Pi is Irrational/Proof 2

Theorem

Pi ($\pi$) is irrational.

Proof

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

We establish a lemma:

Lemma

Let $n \in \Z_{> 0}$ be a positive integer.

Let it be supposed that $\pi$ is irrational, so that:

$\pi = \dfrac p q$

where $p$ and $q$ are integers and $q \ne 0$.

Let $A_n$ be defined as:

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

Then:

$A_n = \paren {4 n - 2} q A_{n - 1} - p^2 A_{n - 2}$

is a reduction formula for $A_n$.

$\Box$

We will use the definition of $A_n$ from the lemma.

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

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

\(\ds A_0\)

\(=\)

\(\ds \int_0^\pi \sin x \rd x\)

\(\ds \)

\(=\)

\(\ds \bigintlimits {- \cos x} 0 \pi\)

Primitive of Sine Function

\(\ds \)

\(=\)

\(\ds 2\)

Cosine of $\pi$, Cosine of $0 \degrees$

Although this article appears correct, it's inelegant. There has to be a better way of doing it. In particular: We could set up the long overdue "Area under Arc of Sine Function" at this point, and then link to it. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Improve}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

and

\(\ds A_1\)

\(=\)

\(\ds q \int_0^\pi x \paren {\pi - x} \sin x \rd x\)

\(\ds \)

\(=\)

\(\ds q \paren {\pi \int_0^\pi x \sin x \rd x - \int_0^\pi x^2 \sin x \rd x }\)

Linear Combination of Integrals

\(\ds \)

\(=\)

\(\ds q \paren {\pi \bigintlimits {- x \cos x + \sin x} 0 \pi - \bigintlimits {- x^2 \cos x + 2 x \sin x + 2 \cos x} 0 \pi }\)

Primitive of $x^n \sin a x$

\(\ds \)

\(=\)

\(\ds q \paren {\pi^2 - \pi^2 + 2 + 2}\)

Cosine of $\pi$, Cosine of $0 \degrees$, Sine of $\pi$, Sine of $0 \degrees$

\(\ds \)

\(=\)

\(\ds 4 q\)

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

By our lemma:

$A_k = \paren {4 n - 2} q A_{k - 1} - p^2 A_{k - 2}$

As $n$, $p$ and $q$ are all integers by hypothesis, $A_k$ is also an integer.

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

$\Box$

Let $x \in \closedint 0 \pi$.

From Shape of Sine Function and Real Sine Function is Bounded, we have:

$0 \le \sin x \le 1$

$\Box$

Let:

\(\ds \map f x\)

\(=\)

\(\ds x \paren {\pi - x}\)

\(\ds \leadsto \ \ \)

\(\ds \map {f'} x\)

\(=\)

\(\ds \pi - 2 x\)

\(\ds \leadsto \ \ \)

\(\ds 0\)

\(=\)

\(\ds \pi - 2 x\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \dfrac \pi 2\)

From Derivative at Maximum or Minimum, $\map f x$ is a maximum or minimum at $x = \dfrac \pi 2$.

Then also:

$\map f 0 = \map f \pi = 0$

and:

$\map f {\dfrac \pi 2} = \dfrac {\pi^2} 4$

So on $\closedint 0 \pi$, $x = \dfrac \pi 2$ is in fact a maximum for $x \paren {\pi - x}$.

So we have:

Hence it follows that:

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

$\Box$

Since $A_n$ is clearly positive and the length of the interval is $\pi$ and the integrand is bounded at $\paren {\pi^2 / 4}^n$, we have:

This article, or a section of it, needs explaining. In particular: Why is the integral so bounded? The integrand is, and I suspect we can establish bounds on the integral based on properties of the integrand (long time since I did this so forget the details), but this all needs to be taken care of You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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

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 deduction that $A_n$ is an integer.

This depends on our supposition that $\pi$ is a rational number,

Hence by Proof by Contradiction it follows that $\pi$ is irrational.

$\blacksquare$