The following is an informal description of Euclid s algorithm for finding the greatest common divisor of a pair of numbers:


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1 Divisibility Euclid s algorithm The following is an informal description of Euclid s algorithm for finding the greatest common divisor of a pair of numbers: Divide the smaller number into the larger, and note the remainder. Divide the remainder into the previous divisor, and note the remainder. Keep doing this until you get a remainder of 0. The last remainder you got right before the zero remainder is your GCF. See if the instructions make sense  try an example, then we ll back up and go through all the theory. Example: Use Euclid s algorithm to find gcd(300, 70). Now, the theory. We need to build this up in stages, and so we ll start with an observation. Let a and b be the two numbers in the previous example: a = 300 and b = 70. Let s and t be a various integers, and consider the quantity sa + tb for each of the pairs s and t. Note that s and t are chosen almost at random, but all choices are such that sa + tb > 0. What can you observe?
2 Theorem: Let a and b be integers, not both equal to 0. Then gcd(a, b) isthe smallest positive integer that can be expressed as a sum ma + nb, where m and n are also integers. The set is bounded below, since sa + tb > 0. [Nicodemi, p. 16] The WellOrdering Principle says the set must have a smallest member; call it d. Anything that divides both a and b will also divide d. And d divides both a and b. So d must equal gcd(a, b). Corollary: An integer x is of the form ma + nb, withm, n Z, if and only if x is a multiple of gcd(a, b). [Nicodemi, p. 17] This corollary tells us whether an equation ma + nb = x with a, b, andx given will have integer solutions for m and n. Example: Will the equation 25m +70n = 15 have integer solutions for m and n? Will the equation 25m +70n = 7 have integer solutions for m and n?
3 Relatively prime Let a and b be integers, not both zero. We say that a and b are relatively prime if gcd(a, b) =1. Example: Which pairs of numbers are relatively prime? 5and7 9and15 9and35 Proposition: Suppose that gcd(a, b) =d. Thengcd ³ a, b d d =1. For convenience in writing, let gcd(a, b) = d. By definition, d a and d b (it s a divisor of both). This means a d and b d are both integers. Also, by the previous theorem, d = ma + nb for some integers m and n. Divide everybody through by d: d d = ma d So 1 = m ³ a d + n µ bd. + nb d. You aren t going to find a postive integer smaller than 1, so 1 is the smallest positive integer that can be expressed as m ³ µ a + n bd. d And by that previous theorem, that makes 1 = gcd ³ a,. b d d
4 Euclid s lemma Suppose that gcd(a, b) = 1 and that a bc. Then a c. Euclid s lemma says that if you have two numbers a and b with no factors in common (other than 1), and you know that a divides some number that does have b has a factor...then that number a must be a factor of c, because it certainly can t be a factor of b. Suppose that gcd(a, b) = 1 and that a bc. By previous theorem, there exists integers m and n such that 1 = ma + nb. Multiply through by c: c = mac + nbc a mac (it appears there as a factor). The hypothesis is that a bc, so a nbc. Since a divides both mac and nbc, it divides their sum: a (mac + nbc) Which is equal to c: a c.
5 Euclid s algorithm Almost there  a couple more propositions, and we ll have an explanation of why Euclid s algorithm works. Proposition: If a>0 and a b, then gcd(a, b) = a. Proposition: Let a and b be integers and suppose that a = qb + r. Then gcd(a, b) = gcd(b, r). This is a classic use of the show two sets are equal by showing each is the subset of the other technique. The idea is to consider the set of common divisors of a and b, and the set of common divisors of b and r, and show that they are the same set (and therefore have the same largest element). Let S be the set of common divisors of a and b. LetT be the set of common divisors of b and r. Note a = qb + r, and therefore r = a qb. Show S T : Suppose x S. Then x a and x b. Therefore, x (a qb), and so x r (if x is a divisor of a and b, then it is also a divisor of r). Therefore x T, and S T. Show T S: Suppose x T. Then x b and x r. Therefore, x (bq + r), and so x a (if x is a divisor of b and r, thenx is a divisor of a). Therefore x S, and T S. So S = T, and the largest element of S must be the largest element of T. gcd(a, b) =gcd(b, r).
6 Example: Take the values a = 300 and b = 70. According to the previous proposition, what can you claim? The fact that this process can be repeated gives us Euclid s algorithm. Recall the original example: 300 = gcd(300, 70) = gcd(70, 20). 70 = gcd(70, 20)=gcd(20, 10). 20 = And gcd(20, 10) = 10, since gcd(300, 70) = gcd(70, 20) = gcd(20, 10) = 10
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