12 Greatest Common Divisors. The Euclidean Algorithm

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1 Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 12 Greatest Common Divisors. The Euclidean Algorithm As mentioned at the end of the previous section, we would like to establish a condition on n so that (Z n, ) is a group. This section provides the tool needed for that. We start this section with the following definition. Definition 12.1 Let a and b be two integers, not both zero. A positive integer d is called a greatest common divisor of a and b if the following two conditions hold: (1) d a and d b. (2) If c a and c b then c d. We write d = (a, b) or d = gcd(a, b). Example 12.1 The greatest common divisor of -42 and 56 is 14. The greatest common divisor is useful for writing fractions in lowest term. For example, 42 = where we cancelled 14 = (42, 56). We next discuss a systematic procedure for finding the greatest commom divisor of two integers, known as the Euclid s Algorithm. For that purpose, we need the following result. Theorem 12.1 If a, b, q, and r are integers such that a = bq + r then (a, b) = (b, r). Let d 1 = (a, b) and d 2 = (b, r). We will show that d 1 = d 2. Since d 2 bq and d 2 r then by Theorem 10.2(a)(b), d 2 (bq + r). That is d 2 a. Thus, by Definition 12.1, d 2 d 1. Since d 1 b then by Theorem 10.2(a) d 1 bq. Since d 1 a then by Theorem 10.2(b)d 1 (a bq). That is, d 1 r. Hence, from the definition of 1

2 d 2, we have d 1 d 2. Since d 1 and d 2 are positive then by Theorem 10.2(d), we have d 1 = d 2. The following theorem, establishes the existence and uniqueness of the greatest common divisor and provide an algorithm of how to find it. Theorem 12.2 (The Euclidean Algorithm) If a and b are two integers, not both zero, then there exists a unique positive integer d such that the two conditions (1) and (2) of Definition 12.1 are satisfied. Uniqueness: Let d 1 and d 2 be two positive integers that satisfy conditions (1) and (2). Then by (2) we can write d 1 d 2 and d 2 d 1. Since d 1 and d 2 are both positive then Theorem 10.2 (d) implies that d 1 = d 2. Existence: Without loss of generality, we may assume that b 0. Note that if a = 0 then d = b. Indeed, b 0 and b b. Moreover, if c is a common divisor of a and b then b = cq so that b = cq. That is, c b. So assume that a 0. By the Division algorithm there exist unique integers q 1 and r 1 such that a = bq 1 + r 1, 0 r 1 < b. If r 1 = 0 then as above, one can easily check that d = b. So assume that r 1 0. Using the Division algorithm for a second time to find unique integers q 2 and r 2 such that b = r 1 q 2 + r 2 0 r 2 < r 1. We keep this process going and eventually we will find integers r n and r n+1 such that r n 2 = r n 1 q n + r n, 0 r n < r n 1 and r n 1 = r n q n+1. That is, r n is the last nonzero remainder in the process. By Theorem 12.1, we have (a, b) = (b, r 1 ) = (r 1, r 2 ) = = (r n 1, r n ) = r n. 2

3 Example 12.2 Performing the arithmetic for the Euclidean algorithm we have So (1776, 1492) = = (1)(1492) = (5)(284) = (3)(72) = (1)(68) = 4(17) An alternative proof for the existence of the greatest common divisor which does not provide a systematic way for finding (a, b) is given next. The theorem is important for its theoretical applications. Theorem 12.3 If a and b are two integers, not both zero, then there exist integers m and n such that (a, b) = ma + nb. That is, (a, b) can be expressed as a linear combination of a and b. Without loss of generality, we assume that b 0. Let S = {xa + yb > 0 : x, y Z}. We first show that S. To see this, note that if b > 0 then b = 0a + (1)b S. If b < 0 then b = 0a + ( 1)b S. Note that 0 S. By Theorem 10.1, S has a smallest element d. Thus, d = ma + nb > 0 for some integers m and n. We will show that d = (a, b). Applying the Division algorithm we can find unique integers q and r such that a = dq + r with 0 r < d. From this equation we see that r = a dq = a (ma + nb)q = (1 mq)a + ( nq)b If r > 0 then r S and r < d. This contradicts the definition of d. Therefore, r = 0 and this gives a = dq and hence d a. A similar argument holds for d b. Finally, if c is an integer such that c a and c b then a = cq and b = cq. Thus, d = ma + nb = mcq + ncq = c(mq + nq ) This means that c d. Thus, d = (a, b). This ends a proof of the Theorem. 3

4 Remark 12.1 The integers m and n in Theorem 12.3 are not unique. Indeed, (a, b) = ma + nb = ma + ab + nb ab = (m + b)a + (n a)b = m a + n b Example 12.3 From Example 12.2, we found that (1776, 1492) = 4. Let s write this as a linear combination of 1776 and We use the equations in Example 12.2, beginning with 72 = (1)(68) + 4 and working backward one step at a time. 4 = 72 (1)(68). Solve the equation 284 = (3)(72) + 68 for 68 and substitute in the previous equation and simplify to obtain 4 = 72 (1)( ) = 72 (4) 284 Solve the equation 1492 = (5)(284)+72 for 72 and substitute in the previous equation and simplify to obtain 4 = ( ) = Finally, solve the equation 1776 = for 284 and substitute to obtain 4 = ( ) = Theorem 12.4 If a and b are integers, not both zero, then (a, b) = 1 if and only if ma+nb = 1 for some integers m and n. Suppose first that (a, b) = 1. By Theorem 12.3, there exist integers m and n such that ma + nb = 1. Conversely, suppose that ma+nb = 1 for some integers m and n. If d = (a, b) then d a and d b so d (ma + nb) (Theorem 10.2 (a)(b)). That is, d 1, 1 d and d > 0 then by Theorem 10.2(d) we must have d = 1. 4

5 Definition 12.2 If a and b are integers such that (a, b) = 1 then we say that a and b are relatively prime. The next result, characterizes those elements of Z n that have multiplicative inverses. Theorem 12.5 [a] Z n has a multiplicative inverse if and only if (a, n) = 1. Suppose first that [a] has a multiplicative inverse [b] in Z n. Then [a] [b] = [ab] = [1]. This implies that ab 1(mod n). Therefore, ab = nq + 1 for some integer q. This last equality can be written in the form ba + ( q)n = 1. Suppose that d = (a, n). By Theorem 10.2(a)(b), d 1. Since 1 d then by Theorem 10.2(d), d = (a, n) = 1. Conversely, suppose that (a, n) = 1. Then by Theorem 12.4, there exist integers m and q such that ma + qn = 1. Thus, ma 1 = ( q)n and hence ma 1(mod n). In terms of, we have [m] [a] = [1]. Thus, [a] has a multiplicative inverse in Z n. As a consequence of the above theorem we have Theorem 12.6 Every nonzero element of Z n has a multiplicative inverse if and only if n is a prime number. Thus, (Z n, ) is a group if and only if n is prime. Suppose first that [a] has a multiplicative inverse for all 1 a < n. Then by Theorem 12.5, (a, n) = 1 for 1 a < n. This is true only if n is prime. Conversely, suppose that n is prime then (a, n) = 1 for every 1 a < n. By Theorem 12.5, [a] has a multiplicative inverse for all 1 a < n. 5

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