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  "body": "\n## Proof\n\nWe find injections in both directions and apply the Schröder–Bernstein theorem.\n\n1. Let $x \\in \\Q$ be a [[rational number|0NCJ5WXN6Z]]. Then $x = p/q$ for some $p \\in \\Z$ and some integer $q \\ge 1$.\n2. So the set $D_x = \\{q \\in \\N : q \\ge 1,\\ qx \\in \\Z\\}$ is nonempty. By the [[well-ordering principle|01BX15HSRR]], it has a least element $q_x$.\n3. Define $r \\colon \\Q \\to \\Z \\times \\N$ by $r(x) = (q_x x,\\ q_x)$. Its first entry is an integer by the choice of $q_x$.\n4. The map $r$ is [[injective|0PVG1CRAJN]]. If $r(x) = r(y)$, then $q_x = q_y$ and $q_x x = q_y y$. Dividing by $q_x \\ge 1$ gives $x = y$.\n5. By [[the integers are countable|066RJJVWH9]], there is a bijection $f \\colon \\N \\to \\Z$. Its inverse $f^{-1} \\colon \\Z \\to \\N$ is a [[bijection|01VG2A0BHW]], so it is injective. Hence $\\Z$ is [[countable|0ZG61TFBES]].\n6. The set $\\N$ is countable, through the identity map. By the [[product of countable sets is countable|08G4XH86JC]], there is an injection $s \\colon \\Z \\times \\N \\to \\N$.\n7. The composite $s \\circ r \\colon \\Q \\to \\N$ is [[injective|01FPXMTBX1]].\n8. The inclusion $\\N \\to \\Q$, $n \\mapsto n$, is injective.\n9. By the [[Schröder–Bernstein theorem|0A9V6RKQG6]], steps 7 and 8 give a bijection $\\N \\to \\Q$. So $\\Q$ is [[countably infinite|0SEZ3JGZP7]].\n\n## Idea\n\nCode each rational by its smallest positive denominator and the matching numerator. This gives an injection into the countable set $\\Z \\times \\N$, so $\\Q$ is countable. Since $\\Q$ also contains a copy of $\\N$, the Schröder–Bernstein theorem upgrades this to a bijection with $\\N$.\n",
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