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Unformatted text preview: Section 3 Prime Numbers and Prime Factorisation Definition 3.1 A prime number is an integer p > 1 whose only positive divisors are 1 and p . Example 3.2 The first few prime numbers are: 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , . . . . See the Prime Pages ( http://www.utm.edu/research/primes/ ) for longer lists of primes and much other interesting information. The primes are useful since they form the ‘building blocks’ from which all other integers are made. Theorem 3.3 (Fundamental Theorem of Arithmetic) Any integer n with n > 1 can be written uniquely in the form n = p k 1 1 p k 2 2 . . . p k r r where the p i are prime numbers with p 1 < p 2 < · · · < p r and the k i are positive integers. Example 3.4 180 = 2 2 · 3 2 · 5. Proof of Theorem 3.3: We shall need some auxiliary results before we can establish the uniqueness part, but we can establish the existence part straight away. We proceed by induction on n . If n is a prime (including n = 2, the base case in the induction), then already n is a product of prime powers (namely a single prime) so there is nothing to prove. Assume then that n > 2 and 16 that n factorises, say n = st where 1 < s, t < n . By induction, both s and t can be written as a product of prime powers. Hence, upon multiplying these expressions together, we see that n = st is also a product of prime powers. This establishes existence of a prime decomposition. square For the uniqueness part we prove: Lemma 3.5 Let p be a prime number. (i) If p  ab , then either p  a or p  b . (ii) If p  a 1 a 2 . . . a s , then p  a i for some i . (iii) If p  q 1 q 2 . . . q t where each q i is a prime number, then p = q j for some j . Proof: (i) Assume p  ab . If p  a then the result holds. So assume that p does not divide a . Then gcd( p, a ) = 1 (since the only divisors of p are 1 and p and the latter does not divide a ). Now part (ii) of Theorem 2.6 tells us that 1 = up + va for some u, v ∈ Z . Hence b = ubp + vab. Now p  ab (by assumption), so we deduce p  ( ubp + vab ); that is, p  b , as required. (ii) Proceed by induction on s . If s = 1, then p  a 1 and there is nothing to prove. Assume then that s > 1. We have p  ba s where b = a 1 a 2 . . . a s 1 . Hence, by (i), either p  b or p  a s . If the first holds, that is, p  b , then by induction p divides one of a 1 , a 2 , . . . , a s 1 . Hence we deduce p  a i for some i , completing the inductive step. (iii) Apply (ii). We deduce p  q j for some j . But as q j is prime, its only divisors are 1 and q j . Hence p = q j . square We now return to the uniqueness part of Theorem 3.3. Assume that we have expressed n in the required form in two different ways: n = p k 1 1 p k 2 2 . . . p k r r = q l 1 1 q l 2 2 . . . q l t t . (3.1) Note that...
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 Spring '10
 Hald,OH
 Differential Equations, Linear Algebra, Algebra, Equations, Prime Numbers, Prime number, prime factorisation

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