Editing Extended Euclidean algorithm (section)

==The case of more than two numbers ==
One can handle the case of more than two numbers iteratively.  First we show that <math>\gcd(a,b,c) = \gcd(\gcd(a,b),c)</math>.  To prove this let <math>d=\gcd(a,b,c)</math>. By definition of gcd <math>d</math> is a divisor of <math>a</math> and <math>b</math>.  Thus <math>\gcd(a,b)=k d</math> for some <math>k</math>.  Similarly <math>d</math> is a divisor of <math>c</math> so <math>c=jd</math> for some <math>j</math>.  Let <math>u=\gcd(k,j)</math>. By our construction of <math>u</math>, <math>ud | a,b,c</math> but since <math>d</math> is the greatest divisor <math>u</math> is a [[Unit (ring theory)|unit]].  And since <math>ud=\gcd(\gcd(a,b),c)</math> the result is proven.

So if <math>na + mb = \gcd(a,b)</math>  then there are <math>x</math> and <math>y</math> such that <math>x\gcd(a,b) + yc = \gcd(a,b,c)</math> so the final equation will be

: <math>x(na + mb) + yc = (xn)a + (xm)b + yc =  \gcd(a,b,c).\,</math>

So then to apply to ''n'' numbers we use induction

:<math>\gcd(a_1,a_2,\dots,a_n) =\gcd(a_1,\, \gcd(a_2,\, \gcd(a_3,\dots, \gcd(a_{n-1}\,,a_n))),\dots),</math>

with the equations following directly.