[proofplan] The key algebraic fact is that the matrix $\begin{pmatrix} p_n & p_{n+1} \\ q_n & q_{n+1} \end{pmatrix}$ has determinant $\pm 1$ and hence is unimodular: it is an invertible integer matrix. Any pair $(p, q) \in \mathbb{Z}^2$ can be written uniquely as $(u, v) \mapsto (u p_n + v p_{n+1}, u q_n + v q_{n+1})$ with integer $u, v$. The inequality $|q\theta - p| \geq |q_n \theta - p_n|$ is then analysed by cases on $v$. The case $v = 0$ is immediate; the case $v \neq 0$ uses a sign argument: the constraints $0 < q < q_{n+1}$ force $u$ and $v$ to have opposite signs, while the geometric fact that $\theta$ lies between consecutive convergents forces $q_n\theta - p_n$ and $q_{n+1}\theta - p_{n+1}$ to have opposite signs, so the two contributions align. [/proofplan] [step:Introduce the unimodular change of basis] The matrix \begin{align*} M_n &:= \begin{pmatrix} p_n & p_{n+1} \\ q_n & q_{n+1} \end{pmatrix} \in \mathbb{Z}^{2 \times 2} \end{align*} has determinant \begin{align*} \det M_n &= p_n q_{n+1} - p_{n+1} q_n = -(p_{n+1} q_n - p_n q_{n+1}) = -(-1)^n = (-1)^{n+1}, \end{align*} by the [Determinant Relations](/theorems/1759) applied at index $n + 1$. Since $\det M_n = \pm 1$, $M_n$ is invertible over $\mathbb{Z}$: its inverse \begin{align*} M_n^{-1} &= \frac{1}{\det M_n} \begin{pmatrix} q_{n+1} & -p_{n+1} \\ -q_n & p_n \end{pmatrix} \in \mathbb{Z}^{2 \times 2} \end{align*} has integer entries. Therefore, for any $(p, q) \in \mathbb{Z}^2$, the unique solution $(u, v) \in \mathbb{R}^2$ of the system \begin{align*} u p_n + v p_{n+1} &= p, & u q_n + v q_{n+1} &= q \end{align*} has $u, v \in \mathbb{Z}$. [/step] [step:Dispose of the case $v = 0$] With $(u, v)$ as in Step 1, suppose $v = 0$. Then $u q_n = q$ and $u p_n = p$. Since $q > 0$ and $q_n > 0$ (from the convergent recurrence, $q_n \geq 1$ for $n \geq 0$; see [Convergent Identity and Interpolation](/theorems/1758)), we have $u = q/q_n > 0$, hence $u \geq 1$ (as $u \in \mathbb{Z}$). Then \begin{align*} |q \theta - p| &= |u q_n \theta - u p_n| = |u| \cdot |q_n \theta - p_n| = u \cdot |q_n \theta - p_n| \geq |q_n \theta - p_n|, \end{align*} using $u \geq 1$. This gives the required bound. [/step] [step:Analyse the signs of $u$ and $v$ when $v \neq 0$] Suppose $v \neq 0$. We show $u$ and $v$ have opposite signs (and $u \neq 0$). From the second equation $u q_n + v q_{n+1} = q$, together with $q_n, q_{n+1} > 0$ and $0 < q < q_{n+1}$: If $v \geq 1$, then $v q_{n+1} \geq q_{n+1} > q$, so $u q_n = q - v q_{n+1} \leq q - q_{n+1} < 0$, giving $u < 0$. If $v \leq -1$, then $v q_{n+1} \leq -q_{n+1} < 0$, so $u q_n = q - v q_{n+1} \geq q + q_{n+1} > 0$, giving $u > 0$. In either case $u$ and $v$ have opposite signs, and in particular $u \neq 0$, hence $|u| \geq 1$. [guided] We need to extract sign information about $u$ and $v$ from the equation $u q_n + v q_{n+1} = q$, using the constraint $0 < q < q_{n+1}$. Assume $v \neq 0$. The plan is to examine the two sub-cases $v \geq 1$ and $v \leq -1$ separately (these exhaust the integer possibilities for $v \neq 0$). **Case $v \geq 1$.