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Unformatted text preview: Example of Extended Euclidean Algorithm Recall that gcd(84 , 33) = gcd(33 , 18) = gcd(18 , 15) = gcd(15 , 3) = gcd(3 , 0) = 3 We work backwards to write 3 as a linear combination of 84 and 33: 3 = 18 15 [Now 3 is a linear combination of 18 and 15] = 18 (33 18) = 2(18) 33 [Now 3 is a linear combination of 18 and 33] = 2(84 2 33)) 33 = 2 84 5 33 [Now 3 is a linear combination of 84 and 33] 1 Some Consequences Corollary 2: If a and b are relatively prime, then there exist s and t such that as + bt = 1. Corollary 3: If gcd( a, b ) = 1 and a  bc , then a  c . Proof: Exist s, t Z such that sa + tb = 1 Multiply both sides by c : sac + tbc = c Since a  bc , a  sac + tbc , so a  c Corollary 4: If p is prime and p  n i =1 a i , then p  a i for some 1 i n . Proof: By induction on n : If n = 1: trivial. Suppose the result holds for n and p  n +1 i =1 a i . note that p  n +1 i =1 a i = ( n i =1 a i ) a n +1 . If p  a n +1 we are done. If not, gcd( p, a n +1 ) = 1. By Corollary 3, p  n i =1 a i By the IH, p  a i for some 1 i n . 2 The Fundamental Theorem of Arithmetic, II Theorem 3: Every n &gt; 1 can be represented uniquely as a product of primes, written in nondecreasing size. Proof: Still need to prove uniqueness. We do it by strong induction. Base case: Obvious if n = 2. Inductive step. Suppose OK for n &lt; n . Suppose that n = s i =1 p i = r j =1 q j . p 1  r j =1 q j , so by Corollary 4, p 1  q j for some j . But then p 1 = q j , since both p 1 and q j are prime. But then n/p 1 = p 2 p s = q 1 q j 1 q j +1 q r Result now follows from I.H. 3 Characterizing the GCD and LCM Theorem 6: Suppose a = n i =1 p i i and b = n i =1 p i i , where p i are primes and i , i N . Some i s, i s could be 0. Then gcd( a, b ) = n i =1 p min( i , i ) i lcm( a, b ) = n i =1 p max( i , i ) i Proof: For gcd, let c = n i =1 p min( i , i ) i . Clearly c  a and c  b . Thus, c is a common divisor, so c gcd( a, b ). If q  gcd( a, b ), must have q { p 1 , . . . , p n } Otherwise q 6  a so q 6  gcd( a, b ) (likewise b ) If q = p i , q  gcd( a, b ), must have min( i , i ) E.g., if &gt; i , then p i 6  a Thus, c gcd( a, b ). Conclusion: c = gcd( a, b ). 4 For lcm, let d = n i =1 p max( i , i ) i . Clearly a  d , b  d , so d is a common multiple. Thus, d lcm( a, b ). Suppose lcm( a, b ) = n i =1 p i i ....
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 Spring '07
 SELMAN
 Number Theory

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