[proofplan] We use the continued-fraction best-approximation theorem of the second kind to identify the unique minimal scaled error among all denominators below the next convergent denominator. This immediately forces any strict second-kind best approximant to be a convergent, and gives the converse classification for convergents, including the exceptional condition for $p_0/q_0$. We then prove that the weak full-range convention introduces no new approximants, because irrationality of $\theta$ prevents two distinct integer pairs from having equal absolute scaled error. Finally, we explain why intermediate fractions do not enter the second-kind classification under this convention. [/proofplan]

[step:Use the second-kind best-approximation theorem to force strict best approximants to be convergents]Let $p/q$ be a strict best approximation of the second kind to $\theta$, where $p \in \mathbb{Z}$, $q \in \mathbb{N}$, and $p/q$ is reduced. Choose $n \ge 0$ such that $q_n \le q < q_{n+1}$, with the convention that when $q_0 = q_1 = 1$ and $q = 1$, we choose $n = 1$ rather than $n = 0$. Since $\theta \in \mathbb{R} \setminus \mathbb{Q}$, its simple continued fraction has convergents $(p_n/q_n)_{n\ge0}$ with denominators $(q_n)_{n\ge0}$ in the same normalization as in the cited theorem. We invoke the strengthened interval form of the [Continued-Fraction Best Approximation Theorem of the Second Kind](/theorems/Continued-Fraction%20Best%20Approximation%20Theorem%20of%20the%20Second%20Kind), using the interval-uniqueness version of that theorem. In the precise form needed here, for every $n \ge 0$, every integer pair $(r,s) \in \mathbb{Z} \times \mathbb{N}$ satisfying $q_n \le s < q_{n+1}$ and $r/s \ne p_n/q_n$ satisfies \begin{align*} |q_n \theta - p_n| < |s\theta - r|. \end{align*} In the initial denominator-one ambiguity $q_0=q_1=1$, this strengthened form is read with the convention that the interval containing $s=1$ is assigned to $n=1$. Since $q_n \le q$, the integer pair $(p_n,q_n)$ lies in the comparison range allowed in the definition of strict second-kind best approximation for $p/q$. If $p/q \ne p_n/q_n$, then the displayed inequality contradicts \begin{align*} |q\theta - p| < |q_n\theta - p_n|. \end{align*} Therefore $p/q = p_n/q_n$. Hence every strict best approximation of the second kind is a continued-fraction convergent.[/step]

[guided]Let $p/q$ be a strict best approximation of the second kind to $\theta$. The definition says that among all integer pairs $(r,s)$ with $1 \le s \le q$, except those representing the same rational number as $p/q$, the scaled error $|q\theta-p|$ is strictly smallest. We now locate $q$ between two consecutive convergent denominators. Choose $n \ge 0$ such that $q_n \le q < q_{n+1}$. If $q_0 = q_1 = 1$ and $q = 1$, we choose $n=1$ rather than $n=0$ so that the denominator-one ambiguity at the beginning of the continued fraction is handled by the later convergent. The key external input is the strengthened interval form of the [Continued-Fraction Best Approximation Theorem of the Second Kind](/theorems/Continued-Fraction%20Best%20Approximation%20Theorem%20of%20the%20Second%20Kind), specifically its interval-uniqueness version. Its hypotheses match our setting: $\theta$ is an irrational real number, and $(p_n/q_n)_{n\ge0}$ and $(q_n)_{n\ge0}$ are the simple continued-fraction convergents and denominators in the same normalization used by the theorem. In the form needed here, it says that for every $n \ge 0$, every integer pair $(r,s) \in \mathbb{Z} \times \mathbb{N}$ satisfying $q_n \le s < q_{n+1}$ and $r/s \ne p_n/q_n$ satisfies \begin{align*} |q_n\theta - p_n| < |s\theta - r|. \end{align*} When $q_0=q_1=1$ and $q=1$, we use the stated initial convention assigning this case to $n=1$. Why does this theorem apply here? The denominator $q$ of our alleged best approximant lies in exactly the interval $q_n \le q < q_{n+1}$ required by the theorem. If $p/q$ were not the convergent $p_n/q_n$, the theorem would produce the strictly better integer pair $(p_n,q_n)$. Since $q_n \le q$, this pair is inside the comparison range for the definition of best approximation of $p/q$. Thus strict bestness would require \begin{align*} |q\theta - p| < |q_n\theta - p_n|, \end{align*} while the continued-fraction theorem gives the opposite strict inequality \begin{align*} |q_n\theta - p_n| < |q\theta - p|. \end{align*} These two inequalities are incompatible. Hence the alternative $p/q \ne p_n/q_n$ is impossible, and so $p/q = p_n/q_n$. Therefore every strict best approximation of the second kind is a convergent.[/guided]

