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  "body": "\n## Proof\n\nWe argue by contradiction. Assume that $\\pi$ is not [[irrational|080T2TXVQV]].\n\n1. By the [[definition of $\\pi$|0C1VYVNV2N]], $\\pi/2$ is the only zero of $\\cos$ in $[0, 2]$. From the [[defining series|0WB57D9488]], $\\cos 0 = 1$, so $\\pi \u003e 0$.\n2. So $\\pi = a/b$ for some positive integers $a$ and $b$. In particular $b \\pi = a$.\n3. From the defining series, $\\sin 0 = 0$, and $\\sin$ is [[odd|0XE690QF65]] because its series has only odd powers. By [[this result|01A8JGR4JM]], $\\sin \\pi = 0$ and $\\cos \\pi = -1$.\n4. By the [[derivatives of sine|00R4MWW532]] and [[of cosine|0V2DV29EGT]], both are differentiable, hence [[continuous|07ZMX4ANJS]]. If $\\cos x \\le 0$ for some $0 \u003c x \u003c \\pi/2$, the [[intermediate value theorem|00KAFBR32E]] gives a zero of $\\cos$ in $[0, x]$.\n5. That zero would be at most $x \u003c \\pi/2 \\le 2$, a second zero of $\\cos$ in $[0, 2]$. So $\\cos \u003e 0$ on $[0, \\pi/2)$.\n6. By [[this result|07J5X2T1ZX]], $\\sin$ is strictly increasing on $[0, \\pi/2]$. So $\\sin x \u003e 0$ for $0 \u003c x \\le \\pi/2$, and $\\sin x \\ge \\sin(\\pi/4) \u003e 0$ for $\\pi/4 \\le x \\le \\pi/2$.\n7. By the [[addition formula|028XEQCE88]] and oddness, $\\sin(\\pi - u) = \\sin \\pi \\cos(-u) + \\cos \\pi \\sin(-u) = \\sin u$. With step 6, $\\sin x \u003e 0$ for $0 \u003c x \u003c \\pi$.\n8. So $\\sin x \\ge 0$ on $[0, \\pi]$. Also $\\sin x \\le 1$, since $\\sin^2 x = 1 - \\cos^2 x \\le 1$ by the [[Pythagorean identity|03H7WZ4T3K]].\n9. Now fix a positive integer $n$, and let $f(x) = \\frac{x^n (a - b x)^n}{n!}$. By the [[binomial theorem|0EHWPWGHB2]], $n! \\, f(x) = \\sum_{k=n}^{2n} c_k x^k$ with integers $c_k$.\n10. By the [[derivative of a polynomial|0MH1WK89PP]], applied $j$ times, the [[$j$-th derivative|084NGSTWM7]] is $f^{(j)}(x) = \\sum_{k \\ge j} \\frac{c_k}{n!} \\frac{k!}{(k-j)!} x^{k-j}$. In particular $f^{(j)} = 0$ for $j \u003e 2n$.\n11. At $x = 0$ only the term $k = j$ survives. So $f^{(j)}(0) = 0$ if $j \u003c n$ or $j \u003e 2n$, and $f^{(j)}(0) = c_j \\frac{j!}{n!}$ if $n \\le j \\le 2n$.\n12. For $j \\ge n$, the number $\\frac{j!}{n!} = (n+1)(n+2) \\cdots j$ is an integer. So $f^{(j)}(0)$ is an integer for every $j \\ge 0$.\n13. By step 2, $a - b(\\pi - x) = b x$ and $\\pi - x = \\frac{a - b x}{b}$. So $f(\\pi - x) = \\frac{(a - b x)^n (b x)^n}{b^n \\, n!} = f(x)$ for all real $x$.\n14. Differentiating $j$ times with the [[chain rule|0RXX2PT1MQ]] gives $(-1)^j f^{(j)}(\\pi - x) = f^{(j)}(x)$. At $x = 0$, $f^{(j)}(\\pi) = (-1)^j f^{(j)}(0)$, an integer.\n15. Let $F = \\sum_{i=0}^{n} (-1)^i f^{(2i)}$. By steps 12 and 14, $F(0)$ and $F(\\pi)$ are integers.