Understanding the proof that the number 
is transcendental, i.e. not a root of any polynomial in
integer coefficients, requires substantial knowledge in mathematics.
That 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
, 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 ŤAn Introduction to the
Theory of Numbersť does indeed contain this proof, and Niven published
a proof of the irrationality of 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 = 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:
f is a polynomial with coefficients that are integer, except for a factor of 1/n!
f(x) = f(-x)
0 <= f(x) <= nan/n! for 0 <= x <=
For 0 <= j < n, the j-th derivative of f is zero at 0 und .
For n <= j, the j-th derivative of f is integer at 0 und
(inferred from (1.)).
F(0) and F() are integer (inferred from (4.) and (5.)).
F + F '' = f
(F 'ˇsin - Fˇcos)' = fˇsin (inferred from (7.))
Hence, the integral over fˇsin, taken from 0 to , 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.