# Thread: How show continued fraction (65-sqrt(44))/37

1. ## How show continued fraction (65-sqrt(44))/37

Please Some one give me the specific procedure . I'm confused because it's not the normal form . thanks

2. Hello, wsc810!

I have a solution, but it might not be acceptable.

$\text{Find the continued fraction for: }\:\dfrac{65 - \sqrt{44}}{37}$

$\displaystyle\text{We have: }\:\frac{65}{37} \;=\;1 + \frac{1}{1 + \frac{1}{3 +\frac{1}{9}}}$

. . $\text{ which can be written: }\,\frac{65}{37} \:=\:[1,1,3,9]$

$\displaystyle \text{We have: }\frac{\sqrt{44}}{37} \:=\:\frac{2}{37}\sqrt{11}$

$\displaystyle \text{And: }\:\sqrt{11} \;=\;3 + \frac{1}{3 + \frac{1}{6 + \hdots}}}$

. . $\text{which can be written: }\,\sqrt{11} \:=\:(\overline{3,3,6})$

$\displaystyle \text{Therefore: }\:\frac{64-\sqrt{44}}{37} \;=\;[1,1,3,9] - \tfrac{2}{37}(\overline{3,3,6})$

3. @wsc810 -- Your textbook should have a section that describes how to convert a general irrational number into a continued fraction. If so, just follow the instructions and apply them to your number. I'll give the steps needed to derive the first two partial quotients. You should be able to take it from there.

Let $\xi_0 = \frac{65 - \sqrt{44}}{37}$. Then the first partial quotient, $a_0$, equals the integer part of $\xi_0$, or, using the floor function,

$\xi_0 = \lfloor\frac{65 - \sqrt{44}}{37}\rfloor$

which can easily be found with a calculator to equal 1. Thus, $a_0 = 1$.

We now define $\xi_1 := \frac{1}{\xi_0 - a_0}$ and $a_1 := \lfloor\xi_1\rfloor$. Therefore, since

$\xi_0 - a_0 = \frac{65 - \sqrt{44}}{37} - 1 = \frac{28 - \sqrt{44}}{37}$

and so

$\xi_1 = \frac{37}{28 - \sqrt{44}} = \frac{1036 + 37\sqrt{44}}{740}$

Thus,

$a_1 = \lfloor\xi_1\rfloor = 1$

which is again a simple calculation.

$1 + \frac{1}{1 + \frac{1}{a_2 + \cdots}}$

You should be able to find the remaining partial quotients $a_2, a_3, \cdots$ by recursively defining

$a_i = \lfloor\xi_i\rfloor, \\\ \xi_{i+1} = \frac{1}{\xi_i - a_i}$

4. Q_0=Q
a_0=[(P a)/Q]
P_0=P
it's the starting value and a=Int(sqrt(d))
$P_{ n 1}=Q_n*a_n - P_n$
$Q_{n 1}=
(d-P_{n 1}^2)/Q_n$

$a_{n 1}=
[(P_{n 1} a)/Q_{n 1}]$

[ *] represents Int(*)
a0=1
P0=65
Q0=37
37*a0 - 65=-28 a0=1
-20*a1-(-28)=8 a1=1
1*a2 -28=6 a2=14
why a2 is not right
the actual vaule is 1 . Mathematica work out expression [1,1,1,2;1,2,1,1,1,12,1,1]

5. Originally Posted by wsc810
Q_0=Q
a_0=[(P + a)/Q]
P_0=P
it's the starting value and a=Int(sqrt(d))
P_{ n + 1}=
Q_n*a_n - P_n
Q_{n + 1}=
(d-P_{n + 1}^2)/Q_n
a_{n + 1}=
[(P_{n + 1} + a)/Q_{n + 1}]
[ *] represents Int(*)
a0=1
P0=65
Q0=37
37*a0 - 65=-28 a0=1
-20*a1-(-28)=8 a1=1
1*a2 -28=6 a2=14
why a2 is not right
the actual vaule is 1 . Mathematica work out expression [1,1,1,2;1,2,1,1,1,12,1,1]

A piece of advise: if you're going to write a lot or some rather involved mathematical

expressions, you better take a peek at the forum's section "Math Resources", subsection

"LaTeX help", and then re-write your stuff using LaTex, otherwise it comes out

too messy and it might be that not many people will even take a look at it.

For example, the following part of you message:

P_{ n + 1}=
Q_n*a_n - P_n
Q_{n + 1}=
(d-P_{n + 1}^2)/Q_n

would come out, using LaTeX, as:

$\displaystyle{P_{n+1}=Q_na_n-P_n\,,\,\,Q_{n+1}=\frac{d-P_{n+1}^2}{Q_n}}$ ,and this is way clearer

and thus much easier to read.

Tonio