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  "body": "\n## Proof\n\nWe use the [[pigeonhole principle|0HGVCWQGY3]].\n\n1. For $k = 0, 1, \\dots, N$, put $u_k = k\\alpha - \\floor{k\\alpha}$, where $\\floor{\\cdot}$ is the [[floor function|0YVRNEDSVG]]. Then $0 \\le u_k \u003c 1$, and $u_0 = 0$.\n2. Put also $u_{N+1} = 1$. So we have $N + 2$ indices $0, 1, \\dots, N+1$, and each $u_k$ lies in $[0, 1]$.\n3. Split $[0, 1]$ into the $N + 1$ intervals $I_j = \\left[\\frac{j}{N+1}, \\frac{j+1}{N+1}\\right)$ for $0 \\le j \\le N - 1$ and $I_N = \\left[\\frac{N}{N+1}, 1\\right]$. Each point of $[0, 1]$ lies in exactly one $I_j$.\n4. The map that sends an index $k$ to the $j$ with $u_k \\in I_j$ goes from a set of $N + 2$ elements to a set of $N + 1$ elements. By the pigeonhole principle, there are indices $i \u003c k$ with $u_i$ and $u_k$ in the same interval.\n5. Each interval has length $\\frac{1}{N+1}$, so $\\abs{u_k - u_i} \\le \\frac{1}{N+1}$.\n6. Case $k \\le N$. Then $u_k - u_i = (k - i)\\alpha - (\\floor{k\\alpha} - \\floor{i\\alpha})$. Put $q = k - i$ and $p = \\floor{k\\alpha} - \\floor{i\\alpha}$. Then $1 \\le q \\le N$ and $\\abs{q\\alpha - p} \\le \\frac{1}{N+1}$.\n7. Case $k = N + 1$. Then $\\abs{1 - u_i} \\le \\frac{1}{N+1}$ with $0 \\le i \\le N$.\n8. In step 7, $i \\ne 0$: for $i = 0$ we would get $1 \\le \\frac{1}{N+1}$, which is false since $N \\ge 1$. So $1 \\le i \\le N$.\n9. In step 7, put $q = i$ and $p = \\floor{i\\alpha} + 1$. Then $q\\alpha - p = u_i - 1$, so $\\abs{q\\alpha - p} \\le \\frac{1}{N+1}$ and $1 \\le q \\le N$.\n10. In both cases we found integers $p$ and $q$ with $1 \\le q \\le N$ and $\\abs{q\\alpha - p} \\le \\frac{1}{N+1}$.\n\n## Idea\n\nThe fractional parts of $0, \\alpha, \\dots, N\\alpha$, together with the point $1$, are $N + 2$ points in $[0, 1]$.\nCut $[0,1]$ into $N + 1$ pieces; two points share a piece, so they are close.\nThe difference of two such multiples of $\\alpha$ is then close to an integer.\n",
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