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  "body": "\n## Proof\n\nWe argue by contradiction. We build an integer that lies strictly between $0$ and $1$.\n\n1. By the [[series for $e$|0G4D3ZKTP8]], $e = \\sum_{n=0}^{\\infty} \\frac{1}{n!}$, where [[$e$|0MVCFMW7S4]] is $\\exp(1)$.\n2. Assume that $e$ is not [[irrational|080T2TXVQV]]. Then $e = a/b$ for an integer $a$ and an integer $b \\ge 1$.\n3. Let $x = b! \\left( e - \\sum_{n=0}^{b} \\frac{1}{n!} \\right)$.\n4. $x$ is an integer. First, $b! \\, e = b! \\, a / b = a \\cdot (b-1)!$, an integer. Second, for $0 \\le n \\le b$, the definition of the [[factorial|0WFCGVMRRC]] gives $\\frac{b!}{n!} = (n+1)(n+2) \\cdots b$, an integer.\n5. For $N \u003e b$, $\\sum_{n=0}^{N} \\frac{1}{n!} - \\sum_{n=0}^{b} \\frac{1}{n!} = \\sum_{n=b+1}^{N} \\frac{1}{n!}$. As $N \\to \\infty$, the left side tends to $e - \\sum_{n=0}^{b} \\frac{1}{n!}$, by the [[limit of a sum|0RAM4E07NR]].\n6. Multiply by $b!$ and use the algebra of limits. Then step 5 gives $x = \\sum_{k=1}^{\\infty} t_k$, where $t_k = \\frac{b!}{(b+k)!}$.\n7. All $t_k$ are positive. By the [[result on series of nonnegative terms|0MX1HZBBXJ]], $x$ is the supremum of the partial sums, so $x \\ge t_1 \u003e 0$.\n8. Let $q = \\frac{1}{b+1}$, so $0 \u003c q \\le \\frac12$. For $k \\ge 1$, $t_k = \\frac{1}{(b+1)(b+2) \\cdots (b+k)} \\le q^k$, since each factor is at least $b + 1$.\n9. For $k \\ge 2$, the factor $b + 2$ exceeds $b + 1$. So $t_2 \u003c q^2$.\n10. By the [[geometric series|01X5CVV1Q5]] with first term $q$ and ratio $q$, $\\sum_{k=1}^{\\infty} q^k = \\frac{q}{1 - q} = \\frac{1}{b}$.\n11. Let $d_k = q^k - t_k \\ge 0$. The series $\\sum d_k$ converges, with sum $\\frac{1}{b} - x$, by step 10 and the limit of a sum.\n12. By the result on series of nonnegative terms, $\\sum d_k \\ge d_1 + d_2 \\ge d_2 \u003e 0$, using step 9. So $x \u003c \\frac{1}{b} \\le 1$.\n13. By steps 7 and 12, $x$ is an integer with $0 \u003c x \u003c 1$. No such integer exists. So $e$ is irrational.\n\n## Idea\n\nIf $e = a/b$, multiplying $e$ by $b!$ makes it an integer, and so does multiplying the first $b + 1$ terms of its series. So the scaled tail of the series is an integer. But the factorials make the tail smaller than a geometric series with sum $1/b$, and it is positive, so it cannot be an integer.\n",
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