Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Two definitions (the first is informal) of continued fraction:

  1. This is the basic Continued Fraction algorithm for real numbers. Let $\alpha \in \mathbb{R}$. If $[\alpha]=\alpha$, then we are done. If not, then let $a_0=[\alpha]$ and write $\alpha=a_0+\frac{1}{\alpha_1}$. If $\alpha_1\in \mathbb{N}$ we are done; otherwise compute $a_1=[\alpha_1]$ and write $\alpha_1=a_1+\frac{1}{\alpha_2}$. If $\alpha_2 \in \mathbb{N}$ we are done; otherwise compute $a_2=[\alpha_2]$ and write $\alpha_2=a_2+\frac{1}{\alpha_3}$. Repeat as necessary.

  2. This is the Continued Fraction Algorithm for quadratic irrationals. The following algorithm computes the continued fraction expansion of the quadratic irrational $\frac{P+\sqrt{D}}{Q}$, where $D$ is a nonsquare positive integer and $P$, $Q$ are any integers. Let $P_0=P$, $Q_0=Q$, and $a_0=[\sqrt{D}]$. For $i\geq 1$, define

(A) $P_i=a_{i-1}Q_{i-1}-P_{i-1}$,

(B) $Q_i=\frac{D-P_i^2}{Q_{i-1}}$,

(C) $a_i=[\frac{P_i+\sqrt{D}}{Q_i}]$.

Is it possible to prove $Q_i\cdot Q_{i-1}=D-P_i^2$ solely using definition 1?

It clearly follows from definition 2. I am wondering if it is possible to prove this identity using definition 1 only.

Thanks. I understand that this may seem like a moot point, but from a teaching perspective, not having to introduce definition 2 makes sense for undergrads and time constraints.

share|cite|improve this question
Yes $-$ once you develop the machinery of convergents of a continued fraction. Obviously you must first define the $P_i$ and $Q_i$ in terms of (1), but that’s standard. – Brian M. Scott Oct 18 '12 at 19:50

Your Answer


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

Browse other questions tagged or ask your own question.