5
$\begingroup$

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.

  • 0
    It's considered rude to formulate your question as a command. Please reformulate it as a question, and indicate what you've already thought about.2011-10-24
  • 0
    Thanks for your notification.2011-10-24
  • 2
    It probably won't hurt to remember that $2\log b/\log 2$ is the same as $2\log_2 b$.2011-10-25

2 Answers 2