Lorena Beard

Answered

2022-07-07

Is there a proof that $\pi$ is an irrational number?

Answer & Explanation

Asdrubali2r

Expert

2022-07-08Added 14 answers

Let

then integrate by parts to show that for $n\ge 2$
${\alpha }^{2}{I}_{n}=2n\left(2n-1\right){I}_{n-1}-4n\left(n-1\right){I}_{n-2}.$
Use induction to show that for positive integer $n$ we have
${\alpha }^{2n+1}{I}_{n}\left(\alpha \right)=n!\left(P\left(\alpha \right)\mathrm{sin}\alpha +Q\left(\alpha \right)\mathrm{cos}\alpha \right),$
where $P\left(\alpha \right)$ and $Q\left(\alpha \right)$ are polynomials of degree less than $2n+1$ in $\alpha$ with integral coefficients.
Show that if $\pi /2=b/a,$, where $a$ and $b$ are integers, then
$\frac{{b}^{2n+1}{I}_{n}\left(\pi /2\right)}{n!}\phantom{\rule{1em}{0ex}}\left(1\right)$
would be an integer.
Note that

which results in contradiction since $\left(1\right)$ is supposed to be an integer but we can show that it is as small as one desires.

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