TITLE: Show that the number of steps in the Euclidean algorithm is less than $\frac{2\log b}{\log 2}$
QUESTION [6 upvotes]: Could anyone tell me how to prove this statement? The number of steps in the Euclidean algorithm is less than $\frac{2\log b}{\log 2}$, where  $b$ is the larger of the two numbers whose GCD is being found.

REPLY [2 votes]: I dislike both answers given. Here is a clean and straightforward calculation.
Take $a=q_1 \cdot b + r_1$ and $b=q_2 \cdot r_1 + r_2$. The key inequality will be  $r_2 <{\frac{1}{2} b}$. This follows from the fact that $r_1<b$. Indeed, if $r_1<b/2$ then automatically $r_2<r_1<b/2$; and if $r_1 > b/2$ then $q_2=1$ and $r_2=b-r_1<b/2$; and finally $r_1=b/2 \implies r_2=0<b/2$ as desired.
In the same fashion one shows that $r_4<r_2/2$ and in general $r_{2(n+1)}<r_{2n}/2$. Then we simply apply the inequalities in order to get $r_{2n}<b/2^n$.
So we have a total of $\log_2(b)$ "double-steps", which is equivalent to $2\log_2(b)$ steps.