{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# s5_2 - 2 INTEGERS AND ALGORITHMS 155 2 Integers and...

This preview shows pages 1–3. Sign up to view the full content.

2. INTEGERS AND ALGORITHMS 155 2. Integers and Algorithms 2.1. Euclidean Algorithm. Euclidean Algorithm. Suppose a and b are in- tegers with a b > 0. (1) Apply the division algorithm: a = bq + r , 0 r < b . (2) Rename b as a and r as b and repeat until r = 0. The last nonzero remainder is the greatest common divisor of a and b . The Euclidean Algorithm depends upon the following lemma. Lemma 2.1.1 . If a = bq + r , then GCD( a, b ) = GCD( b, r ) . Proof. We will show that if a = bq + r , then an integer d is a common divisor of a and b if, and only if, d is a common divisor of b and r . Let d be a common divisor of a and b . Then d | a and d | b . Thus d | ( a - bq ), which means d | r , since r = a - bq . Thus d is a common divisor of b and r . Now suppose d is a common divisor of b and r . Then d | b and d | r . Thus d | ( bq + r ), so d | a . Therefore, d must be a common divisor of a and b . Thus, the set of common divisors of a and b are the same as the set of common divisors of b and r . It follows that d is the greatest common divisor of a and b if and only if d is the greatest common divisor of b and r . Discussion The fact that the Euclidean algorithm actually gives the greatest common divi- sor of two integers follows from the division algorithm and the equality in Lemma 2.1.1. Applying the division algorithm repeatedly as indicated yields a sequence of remainders r 1 > r 2 > · · · > r n > 0 = r n +1 , where r 1 < b . Lemma 2.1.1 says that GCD( a, b ) = GCD( b, r 1 ) = GCD( r 1 , r 2 ) = · · · = GCD( r n - 1 , r n ) . But, since r n +1 = 0, r n divides r n - 1 , so that GCD( r n - 1 , r n ) = r n . Thus, the last nonzero remainder is the greatest common divisor of a and b . Example 2.1.1 . Find GCD (1317, 56).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2. INTEGERS AND ALGORITHMS 156 1317 = 56(23) + 29 56 = 29(1) + 27 29 = 27(1) + 2 27 = 2(13) + 1 2 = 1(2) + 0 GCD (1317,56)=1 Example 2.1.1 shows how to apply the Euclidean algorithm. Notice that when you proceed from one step to the next you make the new dividend the old divisor (replace a with b ) and the new divisor becomes the old remainder (replace b with r ). Recall that you can find the quotient q by dividing b into a on your calculator and rounding down to the nearest integer. (That is, q = b a/b c .) You can then solve for r . Alternatively, if your calculator has a mod operation, then r = mod ( a, b ) and q = ( a - r ) /b . Since you only need to know the remainders to find the greatest common divisor, you can proceed to find them recursively as follows: Basis. r 1 = a mod b , r 2 = b mod r 1 . Recursion. r k +1 = r k - 1 mod r k , for k 2. (Continue until r n +1 = 0 for some n . ) 2.2. GCD ’s and Linear Combinations. Theorem 2.2.1 . If d = GCD( a, b ) , then there are integers s and t such that d = as + bt. Moreover, d is the smallest positive integer that can be expressed this way. Discussion Theorem 2.2.1 gives one of the most useful characterizations of the greatest com- mon divisor of two integers. Given integers a and b , the expression as + bt , where s and t are also integers, is called a linear combination of a and b .
This is the end of the preview. Sign up to access the rest of the document.
• Fall '08
• PenelopeKirby
• Natural number, Prime number, Greatest common divisor, Euclidean algorithm, Fundamental theorem of arithmetic

{[ snackBarMessage ]}