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  "body": "\n## Proof\n\nLet $x_n$ and $a_n$ be the complete quotients and partial quotients of the [[continued fraction expansion|0SJ7Z6VKWK]] of $x$: $x_0 = x$, $a_n = \\floor{x_n}$, $x_{n+1} = 1/(x_n - a_n)$.\nLet $p_n$ and $q_n$ be given by the recurrences in the definition of [[convergents|0SRWNR42BQ]].\n\n1. Since $x$ is irrational, its expansion is infinite by [[the termination theorem|0PMY0YQYXX]]. So $x_n$ and $a_n$ exist for all $n \\ge 0$.\n2. Each $x_n$ is irrational. Indeed $x_0 = x$, and if $x_{n+1}$ were rational then $x_n = a_n + 1/x_{n+1}$ would be rational.\n3. So $x_n$ is not an integer, and $x_n \u003e \\floor{x_n} = a_n$ for each $n$. For $n \\ge 1$ this gives $x_n \u003e a_n \\ge 1$.\n4. Claim A: $p_n q_{n-1} - p_{n-1} q_n = (-1)^{n+1}$ for $n \\ge -1$. We argue by [[induction|0ASJVEFA9J]] on $n$.\n5. For $n = -1$: $p_{-1} q_{-2} - p_{-2} q_{-1} = 1 \\cdot 1 - 0 \\cdot 0 = 1$.\n6. For $n \\ge 0$, the recurrences give $p_n q_{n-1} - p_{n-1} q_n = (a_n p_{n-1} + p_{n-2}) q_{n-1} - p_{n-1} (a_n q_{n-1} + q_{n-2})$. This equals $-(p_{n-1} q_{n-2} - p_{n-2} q_{n-1})$, so the sign flips at each step. This proves claim A.\n7. Claim B: $x = \\dfrac{x_{n+1} p_n + p_{n-1}}{x_{n+1} q_n + q_{n-1}}$ for $n \\ge -1$, with a positive denominator. We argue by induction on $n$.\n8. For $n = -1$ the right side is $\\dfrac{x_0 \\cdot 1 + 0}{x_0 \\cdot 0 + 1} = x_0 = x$.\n9. Claim C: $q_n \\ge 1$ for $n \\ge 0$, $q_n \\ge q_{n-1} \\ge 0$ for $n \\ge 0$, and $q_n \\ge n$ for $n \\ge 1$. Indeed $q_0 = a_0 \\cdot 0 + 1 = 1$ and $q_1 = a_1 q_0 + q_{-1} = a_1 \\ge 1$. For $n \\ge 2$, $q_n = a_n q_{n-1} + q_{n-2} \\ge q_{n-1} + 1$, using $a_n \\ge 1$ and $q_{n-2} \\ge 1$.\n10. In claim B the denominators are positive: $x_{n+1} \u003e 0$, $q_n \\ge 1$ and $q_{n-1} \\ge 0$ for $n \\ge 0$.\n11. Induction step for claim B. Assume it for $n - 1$, where $n \\ge 0$: $x = \\dfrac{x_n p_{n-1} + p_{n-2}}{x_n q_{n-1} + q_{n-2}}$. Substitute $x_n = a_n + 1/x_{n+1}$ and multiply the numerator and the denominator by $x_{n+1} \u003e 0$.\n12. The numerator becomes $x_{n+1}(a_n p_{n-1} + p_{n-2}) + p_{n-1} = x_{n+1} p_n + p_{n-1}$. In the same way the denominator becomes $x_{n+1} q_n + q_{n-1}$. This proves claim B.\n13. Let $n \\ge 0$. By claim B, $x - \\dfrac{p_n}{q_n} = \\dfrac{q_n p_{n-1} - p_n q_{n-1}}{q_n (x_{n+1} q_n + q_{n-1})}$, after putting both fractions over a common denominator.\n14. By claim A, the numerator in step 13 is $-(-1)^{n+1} = (-1)^n$. So $\\abs{x - \\dfrac{p_n}{q_n}} = \\dfrac{1}{q_n (x_{n+1} q_n + q_{n-1})}$.\n15. By step 3, $x_{n+1} \u003e a_{n+1}$. Since $q_n \\ge 1$, we get $x_{n+1} q_n + q_{n-1} \u003e a_{n+1} q_n + q_{n-1} = q_{n+1}$.\n16. By steps 14 and 15, $\\abs{x - p_n/q_n} \u003c \\dfrac{1}{q_n q_{n+1}}$.\n17. By claim C, $q_{n+1} \\ge q_n \\ge 1$, so $\\dfrac{1}{q_n q_{n+1}} \\le \\dfrac{1}{q_n^2}$.\n18. By claim C, $q_n \\ge n$ for $n \\ge 1$, so $\\abs{x - p_n/q_n} \u003c 1/n^2$ for $n \\ge 1$. This tends to $0$, so $p_n / q_n \\to x$ as $n \\to \\infty$.\n\n## Idea\n\nThe number $x$ is the convergent formula with the last partial quotient replaced by the exact complete quotient $x_{n+1}$.\nThe determinant identity $p_n q_{n-1} - p_{n-1} q_n = \\pm 1$ then makes the error exactly $1/(q_n(x_{n+1} q_n + q_{n-1}))$.\nSince $x_{n+1} \u003e a_{n+1}$, this is less than $1/(q_n q_{n+1})$, and the denominators grow.\n",
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