Let $b_0,b_1,b_2$,... be the successive remainders computed in the course of Euclid’s algorithm. Prove that $b_{i+2} < b_{i}/2$ for any i ≥ 1.
So we know that $b_i > b_{i+1} > b_{i+2}$ for every i, and $b_i = kb_{i+1} + b_{i+2}$, for integer $k\geq 1$, so $b_i > (k+1)b_{i+2} \geq 2b_{i+2}$, which leads us to the desired result.
Is my proof correct? Now the lecturer proposed that we take two cases, namely $b_{i+1} =< b_{i}/2$ and $b_{i+1} > b_{i}/2$, but i was struggling to prove the result in the second case, so i took this alternative route.