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College Algebra Exam Review 23

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

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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 q r and p 1 p 2 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
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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 ³ 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 :...
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