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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.