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So, as far as I understand any rational number can be expressed as a continued fraction of finite length

for example: $$\frac {7}{17} = \cfrac {1}{2 + \cfrac{1}{2 + \cfrac{1}{3}}}$$ which for convenience I will express as $[0,2,2,3]$

Now I also understand that some irrationals can be written as infinite continued fractions $$\sqrt{2} =1 + \cfrac{1}{2 +\cfrac{1}{2 + \cfrac{1}{\cdots}}}$$ expressable as $[1,\overline{2}]$

Now my question, can all irrational numbers that can be expressed as the root of a polynomial (algebraic irrational) be expressed as a repeating continued fraction? How would we go about proving this if it is true?

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  • $\begingroup$ Every algebraic irrational, yes. $\endgroup$ – barak manos Nov 20 '16 at 17:23
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    $\begingroup$ no. repeating c.f. means a quadratic irrational, real root of $a x^2 + b x + c$ with integers $a,b,c.$ More restrictions if the c.f. is strictly periodic $\endgroup$ – Will Jagy Nov 20 '16 at 17:26
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    $\begingroup$ To simplify every answer here: NO, unless the number happens to be the root of a quadratic. $\endgroup$ – Brevan Ellefsen Nov 20 '16 at 18:02
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If a simple continued fraction repeats but does not terminate then it converges to a second-degree algebraic number. ("Simple" means all of the numertors are $1$.)

First, look at the case where the repeating part begins immediately: $$ x = a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {\ddots \cfrac{}{a_n + \cfrac 1 {a_1 + _{\large\ddots}}}}}} $$ Then you have $$ x = a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {\ddots \cfrac{}{a_n + \cfrac 1 x}}}} $$ This reduces to a quadratic equation in $x$ with integer coefficients.

To to that reduction to a quadratic equation, start with $$ a_{n-1} + \cfrac 1 {a_n + \cfrac 1 x} = a_{n-1} + \frac{x}{a_n x + 1} = \frac{ (a_{n-1} a_n + 1) x + a_n }{a_{n-1} x + 1} $$ So you have $$ a_{n-2} + \cfrac 1 {a_{n-1} + \cfrac{x}{a_n x + 1}} = a_{n-2} + \frac{a_n x + 1}{a_{n-1}(a_n x + 1) + x } = \text{etc.} $$ Keep going until you have $$ x = \frac{\text{a first-degree polynomial in } x}{\text{another first-degree polyonomial in } x}, $$ then multiply both sides by the denominator to get a quadratic equation.

If the repeating part doesn't start at the beginning, you can work on the repeating part, getting it to be the solution of a quadratic equation, then repeatedly rationalize denominators until you're done.

Here's a concrete example: $$ x = 3 + \cfrac 1 {2 + \cfrac 1 {5 + \cfrac 1 {2 + \cfrac 1 {5 + \cdots}}}} $$ with $2,5$ repeating.

One can write $$ y = 3 + \cfrac 1 x = 3 + \cfrac 1 {2 + \cfrac 1 {5 + \cfrac 1 x}} $$ So $$ x = 2 + \cfrac 1 {5 + \cfrac 1 x} = 2 + \frac x {5x+1} = \frac{11x+2}{5x+1} $$ $$ x = \frac{11x+2}{5x+1} $$ $$ 5x^2 + x = 11x+2 $$ $$ 5x^2 - 10 x - 2 = 0. $$ $$ x = \frac{10 \pm \sqrt{140}}{10} = \frac{5 \pm \sqrt{35}} 5. $$ Since the value we seek is positive, we have $$ x= \frac{5 + \sqrt{35}} 5. $$

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