[step:Apply the endpoint form of the second-kind theorem to identify which convergents are strict]Since $\theta \in \mathbb{R} \setminus \mathbb{Q}$, the simple continued-fraction convergents $(p_n/q_n)_{n\ge0}$ and denominators $(q_n)_{n\ge0}$ are in the same normalization as in the cited theorem. The strengthened endpoint and full-range form of the [Continued-Fraction Best Approximation Theorem of the Second Kind](/theorems/Continued-Fraction%20Best%20Approximation%20Theorem%20of%20the%20Second%20Kind) states that every convergent $p_n/q_n$ with $n \ge 1$ satisfies the endpoint-minimality inequality \begin{align*} |q_n\theta - p_n| < |s\theta - r| \end{align*} for every $(r,s) \in \mathbb{Z} \times \mathbb{N}$ such that $1 \le s \le q_n$ and $r/s \ne p_n/q_n$. This is exactly the full comparison range in the strict second-kind definition, including competitors with endpoint denominator $s=q_n$. Hence each $p_n/q_n$ with $n \ge 1$ is a strict best approximation of the second kind. For $p_0/q_0$, we have $q_0 = 1$ and $p_0 = a_0$. The only competitors with denominator $s \le 1$ are integers $r/1$ with $r \in \mathbb{Z}$ and $r \ne a_0$. Since \begin{align*} 0 < \theta - a_0 = \frac{1}{[a_1;a_2,a_3,\dots]} < 1, \end{align*} the nearest integer competitors are $a_0-1$ and $a_0+1$, and the strict condition is equivalent to \begin{align*} \theta - a_0 < a_0 + 1 - \theta. \end{align*} This is equivalent to \begin{align*} \theta - a_0 < \frac{1}{2}. \end{align*} Writing $\alpha := [a_1;a_2,a_3,\dots]$, so that $\theta - a_0 = 1/\alpha$, this becomes $\alpha > 2$. Since $\alpha = a_1 + \beta$, where $\beta := [0;a_2,a_3,\dots]$ satisfies $0 < \beta < 1$, the inequality $\alpha > 2$ holds exactly when $a_1 \ge 2$. Thus $p_0/q_0$ is a strict second-kind best approximation exactly when $a_1 \ge 2$.[/step]

