# Pi is Irrational

## Proof 1

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

$\exists a \in \Z, b \in \Z_{>0}: \pi = \dfrac a b$

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

We define the polynomial function:

$\forall x \in \R: \map f x = \dfrac {x^n \paren {a - b x}^n} {n!}$

We differentiate this $2 n$ times, and then we build:

$\ds \map F x = \sum_{j \mathop = 0}^n \paren {-1}^j \map {f^{\paren {2 j} } } x = \map f x + \cdots + \paren {-1}^j \map {f^{\paren {2 j} } } x + \cdots + \paren {-1}^n \map {f^{\paren {2 n} } } x$

That is, $\map F x$ is the alternating sum of $f$ and its first $n$ even derivatives.

First we show that:

$(1): \quad \map F 0 = \map F \pi$

From the definition of $\map f x$, and our supposition that $\pi = \dfrac a b$, we have that:

$\forall x \in \R: \map f x = b^n \dfrac {x^n \paren {\pi - x}^n} {n!} = \map f {\pi - x}$

Using the Chain Rule for Derivatives, we can apply the Principle of Mathematical Induction to show that, for all the above derivatives:

$\forall x \in \R: \map {f^{\paren j} } x = \paren {-1}^j \map {f^{\paren j} } {\pi - x}$

In particular, we have:

$\forall j \in \set {1, 2, \ldots, n}: \map {f^{\paren {2 j} } } 0 = \map {f^{\paren {2 j} } } \pi$

From the definition of $F$, it follows that:

$\map F 0 = \map F \pi$

Next we show that:

$(2): \quad \map F 0$ is an integer.

We use the Binomial Theorem to expand $\paren {a - b x}^n$:

$\ds \paren {a - b x}^n = \sum_{k \mathop = 0}^n \binom n k a^{n - k} \paren {-b}^k x^k$

By substituting $j = k + n$, we obtain the following expression for $f$:

$\ds \map f x = \frac 1 {n!} \sum_{j \mathop = n}^{2 n} \binom n {j - n} a^{2 n - j} \paren {-b}^{j - n} x^j$

Note the following:

The coefficients of $x^0, x^1, \ldots, x^{n - 1}$ are all zero
The degree of the polynomial $f$ is at most $2 n$.

So we have:

$\forall j < n: \map {f^{\paren j} } 0 = 0$
$\forall j > 2 n: \map {f^{\paren j} } 0 = 0$

But for $n \le j \le 2 n$, we have:

$\map {f^{\paren j} } 0 = \dfrac {j!} {n!} \dbinom n {j - n} a^{2 n - j} \paren {-b}^{j - n}$

Because $j \ge n$, it follows that $\dfrac {j!} {n!}$ is an integer.

So is the binomial coefficient $\dbinom n {j - n}$ by its very nature.

As $a$ and $b$ are both integers, then so are $a^{2 n - j}$ and $\paren {-b}^{j - n}$.

So $\map {f^{\paren j} } 0$ is an integer for all $j$, and hence so is $\map F 0$.

Next we show that:

$(3): \quad \ds \dfrac 1 2 \int_0^\pi \map f x \sin x \rd x = \map F 0$

As $\map f x$ is a polynomial function of degree $n$, it follows that $f^{\paren {2 n + 2} }$ is the null polynomial.

This means:

$F'' + F = f$

Using the Product Rule for Derivatives and the derivatives of sine and cosine, we get:

$\paren {\map {F'} x \sin x - \map F x \cos x}' = \map f x \sin x$

By the Fundamental Theorem of Calculus, this leads us to:

$\ds \frac 1 2 \int_0^\pi \map f x \sin x \rd x = \frac 1 2 \bigintlimits {\paren {\map {F'} x \sin x - \map F x \cos x} } {x \mathop = 0} {x \mathop = \pi}$

From Sine and Cosine are Periodic on Reals, we have that $\sin 0 = \sin \pi = 0$ and $\cos 0 = - \cos \pi = 1$.

So, from $\map F 0 = \map F \pi$ (see $(1)$ above), we have:

$\ds \frac 1 2 \int_0^\pi \map f x \sin x \rd x = \map F 0$

The final step:

On the interval $\openint 0 \pi$, we have from Sine and Cosine are Periodic on Reals that $\sin x > 0$.

So from $(2)$ and $(3)$ above, we have that $\map F 0$ is a positive integer.

Now, we have that:

$\paren {x - \dfrac \pi 2}^2 = x^2 - \pi x + \paren {\dfrac \pi 2}^2$

and so:

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

Hence:

$\forall x \in \R: x \paren {\pi - x} \le \paren {\dfrac \pi 2}^2$

Also, from Real Sine Function is Bounded, $0 \le \sin x \le 1$ on the interval $\openint 0 \pi$.

So, by the definition of $f$:

$\ds \frac 1 2 \int_0^\pi \map f x \sin x \rd x \le \frac {b^n} {n!} \paren {\frac \pi 2}^{2 n + 1}$

But this is smaller than $1$ for large $n$, from Radius of Convergence of Power Series over Factorial.

Hence, for these large $n$, we have $\map F 0 < 1$, by $(3)$.

This is impossible for the (strictly) positive integer $\map F 0$.

So our assumption that $\pi$ is rational must have been false.

$\blacksquare$

## Proof 2

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.

$\blacksquare$

## Proof 3

From Rational Points on Graph of Sine Function, the only rational point on the graph of the sine function in the real Cartesian plane $\R^2$:

$f := \left\{ {\left({x, y}\right) \in \R^2: y = \sin x}\right\}$

is the point $\left({0, 0}\right)$.

But $\left({\pi, 0}\right)$ is also on $f$.

Hence $\pi$ cannot be rational.

$\blacksquare$

## Historical Note

The proof that $\pi$ (pi) is irrational was demonstrated in $1761$ by Johann Heinrich Lambert.