Pi Squared is Irrational/Proof 3

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Theorem

Pi squared ($\pi^2$) is irrational.


Proof

Aiming for a contradiction, suppose $\pi^2$ is rational.


We establish a lemma:

Lemma

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


Let it be supposed that $\pi^2$ is irrational, so that:

$\pi^2 = \dfrac p q$

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


Let $A_n$ be defined as:

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

Then:

$A_n = \paren {16 n - 8} q A_{n - 1} - 16 p q A_{n - 2}$

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:

\(\ds A_0\) \(=\) \(\ds \frac \pi 2 \frac {p^0} {0!} \int_0^1 \paren {1 - x^2 }^0 \map \cos {\dfrac {\pi x} 2} \rd x\)
\(\ds \) \(=\) \(\ds \frac \pi 2 \int_0^1 \map \cos {\dfrac {\pi x} 2} \rd x\) Zeroth Power of Real Number equals One, Factorial of Zero
\(\ds \) \(=\) \(\ds \frac \pi 2 \times \frac 2 \pi \times \bigintlimits {\map \sin {\dfrac {\pi x} 2} } 0 1\) Primitive of Cosine Function: Corollary
\(\ds \) \(=\) \(\ds 1\) Sine of $0 \degrees$, Sine of $90 \degrees$

and

\(\ds A_1\) \(=\) \(\ds \frac \pi 2 \frac {p^1} {1!} \int_0^1 \paren {1 - x^2 }^1 \map \cos {\dfrac {\pi x} 2} \rd x\)
\(\ds \) \(=\) \(\ds p \paren {\frac \pi 2 \int_0^1 \map \cos {\dfrac {\pi x} 2} \rd x - \frac \pi 2 \int_0^1 x^2 \map \cos {\dfrac {\pi x} 2} \rd x }\) Linear Combination of Integrals
\(\ds \) \(=\) \(\ds p \paren {\frac \pi 2 \times \frac 2 \pi \times \bigintlimits {\map \sin {\dfrac {\pi x} 2} } 0 1 - \frac \pi 2 \bigintlimits {\frac 2 \pi x^2 \map \sin {\dfrac {\pi x} 2} + \paren {\frac 4 {\pi^2} } 2 x \map \cos {\dfrac {\pi x} 2} - 2 \paren {\frac 2 \pi }^3 \map \sin {\dfrac {\pi x} 2} } 0 1 }\) Primitive of Cosine Function: Corollary, Primitive of $x^n \cos a x$
\(\ds \) \(=\) \(\ds p \paren {1 - \paren {1 - \frac 8 {\pi^2} } }\) Sine of $0 \degrees$, Sine of $90 \degrees$, Cosine of $0 \degrees$, Cosine of $90 \degrees$
\(\ds \) \(=\) \(\ds p \paren {\frac 8 {p / q} }\) assuming $\pi^2 = \dfrac p q$
\(\ds \) \(=\) \(\ds 8 q\)


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

By our lemma:

$A_k = \paren {16 n - 8} q A_{k - 1} - 16 p q A_{k - 2}$


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 $x \in \closedint 0 1$.

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

$0 \le \map \cos {\dfrac {\pi x} 2} \le 1$

From Derivative at Maximum or Minimum, we have:

\(\ds \map f x\) \(=\) \(\ds \paren {1 - x^2}^n\)
\(\ds \leadsto \ \ \) \(\ds \map {f'} x\) \(=\) \(\ds n \paren {1 - x^2}^{n - 1} \paren {-2 x}\)
\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds n \paren {1 - x^2}^{n - 1} \paren {-2 x}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds 0\) $0$ is the maximum $\map f 0 = 1$ and $\pm 1$ is the minimum $\map f {\pm 1} = 0$

Therefore, plugging $0$ into $\map f x$, we obtain:

$0 \le \paren {1 - x^2}^n \le 1$

Since $A_n$ is clearly positive and the length of the interval is $1$ and the integrand is bounded at $\paren {1}^n$, from Darboux's Theorem, we have:

$0 < A_n < 1 \times \dfrac \pi 2 \times \dfrac {p^n} {n!} \times \paren {1}^n$

From Power over Factorial:

$\ds \lim_{n \mathop \to \infty} \dfrac {p^n} {n!} = 0$

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 $\pi^2$ is a rational number,

Hence by Proof by Contradiction it follows that $\pi^2$ is irrational.

$\blacksquare$

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