Greatest Commom Divisor In my lecture notes there is the following: 
$$\newcommand{\gcd}{\operatorname{gcd}}
a, b \in \mathbb{N}, \gcd(a, b)=d \implies \exists s, t \in \mathbb{Z} : sa+tb=d \implies tb \equiv d \pmod a$$ 
If $\gcd(a, b)=1$ then $tb \equiv 1 \pmod a \ $ $ \ \ \ [b]_a \in \mathbb{Z}_a^{\star}$ and $[b]_a^{-1}=[t]_a$ 
$$r_0=a, r_1=b \\ r_0=q_1 r_1+r_2 , 0 < r_2 <r_1 \\ r_1=q_2 r_2 + r_3, 0<r_3 < r_2 \\ \dots \\ r_{l-1}=q_{l-1}r_{l-1}+r_l, 0<r_l < r_{l-1} \\ r_{l-1}=q_l r_l+0$$ 
$\gcd(a, b)=r_l$ 
$$b, b-1, b-2, \dots , 0 \\ l \leq b$$ 
Because there isn't any explanation what $b, b-1, b-2, \dots , 0$ means, do you have any idea about what it could mean? What does this represent?

The point here seems to be to give a simple argument why the Euclidean algorithm terminates. 
The sequences of remainders $r_i$ is strictly decreasing and $r_1 = b$. So it will stop at at most $r_b$, as the worst case would be $r_2 = b-1$, $r_3 = b-2$ and so on $r_b = b - (b-1)=1$, so then the following remainder must be $0$. 
Note that there are somewhat more sophisticated arguments that give a much better  bound for the number of necessary steps. 
