Pi is irrational

Understanding the proof that the number π is transcendental, i.e. not a root of any polynomial in integer coefficients, requires substantial knowledge in mathematics. That π is irrational, i.e. not the quotient of two integers, is much easier to prove, only a good math education of a grammar school is needed. The first proof of the irrationality of π, given by J.H. Lambert in 1768, is more complicated than the one below but also more general.

I found the proof below in my notes of my beginner's analysis lecture. In these notes, it does not carry the name of an author. I have, however, been informed by Usenet participants that this proof is due to Ivan Niven. Niven/Zuckerman's book «An Introduction to the Theory of Numbers» does indeed contain this proof, and Niven published a proof of the irrationality of π in 1947 (Bull.Amer.Math.Soc. 53(1947),509). I have hints that the 1947 version has a similar idea as the proof given below but is less elegant; I have not yet seen the 1947 proof, though. - The rumour spread in a previous version of this Web article that "it [is] due to a Japanese mathematician and was pretty recent [...] in the late 60ies" is either unsubstantiated or this still unidentified "Japanese mathematician" is the person who gave Niven's original proof its present form.

Assume π = a/b with positive integers a und b.

Now, for some natural number n define the functions f and F as follows. Strictly speaking, f and F should each have n as an index as they depend on n but this would render things unreadable; remember that n is always the same constant throughout this proof.

Let

f(x) = xⁿ(a-bx)ⁿ/n!

and let

F(x) = f(x) + ... + (–1)^jf^(2j)(x) + ... + (–1)ⁿf⁽²ⁿ⁾(x)

where f^(2j) denotes the 2j-th derivative of f.

Then f and F have the following properties:

1. f is a polynomial with coefficients that are integer, except for a factor of 1/n!

2. f(x) = f(π–x)

3. 0 ≦ f(x) ≦ πⁿaⁿ/n! for 0 ≦ x ≦ π

4. For 0 ≦ j < n, the j-th derivative of f is zero at 0 und π.

5. For n ≦ j, the j-th derivative of f is integer at 0 und π (inferred from (1.)).

6. F(0) and F(π) are integer (inferred from (4.) and (5.)).

7. F + F '' = f

8. (F '·sin - F·cos)' = f·sin (inferred from (7.))

Hence, the integral over f·sin, taken from 0 to π, is integer.

For sufficiently large n, however, inequality (3.) tells us that this integral must be between 0 an 1. Hence, we could have chosen n such that the assumption is led ad absurdum.