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Source. Laczkovich (1984), printed p. 113 (PDF p. 5), equations (7)–(9) and the intervening recurrence. The source invokes standard regular continued-fraction theory. That theory is the explicit external input here; its general proof is not reconstructed on this page.

Let $\alpha$ be irrational and let $[a_0;a_1,a_2,\ldots]$ be its regular continued fraction, where $a_0\in\mathbb Z$ and $a_i$ is a positive integer for $i\ge1$. For its convergents $p_i/q_i$, use

$$p_{-1}=1,\quad p_0=a_0,\qquad q_{-1}=0,\quad q_0=1,$$

and, for $i\ge0$,

$$p_{i+1}=a_{i+1}p_i+p_{i-1},\qquad q_{i+1}=a_{i+1}q_i+q_{i-1}. \tag{1}$$

In particular $q_1=a_1$. The denominators are positive and unbounded; $q_{i+1}\ge q_i$, with strict inequality for $i\ge1$. The approximation errors satisfy

$$0<q_i\alpha-p_i<\frac1{q_{i+1}}\quad(i\text{ even}), \qquad -\frac1{q_{i+1}}<q_i\alpha-p_i<0\quad(i\text{ odd}). \tag{2}$$

If $a_i\le K$ for every $i\ge1$, with $K\ge1$, then the recurrence and $q_{i-1}\le q_i$ give

$$q_{i+1}\le(K+1)q_i\quad(i\ge0). \tag{3}$$

For $i=0$, use $q_1=a_1\le Kq_0$ directly. Thus (3) also gives $q_{j-1}\ge q_j/(K+1)$ when $j=1$; the possible equality $q_0=q_1=1$ causes no exception.

These are exactly the facts used in Lemma 2. For the application on the whole real line, $\sqrt2=[1;2,2,\ldots]$ supplies an irrational with bounded partial quotients, as the source notes on p. 110.

Source precision. The printed recurrence writes $q_{i+1}=a_iq_i+q_{i-1}$. With its indexing $q_0=1$ and $q_1=a_1$, the coefficient is $a_{i+1}$ as in (1). This is a transcription of the classical input with a corrected index, not an author-issued erratum. The separate two-denominator calculation in Lemma 2 is addressed on that result's page.

Bears on. Problem 1125, through Lemma 2 and the two monotonicity theorems.