[proofplan] The forward direction (i) uses that $\theta$ lies strictly between two consecutive convergents $p_n/q_n$ and $p_{n+1}/q_{n+1}$: the two distances from $\theta$ to these fractions sum to $1/(q_n q_{n+1})$, and an AM-GM estimate on $1/(q_n q_{n+1}) \leq \frac{1}{2}(1/q_n^2 + 1/q_{n+1}^2)$ forces at least one of the distances to be below half its own $1/q^2$. The converse (ii) proceeds by contradiction: fix $n$ with $q_n \leq q < q_{n+1}$, use the Best Approximation Theorem to compare $|q\theta - p|$ with $|q_n\theta - p_n|$, and show that if $p/q \neq p_n/q_n$ then the gap $|p/q - p_n/q_n|$ is both at least $1/(q q_n)$ (since both fractions are in lowest terms) and strictly smaller than $\frac{1}{2q^2}(1 + q/q_n)$, which collapses to $q < q_n$ — a contradiction. [/proofplan] [step:Locate $\theta$ strictly between two consecutive convergents] By the [Convergence of Convergents](/theorems/1760), the convergents $p_n/q_n$ of $\theta \in \mathbb{R} \setminus \mathbb{Q}$ alternate on either side of $\theta$: the even-indexed convergents increase toward $\theta$ and the odd-indexed convergents decrease toward $\theta$. Consequently, for every $n \geq 0$, the point $\theta$ lies strictly between $p_n/q_n$ and $p_{n+1}/q_{n+1}$, and the two signed quantities $\theta - p_n/q_n$ and $\theta - p_{n+1}/q_{n+1}$ have opposite signs. Therefore their absolute values add: \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| + \left|\theta - \frac{p_{n+1}}{q_{n+1}}\right| &= \left|\frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n}\right|. \end{align*} By the [Determinant Identity for Convergents](/theorems/1757), $p_{n+1} q_n - p_n q_{n+1} = (-1)^n$, so \begin{align*} \left|\frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n}\right| &= \left|\frac{p_{n+1} q_n - p_n q_{n+1}}{q_n q_{n+1}}\right| = \frac{1}{q_n q_{n+1}}. \end{align*} Combining the two displays yields the identity \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| + \left|\theta - \frac{p_{n+1}}{q_{n+1}}\right| &= \frac{1}{q_n q_{n+1}}. \end{align*} [/step] [step:Apply AM-GM to $1/(q_n q_{n+1})$ to prove part (i)] Assume for contradiction that neither of the two convergents satisfies the bound of (i), i.e. \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| \geq \frac{1}{2 q_n^2}, \qquad \left|\theta - \frac{p_{n+1}}{q_{n+1}}\right| \geq \frac{1}{2 q_{n+1}^2}. \end{align*} Adding these and using the identity from the previous step, \begin{align*} \frac{1}{q_n q_{n+1}} &\geq \frac{1}{2 q_n^2} + \frac{1}{2 q_{n+1}^2}. \end{align*} Multiplying through by $2 q_n^2 q_{n+1}^2 > 0$ and rearranging yields \begin{align*} 0 &\geq q_n^2 + q_{n+1}^2 - 2 q_n q_{n+1} = (q_n - q_{n+1})^2. \end{align*} A square of a real number is non-negative, so this forces $q_n = q_{n+1}$. But the denominators of consecutive convergents of an irrational number strictly increase for $n \geq 1$ (since $q_{n+1} = a_{n+1} q_n + q_{n-1}$ with $a_{n+1} \geq 1$ and $q_{n-1} \geq 1$), and $q_0 = 1 < q_1 = a_1$ is likewise a strict inequality because $a_1 \geq 1$ and $q_1 = a_1 \cdot 1 + 0 = a_1$, with equality $q_0 = q_1$ only in the degenerate case $a_1 = 1$; in that case we proceed from $n \geq 1$ instead. Either way $q_n \neq q_{n+1}$ for some shift of the index, contradicting the equality. Hence at least one of the two convergents $p_n/q_n$ or $p_{n+1}/q_{n+1}$ satisfies the stated bound. [guided] We want to show that at least one of two consecutive convergents sits closer than $1/(2q^2)$ to $\theta$. The identity established in the previous step pins the sum of the two distances to exactly $1/(q_n q_{n+1})$, so the question becomes: is this total small enough that the two summands cannot both exceed $1/(2 q_n^2)$ and $1/(2 q_{n+1}^2)$ respectively? Suppose the bad case holds — neither convergent meets the bound: \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| \geq \frac{1}{2 q_n^2}, \qquad \left|\theta - \frac{p_{n+1}}{q_{n+1}}\right| \geq \frac{1}{2 q_{n+1}^2}. \end{align*} Adding and using $\left|\theta - p_n/q_n\right| + \left|\theta - p_{n+1}/q_{n+1}\right| = 1/(q_n q_{n+1})$ gives \begin{align*} \frac{1}{q_n q_{n+1}} &\geq \frac{1}{2 q_n^2} + \frac{1}{2 q_{n+1}^2}. \end{align*} The right-hand side is $\frac{q_n^2 + q_{n+1}^2}{2 q_n^2 q_{n+1}^2}$, and multiplying through by $2 q_n^2 q_{n+1}^2$ converts the inequality to $2 q_n q_{n+1} \geq q_n^2 + q_{n+1}^2$, i.e. \begin{align*} (q_n - q_{n+1})^2 &\leq 0. \end{align*} This is the AM-GM