finding the quadratic irratonality of simple continued fractions

Question

For instance: find the quadratic irrationality of the simple continued fraction [1;2,3].

I have a handful of these problems to do, so any walk-through of one problem should give me the general idea of how to approach the others. I don't quite understand where the quadratic form comes from; the examples in the book derive it out of thin air.

Answer 1

Thanks Andre and Gerry, I have used your approach to find the answer as such:

$[1;2,3,3,3,3,...]$

Let x = $[1;2,y]$ and y = $[3;y]$. Then $y = 3 + 1/y = (3y+1)/y$ which is to say $y^2 - 3y - 1 = 0$. Solving using the quadratic formula we have $y = (3+\sqrt13)/2$, discarding the negative solution since we know $y$ is positive. Since $x = [1;2,y] = 1+(1/(2+1/y))$ we can substitute in our $y$ value and obtain:

$x = (5+\sqrt13)/2$

which is to say:

$[1;2,3,3,3,3,...] = (5+\sqrt13)/2$

Answer 2

The question seems to be about $[1;2,3,3,3,3,3,\dots]$. A simple approach, assuming that the calculation makes sense, is to let $\alpha=[0;3,3,3,3,3,\dots]$.

If you look at this continued fraction, you can see that $\alpha=\frac{1}{3+\alpha}$. Rewrite as $\alpha^2+3\alpha-1=0$ and solve for $\alpha$. Now it is easy to compute $[1;2,3,3,3,3,3,3,\dots]$. For $$[1;2,3,3,3,3,3,3,\dots]=1+\cfrac{1}{2+\alpha}.$$

Details: So $\alpha=\frac{-3+\sqrt{13}}{2}$, and our original continued fraction is $\frac{3+\alpha}{2+\alpha}$. After a bit of fiddling, this turns out to be $\frac{3+\sqrt{13}}{1+\sqrt{13}}$. That is a correct answer, but you probably don't want a square root in the denominator, so multiply numerator and denominator by $\sqrt{13}-1$.

Remark: A similar process applies, in principle, to any ultimately cycling continued fraction. In all cases we get a quadratic irrationality.