Hello!!!

I am looking at the following exercise:

Let $b=r_0, r_1, r_2, \dots$ be the successive remainders in the Euclidean algorithm applied to $a$ and $b$. Show that after every two steps, the remainder is reduced by at least one half. In other words, verify that $$r_{i+2}< \frac{1}{2} r_i \ ,\text{ for every } i=0,1,2, \dots$$

Conclude that the Euclidean algorithm terminates in at most $2 \log_2{(b)}$ steps, where $\log_{2}$ is the logarithm to the base $2$. In particular, show that the number of steps is at most seven times the number of digits in $b$. [ Hint: What is the value of $\log_2{10}$ ?]

I have tried the following:

The general formula for $r_{i-1}$ in the Euclidean algorithm is the following:

$$r_{i-1}=q_{i+1} r_i+r_{i+1}$$

So we have: $r_i=q_{i+2} r_{i+1}+r_{i+2} \\=q_{i+2}(q_{i+3} r_{i+2}+r_{i+3})+r_{i+2} \\=q_{i+2} q_{i+3} r_{i+2} +q_{i+2} r_{i+3}+r_{i+2}\\ \geq r_{i+2}+q_{i+2} r_{i+3}+r_{i+2} \\=2r_{i+2}+q_{i+2} r_{i+3}> 2r_{i+2}$

So it follows that $r_{i+2}<\frac{1}{2} r_i$.

As for the number of steps, I have thought the following:

We have that $b=r_0>2r_2> 4r_4>8r_6> \dots> 2^m r_{2m}$

The Euclidean algorithm terminates when we find a remainder that is equal to zero.

So we check when $r_{2m}<1$.

$r_{2m}<1 \Rightarrow 2^m r_{2m}<2^m$.

We also have that $2^m r_{2m}<b$.

But from these two inequalities, we cannot find a relation between $m$ and $b$, can we?