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  "body": "\n## Statement\n\nLet $n$ be a natural number with $n \\ge 2$.\nThen there is a [[prime number|0SN0MAEKYX]] $p$ such that $p \\mid n$.\n\n## Remarks\n\nOne proof takes the smallest divisor $d$ of $n$ with $d \u003e 1$. Each divisor $e$ of $d$ with $e \u003e 1$ also divides $n$, so $e = d$. Thus $d$ is prime.\nThis smallest divisor is the [[smallest prime factor|0S2SYHTWPY]] of $n$.\nThe result is also true for $n = 0$, since each prime divides $0$. It is false for $n = 1$.\n\n## Explanations\n\n### L1 human\n\nSplit $n$ into two smaller factors when you can, and repeat with one of the factors.\nThe numbers get smaller each time, so the process stops at a number that you cannot split. This number is a prime, and it divides $n$.\nFor example, $60 = 2 \\cdot 30$, so $2$ is a prime divisor of $60$.\n",
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  "name": "Existence of a prime divisor",
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