# Pi is Irrational/Proof 2

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## Proof

Aiming for a contradiction, suppose $\pi$ is rational.

Then $\pi = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$.

Now let:

$\ds A_n = \frac {q^n} {n!} \int_0^\pi \paren {x \paren {\pi - x} }^n \sin x \rd x$

Integration by Parts twice gives:

 $\ds A_n$ $=$ $\ds \frac {q^n} {n!} \int_0^\pi n \paren {x \paren {\pi - x} }^{n - 1} \paren {\pi - 2 x} \cos x \rd x$ $\ds$ $=$ $\ds \frac {q^n} {n!} \int_0^\pi 2 n \paren {x \paren {\pi - x} }^{n - 1} \sin x - n \paren {n - 1} \paren {x \paren {\pi - x} }^{n - 2} \paren {\pi - 2 x}^2 \sin x \rd x$

Writing:

$\paren {\pi - 2 x}^2 = \pi^2 - 4 x \paren {\pi - x}$

gives the result:

$(1): \quad A_n = \paren {4 n - 2} q A_{n - 1} - \paren {q \pi}^2 A_{n - 2}$

We will deduce that $A_n$ is an integer for all $n$.

First confirm by direct integration that $A_0$ and $A_1$ are integers:

$\ds A_0 = \int_0^\pi \sin x \rd x = 2$
$\ds A_1 = q \int_0^\pi x \paren {\pi - x} \sin x \rd x = 4 q$

Suppose $A_{k - 2}$ and $A_{k - 1}$ are integers.

Then by $(1)$ and the assumption that $q$ and $q \pi$ are integers, $A_k$ is also an integer.

So $A_n$ is an integer for all $n$ by Second Principle of Mathematical Induction.

For $x \in \closedint 0 \pi$, we have:

$0 \le \sin x \le 1$

and:

$0 \le x \paren {\pi - x} \le \pi^2 / 4$

hence:

$0 < A_n < \pi \dfrac {\paren {q \pi^2 / 4}^n} {n!}$

From Power over Factorial:

$\ds \lim_{n \mathop \to \infty} \frac {\paren {q \pi^2 / 4}^n} {n!} = 0$

It follows from the Squeeze Theorem that:

$\ds \lim_{n \mathop \to \infty} A_n = 0$

Hence for sufficiently large $n$, $A_n$ is strictly between $0$ and $1$.

This contradicts the supposition that $A_n$ is an integer.

It follows that $\pi$ is irrational.

$\blacksquare$