College Algebra Exam Review 56

College Algebra Exam Review 56 - 2 ³³³ p i r divides n ,...

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66 1. ALGEBRAIC THEMES Proof. A natural number j less than p k is relatively prime to p k if, and only if, p does not divide j . The number of natural numbers j ± p k such that p does divide j is p k ± 1 , so the number of natural numbers j ± n such that p does not divide j is p k ² p k ± 1 . n Let n be a natural number with prime factorization n D p k 1 1 ³³³ p k s s . A natural number is relatively prime to n if it is not divisible by any of the p i appearing in the prime factorization of n . Let us take our “universal set” to be the set of natural numbers less than or equal to n . For each i ( 1 ± i ± s ), let A i be the set of natural numbers less than or equal to n that are divisible by p i . Then '.n/ D j A 0 1 \ A 0 2 \ ³³³ \ A 0 s j . In order to use the inclusion-exclusion formula, we need to know j A i 1 \ ³³³ \ A i r j for each choice of r and of f i 1 ;:::;i r g . In fact, A i 1 \ ³³³ \ A i r is the set of natural numbers less than or equal to n that are divisible by each of p i 1 ;p i 2 ;:::;p i r and thus by p i 1 p i 2 ³³³ p i r . Since p i 1 p i
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Unformatted text preview: 2 ³³³ p i r divides n , the number of natural numbers a ± n such that p i 1 p i 2 ³³³ p i r divides a is n p i 1 p i 2 ³³³ p i r . Thus we have the formula '.n/ D j A 1 \ A 2 \ ³³³ \ A n j D j U j ² X i j A i j C X i<j j A i \ A j j ² X i<j<k j A i \ A j \ A k j C ³³³ C . ² 1/ s j A 1 \ ³³³ \ A s j D n ² X i n p i C X i<j n p i p j ² X i<j<k n p i p j p k C ³³³ C . ² 1/ s n p 1 p 2 ³³³ p s : The very nice feature of this formula is that the right side factors, '.n/ D n.1 ² 1 p 1 /.1 ² 1 p 2 / ³³³ .1 ² 1 p s /: Proposition 1.9.18. Let n be a natural number with prime factorization n D p k 1 1 ³³³ p k s s . Then (a) '.n/ D n.1 ² 1 p 1 /.1 ² 1 p 2 / ³³³ .1 ² 1 p s /: (b) '.n/ D '.p k 1 1 / ³³³ '.p k s s /:...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

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