 Solution

For a real number $\alpha$, set ${\alpha }_0=\alpha$ and $a_j=\lfloor {\alpha }_j\rfloor$. If ${\alpha }_j-a_j\ne 0$, define ${\alpha }_{j+1}=({\alpha }_j-a_j{)}^{-1}$. The resulting simple continued fraction is $[a_0;a_1,a_2,\dots ]$, where $a_0$ is an integer and subsequent $a_j$ are positive integers. It terminates when a complete quotient is integral; otherwise its convergents tend to $\alpha$.

For the continued fraction of the square root of nineteen, the complete quotients after $a_0=4$ areTheir integer parts are $2,1,3,1,2,8$. Taking the reciprocal after subtracting $8$ from the last quotient returns the first one, so