Why is this value a reduced quadratic irrational

Question

The paper: http://web.math.princeton.edu/mathlab/jr02fall/Periodicity/alexajp.pdf

Note: The paper defines a reduced quadratic irrational $\alpha$ as:

• $\alpha > 1$
• Its the root of a quadratic with integral coefficients
• Its conjugate $-1 < \overset{\sim}{\alpha} < 0$

In the paper (page 3, Proof of Theorem 3.1), it says $\alpha$ is a reduced quadratic irrational ($\alpha = \frac{P + \sqrt{D}}{Q}$), and then says $\alpha$ can be expressed as $\alpha = a_{1} + \frac{1}{\alpha_{1}}$, with $a_{1} = \lfloor\alpha\rfloor$. Then it rearanges to give $\alpha_{1}=\frac{1}{\alpha - a_{1}}$

This all makes sense.

But then it says (without any explanation) that $\alpha_{1}=\frac{1}{\alpha - a_{1}} = \frac{P_{1} + \sqrt{D}}{Q_{1}} > 1$

I agree with the $>1$ ($\alpha_{1}$ is the reciprocal of the fractional part of $\alpha$)

However I don't see how you can go from $\frac{1}{\alpha - a_{1}}$ to $\frac{P_{1} + \sqrt{D}}{Q_{1}}$ so simply.

I worked out the value of $\alpha_{1}$ in terms of $P$, $Q$, $D$ and $a_{1}$: $\alpha_{1} = \frac{(a_{1}Q - P) + \sqrt{D}}{2Pa_{1} - Qa^{2}_{1} + \frac{D - P^{2}}{Q}}$ and because $\frac{P + \sqrt{D}}{Q}$ is the root of a quadratic with integer coefficients we can work out that the $\frac{D - P^{2}}{Q} = -2c \in \Bbb{Z}$ (if the quadratic it was the root of was $ax^{2} + bx + c$)

But as nothing like this is included, I assume that there's a more obvious reason that we know that $\alpha_{1}$ is a reduced quadratic irrational.

In addition, Definition 2.4 also requires that the conjugate root $-1 < \overset{\sim}{\alpha} < 0 $, however I see no reason why this is true.

UPDATE:

I tried setting $\alpha=\frac{P + \sqrt{D}}{Q}$ and therefore $\alpha - \lfloor\alpha\rfloor = \frac{P' + \sqrt{D}}{Q}$.

With this I was able to substitute to give:

• $\alpha_1 = (\frac{P' + \sqrt D}{Q})^{-1} = \frac{Q(\sqrt D - P')}{D - (P')^2} = \frac{-P' + \sqrt D}{\frac{D - (P')^2}{Q}} > 1$
• $\overset{\sim}{\alpha_{1}} = \frac{-P' - \sqrt{D}}{\frac{D - (P')^{2}}{Q}} > 1 - \frac{2\sqrt D}{\frac{D - (P')^2}{Q}} = 1 - \frac{2Q\sqrt D}{D - (P')^2}$

Then working backwards (to prove $\overset{\sim}{\alpha_1} > -1$):

$1 - \frac{2Q\sqrt D}{D - (P')^2} > -1$

$\frac{2Q\sqrt D}{D - (P')^2} < 2$

$\frac{Q\sqrt D}{D - (P')^2} < 1$

$Q\sqrt{D} < D - (P')^2$

$Q\sqrt{D} < (\sqrt D + P')(\sqrt D - P')$

Note that $\alpha - \lfloor\alpha\rfloor = \frac{P' + \sqrt D}{Q}$ is the fractional part of $\alpha$ and therefore $\frac{P' + \sqrt D}{Q} < 1 \Rightarrow P' + \sqrt D < Q$

Therefore we can perform the following substitution:

$Q\sqrt D < Q(\sqrt D - P')$

$\sqrt D < \sqrt D - P'$

$0 < - P'$

$P' < 0$

$\therefore \overset{\sim}{\alpha_1} > -1 \iff P' < 0 $

However I can't figure out how to prove $P' < 0$

Answer 1

From the variables used this is a continuation of various studies of continued fractions and the fact that a 'reduced quadratic irrational' is a continued fraction with no acyclic member (the entire continued fraction repeats).

It is proven in continued fraction if you reverse the continued fraction of a reduced quadratic irrational and call that new value beta, then -1/beta is the other root of the quadratic equation of the original reduced quadratic irrational.

Since alpha is all positive integers, so is beta, so beta > 1, so -1/beta is between -1 and 0.

A good source on proving the above facts is in an older book called "Continued Fractions" by C. D. Olds that is book 9 of "New Mathematical Library".

Where you say "But then it says (without any explanation)" is also a fact determined about reduced quadratic irrationals in the book above. In addition, the book proves that there is a finite number of P and Q values of that form for any given positive non-square D. In particular, P < $\sqrt{D}$ and Q < 2 $\sqrt{D}$.