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As we know $\sqrt{2},\sqrt{3}$ are irrational numbers. And I see some proofs on the net.

So I doubt that how $e,\pi$ or already known irrational numbers are proved to be irrational.

In fact, I got interested in Riemann zeta function $$\zeta(s)=\sum_{n=0}^{\infty} \frac{1}{n^s},$$

we know $\zeta(2)=\pi^2/6$ from Euler, 1737.

One mathematician (sorry to forgot his name) proved $\zeta(3)$ to be also irrational 40 years before.

Can somebody explain how he could do with that? To understand Apéry's theorem, is it very hard?

An question raises that could one real number make up of two different irrationals (for example: $e,\pi$), $e\pi$, or others can be rational?

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More information about $\zeta(3)$ being irrational: en.wikipedia.org/wiki/Apéry%27s_theorem –  user67258 May 21 '13 at 14:58
    
For reference: proof $e$ is irrational and proof $\pi$ is irrational. Are you saying, to begin with, you doubt these are right? –  al0 May 21 '13 at 15:02
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Roger Apéry showed that $\zeta(3)$ is irrational in 1978 (Apéry's Theorem). –  robjohn May 21 '13 at 15:03
    
Internet searches easily lead to answers to many of these questions. In particular, you'd probably find some proofs of irrationality on Wikipedia. Regarding the last question, see the related questions math.stackexchange.com/q/159350 and math.stackexchange.com/q/28243. –  Jonas Meyer May 21 '13 at 15:14
    
I don't doubt they are right. But wonder how to do that. –  Jimmy Wang May 21 '13 at 15:22
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2 Answers 2

It is known to be very difficult to find the arithmetic nature of most real numbers such as the constants that appear in mathematical analysis, e.g. $\zeta (5)=\sum_{n\geq 1}\frac{1}{n^{5}}$.

Roger Apéry proved directly that $\zeta(2)$ is irrational. And he was the first to prove that $\zeta(3)$ is irrational too. He constructed two sequences $(a_n),(b_n)$ $[1]$

$$\begin{equation*} a_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}c_{n,k}, \qquad b_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2},\end{equation*}$$

where

$$\begin{equation*} c_{n,k}=\sum_{m=1}^{n}\frac{1}{m^{3}}+\sum_{m=1}^{k}\frac{\left( -1\right) ^{m-1}}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\quad k\leq n. \end{equation*}$$

The ratio $a_n/b_n\to\zeta(3)$ and has the following properties:

  1. $2(b_{n}\zeta (3)-a_{n})$ satisfies $\lim\sup \left\vert 2(b_{n}\zeta (3)-a_{n})\right\vert^{1/n}\le(\sqrt{2}-1)^4 $.
  2. $b_{n}\in \mathbb{Z},2(\operatorname{lcm}(1,2,\ldots ,n))^{3}a_{n}\in \mathbb{Z}$.
  3. $\left\vert b_{n}\zeta (3)-a_{n}\right\vert >0$.

This is enough to prove the irrationality of $\zeta (3)$ by contradiction. $[2]$.

There is The Tricki entry To prove that a number is irrational, show that it is almost rational that gives two examples and explains the principle of some proofs of the irrationality of numbers.

References.

$[1]$ Poorten, Alf., A Proof that Euler Missed…, Apéry’s proof of the irrationality of $\zeta(3)$. An informal report, Math. Intelligencer 1, nº 4, 1978/79, pp. 195-203.

$[2]$ Fischler, Stéfane, Irrationalité de valeurs de zêta (d’ après Apéry, Rivoal, …), Séminaire Bourbaki 2002-2003, exposé nº 910 (nov. 2002), Astérisque 294 (2004), 27-62

$[3]$ Apéry, Roger (1979), Irrationalité de $\zeta2$ et $\zeta3$, Astérisque 61: 11–13

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One last thing: Is there a general method for proving irrationality? –  Vladimir Putin May 21 '13 at 16:25
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@GustavoBandeira No! The test above or a similar one provides a method but the difficulty is on how to construct such sequences $(a_n),(b_n)$. –  Américo Tavares May 21 '13 at 16:29
    
Thanks for the answer. –  Vladimir Putin May 21 '13 at 16:32
    
@Gustavo Bandeira You are welcome. –  Américo Tavares May 21 '13 at 16:39
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Some combinations of irrationals can be proved to be irrational in a more or less systematic way. For example, $(\sqrt{2})^\pi$ is irrational (and in fact transcendental) by Gelfond's theorem (see http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem).

One approach to proving the irrationality of a number is to show that it is approximated "too well" by rational numbers $p/q$ where "too well" is assigned a specific meaning in terms of the denominator $q$. From this point of view, $e$ is irrational because the terms in the familiar power series defining it: $e=\sum 1/n!$ tend to zero rapidly, so that the partial sums give extremely good approximations of $e$.

To get more details, see Spivak's book "Calculus" where he proves irrationality of $\pi$ on page 307, and that $e$ is transcendental on page 409.

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