Difference between revisions of "Euclidean algorithm"

Revision as of 13:18, 27 August 2006

The Euclidean algorithm allows us to find the greatest common divisor of any two nonnegative integers.

Main idea and informal description

If we have two non-negative integers $a,b$ with $a\ge b$ and $b=0$ , then the greatest common divisor is $\displaystyle{a}$ . If $a\ge b>0$ , then the set of common divisors of $\displaystyle{a}$ and $b$ is the same as the set of common divisors of $b$ and $r$ where $r$ is the remainder of division of $\displaystyle{a}$ by $b$ . Indeed, we have $a=mb+r$ with some integer $m$ , so, if $\displaystyle{d}$ divides both $\displaystyle{a}$ and $b$ , it must divide both $\displaystyle{a}$ and $mb$ and, thereby, their difference $r$ . Similarly, if $\displaystyle{d}$ divides both $b$ and $r$ , it should divide $\displaystyle{a}$ as well. Thus, the greatest common divisors of $\displaystyle{a}$ and $b$ and of $b$ and $r$ coincide: $GCD(a,b)=GCD(b,r)$ . But the pair $(b,r)$ consists of smaller numbers than the pair $(a,b)$ ! So, we reduced our task to a simpler one. And we can do this reduction again and again until the smaller number becomes $0$

Steps

Start with two nonnegative integers, ${a}$ and $b$ .

If $b=0$ , then $\gcd(a,b)=a$ .
Otherwise take the remainder when ${a}$ is divided by $b$ ( ${a}\pmod {b}$ ), and find $\gcd(b,a\pmod {b})$ .

Repeat this until $b=0$ .

Simple Example

To see how it works, just take an example. Say $\displaystyle a=112,b=42$ . We have $112\equiv 28\pmod {42}$ , so $\displaystyle{\gcd(112,42)}=\displaystyle\gcd(42,28)$ . Similarly, $42\equiv 14\pmod {28}$ , so $\displaystyle\gcd(42,28)=\displaystyle\gcd(28,14)$ . Then $28\equiv {0}\pmod {14}$ , so ${\displaystyle \gcd(28,14)}={\displaystyle \gcd(14,0)} = 14$ . Thus $\displaystyle\gcd(112,42)=14$ .

Usually the Euclidean algorithm is written down just as a chain of divisions with remainder:

$112 = 2 \cdot 42 + 28 \qquad (1)$
$42 = 1\cdot 28+14\qquad (2)$
$28 = 2\cdot 14+0\qquad (3)$

Linear Representation

An added bonus of the Euclidean algorithm is the "linear representation" of the greatest common divisor. This allows us to write $\gcd(a,b)=ax+by$ , where $x,y$ are some integer constants that can be determined using the algorithm.

In the example, we can rewrite equation $(2)$ from above as

$14 = 42-1\cdot 28$
$14 = 42-1\cdot (112-2\cdot 42)$
$14 = 3\cdot 42-1\cdot 112.$

Examples

AIME 1985/13

1959 IMO Problems/Problem 1

@@ Line 16: / Line 16: @@
 == Simple Example ==
-To see how it works, just take an example. Say <math>\displaystyle a=112,b=42</math>. We have <math>112\equiv 28\pmod {42}</math>, so <math>\displaystyle{\gcd(112,42)}=\displaystyle\gcd(42,28)</math>. Similarly, <math>42\equiv 14\pmod {42}</math>, so <math>\displaystyle\gcd(42,28)=\displaystyle\gcd(28,14)</math>. Then <math>28\equiv {0}\pmod {14}</math>, so <math>{\displaystyle \gcd(28,14)}={\displaystyle \gcd(14,0)} = 14</math>. Thus <math>\displaystyle\gcd(112,42)=14</math>.
+To see how it works, just take an example. Say <math>\displaystyle a=112,b=42</math>. We have <math>112\equiv 28\pmod {42}</math>, so <math>\displaystyle{\gcd(112,42)}=\displaystyle\gcd(42,28)</math>. Similarly, <math>42\equiv 14\pmod {28}</math>, so <math>\displaystyle\gcd(42,28)=\displaystyle\gcd(28,14)</math>. Then <math>28\equiv {0}\pmod {14}</math>, so <math>{\displaystyle \gcd(28,14)}={\displaystyle \gcd(14,0)} = 14</math>. Thus <math>\displaystyle\gcd(112,42)=14</math>.
 Usually the Euclidean algorithm is written down just as a chain of divisions with remainder:

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