π and π2 are irrational

Theorem 1.

π and π2 are irrational.


For any strictly positive integer n ,x(0,1) we define:


where cm are integers. For 0<x<1 we have

0<f(x)<1n! (1)

For a contradictionMathworldPlanetmathPlanetmath, suppose π2 is rational, so that π2=ab, where a,b are positive integers.

For x(0,1) let us define


We have that f(0)=0 and f(m)(0)=0 if m<n or m>2n. But, if nm2n, then


an integer. Hence f(x) and all its derivates take integral values at x=0.Since f(1-x)=f(x), the same is true at x=1

so that G(0) and G(1) are integers. We have

ddx[G(x)sinπx-πG(x)cosπx] = [G′′(x)+π2G(x)]sinπx
= bnπ2n+2f(x)sinπx
= π2ansinπxf(x).



witch is an integer. But by equation 1,


For a large enough n, we obtain a contradiction.

For any integer n, if an is irrational then a is irrational http://planetmath.org/?op=getobj&from=objects&id=5779(proof), and since π2 is irrational π2=π is also irrational. ∎

The irrationality of π was Proved by Lambert in 1761. The above proof is not the original proof due to Lambert.


  • 1 G.H.Hardy and E.M.Wright An Introduction to the Theory of Numbers, Oxford University Press, 1959

See also

  • The MacTutor History of Mathematics Archive, http://www-gap.dcs.st-and.ac.uk/ history/HistTopics/Pi_through_the_ages.htmlA history of Pi

  • The MacTutor History of Mathematics Archive, http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Lambert.htmlJohann Heinrich Lambert

  • http://numbers.computation.free.fr/Constants/Miscellaneous/irrationality.htmlIrrationality proofs

Title π and π2 are irrational
Canonical name piAndpi2AreIrrational
Date of creation 2013-03-22 14:44:00
Last modified on 2013-03-22 14:44:00
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 15
Author mathcam (2727)
Entry type Theorem
Classification msc 51-00
Classification msc 11-00