Greatest Common Divisor MATH Greatest Common Divisor. Benjamin V.C. Collins, James A. Swenson MATH 2730

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1 MATH 2730 Greatest Common Divisor Benjamin V.C. Collins James A. Swenson

2 The world s least necessary definition Definition Let a, b Z, not both zero. The largest integer d such that d a and d b is called the greatest common divisor of a and b, and is denoted by gcd(a, b). Existence We know that there is at least one common divisor of a and b; namely, 1. Because (a, b) (0, 0), at least one of {a, b} has only a finite number of divisors. This shows that the set of common divisors of a and b is nonempty and finite, so it has a greatest element.

3 Examples Example gcd(12, 40) = 4 gcd( 625, 100) = 25 gcd(81, 70) = 1 gcd(220, 8294) = = =

4 Euclid s algorithm Euclid(a,b) Input: positive integers a and b Output: gcd(a, b) Method: assign c = a mod b if c = 0: return b else: return Euclid(b,c)

5 Euclid s algorithm assign c = a mod b if c = 0: return b else: return Euclid(b,c) Input Division Algorithm mod operation Output (220, 8294) 220 = mod 8294 = 220 Euclid(8294, 220) (8294, 220) 8294 = mod 220 = 154 Euclid(220, 154) (220, 154) 220 = mod 154 = 66 Euclid(154, 66) (154, 66) 154 = mod 66 = 22 Euclid(66, 22) (66, 22) 66 = mod 22 = 0 22

6 Why does Euclid s algorithm work? Lemma Let a Z and b Z +. If c = a mod b, then gcd(a, b) = gcd(b, c). Proof. Let a Z and b Z +, and let c = a mod b. Let d = gcd(a, b) and e = gcd(b, c). We have a = qb + c for some q Z. We know that d a and d b, so d (a qb). That is, d c. Since d b and d c, d is a common divisor of b and c, so d gcd(b, c) = e. On the other hand, e b and e c, so e (qb + c). That is, e a. Since e a and e b, e is a common divisor of a and b. Therefore, e gcd(a, b) = d. Therefore d = e. That is, gcd(a, b) = gcd(b, c).

7 Euclid s algorithm works! Proposition Euclid s algorithm correctly computes gcd(a, b) for all a, b Z +. Proof. Assume wlog that b a. The proof is by strong induction on a. Base case: Suppose a = 1. Then b = 1, so gcd(a, b) = 1. But c = a mod b = 0, so Euclid s algorithm returns b = 1. Strong induction hypothesis: Suppose that Euclid s algorithm correctly computes gcd(a, b) whenever 1 a k and b a. Apply Euclid s algorithm to a = k + 1 and any b k + 1. If (k + 1) mod b = 0, then b = gcd(k + 1, b), and Euclid s algorithm returns b. If (k + 1) mod b 0, then 1 b k. By induction, Euclid s algorithm returns gcd(b, c). By the lemma, that s gcd(k + 1, b). Either way, Euclid s algorithm computes gcd(k + 1, b) correctly.

8 Linear combinations and the GCD Theorem Let a and b be integers, not both zero. The smallest positive integer of the form ax + by (x, y Z) is gcd(a, b). Example Find integers x and y such that 12x + 40y = 4. (x=-3, y=1) Find integers x and y such that ( 625)x + ( 100)y = 25. (x=-1, y=6) Find integers x and y such that 8294x + 220y = 22.

9 Finding the linear combination Find integers x and y such that 8294x + 220y = 22. Equation Rearranged Substituted 154 = (2) = 154(1) + 66( 2) 22 = 154(1) + 66( 2) 220 = (1) = 220(1) + 154( 1) 22 = 154(1) + [220(1) + 154( 1)]( 2) 22 = 154(3) + 220( 2) 8294 = (37) = 8294(1) + 220( 37) 22 = [8294(1) + 220( 37)](3) + 220( 2) 22 = 8294(3) + 220( 113)

10 Proof of the theorem Theorem Let a and b be integers, not both zero. The smallest positive integer of the form ax + by (x, y Z) is gcd(a, b). Proof. Let D = {k Z + : x, y Z, k = ax + by}. Using the well-ordering principle, let c = au + bv be the least element of D. Using the division algorithm, write a = qc + r, with 0 r < c. a = q(au + bv) + r r = a(1 qu) + b( qv) Now r is of the form ax + by, but r < c, so r is not positive. But r 0, so r = 0. So a = qc, which implies that c a. A similar argument shows that c b. So c is a common divisor of a and b.

11 Proof of the theorem (continued) Theorem Let a and b be integers, not both zero. The smallest positive integer of the form ax + by (x, y Z) is gcd(a, b). Proof (continued).... So c is a common divisor of a and b. Now, suppose that d is any common divisor of a and b. That is, a = dm and b = dn for some m, n Z. Then we have: c = au + bv = dmu + dnv = d(mu + nv). Therefore, d c. But c > 0, so d c. By the definition of gcd, c = gcd(a, b).

12 Relatively prime Definition Let a and b be integers. We say a and b are relatively prime provided that gcd(a, b) = 1. Example 70 is relatively prime to is not relatively prime to Any integer x is relatively prime to 1.

13 Using the theorem Greatest Common Divisor Corollary Given a, b Z, a and b are relatively prime if and only if there exist integers x and y such that ax + by = 1. Proof. Let a, b Z. Suppose a and b are relatively prime. By the definition of relatively prime, gcd(a, b) = 1. So, by the theorem, there exist integers x and y such that ax + by = 1. On the other hand, suppose that there exist integers x and y such that ax + by = 1. Then 1 is the smallest positive integer of the form ax + by (x, y Z). So, by the theorem, 1 = gcd(a, b). By the definition of relatively prime, a and b are relatively prime.

14 Using the corollary Theorem If m and n are relatively prime, and j Z, then m + jn and n are relatively prime. Proof. Suppose that m and n are relatively prime, and let j Z. By the corollary, there exist x, y Z such that: mx + ny = 1 mx + jnx + ny jnx = 1 (m + jn)x + n(y jx) = 1 So there exist z, w Z such that (m + jn)z + nw = 1; namely, z = x and w = y jx. Therefore m + jn and n are relatively prime.

15 Another example Greatest Common Divisor Proposition Let a, b, n be integers with n > 0. If ab 1 (mod n), then both a and b are relatively prime to n. Proof. Let a, b, n be integers with n > 0, and suppose ab 1 (mod n). By definition of, n (ab 1). So ab 1 = nk for some k Z. ab + n( k) = 1 Since b Z and k Z, the corollary implies that a and n are relatively prime. On the other hand, since a Z and k Z, the corollary implies that b and n are relatively prime. So both a and b are relatively prime to n.