[guided]For convergents after the initial one, the strengthened endpoint and full-range version of the [Continued-Fraction Best Approximation Theorem of the Second Kind](/theorems/Continued-Fraction%20Best%20Approximation%20Theorem%20of%20the%20Second%20Kind) supplies exactly the needed comparison. The theorem applies because $\theta$ is irrational and the symbols $p_n/q_n$ and $q_n$ denote the simple continued-fraction convergents and their denominators in the theorem's normalization. It says that for every $n \ge 1$ and every competitor $(r,s) \in \mathbb{Z} \times \mathbb{N}$ with $1 \le s \le q_n$ and $r/s \ne p_n/q_n$, one has \begin{align*} |q_n\theta - p_n| < |s\theta-r|. \end{align*} This is precisely the strict second-kind condition for $p_n/q_n$, because the definition compares against the full denominator range $1 \le s \le q_n$, including the endpoint $s=q_n$. The initial convergent needs a separate calculation because $q_0=1$ and the endpoint theorem is being used only for $n \ge 1$. We have $p_0=a_0$ and $q_0=1$. Thus the only allowed competitors have the form $r/1$ with $r \in \mathbb{Z}$ and $r \ne a_0$. Since \begin{align*} 0 < \theta-a_0 = \frac{1}{[a_1;a_2,a_3,\dots]} < 1, \end{align*} the closest competing integer errors occur for $r=a_0-1$ and $r=a_0+1$. All other integers are farther away from $\theta$ than one of these two nearest competitors. The strict condition is therefore equivalent to requiring the error to $a_0$ to be smaller than the error to $a_0+1$: \begin{align*} \theta-a_0 < a_0+1-\theta. \end{align*} Equivalently, \begin{align*} \theta-a_0 < \frac{1}{2}. \end{align*} Define $\alpha := [a_1;a_2,a_3,\dots]$. Then $\theta-a_0=1/\alpha$, so the preceding inequality is equivalent to $\alpha>2$. Define also $\beta := [0;a_2,a_3,\dots]$. Then $0<\beta<1$ and $\alpha=a_1+\beta$. Therefore $\alpha>2$ holds exactly when $a_1\ge2$. Hence $p_0/q_0$ is strict exactly in that case.[/guided]

[step:Show that equality of scaled errors cannot occur for distinct competitors]Let $(p,q),(r,s) \in \mathbb{Z} \times \mathbb{N}$ be two integer pairs with \begin{align*} |q\theta - p| = |s\theta - r|. \end{align*} Then either \begin{align*} q\theta - p = s\theta - r \end{align*} or \begin{align*} q\theta - p = -(s\theta - r). \end{align*} In the first case, \begin{align*} (q-s)\theta = p-r. \end{align*} If $q \ne s$, then \begin{align*} \theta = \frac{p-r}{q-s}, \end{align*} contradicting $\theta \notin \mathbb{Q}$. If $q=s$, then $p=r$, so the integer pairs are identical. In the second case, \begin{align*} (q+s)\theta = p+r. \end{align*} Since $q,s \in \mathbb{N}$, we have $q+s>0$, and hence \begin{align*} \theta = \frac{p+r}{q+s}, \end{align*} again contradicting $\theta \notin \mathbb{Q}$. Therefore two distinct integer pairs cannot have the same absolute scaled error.[/step]

[guided]We now prove the key fact needed for the weak convention: for irrational $\theta$, ties in absolute scaled error cannot occur between distinct integer pairs. Take two integer pairs $(p,q),(r,s) \in \mathbb{Z} \times \mathbb{N}$ and suppose \begin{align*} |q\theta - p| = |s\theta - r|. \end{align*} Equality of absolute values means that the two [real numbers](/page/Real%20Numbers) are either equal or negatives of each other. First suppose \begin{align*} q\theta - p = s\theta - r. \end{align*} Rearranging gives \begin{align*} (q-s)\theta = p-r. \end{align*} If $q \ne s$, then division by the nonzero integer $q-s$ gives \begin{align*} \theta = \frac{p-r}{q-s}, \end{align*} which is rational because $p-r$ and $q-s$ are integers. This contradicts the hypothesis $\theta \in \mathbb{R} \setminus \mathbb{Q}$. Therefore, in this first-sign case, we must have $q=s$. Substituting $q=s$ into $(q-s)\theta=p-r$ gives $p-r=0$, so $p=r$. Thus the two integer pairs are actually the same. Now suppose the signs are opposite: \begin{align*} q\theta - p = -(s\theta-r). \end{align*} Then \begin{align*} (q+s)\theta = p+r. \end{align*} Here $q+s$ is a positive integer because $q,s \in \mathbb{N}$. Dividing by $q+s$ gives \begin{align*} \theta = \frac{p+r}{q+s}, \end{align*} again contradicting irrationality of $\theta$. Hence the second-sign case cannot occur at all. The conclusion is that equal absolute scaled errors force equality of the integer pairs. In particular, no distinct allowed competitor can tie the scaled error of a candidate.[/guided]

