If n is an integer greater than one, one defines

When n is even, is irrational iff is irrational since

and , the n th Bernoulli number, is rational.

The nature of for odd n remained a mystery until 1978, when it was shown by Roger Apéry that is irrational. (The mystery remains for odd n>3).

As before we use the fact that

x is irrational if there exist integers h and k such that for any given real number

Apéry [1] described sequences that satisfy this condition for sufficiently large n when . In 1979 Beukers described [2] a simpler method of arriving at the same result. This works as follows (to keep the post, and my editing time, short, I haven’t included proofs of the lemmas. Instead, the full article with lemmas is available as a pdf here)

Firstly we note the following formulas (see Lemma 1)

When i>j

Suppose is some polynomial in x of degree n with integer coefficients. Then we can represent as

with the a_{ij} integers.

Combining the three expressions above we can obtain

Let d_{n} be the least integer that is divisible by each of the integers 1,2, .. ,n. Then it is not hard to see that the denominators of all the fractions on the right hand side of the above expression divide (d_{n})^{3}. Therefore we have

with k_{n}, h_{n} integers. To complete the proof we need to find an such that for large enough n, we can make the absolute value of the left hand side as close as we like to zero, without actually being zero.

It was established in connection with the proof of the Prime Number Theorem that [3] d_{n}<e^{(1+δ)n} for sufficiently large n and any given positive real number δ. Consequently (d_{n})^{3}<e^{(3+δ)n} for sufficiently large n.

In an earlier post we made use of the polynomial

This is a polynomial of degree n with integer coefficients, and each of its coefficients is divisible by n! (Lemma 2). We take

It is not too difficult to establish (Lemma 3) that

Consequently

And the result is proved.

[1] Roger Apéry, ‘Irrationalité de ζ(2)et ζ(3)’, Astérisque 61, 1979, pp 11-13; and ‘Interpolation de fractions continues et irrationalite de certaines constantes’, Bulletin de la Section des Sciences du C.T.H.S. III, 1981, pp 37-53.

[2] Frits Beukers, ‘A Note on the Irrationality of ζ(2)and ζ(3)’, Bulletin of the London Mathematical. Society, 11, 1979, pp 268-272.

[3] In analytical number theory, the function ln d_{n} is usually called ψ(n). This function was introduced around 1850 by Chebyshev who showed that the conjecture lim_{n→∞}ψ(n)/n=1 (if such a limit exists) implies the prime number theorem. (Chebyschev also obtained upper and lower bounds on ψ(n)/n and showed that if a limit exists, it must be 1). Most proofs of the Prime Number Theorem whether by way of Riemann’s zeta function or by way of the ‘elementary proof’ proceed by proving the limit exists. Consequently ln d_{n}/n<1+δ for sufficiently large n and any given real positive δ.

April 27, 2011 at 8:29 am |

Dear Cliffbott,

Thank you for this post. I always wondered just how accessible the proof of the irrationality of zeta(3) would be. From the pdf that I’ve downloaded, it seems like I’ll be able to follow it.

My interest in irrationals and transcendentals stems from Niven’s delightful first book (No. 1 in the NML series). Will move on to his later (Carus Monograph) book as well as Gelfond’s book from the same age (half a century ago).

Hope you keep enjoying your Mathematics!

Sincerely,

Andrew

April 27, 2011 at 8:48 am |

Hi once again!

I’m sorry for not realising that your name is Cliff Bott in itself (I thought that the bott was a truncation of our surname).

I’d like to ask you something that occurred to me while perusing Sierpinski’s “Pythagorean Triangles”:

The T-ratios of any Pyth. triangle are rational. So, we can say that the acute angles are necessarily irrational. But can we say that they are irrational multiples of pi? Or what can one say, at best?

Sincerely,

Andrew

April 27, 2011 at 12:39 pm |

Thanks Andrew. I have now fixed the way my name displays. I’m a mathematical hobbyist, not a prefessional mathematician, so I’m not sure I can help you with the question, but I’ll try to have a look at it and think about it over the next couple of weeks. I’ll try to email separately in case you don’t see this comment.