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  "body": "\n## Proof\n\nWe use the notation $x_n$, $a_n$ of the [[continued fraction expansion|0SJ7Z6VKWK]] of $x$: $x_0 = x$, $a_n = \\floor{x_n}$, and $x_{n+1} = 1/(x_n - a_n)$ when $x_n \\ne a_n$.\n\n1. Suppose the expansion is finite. Then the process stops at some $n$, so $x_n = a_n$ is an integer.\n2. For $k \u003c n$, the process did not stop at $k$, so $x_{k+1} = 1/(x_k - a_k)$. Solving gives $x_k = a_k + 1/x_{k+1}$.\n3. If $x_{k+1}$ is rational, then so is $x_k = a_k + 1/x_{k+1}$, since $a_k$ is an integer.\n4. By step 1 and step 3, applied for $k = n-1, n-2, \\dots, 0$ in turn, each $x_k$ is rational. In particular $x = x_0$ is rational.\n5. Conversely, let $x = r/s$ with integers $r$ and $s$, $s \u003e 0$. We show that the process stops.\n6. Claim: if the process has not stopped before step $k$, then $x_k = r_k / s_k$ with integers $r_k$ and $s_k$, and $1 \\le s_k \\le s - k$.\n7. For $k = 0$, take $r_0 = r$ and $s_0 = s$.\n8. Suppose the claim holds for $k$, and the process does not stop at $k$, so $x_k \\ne a_k$. Since $a_k$ is the [[floor|0YVRNEDSVG]] of $x_k$, we have $0 \u003c x_k - a_k \u003c 1$.\n9. Put $t_k = r_k - a_k s_k$, an integer. Then $x_k - a_k = t_k / s_k$, so $0 \u003c t_k \u003c s_k$ by step 8.\n10. Then $x_{k+1} = 1/(x_k - a_k) = s_k / t_k$. Put $r_{k+1} = s_k$ and $s_{k+1} = t_k$. We get $1 \\le s_{k+1} \\le s_k - 1 \\le s - (k+1)$. This proves the claim by [[induction|0ASJVEFA9J]] on $k$.\n11. If the process had not stopped before step $s$, the claim would give $1 \\le s_s \\le 0$, which is impossible. So the process stops at some $k \u003c s$, and the expansion of $x$ is finite.\n12. So the expansion of $x$ is finite if and only if $x$ is rational.\n\n## Idea\n\nFor a fraction $r/s$, each step of the algorithm replaces the denominator by a smaller positive integer, exactly as in the Euclidean algorithm.\nA decreasing sequence of positive integers must stop.\nConversely, a finite expansion can be folded back from its last term, and each step keeps the number rational.\n",
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