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  "body": "\n## Proof\n\nWe follow the argument of Hardy and Wright.\nFor integers $c_0$ and $c_1, \\dots, c_m \\ge 1$, let $P_k$ and $Q_k$ ($-2 \\le k \\le m$) be given by the recurrences in the definition of [[convergents|0SRWNR42BQ]] with terms $c_k$.\n\n1. Fact 1: $P_k Q_{k-1} - P_{k-1} Q_k = (-1)^{k+1}$ for $-1 \\le k \\le m$. For $k = -1$ the left side is $1$. For $k \\ge 0$, inserting the recurrences shows that the left side is minus its value at $k - 1$.\n2. Fact 2: $Q_k \\ge Q_{k-1} \\ge 0$ and $Q_k \\ge 1$ for $0 \\le k \\le m$. Indeed $Q_{-1} = 0$, $Q_0 = 1$, and $Q_k = c_k Q_{k-1} + Q_{k-2} \\ge Q_{k-1}$ for $k \\ge 1$, by [[induction|0ASJVEFA9J]].\n3. Fact 3: let $y_{m+1} \u003e 0$ be real and put $y_k = c_k + 1/y_{k+1}$ for $k = m, m-1, \\dots, 0$. Then $y_0 = \\dfrac{y_{k+1} P_k + P_{k-1}}{y_{k+1} Q_k + Q_{k-1}}$ for $-1 \\le k \\le m$.\n4. Proof of fact 3, by induction on $k$. For $k = -1$ the right side is $y_0$. For the step, substitute $y_k = c_k + 1/y_{k+1}$ into the formula for $k - 1$ and multiply above and below by $y_{k+1}$. The recurrences turn the result into the formula for $k$.\n5. Fact 4: in fact 3, assume $y_{m+1} \u003e 1$. Then the [[continued fraction expansion|0SJ7Z6VKWK]] of $y_0$ has at least $m + 2$ terms, and its first $m + 1$ terms are $c_0, \\dots, c_m$.\n6. Proof of fact 4. First, $y_k \u003e 1$ for $1 \\le k \\le m + 1$: this holds for $k = m + 1$, and $y_k = c_k + 1/y_{k+1} \u003e c_k \\ge 1$ for $1 \\le k \\le m$.\n7. For $0 \\le k \\le m$ we have $0 \u003c 1/y_{k+1} \u003c 1$. So the [[floor|0YVRNEDSVG]] of $y_k = c_k + 1/y_{k+1}$ is $c_k$, and $y_k \\ne c_k$.\n8. Let $x_k$, $a_k$ be the complete and partial quotients of $y_0$. By induction on $k$, step 7 gives $x_k = y_k$ and $a_k = c_k$ for $0 \\le k \\le m$, and the process does not stop before step $m + 1$. This proves fact 4.\n9. Fact 5: by fact 4, the $m$-th convergent of $y_0$ is $P_m / Q_m$, because $P_m$ and $Q_m$ depend only on $c_0, \\dots, c_m$.\n10. Now the proof. Write $p/q = p'/q'$ in lowest terms with $q' \\ge 1$. Then $q' \\le q$, so $\\abs{\\alpha - p'/q'} \u003c \\frac{1}{2q^2} \\le \\frac{1}{2q'^2}$. So we may assume that $\\gcd(p, q) = 1$.\n11. The [[expansion|0SJ7Z6VKWK]] of $p/q$ is finite by [[the termination theorem|0PMY0YQYXX]]. Say it is $[a_0; a_1, \\dots, a_n]$, a [[simple continued fraction|06XF7BCFFJ]], and the process stops with $x_n = a_n$.\n12. By fact 3 with $m = n - 1$ and $y_n = x_n = a_n$, we get $p/q = \\dfrac{a_n p_{n-1} + p_{n-2}}{a_n q_{n-1} + q_{n-2}} = p_n / q_n$. For $n = 0$ this reads $p/q = a_0 = p_0/q_0$.\n13. If $n \\ge 1$, then $a_n = x_n = 1/(x_{n-1} - a_{n-1}) \u003e 1$, so $a_n \\ge 2$.\n14. Consider also the list $a_0, \\dots, a_{n-1}, a_n - 1, 1$ of length $n + 2$. Its terms after the first are positive integers by step 13. Its numerators $p'_k$ satisfy $p'_k = p_k$ for $k \\le n - 1$, $p'_n = p_n - p_{n-1}$ and $p'_{n+1} = p'_n + p_{n-1} = p_n$. In the same way $q'_{n+1} = q_n$.\n15. So we have two lists $c_0, \\dots, c_m$, with $m = n$ and $m = n + 1$, whose terms after the first are positive. For both, $P_m / Q_m = p_n / q_n = p/q$.\n16. By fact 1, $\\gcd(P_m, Q_m) = 1$, and $Q_m \\ge 1$ by fact 2. Since also $\\gcd(p, q) = 1$ and $q \\ge 1$, we get $P_m = p$ and $Q_m = q$ for both lists.\n17. If $\\alpha = p/q$, then by step 11 the expansion of $\\alpha$ is $[a_0; \\dots, a_n]$, and its $n$-th convergent is $p_n / q_n = p/q$. We are done in this case.\n18. Otherwise write $\\alpha - p/q = \\sigma \\theta / q^2$ with $\\sigma \\in \\set{1, -1}$ and $0 \u003c \\theta \u003c \\frac{1}{2}$. This is possible by the hypothesis.\n19. The two values of $m$ in step 15 differ by one. Choose the list with $(-1)^m = \\sigma$.\n20. Put $\\omega = \\dfrac{1}{\\theta} - \\dfrac{Q_{m-1}}{Q_m}$. By fact 2, $0 \\le Q_{m-1}/Q_m \\le 1$. Since $1/\\theta \u003e 2$, we get $\\omega \u003e 1$.\n21. Put $\\beta = \\dfrac{\\omega P_m + P_{m-1}}{\\omega Q_m + Q_{m-1}}$. By the choice of $\\omega$, $\\omega Q_m + Q_{m-1} = Q_m / \\theta \u003e 0$.\n22. Over a common denominator, $\\beta - \\dfrac{P_m}{Q_m} = \\dfrac{Q_m P_{m-1} - P_m Q_{m-1}}{Q_m (\\omega Q_m + Q_{m-1})}$. By fact 1 the numerator is $(-1)^m = \\sigma$.\n23. With step 21 and $Q_m = q$, this gives $\\beta - p/q = \\sigma \\theta / q^2 = \\alpha - p/q$. So $\\beta = \\alpha$.\n24. By fact 3 with $y_{m+1} = \\omega$, the number $y_0$ of fact 3 equals $\\beta = \\alpha$.\n25. Since $\\omega \u003e 1$, fact 5 shows that the $m$-th convergent of $\\alpha$ is $P_m / Q_m = p/q$.\n\n## Idea\n\nWrite $p/q$ as a finite continued fraction; there are two ways, of lengths differing by one, so the parity of the length can be chosen.\nThen $\\alpha$ is that finite continued fraction with one extra real tail $\\omega$.\nThe bound $\\frac{1}{2q^2}$ makes $\\omega \u003e 1$, and a tail bigger than $1$ does not change the earlier partial quotients.\nSo $p/q$ appears among the convergents of $\\alpha$.\n",
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