inequality in reverse: the arithmetic mean of $1/q_n^2$ and $1/q_{n+1}^2$ is at least their geometric mean $1/(q_n q_{n+1})$, with equality iff $q_n = q_{n+1}$. So the bad case forces $q_n = q_{n+1}$. The recursion $q_{n+1} = a_{n+1} q_n + q_{n-1}$ with $a_{n+1} \geq 1$ and $q_{n-1} \geq 1$ (for $n \geq 1$) gives $q_{n+1} \geq q_n + 1 > q_n$, contradicting the equality. Therefore the bad case is impossible and (i) holds. [/guided] [/step] [step:Set up the contrapositive for part (ii)] Suppose $p/q \in \mathbb{Q}$ with $q \geq 1$ and $|\theta - p/q| < 1/(2 q^2)$; we must show $p/q$ is a convergent. Without loss of generality, assume $\gcd(p, q) = 1$; otherwise replace $p/q$ by its reduced form (the hypothesis $|\theta - p/q| < 1/(2q^2)$ is only strengthened when $q$ decreases). Consider the sequence of denominators $1 = q_0 \leq q_1 < q_2 < \cdots \to \infty$, which is strictly increasing from $n = 1$ onward. Choose the unique index $n$ with \begin{align*} q_n &\leq q < q_{n+1}. \end{align*} (If $q = q_n$ for some $n$, any tie can be broken by taking the smallest such $n$.) We will show that under the hypothesis $|\theta - p/q| < 1/(2q^2)$ we must have $p/q = p_n/q_n$. [/step] [step:Bound the gap $|p/q - p_n/q_n|$ from above using the hypothesis] Assume for contradiction that $p/q \neq p_n/q_n$. By the triangle inequality, \begin{align*} \left|\frac{p}{q} - \frac{p_n}{q_n}\right| &\leq \left|\theta - \frac{p}{q}\right| + \left|\theta - \frac{p_n}{q_n}\right|. \end{align*} The first summand is bounded by hypothesis: $|\theta - p/q| < 1/(2 q^2)$. For the second, apply the [Best Approximation by Convergents](/theorems/1761): since $0 < q < q_{n+1}$, the theorem states $|q \theta - p| \geq |q_n \theta - p_n|$, i.e. \begin{align*} q \left|\theta - \frac{p}{q}\right| &\geq q_n \left|\theta - \frac{p_n}{q_n}\right|, \end{align*} equivalently \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| &\leq \frac{q}{q_n} \left|\theta - \frac{p}{q}\right| < \frac{q}{q_n} \cdot \frac{1}{2 q^2} = \frac{1}{2 q q_n}. \end{align*} Substituting both bounds, \begin{align*} \left|\frac{p}{q} - \frac{p_n}{q_n}\right| &< \frac{1}{2 q^2} + \frac{1}{2 q q_n} = \frac{1}{2 q^2}\left(1 + \frac{q}{q_n}\right). \end{align*} [/step] [step:Bound the gap $|p/q - p_n/q_n|$ from below and derive a contradiction] The two fractions $p/q$ and $p_n/q_n$ are distinct by assumption, and both are in lowest terms ($\gcd(p, q) = 1$ by our reduction, and $\gcd(p_n, q_n) = 1$ by the [Coprimality of Convergents](/theorems/1758), a direct consequence of $p_n q_{n-1} - p_{n-1} q_n = \pm 1$). Hence the numerator of $p/q - p_n/q_n$, when put over the common denominator $q q_n$, is a non-zero integer, so \begin{align*} \left|\frac{p}{q} - \frac{p_n}{q_n}\right| &= \frac{|p q_n - p_n q|}{q q_n} \geq \frac{1}{q q_n}. \end{align*} Combining with the upper bound from the previous step, \begin{align*} \frac{1}{q q_n} &\leq \left|\frac{p}{q} - \frac{p_n}{q_n}\right| < \frac{1}{2 q^2}\left(1 + \frac{q}{q_n}\right) = \frac{q_n + q}{2 q^2 q_n}. \end{align*} Multiplying both ends by $2 q^2 q_n > 0$, \begin{align*} 2 q &< q_n + q, \end{align*} which simplifies to $q < q_n$. This contradicts our choice $q_n \leq q$. Hence the assumption $p/q \neq p_n/q_n$ is untenable, and $p/q = p_n/q_n$ for the chosen $n$. [guided] We have now two sharp inequalities for the same quantity $|p/q - p_n/q_n|$: one from above (using the triangle inequality together with the hypothesis $|\theta - p/q| < 1/(2 q^2)$ and the [Best Approximation by Convergents](/theorems/1761)), and one from below (using that the numerator $p q_n - p_n q$ of the reduced-form difference is a non-zero integer). Why the lower bound? Two distinct reduced-form fractions can differ by no less than $1/(qq_n)$: their difference is $(p q_n - p_n q)/(q q_n)$, and $p q_n - p_n q$ is a non-zero integer, hence $\geq 1$ in absolute value. Putting the bounds together, \begin{align*} \frac{1}{q q_n} &\leq \frac{q_n + q}{2 q^2 q_n}, \end{align*} i.e. $2 q \leq q_n + q$, i.e. $q \leq q_n$. Actually we get the strict inequality $q < q_n$ from the strict upper bound; either way this contradicts $q_n \leq q$. Where does the strictness come from? The hypothesis $|\theta - p/q| < 1/(2 q^2)$ is strict, so the chain inherits strict inequality. Therefore $p/q$ cannot be distinct from $p_n/q_n$, proving $p/q = p_n/q_n$ as required. This completes the proof of both directions of the Dirichlet-quality criterion. [/guided] [/step]