# Proof consider the set s of non negative integers of

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Proof. Consider the set S of non-negative integers of the form a - xb with x Z . This set is clearly non-empty, and so contains a minimum. Let r = a - qb be the smallest integer in this set. By definition, we have r 0. Also, we must have r < b , since otherwise, we would have r - b S , contradicting the minimality of r . That proves the existence of r and q . For uniqueness, suppose that a = bq + r and a = bq 0 + r 0 , where 0 r, r 0 < b . Then subtracting these two equations and rearranging terms, we obtain r 0 - r = b ( q - q 0 ) . (1.1) Now observe that by assumption, the left-hand side of (1.1) is less than b in absolute value. However, if q 6 = q 0 , then the right-hand side of (1.1) would be at least b in absolute value; therefore, we must have q = q 0 . But then by (1.1), we must have r = r 0 . 2 In the above theorem, it is easy to see that q = b a/b c , the greatest integer less than or equal to a/b . We shall write r = a rem b . For a Z and a positive integer b , it is clear that b | a if and only if a rem b = 0. 1.2 Ideals and Greatest Common Divisors To carry on with the proof of Theorem 1.2, we introduce the notion of an ideal in Z , which is a non- empty set of integers that is closed under addition and subtraction, and closed under multiplication by integers. That is, a non-empty set I Z is an ideal if and only if for all a, b I and all z Z , we have a + b I, a - b I, and az I. Note that in fact closure under addition and subtraction already implies closure under multiplication by integers, and so the definition is a bit redundant. For a 1 , . . . , a k Z , define a 1 Z + · · · + a n Z := { a 1 z 1 + · · · a k z k : z 1 , . . . , z k Z } . We leave it to the reader to verify that a 1 Z + · · · + a n Z is an ideal, and this ideal clearly contains a 1 , . . . , a k . An ideal of the form a Z is called a principal ideal . Theorem 1.4 For any ideal I Z , there exists a unique non-negative integer d such that I = d Z . 2

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Proof. We first prove the existence part of the theorem. If I = { 0 } , then d = 0 does the job, so let us assume that I 6 = { 0 } . Since I contains non-zero integers, it must contain positive integers, since if x I then so is - x . Let d be the smallest positive integer in I . We want to show that I = d Z . We first show that I d Z . To this end, let c be any element in I . It suffices to show that d | c . Using the Division with Remainder Property, write c = qd + r , where 0 r < d . Then by the closure properties of ideals, one sees that r = c - qd is also an element of I , and by the minimality of the choice of d , we must have r = 0. Thus, d | c . We next show that d Z I . This follows immediately from the fact that d I and the closure properties of ideals. That proves the existence part of the theorem. As for uniqueness, note that if d Z = d 0 Z , we have d | d 0 and d 0 | d , from which it follows that d 0 = ± d . 2 For a, b Z , we call d Z a common divisor of a and b if d | a and d | b ; moreover, we call d the greatest common divisor of a and b if d is non-negative and all other common divisors of a and b divide d . It is immediate from the definition of a greatest common divisor that it is unique if it exists at all.
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• Spring '13
• MRR

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