1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

$$\begin{gathered} \frac{{113}}{{50}} = \frac{{100}}{{50}} + \frac{{13}}{{50}} = 2 + \frac{1}{{\frac{{39}}{{13}} + \frac{{11}}{{13}}}} = 2 + \frac{1}{{3 + \frac{{11}}{{13}}}} \hfill \\ = 2 + \frac{1}{{3 + \frac{{11}}{{11 + 2}}}} = 2 + \frac{1}{{3 + \frac{1}{{1 + \frac{2}{{10 + 1}}}}}} \hfill \\ = 2 + \frac{1}{{3 + \frac{1}{{1 + \frac{1}{{5 + \frac{1}{2}}}}}}} = [2;3,1,5,2] \hfill \\ \end{gathered}$$

$\endgroup$
  • $\begingroup$ So I was right? $\endgroup$ – Jessie Mar 21 '15 at 21:37
  • $\begingroup$ Yes! Can't follow your way of calculation exactly, so I wanted to show my way! $\endgroup$ – Frieder Mar 21 '15 at 21:41
1
$\begingroup$

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.

$$ \begin{array}{cccccccccccc} & & 2 & & 3 & & 1 & & 5 & & 2 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{2}{1} & & \frac{7}{3} & & \frac{9}{4} & & \frac{52}{23} & & \frac{113}{50} \end{array} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.