Pi Squared is Irrational

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Theorem

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


Decimal Expansion

The decimal expansion of Pi squared ($\pi^2$) begins:

$9 \cdotp 86960 \, 44010 \, 89358 \ldots$


Proof 1

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


We establish a lemma:

Lemma

Let $n \in \Z_{> 0}$ 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:

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

Then:

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

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:

\(\ds A_0\) \(=\) \(\ds \int_0^\pi \sin x \rd x\)
\(\ds \) \(=\) \(\ds 2\) Area under Arc of Sine Function

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:

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


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 $x \in \closedint 0 \pi$.

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

$0 \le \sin x \le 1$

From Derivative at Maximum or Minimum, we have:

\(\ds \map f x\) \(=\) \(\ds x \paren {\pi - x}\)
\(\ds \leadsto \ \ \) \(\ds \map {f'} x\) \(=\) \(\ds \pi - 2 x\)
\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds \pi - 2 x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac \pi 2\)

Therefore, plugging $\dfrac \pi 2$ into $\map f x$, we obtain:

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

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:

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

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 $\pi^2$ is a rational number,

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

$\blacksquare$


Proof 2

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

Then $\pi^2 = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$.

Let us define a polynomial:

$\ds \map f x = \frac {\paren {1 - x^2}^n} {n!} = \sum_{m \mathop = n}^{2 n} \frac {c_m} {n!} x^m$

for $c_m \in \Z$.

$\map f x = \map f {-x}$ and so we get:

\(\ds \map {f^{\paren k} } x = \paren {-1}^k \map {f^{\paren k} } {-x}\) \(=\) \(\ds \begin {cases} \ds \sum_{m \mathop = n}^{2 n} \frac{k!} {n!} \binom m k c_m x^{m - k} & : 0 \le k \le n - 1 \\ \ds \sum_{m \mathop = k}^{2 n} \frac {k!} {n!} \binom m k c_m x^{m - k} & :n \le k \le 2 n \end {cases}\)
\(\ds \map {f^{\paren k} } 1 = \paren {-1}^k \map {f^{\paren k} } {-1}\) \(=\) \(\ds \begin {cases} 0 & : 0 \le k \le n - 1 \\ \dfrac {k!} {n!} c_k & : n \le k \le 2 n \end {cases}\)

hence they are all integers.


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Now let:

$\ds A_n = \int_{-1}^1 \map f x \map \cos {\dfrac {\pi x} 2} \rd x = 2 \int_0^1 \map f x \map \cos {\dfrac {\pi x} 2} \rd x$

Repeated integration by Parts gives:

\(\ds A_n\) \(=\) \(\ds \frac {2^2} \pi \intlimits {\map {f^{\paren 0} } x \map \sin {\dfrac {\pi x} 2} } 0 1 + \frac {2^3} {\pi^2} \intlimits {\map {f^{\paren 1} } x \map \cos {\dfrac {\pi x} 2} } 0 1 - \cdots \pm \frac {2^{2 n + 2} } {\pi^{2 n + 1} } \intlimits {\map {f^{\paren {2 n} } } x \map \sin {\dfrac {\pi x} 2} } 0 1 \pm \frac {2^{2 n + 2} } {\pi^{2 n + 1} } \int_0^1 \map {f^{\paren {2 n + 1} } } x \map \sin {\dfrac {\pi x} 2} \rd x\)
\(\ds \) \(=\) \(\ds \frac {2^2} \pi \intlimits {\map {f^{\paren 0} } x \map \sin {\dfrac {\pi x} 2} } 0 1 - \frac {2^4} {\pi^3} \intlimits {\map {f^{\paren 2} } x \map \sin {\dfrac {\pi x} 2} } 0 1 + \frac {2^6} {\pi^5} \intlimits {\map {f^{\paren 4} } x \map \sin {\dfrac {\pi x} 2} } 0 1 - \cdots + \paren {-1}^n \frac {2^{2 n + 2} } {\pi^{2 n + 1} } \intlimits {\map {f^{\paren {2 n} } } x \map \sin {\dfrac {\pi x} 2} } 0 1\)
\(\ds \) \(=\) \(\ds \frac {4} \pi \map {f^{\paren 0} } 1 - \frac {4^2} {\pi^3} \map {f^{\paren 2} } 1 + \frac {4^3} {\pi^5} \map {f^{\paren 4} } 1 - \cdots + \paren {-1}^n \frac {4^{n + 1} } {\pi^{2 n + 1} } \map {f^{\paren {2 n} } } 1\)
\(\ds \leadsto \ \ \) \(\ds \frac \pi 4 A_n\) \(=\) \(\ds \sum_{k \mathop = 0}^n \paren {-\frac 4 {\pi^2} }^k \map {f^{\paren {2 k} } } 1\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \paren {-\frac {4 q} p}^k \map {f^{\paren {2 k} } } 1\)

Now let:

$\ds N_n = \frac \pi 4 p^n A_n = \sum_{k \mathop = 0}^n p^{n - k} \paren {-4 q}^k \map {f^{\paren {2 k} } } 1$

Since $\map f x >0$ and $\map \cos {\dfrac {\pi x} 2} > 0$ on the open interval $\openint {-1} 1$:

$A_n > 0$

Hence $N_n$ is a positive integer.


Nevertheless, for $x \in \closedint {-1} 1$:

$0 \le \map f x \le \dfrac 1 {n!}$

and:

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

Hence:

$0 < A_n < \dfrac 2 {n!}$

Therefore:

$0 < N_n < \dfrac \pi 2 \dfrac {p^n} {n!}$

From Power over Factorial:

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

It follows from the Squeeze Theorem that:

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

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

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

It follows that $\pi^2$ is irrational.

$\blacksquare$


Also see


Historical Note

$\pi^2$ was demonstrated to be irrational in $1794$ by Joseph Louis Lagrange.


Sources

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