Expansion of Associated Reduced Quadratic Irrational

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Lemma

Let $\alpha$ be a reduced quadratic irrational which is associated to $n$.

Let $\alpha$ be transformed into its integer part and fractional part via:

$\alpha = \left \lfloor {\alpha} \right \rfloor + \dfrac 1{\alpha'}$


Then the resulting quadratic irrational $\alpha'$ is also reduced and associated to $n$.


Proof

Let $\alpha = \dfrac{\sqrt n + P} Q$ and $X = \left \lfloor {\alpha} \right \rfloor$

Then we have:

$\dfrac 1 {\alpha'} = \dfrac {\sqrt n - \left({Q X - P}\right)} Q$

Because $\sqrt n$ is irrational, we must have:

$\dfrac 1 {\alpha'} > 0$

Because $\dfrac 1 {\alpha'}$ is the fractional part:

$0 < \dfrac 1 {\alpha'} < 1 \implies \alpha' > 1$

Transforming:

\(\ds \alpha'\) \(=\) \(\ds \frac Q {\sqrt n - \left({Q X - P}\right)} \cdot \frac {\sqrt n + \left({Q X - P}\right)} {\sqrt n + \left({Q X - P}\right)}\)
\(\ds \) \(=\) \(\ds \frac {\sqrt n + \left({Q X - P}\right)} {\frac 1 Q \left({n - \left({Q X - P}\right)^2}\right)}\)

we have:

$P' = Q X - P$

and:

$Q' = \dfrac 1 Q \left({n - \left({Q X - P}\right)^2}\right)$


To show $Q' \in \Z$:

We have that:

$n - \left({Q X - P}\right)^2 \equiv n - P^2 \pmod Q$

Because $\alpha$ is associated to $n$, $Q$ must divide this quantity.

Hence $Q'$ is an integer as defined.


Because $X = \left \lfloor{\dfrac{\sqrt n + P} Q}\right \rfloor$ is an integer and $\alpha$ is irrational:

$X < \dfrac {\sqrt n + P} Q$

Hence:

$P' = QX - P < \sqrt n $

forcing:

$\tilde{\alpha}' < 0$


Because $\alpha > 1$:

$X \ge 1 \iff 0 \le X - 1$

Thus:

\(\ds \tilde \alpha = \frac {P - \sqrt n} Q\) \(<\) \(\ds 0 \le X - 1\)
\(\ds \leadsto \ \ \) \(\ds Q\) \(<\) \(\ds \sqrt n + \left({Q X - P}\right)\)
\(\ds \leadsto \ \ \) \(\ds Q \left({\sqrt n - \left({Q X - P}\right)}\right)\) \(<\) \(\ds n - \left({Q X - P}\right)^2\)
\(\ds \leadsto \ \ \) \(\ds -\tilde \alpha' = \frac {\sqrt n - \left({Q X - P}\right)} {\frac 1 Q \left({n - \left({Q X - P}\right)^2}\right)}\) \(<\) \(\ds 1\)

Hence $\tilde \alpha' > -1$ and $\alpha'$ is reduced.


Because $Q' = \dfrac 1 Q \left({n - \left({P'}\right)^2}\right)$:

$n - \left({P'}\right)^2 \equiv Q Q' \equiv 0 \pmod {Q'}$

Hence $\alpha'$ is associated to $n$.


Thus $\alpha'$ is both reduced and associated to $n$.

$\blacksquare$

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