Continued Fraction Algorithm for 113/50 The numbers $a_k$ can be found for $\frac{113}{50}$ by using a continued fraction algorithm. Note that $\frac{113}{50}$ is rational, and as a result it will have to terminate. Can anyone help me determine the numbers $a_k$ for $\frac{113}{50}$?
What I have Done:
$\frac{113}{50} = 2.26$ 
$i=2$ 
$2.26 - 2 = .26$ 
$\frac{26}{100}$  
$\frac{100}{26} = 3+ \frac{22}{26} $ 
3 is  $a_1$  
$\frac{26}{22} = 1 + \frac{4}{22}  $
1 is  $a_2$ 
$\frac{22}{4} = 5 + \frac{2}{4}$  
$a_3 = 5$  
$\frac{4}{2} = 2 $ 
$a_4 = 2 $
Is this correct, and what do I do next?

Meanwhile, here is the calculation of the "convergents." It begins with the formal fractions $0/1$ and the fake (but necessary) $1/0.$ Then, separately for numerator and denominator, for each "digit" or "partial quotient" $a_k,$ you get then next number from the evident rule, now visible: for example, $0 + 2 \cdot 1 = 2,$ then $1 + 3 \cdot 2 = 7,$ then $2 + 1 \cdot 7 = 9,$ and so on. Similar for denominators. I have found this litle grid way of presenting the convergents very helpful. For on thing, you can clearly see how the "cross product" of two consecutive convergents is $\pm 1.$ As in: $1 \cdot 1 - 2 \cdot 0 = 1, \; \;$  $2 \cdot 3 - 7 \cdot 1 = -1, \; \;$ $7 \cdot 4 - 9 \cdot 3 = 1, \; \;$  $9 \cdot 23 - 52 \cdot 4 = -1, \; \;$ and so on.
$$  
\begin{array}{cccccccccccc}
 & & 2 & & 3 & & 1 & & 5 & & 2 &  \\
\frac{0}{1} & \frac{1}{0} & & \frac{2}{1} &  &  \frac{7}{3} & &  \frac{9}{4}  & & \frac{52}{23} & &  \frac{113}{50}   
\end{array}
$$
