User:Marco Polo

Theorem: ${\displaystyle \pi }${\displaystyle \pi } is irrational

Proof: Suppose ${\displaystyle \pi ={\frac {p}{q}}}${\displaystyle \pi ={\frac {p}{q}}}, where ${\displaystyle p}${\displaystyle p} and ${\displaystyle q}${\displaystyle q} are integers. Consider the functions ${\displaystyle f_{n}(x)\!}${\displaystyle f_{n}(x)\!} defined on ${\displaystyle [0,\pi ]\!}${\displaystyle [0,\pi ]\!} by

${\displaystyle f_{n}(x)={\frac {q^{n}x^{n}(\pi -x)^{n}}{n!}}={\frac {x^{n}(p-qx)^{n}}{n!}}}${\displaystyle f_{n}(x)={\frac {q^{n}x^{n}(\pi -x)^{n}}{n!}}={\frac {x^{n}(p-qx)^{n}}{n!}}}

Clearly ${\displaystyle f_{n}(0)=f_{n}(\pi )=0\!}${\displaystyle f_{n}(0)=f_{n}(\pi )=0\!} for all ${\displaystyle n}${\displaystyle n}. Let ${\displaystyle f_{n}^{(m)}(x)}${\displaystyle f_{n}^{(m)}(x)} denote the ${\displaystyle m}${\displaystyle m}-th derivative of ${\displaystyle f_{n}(x)\!}${\displaystyle f_{n}(x)\!}. Note that

${\displaystyle f_{n}^{(m)}(0)=-f_{n}^{(m)}(\pi )=0{\mbox{ for }}m<=n{\mbox{ or for }}m>2n}$2n}"/>; otherwise some integer

${\displaystyle \operatorname {max} \ f_{n}(x)=f_{n}\left({\frac {\pi }{2}}\right)={\frac {q^{n}\left({\frac {\pi }{2}}\right)^{2n}}{n!}}}${\displaystyle \operatorname {max} \ f_{n}(x)=f_{n}\left({\frac {\pi }{2}}\right)={\frac {q^{n}\left({\frac {\pi }{2}}\right)^{2n}}{n!}}}

By repeatedly applying integration by parts, the definite integrals of the functions ${\displaystyle f_{n}(x)\sin x\!}${\displaystyle f_{n}(x)\sin x\!} can be seen to have integer values. But ${\displaystyle f_{n}(x)\sin x\!}${\displaystyle f_{n}(x)\sin x\!} are strictly positive, except for the two points 0 and ${\displaystyle \pi }${\displaystyle \pi }, and these functions are bounded above by ${\displaystyle {\frac {1}{\pi }}}${\displaystyle {\frac {1}{\pi }}} for all sufficiently large n. Thus for a large value of n, the definite integral of ${\displaystyle f_{n}(x)\sin x\!}${\displaystyle f_{n}(x)\sin x\!} is some value strictly between 0 and 1, a contradiction.