Pi is Irrational

From ProofWiki
Jump to navigation Jump to search

Theorem

Pi ($\pi$) is irrational.


Proof 1

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

Then from Existence of Canonical Form of Rational Number:

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

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

We define the polynomial function:

$\forall x \in \R: f \left({x}\right) = \dfrac {x^n \left({a - b x}\right)^n} {n!}$

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

$\displaystyle F \left({x}\right) = \sum_{j \mathop = 0}^n \left({-1}\right)^j f^{\left({2 j}\right)} \left({x}\right) = f \left({x}\right) + \cdots + \left({-1}\right)^j f^{\left({2 j}\right)} \left({x}\right) + \cdots + \left({-1}\right)^n f^{(2 n)} \left({x}\right)$

That is, $F \left({x}\right)$ is the alternating sum of $f$ and its first $n$ even derivatives.


First we show that:

$(1): \quad F \left({0}\right) = F \left({\pi}\right)$

From the definition of $f \left({x}\right)$, and our supposition that $\pi = \dfrac a b$, we have that:

$\forall x \in \R: f \left({x}\right) = b^n \dfrac {x^n \left({\pi - x}\right)^n} {n!} = f \left({\pi - x}\right)$


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

$\forall x \in \R: f^{\left({j}\right)} \left({x}\right) = \left({-1}\right)^j f^{\left({j}\right)} \left({\pi - x}\right)$

In particular, we have:

$\forall j \in \left\{{1, 2, \ldots, n}\right\}: f^{\left({2 j}\right)} \left({0}\right) = f^{\left({2 j}\right)} \left({\pi}\right)$

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

$F \left({0}\right) = F \left({\pi}\right)$


Next we show that:

$(2): \quad F \left({0}\right)$ is an integer.

We use the Binomial Theorem to expand $\left({a - bx}\right)^n$:

$\displaystyle \left({a - bx}\right)^n = \sum_{k \mathop = 0}^n \binom n k a^{n - k} \left({-b}\right)^k x^k$

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

$\displaystyle f \left({x}\right) = \frac 1 {n!} \sum_{j \mathop = n}^{2 n} \binom n {j - n} a^{2 n - j} \left({-b}\right)^{j - n} x^j$

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:

$\forall j < n: f^{\left({j}\right)} \left({0}\right) = 0$
$\forall j > 2 n: f^{\left({j}\right)} \left({0}\right) = 0$

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

$\displaystyle f^{\left({j}\right)} \left({0}\right) = \frac {j!} {n!} \binom n {j - n} a^{2 n - j} \left({-b}\right)^{j - n}$

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 $\left({-b}\right)^{j - n}$.

So $f^{\left({j}\right)} \left({0}\right)$ is an integer for all $j$, and hence so is $F \left({0}\right)$.


Next we show that:

$(3): \quad \displaystyle \dfrac 1 2 \int_0^\pi f \left({x}\right) \sin x \rd x = F \left({0}\right)$

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

This means:

$F'' + F = f$

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

$\left({F' \left({x}\right) \sin x - F \left({x}\right) \cos x}\right)' = f \left({x}\right) \sin x$

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

$\displaystyle \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \rd x = \frac 1 2 \left[{\left({F' \left({x}\right) \sin x - F \left({x}\right) \cos x}\right)}\right]_{x \mathop = 0}^{x \mathop = \pi}$

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

So, from $F \left({0}\right) = F \left({\pi}\right)$ (see $(1)$ above), we have:

$\displaystyle \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \rd x = F \left({0}\right)$


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 $F \left({0}\right)$ is a positive integer.

Now, we have that:

$\left({x - \dfrac \pi 2}\right)^2 = x^2 - \pi x + \left({\dfrac \pi 2}\right)^2$

and so:

$x \left({\pi - x}\right) = \left({\dfrac \pi 2}\right)^2 - \left({x - \dfrac \pi 2}\right)^2$

Hence:

$\forall x \in \R: x \left({\pi - x}\right) \le \left({\dfrac \pi 2}\right)^2$

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

So, by the definition of $f$:

$\displaystyle \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \rd x \le \frac {b^n} {n!} \left({\frac \pi 2}\right)^{2 n + 1}$

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 $F \left({0}\right) < 1$, by $(3)$.

This is impossible for the positive integer $F \left({0}\right)$.

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

$\blacksquare$


Proof 2

For any integer $q$, let:

$\displaystyle A_n = \frac {q^n} {n!} \int_0^\pi \left[{x \left({\pi - x}\right)}\right]^n \sin x \, \mathrm d x $

Integration by Parts twice gives:

\(\displaystyle A_n\) \(=\) \(\displaystyle \frac {q^n} {n!} \int_0^\pi n \left[{x \left({\pi - x}\right)}\right]^{n - 1} \left({\pi - 2 x}\right) \cos x \, \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {q^n} {n!} \int_0^\pi 2 n \left[{x \left({\pi - x}\right)}\right]^{n - 1} \sin x - n \left({n - 1}\right) \left[{x \left({\pi - x}\right)}\right]^{n - 2} \left({\pi - 2 x}\right)^2 \sin x \, \mathrm d x\)

Writing:

$\left({\pi - 2 x}\right)^2 = \pi^2 - 4 x \left({\pi - x}\right)$

gives the result:

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


Suppose $\pi$ is rational.

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

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:

$\displaystyle A_0 = \int_0^\pi \sin x \, \mathrm d x = 2$
$\displaystyle A_1 = q \int_0^\pi x \left({\pi - x}\right) \sin x \, \mathrm d x = 4 q$

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 $x \in \left[{0 \,.\,.\, \pi}\right]$, we have:

$0 \le \sin x \le 1$

and:

$0 \le x \left({\pi - x}\right) \le \pi^2 / 4$

hence:

$0 < A_n < \pi \dfrac {\left({q \pi^2 / 4}\right)^n} {n!}$

From Power over Factorial:

$\displaystyle \lim_{n \to \infty} \frac {\left({q \pi^2 / 4}\right)^n} {n!} = 0$

It follows from Squeeze Theorem that:

$\displaystyle \lim_{n \to \infty} A_n = 0$

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

This contradicts 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.


Sources