The Euclidean algorithm stops in $\hbox{$O\!\left(\log(N)\right)$}$ steps

Let F_n be the largest Fibonacci number less than or equal to N. By Proposition 1, we need to show that $n=\hbox{$O\!\left(\log N\right)$}$. We have  

$$n \log\tau=\log\tau^n\le\log F_n\le\log N \quad\hbox{$\Rig... ...row$}\quad n\le \log N/\log\tau=\hbox{$O\!\left(\log N\right)$}$$ (4)

as required. Thus, the Euclidean algorithm stops in $\hbox{$O\!\left(\log N\right)$}$ steps as advertized!