Proving that Pi is Irrational

In this post, we prove that $\pi$ is irrational.

Theorem 1. $\pi$ is irrational.

We will prove this by following the proof provided by Niven (1947). Let’s first establish a few lemmas.

Lemma 1. For integers $a, b, n > 0$ define the functions

$\displaystyle \begin{aligned} f(x)& := \frac{x^n (a-bx)^n}{n!}, \quad F(x) := \sum_{k=0}^n {(-1)}^k f^{(2k)}(x).\end{aligned}$

Then $F(0)$ and $F(a/b)$ are integers.

Proof of Lemma 1. Since $f(x) = f(a/b-x)$ , $f^{(2k)}(a/b) = f^{(2k)}(0)$ . It therefore suffices to check that $f^{(k)}(0)$ is an integer for any $k$ , which is straightforward upon using the binomial theorem:

$\displaystyle n! f(x) = x^n (a-bx)^n = \sum_{j=0}^n {n \choose j} a^{n-j} b^j x^{n+j}.$

Observe that differentiating $n-1$ times and setting $x = 0$ yields $f^{(k)}(0) = 0$ for $k = 0,1,\dots,n-1$ . Differentiating one more time yields

$\displaystyle n! f^{(n)}(x) = \sum_{j=0}^n {n \choose j} \cdot \frac{(n+j)!}{j!} a^{n-j} b^j x^j.$

Dividing by $n!$ on both sides,

$\displaystyle f^{(n)}(x) = \sum_{j=0}^n {n \choose j} {n + j \choose j} \cdot a^{n-j} b^j x^j.$

This yields integer values of $f^{(k)}(0)$ for $k =n,\dots, 2n$ .

We also need a lemma involving integrals.

Lemma 2. Following the notations in Lemma 1,

$\displaystyle \int_{0}^{\pi} f(x) \sin(x)\, \mathrm dx = F(\pi) + F(0).$

Proof of Lemma 2. A quick observation yields

$\displaystyle \begin{aligned} F(x) + F''(x) &= \sum_{k=0}^n {(-1)}^k f^{(2k)}(x) + \sum_{k=0}^n {(-1)}^k f^{(2k+2)}(x)\\ &= \sum_{k=0}^n {(-1)}^k f^{(2k)}(x) - \sum_{k=1}^n {(-1)}^{k} f^{(2k)}(x) = f(x). \end{aligned}$

Thus, integrating yields

$\displaystyle \begin{aligned} \int_{0}^{\pi} f(x) \sin(x)\, \mathrm dx &= \int_{0}^{\pi} (F(x) + F''(x)) \sin(x)\, \mathrm dx \\ &= \underbrace{\int_0^{\pi} F(x) \sin(x)\, \mathrm dx}_{I_1} + \underbrace{\int_0^{\pi} F''(x) \sin(x)\,\mathrm dx}_{I_2}. \end{aligned}$

Both integrals can be simplified using integration by parts. For $I_1$ ,

$\displaystyle \begin{aligned} I_1 = \int_0^{\pi} F(x) \sin(x)\, \mathrm dx &= [\underbrace{-{\cos(x)}}_{\text I} \underbrace{F(x)}_{\text S} ]_0^\pi - \int_0^{\pi} \underbrace{-{\cos(x)} }_{\text I} \underbrace{ F'(x) }_{\text D}\, \mathrm dx \\ &= F(\pi) + F(0) + \int_0^{\pi} F'(x)\cos(x)\, \mathrm dx. \end{aligned}$

For $I_2$ ,

$\displaystyle \begin{aligned} I_2 = \int_0^{\pi} F''(x) \sin(x)\, \mathrm dx &= [\underbrace{F'(x)}_{\text I} \underbrace{\sin(x)}_{\text S} ]_0^\pi - \int_0^{\pi} \underbrace{F'(x) }_{\text I} \underbrace{ \cos(x) }_{\text D}\, \mathrm dx \\ &= 0 -\int_0^{\pi} F'(x)\cos(x)\, \mathrm dx \\ &= -\int_0^{\pi} F'(x)\cos(x)\, \mathrm dx. \end{aligned}$