Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can one find a quadratic irrational when knowing its periodic continued fraction?

For example(using Wikipedia notion), how can one find the quadratic irrational that its continued fraction is $[0; \overline{1,4,1}]$ ?

share|cite|improve this question
It is not a "fraction," it is a quadratic irrationality. – André Nicolas Feb 23 '14 at 20:18
If periodic (therefore infinite) it is a "quadratic irrational" – Will Jagy Feb 23 '14 at 20:19
Sorry about that, changed it. – Eliran Koren Feb 23 '14 at 20:32
up vote 7 down vote accepted

Let $a$ be a quadratic irrational with periodic continued fraction of the form $[0; \overline{1,4,1}]$.

$a=\cfrac{1}{1+\cfrac{1}{4+\cfrac{1}{1+a}}} =\cfrac{1}{1+\cfrac{1}{\cfrac{5+4a}{1+a}}} = \cfrac{1}{1+\cfrac{1+a}{5+4a}}=\cfrac{1}{\cfrac{6+5a}{5+4a}}=\cfrac{5+4a}{6+5a} \Rightarrow$

$6a+5a^2=5+4a \Rightarrow$ $5a^2+2a-5=0 \Rightarrow$ $a=\frac{-1 \pm \sqrt{26}}{5} $
We know that $a>0$ ,thus:

$a=\frac{ \sqrt{26}-1}{5}$

share|cite|improve this answer

You might consider taking it over to Wolfram and seeing what it says

$\frac{\sqrt{26}-1}{5} $

Now you can try to prove it.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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