Let $x$ be irrational with $n$th continued fraction convergent

$$\frac{p_{n}}{q_{n}}=a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}+\frac{1}{a_{n-1}+\frac{1}{a_{n}}}}}}$$

Show that

$$\left(\begin{array}{cc} p_{n} & p_{n-1} \\ q_{n} & q_{n-1} \end{array}\right)=\left(\begin{array}{cc} a_{0} & 1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} a_{1} & 1 \\ 1 & 0 \end{array}\right) \cdots\left(\begin{array}{cc} a_{n-1} & 1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} a_{n} & 1 \\ 1 & 0 \end{array}\right)$$

and deduce that

$$\left|\frac{p_{n}}{q_{n}}-x\right| \leqslant \frac{1}{q_{n} q_{n+1}} .$$

[You may quote the result that $x$ lies between $p_{n} / q_{n}$ and $p_{n+1} / q_{n+1} \cdot$ ]

We say that $y$ is a quadratic irrational if it is an irrational root of a quadratic equation with integer coefficients. Show that if $y$ is a quadratic irrational, we can find an $M>0$ such that

$$\left|\frac{p}{q}-y\right| \geqslant \frac{M}{q^{2}}$$

for all integers $p$ and $q$ with $q>0$.

Using the hypotheses and notation of the first paragraph, show that if the sequence $\left(a_{n}\right)$ is unbounded, $x$ cannot be a quadratic irrational.