In how many steps Euclidean Algorithm will find the GCD for two integers
It can be observed that the product $ab$ drops by a factor of at least two for each iteration.

Prior to each iteration, we have the pair $(a, b)$ such that $a &lt; b$, which is replaced by the pair $(r, a)$ at the end of each iteration in preperation for the next, where $r = b \bmod a$. We then have $r &lt; a$ and $a + r ≤ b$. Hence, $b ≥ a + r &gt; 2r$. So, $ar &lt; \frac{1}{2}ab$.

Supposing it takes N steps to compute $gcd(a, b)$ using the Euclidean algorithm, we then have $ar ≤ \frac{ab}{2^N}$ after N steps. It follows that $ab ≥ 2^N$.

Hence, $N ≤ \log_2 ab = \log_2 a + \log_2 b$.

Therefore, the number of steps it takes to compute $gcd(a, b)$ using the Euclidean algorithm is at most $\log_2 a + \log_2 b$.

Furthermore, on Wikipedia, you can observe that the worst case is $N ≤ 5 \log_{10} a$, i.e. five times the number of (base-10) digits of $min(a, b) = a$, where $(a, b)$ is a pair of consecutive Fibonacci numbers. This relationship is useful because $gcd(F_{n+1}, F_{n+2}) = 1$ (i.e. all consecutive Fibonacci numbers are coprime).

Given these upper-bounds, the asymptotic computational complexity of the Euclidean algorithm can be expressed using Big-O notation as $\mathcal{O}(\log b)$.

Since, 

$$\mathcal{O}(\log b) = \mathcal{O}(\log_2 a + \log_2 b) = \mathcal{O}(5\log_{10} a)$$

where $b ≥ a$.