Ref:

• “A simple proof that ${ \pi }$ is irrational” by Niven. Link to paper: Link.

• Math StackExchange posts here, here.

Updated: 15/3/26

Belated Happy ${ \pi }$ Day!

[Inner product of a polynomial with ${ \sin(x) }$]

Let ${ P(x) }$ be a polynomial. Note that

\[{ \small {\begin{aligned} &\, \underline{\int P(x) \sin(x) \, dx} \\ = &\, - P(x) \cos(x) + \int P ^{’} (x) \cos(x) \, dx \\ = &\, - P(x) \cos(x) + P ^{(1)} (x) \sin(x) - \underline{ \int P ^{(2)} (x) \sin(x) \, dx } \\ = &\, - P(x) \cos(x) + P ^{(1)} (x) \sin(x) + P ^{(2)} (x) \cos(x) - P ^{(3)} (x) \sin(x) + \int P ^{(4)} (x) \sin(x) \, dx . \end{aligned}} }\]

Suppose ${ \deg(P) = 2N . }$ Note that

\[{ \small {\begin{aligned} &\, \int _0 ^{\pi} P(x) \sin(x) \, dx \\ = &\, (P(0) + P(\pi)) - (P ^{(2)} (0) + P ^{(2)} (\pi)) + \int _0 ^{\pi} P ^{(4)} (x) \sin(x) \, dx \\ = &\, \sum _{j=0} ^{\infty} (-1) ^{j} (P ^{(2j)} (0) + P ^{(2j)} (\pi) ) \\ = &\, \sum _{j=0} ^{N} (-1) ^j (P ^{(2j)} (0) + P ^{(2j)} (\pi)) . \end{aligned}} }\]

Hence if ${ \deg(P) = 2N, }$ the inner product

\[{ \int _0 ^{\pi} P(x) \sin(x) \, dx = F(0) + F(\pi) , }\]

where

\[{ F(x) := \sum _{j=0} ^{N} (-1) ^j P ^{(2j)} (x) . }\]

[${ \pi }$ is irrational]

Suppose to the contrary ${ \pi = \frac{a}{b} , }$ where ${ a, b }$ are positive integers.

We will cleverly choose a polynomial ${ P(x) }$ (of degree ${ 2N }$), which is “small over ${ (0, \pi) }$” and such that the inner product

\[{ \int _0 ^{\pi} P(x) \sin(x) \, dx = F(0) + F(\pi) }\]

is an integer (!).

Suppose ${ P(x) }$ is of the form

\[{ \text{Suppose:} \quad P(x) = \frac{1}{N!} (c _N x ^N + \ldots + c _{2N} x ^{2N}) , \quad c _j \in \mathbb{Z} . }\]

Note that all ${ P ^{(j)} (0) }$ are integers.

Suppose further

\[{ \text{Suppose:} \quad P(x) = P \left( \frac{a}{b} - x \right) . }\]

Note that all ${ P ^{(j)} \left( \frac{a}{b} \right) }$ are integers.

Hence to satisfy above constraints a clever choice of ${ P }$ is

\[{ \boxed{ \text{Choice:} \quad P(x) = \frac{b ^N}{N!} x ^N \left( \frac{a}{b} - x \right) ^N } . }\]

By the choice of the polynomial the inner product

\[{ \int _0 ^{\pi} P(x) \sin(x) \, dx }\]

is an integer.

But we have the bound

\[{ 0 < P(x) \sin(x) < \frac{\pi ^N a ^N}{N!} \quad \text{ for } x \in (0, \pi) }\]

Note that for ${ x \in (0, \pi) , }$

\[{ P(x) = \frac{1}{N!} {\color{red}{x ^N}} {\color{blue}{(a - bx) ^N}} < \frac{1}{N!} {\color{red}{\pi ^N}} {\color{blue}{a ^N }} . }\]

Hence the inner product

\[{ \int _0 ^{\pi} P(x) \sin(x) \, dx }\]

can get arbitrarily small by choosing large enough ${ N . }$

A contradiction.

Hence ${ \pi }$ is irrational, as needed. ${ \blacksquare }$