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  "body": "\n## Proof\n\nWe check that the given map is injective and surjective.\n\n1. Each $n \\in \\N$ is either $2k$ or $2k + 1$ for exactly one $k \\in \\N$, by the [[division algorithm|0CHQKZFK5S]] with divisor $2$. So $f(2k) = k$ and $f(2k+1) = -(k+1)$ define a function $f \\colon \\N \\to \\Z$.\n2. For even $n = 2k$, $f(n) = k \\ge 0$. For odd $n = 2k + 1$, $f(n) = -(k + 1) \\le -1$.\n3. The map $f$ is [[injective|0PVG1CRAJN]]. Let $f(n) = f(n')$. By step 2, $n$ and $n'$ are both even or both odd.\n4. If $n = 2k$ and $n' = 2k'$, then $k = k'$, so $n = n'$. If $n = 2k + 1$ and $n' = 2k' + 1$, then $-(k+1) = -(k'+1)$, so $k = k'$ and $n = n'$.\n5. The map $f$ is [[surjective|0XKH881BZ7]]. Let $m \\in \\Z$. If $m \\ge 0$, then $f(2m) = m$.\n6. If $m \u003c 0$, put $k = -m - 1$. Then $k \\in \\N$, and $f(2k + 1) = -(k + 1) = m$.\n7. So $f$ is a [[bijection|0SNQEF7Z53]] from $\\N$ to $\\Z$, and $\\Z$ is [[countably infinite|0SEZ3JGZP7]].\n\n## Idea\n\nList the integers as $0, -1, 1, -2, 2, \\dots$. The even positions hold the nonnegative integers and the odd positions hold the negative ones. Each integer appears exactly once in this list.\n",
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