College Algebra Exam Review 23

College Algebra Exam Review 23 - n D 1 ; 1 cannot be...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
1.6. DIVISIBILITY IN THE INTEGERS 33 Proposition 1.6.16. Let p be a prime number, and a and b nonzero inte- gers. If p divides ab , then p divides a or p divides b . Proof. If p does not divide a , then a and p are relatively prime, so 1 D ˛a C ˇp for some integers ˛ and ˇ , by Proposition 1.6.13 . Multiplying by b gives b D ˛ab C ˇpb , which shows that b is divisible by p . n Corollary 1.6.17. Suppose that a prime number p divides a product a 1 a 2 :::a r of nonzero integers. Then p divides one of the factors. Proof. Exercise 1.6.12 . n Theorem 1.6.18. The prime factorization of a natural number is unique. Proof. We have to show that for all natural numbers n , if n has factoriza- tions n D q 1 q 2 :::q r ; n D p 1 p 2 :::p s ; where the q i ’s and p j ’s are prime and q 1 ± q 2 ± ²²² ± q r and p 1 ± p 2 ± ²²² ± p s , then r D s and q i D p i for all i . We do this by induction on n . First check the case
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n D 1 ; 1 cannot be written as the product of any nonempty collection of prime numbers. So consider a natural number n 2 and assume inductively that the assertion of unique factorization holds for all natural numbers less than n . Consider two factorizations of n as before, and assume without loss of generality that q 1 p 1 . Since q 1 divides n D p 1 p 2 :::p s , it follows from Proposition 1.6.17 that q 1 divides, and hence is equal to, one of the p i . Since also q 1 p 1 p k for all k , it follows that p 1 D q 1 . Now dividing both sides by q 1 , we get n=q 1 D q 2 :::q r ; n=q 1 D p 2 :::p s :...
View Full Document

This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

Ask a homework question - tutors are online