= Proof that π is irrational
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= Irrationality of pi
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<Ivan Niven>'s 1947 <proof by contradiction> constructs an <integer> strictly between zero and one.<A simple proof that pi is irrational by Ivan Niven>[Page 509]{ref} Suppose $\pi=a/b$ for <positive integers> $a,b$. For a <positive integer> $n$, put
$$
f_n(x)=\frac{x^n(a-bx)^n}{n!},\qquad F_n=f_n-f_n''+f_n^{(4)}-\cdots+(-1)^nf_n^{(2n)}.
$$
Every <derivative> of $f_n$ at zero is an <integer>: its coefficients have denominator $n!$, its lowest <polynomial degree> is $n$, and differentiation contributes the divisible <factorial>. The symmetry $f_n(\pi-x)=f_n(x)$ gives the same assertion at $\pi$.
Cancellation gives $F_n+F_n''=f_n$, hence
$$
I_n=\int_0^\pi f_n(x)\sin x\,dx=\left[F_n'(x)\sin x-F_n(x)\cos x\right]_0^\pi=F_n(\pi)+F_n(0)\in\mathbb Z.
$$
The integrand is positive in $(0,\pi)$, but $x(a-bx)=bx(\pi-x)\leq b\pi^2/4$, so
$$
0<I_n\leq\frac{\pi(b\pi^2/4)^n}{n!}\longrightarrow0.
$$
For large $n$ this is impossible for an <integer>. Therefore <Pi> is an <irrational number>.
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