This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Euclidean algorithm and greatest common divisor So you have two positive integers, a and b , and you want to find d = gcd ( a, b ) and s, t such that as + bt = d . Well, here’s how you do it. Suppose a ≥ b . Then, by the division algorithm, there are unique q 1 , r 1 such that a = q 1 b + r 1 . Well, unless r 1 is equal to 0, we aren’t done. Remember the division algorithm, we’re guaranteed that r 1 < b . So we can do the division algorithm on those two, so that b = q 2 r 1 + r 2 . Now we are guaranteed that r 2 < r 1 . If r 2 = 0, we’re done, and the gcd ( a, b ) = r 1 . Otherwise, we keep doing this (getting q i and r i until, at long last, r i = 0. Then the gcd ( a, b ) = r i 1 , the previous remainder. But that’s only half the battle. We need to find s, t such that as + bt = d . So here’s what we do. Notice that the remainder in each equation is used in the next line as a divisor. So what we can do is solve each equation for r i . So, from the first equation, we get that....
View
Full
Document
This note was uploaded on 08/11/2008 for the course MATH 240 taught by Professor Miller during the Fall '08 term at University of Wisconsin.
 Fall '08
 Miller
 Math, Division, Integers

Click to edit the document details