\n16. By [[linearity of the derivative|0M0TKFRX2A]], $F'' = \\sum_{i=0}^{n} (-1)^i f^{(2i+2)} = -\\sum_{i=1}^{n+1} (-1)^i f^{(2i)}$. Since $f^{(2n+2)} = 0$, this gives $F + F'' = f^{(0)} = f$.\n17. Let $G(x) = F'(x) \\sin x - F(x) \\cos x$. By the [[product rule|0V9A43H7WV]], $G' = F'' \\sin + F' \\cos - F' \\cos + F \\sin = (F'' + F) \\sin = f \\sin$.\n18. Let $g(x) = f(x) \\sin x$. It is [[continuous|0VGMHCF4SC]], since [[polynomials are continuous|0C5TP6RYSV]]. So $g$ is [[Riemann integrable|06SESV7GQY]] on every closed interval.\n19. By the [[fundamental theorem of calculus|0NHGY0N0JB]], $J = \\int_0^{\\pi} g(x) \\, dx = G(\\pi) - G(0)$.\n20. By steps 1 and 3, $G(\\pi) = F'(\\pi) \\cdot 0 - F(\\pi) \\cdot (-1) = F(\\pi)$ and $G(0) = -F(0)$. So $J = F(\\pi) + F(0)$ is an integer.\n21. Let $0 \\le x \\le \\pi$. Then $0 \\le x^n \\le \\pi^n$, and $0 \\le a - b x = b(\\pi - x) \\le a$. So $0 \\le f(x) \\le \\frac{\\pi^n a^n}{n!}$.\n22. By step 8, $0 \\le g(x) \\le \\frac{(\\pi a)^n}{n!}$ on $[0, \\pi]$. A constant $C$ has $\\int_\\alpha^\\beta C \\, dx = C(\\beta - \\alpha)$, by the fundamental theorem with $G(x) = C x$.\n23. By [[monotonicity of the integral|0MBWQKM8QR]], $J \\le \\pi \\frac{(\\pi a)^n}{n!}$.\n24. Let $\\pi/4 \\le x \\le \\pi/2$. Then $x \\ge \\pi/4$, $a - b x = b(\\pi - x) \\ge b \\pi / 2$, and $\\sin x \\ge \\sin(\\pi/4) \u003e 0$ by step 6.\n25. So $g(x) \\ge m$ on $[\\pi/4, \\pi/2]$, where $m = \\frac{(\\pi/4)^n (b \\pi / 2)^n}{n!} \\sin(\\pi/4) \u003e 0$.\n26. By [[additivity over intervals|05RCDKHZM6]], $J$ is the sum of the integrals of $g$ over $[0, \\pi/4]$, $[\\pi/4, \\pi/2]$ and $[\\pi/2, \\pi]$.\n27. Since $g \\ge 0$, monotonicity makes the first and last integrals at least $0$, and the middle one at least $m \\pi / 4 \u003e 0$. So $J \u003e 0$, and $J \\ge 1$ as an integer.\n28. Let $C = \\pi a$ and $t_n = \\frac{C^n}{n!} \u003e 0$. Then $\\frac{t_{n+1}}{t_n} = \\frac{C}{n+1} \\to 0$. By the [[ratio test|0QNY6ZBXPN]], $\\sum t_n$ converges.\n29. By the [[term test|0VVY4H92G2]], $t_n \\to 0$. So there is a positive integer $n$ with $\\pi t_n \u003c 1$.\n30. For this $n$, steps 23 and 27 give $1 \\le J \\le \\pi t_n \u003c 1$, a contradiction. So $\\pi$ is irrational.\n\n## Idea\n\nIf $\\pi = a/b$, Niven builds a polynomial $f$ whose derivatives at $0$ and $\\pi$ are all integers. Two integrations by parts, packed into the function $G$, turn $\\int_0^\\pi f(x) \\sin x \\, dx$ into an integer. This integral is positive but tends to $0$ as $n$ grows, since $n!$ beats $(\\pi a)^n$, and no integer lies strictly between $0$ and $1$.\n",
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