[step:Deduce that the weak full-range convention gives the same approximants]Let $p/q$ be a weak best approximation of the second kind under the full-range convention. Thus \begin{align*} |q\theta - p| \le |s\theta-r| \end{align*} for every $(r,s) \in \mathbb{Z} \times \mathbb{N}$ with $1 \le s \le q$ and $r/s \ne p/q$. By the previous step, equality cannot occur for any distinct integer pair $(r,s)$ in this comparison range. Therefore the weak inequality is automatically strict for every allowed competitor: \begin{align*} |q\theta - p| < |s\theta-r|. \end{align*} Hence every weak best approximation under the full-range convention is strict. The converse implication from strict to weak is immediate because a strict inequality implies the corresponding weak inequality. Therefore the weak and strict second-kind best approximations coincide, and the classification already proved for strict best approximations applies.[/step]

[guided]Let $p/q$ be a weak best approximation of the second kind under the full-range convention. This means that $p \in \mathbb{Z}$, $q \in \mathbb{N}$, and every allowed competitor $(r,s) \in \mathbb{Z} \times \mathbb{N}$ with $1 \le s \le q$ and $r/s \ne p/q$ satisfies \begin{align*} |q\theta - p| \le |s\theta-r|. \end{align*} The only difference between this condition and the strict condition is the possible equality case. The preceding step removes exactly that possible equality case. If equality held for an allowed competitor, then the two integer pairs $(p,q)$ and $(r,s)$ would have the same absolute scaled error. Since $\theta$ is irrational, the previous step proves that equal absolute scaled errors force the integer pairs themselves to be equal. But equality of the pairs gives $r/s=p/q$, which is excluded from the comparison range. Therefore equality cannot occur for any allowed competitor, and the weak inequality improves to \begin{align*} |q\theta - p| < |s\theta-r|. \end{align*} Thus every weak full-range best approximation is strict. Conversely, every strict best approximation is weak because a strict inequality implies the corresponding non-strict inequality. Hence the weak and strict second-kind best approximations coincide under this convention, and the strict classification already proved gives the stated weak classification.[/guided]

[step:Separate intermediate fractions from second-kind best approximation] For $n \ge 0$ and $1 \le m < a_{n+1}$, define the integers $P_{n,m} := m p_n + p_{n-1}$ and $Q_{n,m} := m q_n + q_{n-1}$. The corresponding intermediate fraction is \begin{align*} \frac{P_{n,m}}{Q_{n,m}} = \frac{m p_n + p_{n-1}}{m q_n + q_{n-1}}. \end{align*} These fractions are continued-fraction intermediate candidates, but they are not additional strict second-kind best approximants under the present convention, because the strict second-kind best approximants have already been shown to be exactly the listed convergents. Whether the displayed integer pair $(P_{n,m},Q_{n,m})$ is first reduced is irrelevant for this exclusion: any rational number represented by it that were a strict second-kind best approximant would already have to equal one of the convergents in the classification proved above. Their relevance belongs to first-kind approximation. In first-kind approximation the quantity compared is \begin{align*} \left|\theta - \frac{p}{q}\right| = \frac{|q\theta-p|}{q}, \end{align*} so changing the denominator changes the comparison by dividing the scaled error by $q$. Therefore a rational number with a larger scaled error $|q\theta-p|$ may still have a smaller ordinary error $|\theta-p/q|$ because its denominator is larger. This denominator normalization is absent from the second-kind comparison, which uses $|q\theta-p|$ directly. Hence the intermediate fractions are not additional second-kind best approximants under the full-range convention of this theorem. [/step]