# 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?

$$\begin{gathered} \frac{{113}}{{50}} = \frac{{100}}{{50}} + \frac{{13}}{{50}} = 2 + \frac{1}{{\frac{{39}}{{13}} + \frac{{11}}{{13}}}} = 2 + \frac{1}{{3 + \frac{{11}}{{13}}}} \hfill \\ = 2 + \frac{1}{{3 + \frac{{11}}{{11 + 2}}}} = 2 + \frac{1}{{3 + \frac{1}{{1 + \frac{2}{{10 + 1}}}}}} \hfill \\ = 2 + \frac{1}{{3 + \frac{1}{{1 + \frac{1}{{5 + \frac{1}{2}}}}}}} = [2;3,1,5,2] \hfill \\ \end{gathered}$$
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.
$$\begin{array}{cccccccccccc} & & 2 & & 3 & & 1 & & 5 & & 2 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{2}{1} & & \frac{7}{3} & & \frac{9}{4} & & \frac{52}{23} & & \frac{113}{50} \end{array}$$