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Article | Pi is irrational |

Author | Helmut Richter |

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Article URL | https://hhr-m.de/pi-irrat/ |

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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)` = `x ^{n}`(

and let

`F(x)` = `f(x)` + ... + (–1)^{j}`f`^{(2j)}`(x)` + ... + (–1)^{n}`f`^{(2n)}`(x)`

where `f`^{(2j)} denotes the 2`j`-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`(π–`x`)0 ≦

`f(x)`≦ π/^{n}a^{n}`n`! for 0 ≦`x`≦ πFor 0 ≦

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

`n`≦`j`, the`j`-th derivative of`f`is integer at 0 und π (inferred from (1.)).`F`(0) and`F(π)`are integer (inferred from (4.) and (5.)).`F`+`F`'' =`f`(

`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*.