This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View Full
Document
 Spring '07
 SELMAN

Click to edit the document details