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clickbait: 'The number pi is famously not rational, in spite of joking attempts at legislation to fix its value at 3 or 22/7.',
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text: 'The number [49r $\\pi$] is not [4zq rational].\n\n# Proof\n\nFor any fixed real number $q$, and any [-45h] $n$, let $$A_n = \\frac{q^n}{n!} \\int_0^{\\pi} [x (\\pi - x)]^n \\sin(x) dx$$\nwhere $n!$ is the [-5bv] of $n$, $\\int$ is the [-definite_integral], and $\\sin$ is the [-sin_function].\n\n## Preparatory work\n\nExercise: $A_n = (4n-2) q A_{n-1} - (q \\pi)^2 A_{n-2}$.\n%%hidden(Show solution):\nWe use [-integration_by_parts].\n\n[todo: show this]\n%%\n\nNow, $$A_0 = \\int_0^{\\pi} \\sin(x) dx = 2$$\nso $A_0$ is an integer.\n\nAlso $$A_1 = q \\int_0^{\\pi} x (\\pi-x) \\sin(x) dx$$ which by a simple calculation is $4q$.\n%%hidden(Show calculation):\nExpand the integrand and then integrate by parts repeatedly:\n$$\\frac{A_1}{q} = \\int_0^{\\pi} x (\\pi-x) \\sin(x) dx = \\pi \\int_0^{\\pi} x \\sin(x) dx - \\int_0^{\\pi} x^2 \\sin(x) dx$$\n\nThe first integral term is $$[-x \\cos(x)]_0^{\\pi} + \\int_0^{\\pi} \\cos(x) dx = \\pi$$\n\nThe second integral term is $$[-x^2 \\cos(x)]_{0}^{\\pi} + \\int_0^{\\pi} 2x \\cos(x) dx$$\nwhich is $$\\pi^2 + 2 \\left( [x \\sin(x)]_0^{\\pi} - \\int_0^{\\pi} \\sin(x) dx \\right)$$\nwhich is $$\\pi^2 -4$$\n\nTherefore $$\\frac{A_1}{q} = \\pi^2 - (\\pi^2 - 4) = 4$$\n%%\n\nTherefore, if $q$ and $q \\pi$ are integers, then so is $A_n$ [5fz inductively], because $(4n-2) q A_{n-1}$ is an integer and $(q \\pi)^2 A_{n-2}$ is an integer.\n\nBut also $A_n \\to 0$ as $n \\to \\infty$, because $\\int_0^{\\pi} [x (\\pi-x)]^n \\sin(x) dx$ is in modulus at most $$\\pi \\times \\max_{0 \\leq x \\leq \\pi} [x (\\pi-x)]^n \\sin(x) \\leq \\pi \\times \\max_{0 \\leq x \\leq \\pi} [x (\\pi-x)]^n = \\pi \\times \\left[\\frac{\\pi^2}{4}\\right]^n$$\nand hence $$|A_n| \\leq \\frac{1}{n!} \\left[\\frac{\\pi^2 q}{4}\\right]^n$$\n\nFor $n$ larger than $\\frac{\\pi^2 q}{4}$, this expression is getting smaller with $n$, and moreover it gets smaller faster and faster as $n$ increases; so its limit is $0$.\n%%hidden(Formal treatment):\nWe claim that $\\frac{r^n}{n!} \\to 0$ as $n \\to \\infty$, for any $r > 0$.\n\nIndeed, we have $$\\frac{r^{n+1}/(n+1)!}{r^n/n!} = \\frac{r}{n+1}$$\nwhich, for $n > 2r-1$, is less than $\\frac{1}{2}$.\nTherefore the ratio between successive terms is less than $\\frac{1}{2}$ for sufficiently large $n$, and so the sequence must shrink at least geometrically to $0$.\n%%\n\n## Conclusion\n\nSuppose (for [46z contradiction]) that $\\pi$ is rational; then it is $\\frac{p}{q}$ for some integers $p, q$.\n\nNow $q \\pi$ is an integer (indeed, it is $p$), and $q$ is certainly an integer, so by what we showed above, $A_n$ is an integer for all $n$.\n\nBut $A_n \\to 0$ as $n \\to \\infty$, so there is some $N$ for which $|A_n| < \\frac{1}{2}$ for all $n > N$; hence for all sufficiently large $n$, $A_n$ is $0$.\nWe already know that $A_0 = 2$ and $A_1 = 4q$, neither of which is $0$; so let $N$ be the first integer such that $A_n = 0$ for all $n \\geq N$, and we can already note that $N > 1$.\n\nThen $$0 = A_{N+1} = (4N-2) q A_N - (q \\pi)^2 A_{N-1} = - (q \\pi)^2 A_{N-1}$$\nwhence $q=0$ or $\\pi = 0$ or $A_{N-1} = 0$.\n\nCertainly $q \\not = 0$ because $q$ is the denominator of a fraction; and $\\pi \\not = 0$ by whatever definition of $\\pi$ we care to use.\nBut also $A_{N-1}$ is not $0$ because then $N-1$ would be an integer $m$ such that $A_n = 0$ for all $n \\geq m$, and that contradicts the definition of $N$ as the *least* such integer.\n\nWe have obtained the required contradiction; so it must be the case that $\\pi$ is irrational.',
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