** Then $v q_{n+1} \geq q_{n+1}$, since $q_{n+1} > 0$. The assumption $q < q_{n+1}$ gives \begin{align*} u q_n &= q - v q_{n+1} \leq q - q_{n+1} < 0. \end{align*} Dividing by $q_n > 0$ gives $u < 0$. **Case $v \leq -1$.** Then $-v \geq 1$, so $-v q_{n+1} \geq q_{n+1} > 0$, whence $v q_{n+1} \leq -q_{n+1} < 0$. Then \begin{align*} u q_n &= q - v q_{n+1} \geq q + q_{n+1} > q_{n+1} > 0. \end{align*} Dividing by $q_n > 0$ gives $u > 0$. In both cases $u \neq 0$ and $u$, $v$ have opposite signs. This is the crucial observation: had $u$ and $v$ had the same sign, the two contributions $u(q_n \theta - p_n)$ and $v(q_{n+1}\theta - p_{n+1})$ below would cancel rather than reinforce, and the proof would collapse. [/guided] [/step] [step:Analyse the signs of $q_n \theta - p_n$ and $q_{n+1} \theta - p_{n+1}$] We show that $q_n \theta - p_n$ and $q_{n+1} \theta - p_{n+1}$ have opposite signs. From Step 2 of the proof of [Convergents Converge](/theorems/1760), $\theta$ lies strictly between $p_n/q_n$ and $p_{n+1}/q_{n+1}$. Since $q_n, q_{n+1} > 0$, the signs of $q_n \theta - p_n$ and $q_{n+1} \theta - p_{n+1}$ agree with the signs of $\theta - p_n/q_n$ and $\theta - p_{n+1}/q_{n+1}$ respectively. Strict betweenness of $\theta$ means $\theta - p_n/q_n$ and $\theta - p_{n+1}/q_{n+1}$ have opposite signs. Therefore $q_n \theta - p_n$ and $q_{n+1} \theta - p_{n+1}$ have opposite signs, and both are nonzero (as $\theta$ is irrational). [/step] [step:Combine the sign information to establish the lower bound when $v \neq 0$] Starting from the equations in Step 1, \begin{align*} q \theta - p &= (u q_n + v q_{n+1}) \theta - (u p_n + v p_{n+1}) \\ &= u (q_n \theta - p_n) + v (q_{n+1} \theta - p_{n+1}). \end{align*} From Step 3, $u$ and $v$ have opposite signs. From Step 4, $q_n \theta - p_n$ and $q_{n+1} \theta - p_{n+1}$ have opposite signs. Therefore the two products \begin{align*} u (q_n \theta - p_n) \quad \text{and} \quad v (q_{n+1} \theta - p_{n+1}) \end{align*} have the **same** sign: the sign flip of $u$ relative to $v$ is compensated by the sign flip of $q_n \theta - p_n$ relative to $q_{n+1} \theta - p_{n+1}$. When two real numbers $A$, $B$ have the same sign (both $\geq 0$ or both $\leq 0$), $|A + B| = |A| + |B|$. Therefore \begin{align*} |q\theta - p| &= |u (q_n \theta - p_n) + v (q_{n+1} \theta - p_{n+1})| = |u| \cdot |q_n \theta - p_n| + |v| \cdot |q_{n+1} \theta - p_{n+1}|. \end{align*} Since $|u| \geq 1$ and $|v| \geq 1$ and both terms on the right are non-negative, \begin{align*} |q\theta - p| &\geq |u| \cdot |q_n \theta - p_n| \geq |q_n \theta - p_n|. \end{align*} [/step] [step:Conclude] Steps 2 and 5 together handle all cases: either $v = 0$ (Step 2) or $v \neq 0$ (Step 5). In both cases, $|q\theta - p| \geq |q_n\theta - p_n|$, completing the proof. [/step]