(x − y). We have (32
9
8
8
9
8
7
9
9
− 1) = (32 + 1) (32 − 1) =
1
0
(32 + 1) (32 + 1) (32 − 1) ・・・ = (32 + 1) (32 + 1) (32 + 1)・・・(32 + 1) (32 +
1) (3 − 1).
By Example 1.5, 2 || 32k+ 1, for positive integers k. Thus the answer is 9 + 2 + 1 = 12.
Corollary 1. : If p is prime number and p|a1a2...an, then p|ak for any k with 1≤ k≤ n
Corollary 2. : If p, q1, q2, ..., qn are prime numbers and p|q1q2...,qn, then p = qk for any k
with 1≤ k≤ n ,
Corollary 3. :For any integer number n > 1 can be written uniquely in the canonical form
is 𝑛 = 𝑝1 𝑘1 𝑝2 𝑘2 … 𝑝𝑟 𝑘𝑟 for i = 1, 2, ... r, ki is a positive integer number and pi
is prime number with p1 ≤p2 ≤ ... ≤ pr.
Theorem 1.2 : Any integer n greater than 1 has a unique representation (up to a
permutation) as a product of primes.
Teorema 1.3 : √2 is an irrasional number.
Proof :
If √2 is a rasional number, suppose √2= a/b with a and b integer positif number and GCD
(a,b) = 1. If both of sides are squared, so we obtain a2 = 2b2 then b | a2.
If b>1, by the fundamental theorem of arithmetic is guaranteed prims p such that p|b. This
result p|a2, dan by theorem 1.1, so we conclude p|a. Since p|a dan p|b, then gcd(a,b)≥p ≥ 1
contradicts the assumption gcd(a, b) = 1.
If b=1, then a2 = 2 , it is not impossible( there is not integer number multiplied by itself is
equal to 2). Thus, the assumption must be wrong, in other words, √ 2 is an irrational
number.