A computation with continued fractions

Question

I am puzzled with the following property on continued fractions. Let's say I denote $x=(n_0, n_1, ...)$ the negative continued fraction (with minuses), where the $n_i > 1$ for $i \geq 1$. If I let $T : x \mapsto x+1$ and $S : x \mapsto -1/x$, then formally $$Tx = (n_0+1, n_1, ...)$$ and $$Sx = (0, n_0, n_1, ...)$$ However these could be improper, because the $n_0$ can be less than $2$. So this has to be fixed in some sense. I read the claims that if $n_0 \leq -1$ then $$Sx = (1, 2, 2, ...2, n_1+1, n_2, ...)$$ where there are $-n_0 - 1$ occurences of $2$. Similarly, if $n_0=1$ and $n_1 \geq 3$, I read that $$Sx = (-1, n_1-1, n_2...)$$ How can we get such results? Could you explain to me what is the strategy, the computations behind?