= Solution
Write the <continued fraction> as $\theta=[a_0;a_1,a_2,\ldots]$ and define its <continued fraction convergent>[convergents] by
$$
p_{-1}=1,\quad p_0=a_0,\quad p_n=a_np_{n-1}+p_{n-2},
$$
$$
q_{-1}=0,\quad q_0=1,\quad q_n=a_nq_{n-1}+q_{n-2}.
$$
Induction using these recurrences gives
$$
p_nq_{n-1}-p_{n-1}q_n
=-(p_{n-1}q_{n-2}-p_{n-2}q_{n-1})
=\boxed{(-1)^{n-1}}.
$$
In particular, consecutive convergents have coprime numerator and denominator. The standard best-approximation consequence of the same determinant identity is that, for $0<q\le q_n$ and $p/q\ne p_n/q_n$,
$$
|q\theta-p|>|q_n\theta-p_n|.
$$
Dividing by $q\le q_n$ gives
$$
\left|\theta-\frac pq\right|
=\frac{|q\theta-p|}{q}
>\frac{|q_n\theta-p_n|}{q_n}
=\left|\theta-\frac{p_n}{q_n}\right|.
$$
The contrapositive proves that any strictly better approximation has $q>q_n$.
For $\sqrt{12}$, the <continued-fraction algorithm> gives
$$
\sqrt{12}=3+(\sqrt{12}-3),
\qquad
\frac1{\sqrt{12}-3}=2+\frac1{\sqrt{12}+3},
$$
and the remainder then repeats. Hence
$$
\boxed{\sqrt{12}=[3;\overline{2,6}]}.
$$
The convergent $[3;2]=7/2$ supplies a solution of the <Pell equation>:
$$
\boxed{x=7,\qquad y=2},
\qquad 7^2-12\cdot2^2=1.
$$
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